Electric Circuit Analysis Complete.pdf

radians to the left. Therefore, points on the solid curve, such as its peaks, occur d> rad, or 4>/w s, earlier than corresponding points on the dashed curve. Accordingly, we shall say that V sin (a>t + d») leads V sin rad (or degrees). In general, the sinusoid m

i>i =

I XAMI'I I 10.1

To determine how much one sinusoid leads or lags another of the same frequency, we must first express both as sine waves or as cosine waves with positive amplitudes. For example, let v = 4 cos (It + 30°) t

and v = - 2 sin (2t + 18°)

m

V„t sin ((ot + a)

2

Then, since -sin wt = sin (air + 180°)

leads the sinusoid

we have

v = V „2 sin (tot + B) 2

i5 = 2 sin (2/ + 18° + 180°) 2

by a - 8. An equivalent expression is that V2 lags Vi by a — B.

= 2 cos (2t + 18° + 180° - 90°) = 2 cos (2t + 108°)

. V sin(w( + m

Comparing this last expression with c i , we see that vi leads v by 30° - 108° = — 78°, which is the same as saying that vi lags v by 78°. 2

2

The sum of a sine wave and a cosine wave of the same frequency is another sinusoid of that frequency. To show this, consider A cos tot + B sin wt = VA

2

+ B

: COS (Ot

2

B VA + B 2

FIGURE 10.2

T w o sinusoids with

different phases

which by Fig. 10.3 may be written A cos tot + B sin (ot = V A + B (cos ait cos 6 + sin ait sin 6)

As an example, consider

2

A cos (ot + B sin (ot = VA

and

2

v

2

= 6 sin (It - 12°)

e = tan'^ A

2

(10.7) C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

+ B cos (tot - 6) 2

(10.9)

where, by Fig. 10.3,

Then V\s v (or v lags i>,) by 30 - (-12) = 42° Thus far we have considered sine functions rather than cosine functions in defining sinusoids. It does not matter which form we use since

310

2

By a formula from trigonometry, this is

t), = 4 sin (2t + 30°)

2

sin (tit 1

(10.10)

A similar result may be established if the sine and cosine terms have phase angles other than zero, indicating that, in general, the sum of two sinusoids of a given frequency is another sinusoid of the same frequency. S e c t i o n 10.1

P r o p e r t i e s of S i n u s o i d s

311

AN HI ( IH< l l l l IXAMIMI n, Icl us hnd the forced compoquatton is / ' ' i Hi A F I G U R E 10.3

( liuptt'i

IIIIMI H I

T r i a n g l e u s e f u l in a d d i n g t w o s i n u s o i d s

It We must be clear on what is meant by (10.10), since some mathematics books take this expression as the principal value of the arctangent and place 6 in a specific quadrant. We mean that the terminal side of the angle 0 is in the quadrant where the point (A, B) is located. EXAMPLE 10.2

V

(10.11)

V„ cos tot

til

let us assume the trial solution

A cos tot + B sin tot

m

coi wf

We have - 5 cos 3? + 12 sin 3t = V 5 + 12 cos |^3f - tan 2

2

F I G U R E 10.4

1

= 13 cos (3t - 112.6°)

Substituting the trial solution into (10.11), we have L(-toA

since tan" (12/—5) is in the second quadrant, because A = - 5 < 0 and B = 12 > 0. ' 1

RL c i r c u i t

sin f

Therefore, equating coefficients of like terms, we must have RA + (oLB = V„ -ooLA + RB = 0

EXERCISES 10.1.1

from which RV R + (o L m

Find the period of the following sinusoids:

2

(a) 4 cos (3f + 33°). (b) cos

10.1.2

10.1.3

(it

+ ^

+ 3 sin ^2t -

(c) 8 sin 2irt. Answer (a) 27r/3; (b) IT; (c) 1 Find the amplitude and phase of the following sinusoids: (a) 6 cos 2f + 8 sin 2t. (b) (4V3 - 3) cos (It + 30°) + ( 3 V I - 4) cos (It + 60°). [Suggestion: In (b), expand both functions and use (10.9).] Answer (a) 10, -53.1°; (b) 5, 36.9° Find the frequency of the following sinusoids: (a) 3 cos (67r< - 10°). (b) 4 sin 377t. Answer (a) 3; (b) 60 Hz

2

(oLV R + to L m

B

^j.

2

2

2

2

The forced response is then RV„ R + (o L 2

2

2

wLV„ ait + — "" ' R + (o L sin tot

; COS

2

2

which by (10.9) and (10.10) may be written as if =

VR

2

+ (o L 2

(at-

tan- Y)

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

1

* > 1012

2

The forced response is, therefore, a sinusoid like the excitation, as we predicted when we chose the trial solution. We may write it in the form if = /„ cos (a.» + )

312

2

S e c t i o n 10.2

A n RL C i r c u i t E x a m p l e

(10.13)

313

where

iNMall 1

I,-

,

V

VR

2

- H I T , we list a lew of these iliw method ot analysis

1

1

~

+
(10.15)

2

and

•'«• " 'I I nilllih) I ii in.I h in- tin- iml and imaginary parts of A, fM|*> iiv»l> I <|iiivnUMly, w# niiv wv ihui 1

= -tan" — 1

(10.14)

a • Re A,

Since the natural response is („ = Aie' " it is clear that after a short time i„ -* 0, and the current settles down to its ac steady state value, given by (10.12). The method we have used is straightforward and conventional but, as the reader might agree, is rather laborious for such a simple problem. For a second-order circuit, the method is more tedious, as was illustrated by the example of (9.34). For very high-order circuits the procedure is, of course, even more complicated. Evidently, we need a better method. One such method is developed in the remainder of this chapter, and its use allows us to treat circuits with storage elements in the same way we treated resistive circuits in Chapters 2, 4, and 5. R

b - 1m A

where H* »ml Im ilanolr ihr rttl /•«"< of and the imaginary part of. I I " • 11111itN i I \K W I I I I I I I .dsn in the polar form.

A

L

A'\A

I

c"

(10.16)

|A|Za

wherr | A \» Ihc magnitude, given by | A | = Va

2

+ b

2

ami II is tin- angle or argument, given by a = tan

I IK-SC

,

b

a relations between rectangular and polar forms are illustrated in Fig. 10.5.

EXERCISES 10.2.1

10.2.2

Im

Find the forced response i, in Fig. 10.4 if L « 60 mH, R = 8 left, Vm = 4 V, and to = 100,000 rad/s. Answer 0.4 cos (lOO.OOOf - 36.9°) mA Find the forced component of v. Answer (RIjVl + \xi R C ) cos (tot - tan"' toRC) V 2

2

A

=a + /A

2

F I G U R E 10.5

I XAMPLE 10.3

G e o m e t r i c a l representation of a c o m p l e x n u m b e r A

Suppose that we have A = 4 + ;'3. Then | A | = V 4 + 3 = 5 and a = tan" 5 = 36.9°. Therefore, the polar form is 2

2

1

EXERCISE

10.2.2

A = 5/36.9° I XAMPLE 10.4

10.3

Consider A = -5 - jl2. Since both a and b are negative, the line segment representing A lies in the third quadrant, as shown in Fig. 10.6, from which we see that

AN ALTERNATIVE M E T H O D USING COMPLEX NUMBERS

314

V 5 + 12 = 13 2

and

The alternative method of analyzing circuits with sinusoidal excitations, which we consider in the remainder of the chapter, relies heavily on the concept of complex numbers. The reader who is unfamiliar with complex numbers, or who needs to review the subject, should consult Appendices C and D, where complex numbers and

Thus we have A = 13/247.4°.

C h a p t e r 10

S e c t i o n 10.3

S i n u s o i d a l Excitation a n d P h a s o r s

2

12 a = 180° + tan"' — = 247.4°

A n Alternative M e t h o d U s i n g C o m p l e x N u m b e r s

315

To solve (Ins equation, wc try ii = Ae which, substituted into (10.20), yields (j(oL + R)Ae'°" = V e'" m

from which A =

V«

~—-rrr R + j(oL

-

12

s

* V/? + (o L 2

2

7

using complex number division. Therefore, we have

A • -5-/12

'

FIGURE 10.6 Complex number with negative real and imaginary parts

VR

2

+

uL 2

2

Let us now observe that

Other useful results are

Re i , = Re

; = 1/90°

; 2

j = -1 = 1/180" 2

and so on. By Euler's formula, which is discussed in Appendix D and which we have used in Chapter 9, we know that

=

2

2

2

2

/ , (oL\ cos I cut - tan — \

t>=Rei,

Therefore, we may say that m

2

which, by (10.12), is the correct forced response of Fig. 10.4. That is,

M

V„ cos tot = Re(V e**)

= g

V„ = V/? + w L

V cos (ot + jV„ sin (ot = V„e m

^-"--'^/») + a> L J

V m

IVR

(10.17)

and V sin tot = lm(V„e*")

(10.21)

We have established for this example the interesting result that if i, is the complex response to the complex forcing function v<, then i = Re i i is the response to v = Re v,. That is, v, yields i , and Re v, = v yields Re i'i = i>. The reason for this is that the describing equation (10.20) contains only real coefficients. Thus, from (10.20), we have f

g

t

m

Rei^Z,^ + Ri^j = R

Returning to the RL circuit example of Fig. 10.4, we know that exponentials are mathematically easier to handle as excitations than sinusoids. Therefore, let us see what happens if we apply the complex excitation oi = V e""

(10.18)

v, = V cos mt = Re u ,

(10.19)

m

instead of the real excitation m

We cannot duplicate such a complex excitation in the laboratory, but there is no reason we cannot consider it abstractly. In this case the forced component of the current, which we call i'i, satisfies L-j

316

C h a p t e r 10

+ Ri, = c i = V e*"

S i n u s o i d a l Excitation a n d P h a s o r s

m

(10.20)

Z,4( at e I'I) + A ( R

R e

ii) " « v

c o s

<»'

and therefore, by (10.11), i = if = Re i ,

(10.22)

Thus we see that it is easier to use the complex forcing function v, to find the complex response i i . Then since the real forcing function is Re « i , the real response is Re I'I. This principle holds for all our circuit analyses, since the describing equations are linear with real coefficients, as may be seen in a development analogous to that leading to (10.22). S e c t i o n 10.3

An Alternative M e t h o d Using C o m p l e x Number s

317

in.I ii I I . I ii

EXERCISES 10.3.1

Replace the real forcing function /„ cos tot in Exercise 10.2.2 by the complex forcing function I e'"", find the resulting complex response t)i, and show that the real response is v = Re v,. Show that, for a real,

i = Re i ,

m

10.3.2

Re

-dt) d,^

a

10.3.3

iipli » irsponse i . . as shown in Fig. 10.8. Then we know from the

n Mill'. .. I ili. |M.. rding section thai the real response of Fig. 10.7 is

=a

I M| li i. .ii I I i "

qumft of the fact that the coefficients in the describing equation are .1 "in previously.

x)

and use this result to establish (10.22). (Suggestion: Let x = f + jg, where/and g are real.) In Exercise 10.2.2, replace the current source by ii = I e'"' A and show that the response t)i has the property that Re v, = v, where v is the original response.

,.-v «><"" m

m

FIGURE 10.8

G e n e r a l circuit with a c o m p l e x excitation

Ihc de=:ribing equation may be solved for the forced response by the method •I i h.ipter 9. That is, since we may write the excitation as

10.4 COMPLEX

(10.26)

v, = V e S»

EXCITATIONS

m

which is a constant times e "', then the trial solution is

Let us now generalize the results using complex excitation functions in the preceding section. The excitation, as well as the forced response, may be a sinusoidal voltage or current. However, to be specific, let us consider the input to be a voltage source and the output to be a current through some element. The other cases may be considered in an analogous way. In general, we know that if v = V cos (ait + 6) s

J

i'i = Ae*

Im cos (tot + ) = Re[Ae'""] which requires that A = Ime * 1

and hence

the forced response is of the form i = Im cos (ait +
(10.24)

as indicated in the general circuit of Fig. 10.7. Therefore, if by some means we can find /„ and , we have our answer, since to, 6, and V are known. To solve for i in Fig. 10.7, let us apply the complex excitation m

V e' m

i, = Ime^e"-

(10.28)

Taking the real part, we have the solution (10.24), of course. IXAMI'll 10.5

Let us find the forced response 1/of ^ + 2 ^ + 8i = 12V2 cos (2t + 15°) dt dt

(10.25)

(M+e)

-

Comparing (10.24) and (10.26), we must have

(10.23)

m

(10.27)

J$

First we replace the real excitation by the complex excitation, FIGURE 10.7

v * V g

m

cos (ut + 0)

G e n e r a l circuit w i t h input a n d output

f^M

I * / „ cos (wr +

l>, = 12V2

)

^ dt

318

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

e ^

m

where for convenience the phase is written in degrees. (This is, of course, an inconsistent mathematical expression, but as long as we interpret it correctly it should present no difficulty.) The complex response U satisfies

S e c t i o n 10.4

2

+ 2 ^ + Si, = 12V2 dt

C o m p l e x Excitations

e**

+m

319

and it must have the general form i, = Ae* Therefore, we must have

PMASORS,

(-4 + j4 + S)Ae ' = 12V2 e»'<•»""

The results obtained in the preceding section may he put in much more compact form by the use of quantities called phasors, which we shall introduce in this see tion. The phasor method of analyzing circuits is credited generally to Charles Proteus Steinmetz (1865-1923), a famous electrical engineer with the General Electric Company in the early part of this century. To begin, let us recall the general sinusoidal voltage,

J2

. l2V2e>> A = . . 4 + j4

r

=

12V2/15

0

= 3/-30°

which gives i'i = ( 3 / - 3 0 ° ) g ^

v = V cos (wt + B)

(10.29)

m

Thus the real answer is if = Re i, = 3 cos (2» - 30°)

which, of course, is the source voltage v of the preceding section. If the frequency to is known, then v is completely specified by its amplitude V and its phase 6. These quantities are displayed in a related complex number, g

m

EXERCISES 10.4.1

(a) From the time-domain equations find the forced response v if v, = 10e^' V (b) Using the result in (a), find the forced response v if v. = 10 cos 8t V Answer (a) 2e '~" '"> V; (b) 2 cos (8/ - 53 1°) V

V = Ve m

je

= V [6

(10.30)

m

M

which is defined as a phasor, or a phasor representation. To distinguish them from other complex numbers, phasors are printed in boldface type, as indicated. The motivation for the phasor definition may be seen from the equivalence, by Euler's formula, of

ion

J

Wv-

V cos (a>t + d) = Re(V e>"e>"") m

EXERCISE 10.4.1

10.4.2

Find the forced response v in Exercise 10.4.1 if sin 8t = Im e'"'.) Answer 2 sin (8/ - 53.1°) V

10.4.3

Using the method of complex excitation, find the forced response i if v = 20 cos 2t V. Answer 2 cos (2t + 36.9°) A J-m.

1 H

80

(10.31)

Therefore, in view of (10.29) and (10.30), we have

= 10 sin 8/ V. (Suggestion

v = Re(Ve*") LE 10.6

(10.32)

Suppose that we have

8

I

m

v = 10 cos (4t + 30°) V The phasor representation is then

VW-

V = 10/30° V since V = 10 and 6 = 30°. Conversely, since oi = 4 rad/s is assumed to be known, v is readily obtained from V. m

EXERCISE 10.4.3

10.4.4

320

Repeat Exercise 10.4.3 if v = 16 cos 4t V. Answer 2 cos 4t A

In an identical fashion we define the phasor representation of the time-domain current ; = /„ cos (a>t + 4>)

g

to be I = l e>* = l j± m

C h a p t e r 10

(10.33)

S i n u s o i d a l Excitation a n d P h a s o r s S e c t i o n 10.5

Phasors

m

(10.34) 321

Thus if we know, for example, that to = 6 rad/s and that I = 2/15" A, then we have

I i a solution, we have jwLU

i = 2 cos (6< + 15°) A

+ RU

m

*

M

\e * h

:

Dividing out the factor e ", we have the phasor equation 1

We have chosen to represent sinusoids and their related phasors on the basis of cosine functions using the fact that cos wt = Re(e ). We could have chosen sine functions just as easily, using sin wt = Im (<>*") (see Exercise 10.4.2). Thus if a function such as

jwLl + Rl = V

iM

v = 8 sin (3t + 30°)

I

is given, we may change it to

=

n—;——r R + Jo>L

. - - / -tan ' — Vj? + w L 1 R 2

l

= 8 cos (3t - 60°) ii

Then the phasor representation is

=

j(M-im-'o,L/K)

VR

2

V = 8 / -60"

„

+

Vm

C 0 S

2

It is important to note that if we can go directly from (10.35) to (10.36), there is a vast saving of time and effort. Also, in the process we have converted the differential equation into an algebraic equation, somewhat like those encountered in resistive circuits. Indeed, the only difference is that the numbers here are complex, whereas in resistive circuits they were real. With the hand calculator as commonly available as it is today, even the complexity of the numbers presents little difficulty. In the remainder of the chapter we shall see how to bypass all the steps between (10.35) and (10.36) by studying the phasor relationships of the circuit elements and considering Kirchhoff's laws as they pertain to phasors. Indeed, as we shall see, we may go directly from the circuit to (10.36), bypassing even the step of writing down the differential equation. In general, the real solutions are time-domain functions, and their phasors are frequency-domain functions (i.e., they are functions of the frequency a>). This is illustrated by the phasor I of Example 10.7. Thus to solve the time-domain problems, we may convert to phasors and solve the corresponding frequency-domain problems, which are generally much easier. Finally, we convert back to the time domain by finding the time function from its phasor representation.

To see how the use of phasors can greatly shorten the work, let us reconsider Fig. 10.4 and its describing equation (10.11), rewritten as dt

+ wV

Taking the real part, we have i = i>, obtained earlier in (10.12).

Had we chosen to base the phasors on sine functions, then we would keep v as is and write its phasor as 8/30°, which, of course, would represent 8 sin (3t + 30°) in the time domain. An illustration using sine-based phasors is given in Example

, <*

2

Substituting this value into the expression for i i , we have

c = 8 cos (3» + 30° - 90°)

EXAMPLE 10.7

(10.36)

Therefore,

°"

(-) 10 35

Following our method, we replace the excitation V„ cos wt by the complex forcing function d = V e* m

which may be written v, = Ve* since 6 = 0, and therefore V = V /0 = V„. Substituting this value and j = i , into (10.35), we have m

+ Ri, = Ve* at whose solution i , is related to the real solution i by i = Re ii Next, trying

EXERCISES 11,1.1

Find the phasor representation of (a) 4 cos (It + 45°), (b) 8 cos 2/ + 15 sin 2f, and (c) - 2 sin (5/ - 65°). Answer (a) 4 /_45°; (b) 17 / - 6 1 . 9 ° ; (c) 2 [2? Ill S.2 Find the time-domain function represented by the phasors (a) 1 0 / - 1 7 ° , (b) 6 + j$, and (c) —j6. In all cases w = 3. Answer (a) 10 cos (3/ - 17°); (b) 10 cos (3f + 53.1°); (c) 6 cos (3/ - 90°)

ii - le*"

322

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

S e c t i o n 10.5

Phasors

323

10.6 V O L T A G E - C U R R E N T RELATIONSHIPS FOR PHASORS

In this section we show that relationships between phasor voltage and phasor current for resistors, inductors, and capacitors are very similar to Ohm's law for resistors. In fact, the phasor voltage is proportional to the phasor current, as in Ohm's law, with the proportionality factor being a constant or a function of the frequency a>. We begin by considering the voltage-current relation for the resistor, v = Ri

(10.37)

where v = V cos (tot + 0) m

III

(10.38) I = Im cos ((Ot + (b) If we apply the complex voltage V„e ' \e complex current which results is I e °" *\ substituted into (10.37) yields

UNI

III III

\i i i i n e n l w a v e f o r m s for a resistor

J{<J +e

m

A

+

Ve m

Ault+e)

=

V - lOcosOOOt + 30°) V

RI e *^ m

Au>,+

(10.41)

.elm . with the polarity indicated in Fig. 10.9(a). Then the

Dividing out the factor e "' results in J

V„e* = RI e'*

V • 10/30° V

(10.39)

m

which, since V„e and Ime'* are the phasors V and I , respectively, reduces to jl>

V = RI

I lie i>li.i .i »i i MI i t u t IN

V R

(10.40)

Thus the phasor or frequency-domain relation for the resistor is exactly like the timedomain relation. The voltage-current relations for the resistor are illustrated in Fig. 10.9.

I I I . i. I . . i .

m

the

lime

10/30° = 2/30° A 5

domain we have i = 2 cos (lOOt + 30°) A

l in i

(10.42)

<>l . ourse, simply the result we would have obtained using Ohm's law.

In the ease of the inductor, substituting the complex current and voltage into tin- nine-domain relation, . di v = Ldt five! the complex relation

(a) FIGURE 10.9

V o l t a g e - c u r r e n t r e l a t i o n s for a resistor

From (10.39) we have V„ = RI„ and 8 = d>. Thus the sinusoidal voltage and current for a resistor have the same phase angle, in which case they are said to be in phase. This phase relationship is shown in Fig. 10.10, where the voltage is represented by the solid line and the current by the dashed line.

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

=

R in t h e (a) t i m e a n d (b) f r e q u e n c y

domains

324

dt

(b)

ja>LI e* +* m

M

)

Again, dividing out the factor e'"" and identifying the phasors, we obtain the phasor relation V = j(oL\)

S e c t i o n 10.6

V o l t a g e - C u r r e n t Relationships for Phasors

325

Thus the phasor voltage V, as in Ohm's law, is proportional to the phasor current I, with the proportionality factor juL. The voltage-current relations for the inductor are shown in Fig. 10.11.

gives the complex relation l *°***) = C—[V„e' '"' "] dt me

l

+<

= jwCVme*"** Again dividing by e*" and identifying the phasors, we obtain the phasor relation v = L di/dl

V =

fuLl

I = jwC\) or (b)

(a)

FIGURE 10.11 V o l t a g e — c u r r e n t r e l a t i o n s for a n i n d u c t o r L in the (a) t i m e a n d (b) frequency domains

If the current in the inductor is given by the second equation of (10.38), then by (10.43), the phasor voltage is V =

(j(oL)(l [^ m

= b>Ll / + 90"

V =

I jaiC

(10.45)

Thus the phasor voltage V is proportional to the phasor current I , with the proportionality factor given by 1/jaiC. The voltage-current relations for a capacitor in the time and frequency domains are shown in Fig. 10.13.

m

since j = 1/90°. Therefore, in the time domain we have

= ;o)CV

i = C dv/dl

t> = (oLI cos (cat +
o +

Comparing this result with the second equation of (10.38), we see that in the case of an inductor the current lags the voltage by 90°. Another expression that is used is that the current and voltage are 90° out of phase. This is shown graphically in Fig. 10.12. Finally, let us consider the capacitor. Substituting the complex current and voltage into the time-domain relation, -r > dx

dt

Ii

—— 1 1

o

(b)

(a) FIGURE 10.13

—

V o l t a g e - c u r r e n t r e l a t i o n s for a c a p a c i t o r in t h e (a) t i m e a n d (b) f r e q u e n c y

domains

In the general case, if the capacitor voltage is given by the first equation of (10.38), then by (10.44), the phasor current is I =

(j(oC)(V [6) m

= (ocvje + 90° Therefore, in the time domain we have i = o>CV cos (wt + 8 + 90°) m

which, by comparison with the first equation of (10.38), indicates that in the case of a capacitor the current and voltage are out of phase with the current leading the voltage by 90°. This is shown graphically in Fig. 10.14.

326

C h a p t e r 10

S i n u s o i d a l Excitation a n d Phasors S e c t i o n 10.6

V o l t a g e - C u r r e n t R e l a t i o n s h i p s for P h a s o r s

327

o + Phasor circuit

1

FIGURE 10.15

G e n e r a l phasor circuit

We define the ratio of the phasor voltage to the phasor current as the impedance of the circuit, which we denote by Z. That is,

FIGURE 10.14

EXAMPLE 1 0 . 9

Z = v

V o l t a g e a n d c u r r e n t w a v e f o r m s for a c a p a c i t o r

If the voltage of (10.41) is applied across a 1 -/iF capacitor, then by (10.44) the phasor current is

which by (10.46) is Z = | Z l/ft, = f/6

I = y(IOO)(IO")(10/30°) A = 1/120° mA

and therefore the current leads the voltage by 90°.

EXERCISES 10.6.1

Using phasors, find the ac steady-state current i if v = 12 cos (1000/ + 30°) V in (a) Fig. 10.9(a) for R = 4 kft, (b) Fig. 10.11(a) for L = 15 mH, and (c) Fig. 10.13(a) forC = \uF. Answer

10.6.2

10.7

|Z|=y=,

e =

e-4>

7

Impedance, as is seen from (10.47), plays the role, in a general circuit, of resistance in resistive circuits. Indeed, (10.47) looks very much like Ohm's law; also like resistance, impedance is measured in ohms, being a ratio of volts to amperes. It is important to stress that impedance is a complex number, being the ratio of two complex numbers, but it is not a phasor. That is, it has no corresponding sinusoidal time-domain function of any physical meaning, as current and voltage phasors have. The impedance Z is written in polar form in (10.48); in rectangular form it is generally denoted by Z = R + jX

(10.49)

where R = Re Z is the resistive component, or simply resistance, and X = Im Z is the reactive component, or reactance. In general, Z = Z(ja>) is a complex function of jto, but R = R{u>) and X = X(u>) are real functions of u>. Both R and X, like Z, are measured in ohms. Evidently, comparing (10.48) and (10.49) we may write | Z | = VR

2

IMPEDANCE A N D ADMITTANCE

Let us now consider a general circuit of phasor quantities with two accessible terminals, as shown in Fig. 10.15. If the time-domain voltage and current at the terminals are given by (10.38), the phasor quantities at the terminals are

(10.48)

2

(a) 3 cos (1000/ + 30°) mA; (b) 0.8 cos (lOOOr - 60°) A;

(c) 6 cos (1000/ + 120°) mA In Exercise 10.6.1, find /' in each case at / = 2 ms. Answer (a) -2.445 mA; (b) 0.464 A; (c) -3.476 mA

-
where | Z | is the magnitude and 6 the angle of Z. Evidently,

The time-domain current is then / = cos (100/ + 120°) mA

(10.47)

+ X

2

' „ ,x ft, = t a n ' ri

and R = | Z | cos 6

Z

X = | Z | sin 6z These relations are shown graphically in Fig. 10.16.

328

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

S e c t i o n 10.7

Impedance and Admittance

329

Since a /., and C are positive, we see that inductive reactance is positive and I hat capacitive reactance is negative. In the general case of (10.49), we may have X • 0, in which case the circuit appears to be resistive; X > 0, in which case its Fell luce appears to be inductive; and X < 0, in which case its reactance appears to he capacitive. These cases are possible when resistance, inductance, and capacitance are all present in the circuit, as we shall see. As an example, the circuit with ini|)odance given by Z = 4 + y3, which we have just considered, has reactance X 3, which is of the inductive type. In all cases of passive circuits, as we shall see in Chapter 12, the resistance R is nonnegative. The reciprocal of impedance, denoted by

FIGURE 10.16 Graphical representation of impedance EXAMPLE 10.10 Suppose in Fig. ,0.15 that V = ,0/56,9° V and I = 2/20° A. Then we have ™ =

10/56.9° 2/20°

=

5

/

3

6

9

°

f t

In rectangular form this is Z = 5(cos 36.9° + j sin 36.9°) = 4 + /3fl The impedances of resistors, inductors, and capacitors are readily found fro, their V - I relations of (10.40), (10.43), and (10.45). Distinguishing the: impedances with subscripts R, L , and C, respectively, we have, from these equ and (10.47),

is called admittance and is analogous to conductance (the reciprocal of resistance) in icsistive circuits. Evidently, since Z is a complex number, then so is Y, the standard representation being Y = G + jB

(10.55)

The quantities G = Re Y and B = Im Y are called conductance and susceptance, respectively, and are related to the impedance components by

Z* = R Z = ja>L = (oL/90° t

Zc =

(10.50)

1 ju>C

= -W-90° o)C (oC

n rte case of a res.stor, the .mpedance is purely resistive, its reactance being z e r o Impedances of tnductors and capacitors are purely reactive h a v i n g ™ ? ! components. The inductive reactance is denoted by ' " ' 8

X = toL L

e S

S

^

g

^

= I = R-TTX

b

The units of Y, G, and B are all Siemens, since in general Y is the ratio of a current to a voltage phasor. To obtain the relation between components of Y and Z we may rationalize the last member of (10.56), which results in

V (

G + jB

(10.51)

R

1 + jX

so that

=

2

Equating real and imaginary parts results in

jX

L

G = (10.52)

Zc = jX

c

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

R

+ X

R2

B = —

R2

a n d thus

330

- jX - jX

jX + X

and the capacitive reactance is denoted by


R R

-

R R2

Zl

( 1 0 5 6 )

(10.53)

2

X + X

(10.57) 2

Therefore, we note that R and G are not reciprocals except in the purely resistive case (X = 0). Similarly, X and B are never reciprocals, but in the purely reactive case (R = 0) they are negative reciprocals. S e c t i o n 10.7

Impedance a n d Admittance

331

are (he phasor voltages around the loop. Thus KVL holds for phasors A similar development will also establish KC'l In circuits having sinusoidal excitations with a common frequency a>, if we are interested only in the forced, or ac steady-state response, we may find the phasor voltages or currents of every element and use Kirchhoff's laws to complete the anal ysis The ac steady-state analysis is therefore identical to the resistive circuit analysis of Chapters 2, 4, and 5, with impedances replacing resistances and phasors replacing time-domain quantities. Once we have found the phasors, we can convert immediately to the time-domain sinusoidal answers.

EXAMPLE 10.11 If we have Z = 4 + ;3 then v =

1

±

= ~ > _ _ 4 + 3 25 4

4 + ;3

3

2

2

,2 25

7

Therefore, G = ^ and B = - 55. Further examples are Yr = G

I 10.12 Consider the circuit of Fig. 10.17, which consists of N impedances connected in series. By KCL for phasors, the single phasor current I flows in each element. Therefore, the voltages shown across each element are

Jft»L Y = ya>C c

V, = Z , I

which are the admittances of a resistor, with R = 1/G, an inductor, and a capacitor,

V = Z I 2

2

EXERCISES 10.7.1

Find the impedance seen at the terminals of the source in Fig. 10.4 in both rectangu lar and polar form. Answer R + ja>L, VR + a> Z. /tan~'

10.7.2

Answer

10.7.3

—

2

—- — / —

—-

—

—

/ — tan

(a) 0.06,

0.08;

(b) 3.2,

-2.4;

(c) - 4 ,

and by KVL around the circuit, V = V, + v + • • • + 2

\

= (Z, + Z + • • • + Z„)I

—

2

Find the conductance and susceptance if Z is (a) 6 - jS, (b) 0.2 + y0.15, and i< 1 V2 Answer

V« = Z„I

2

Since we must also have, from Fig. 10.17, V = Z.,1 where Z*, is the equivalent impedance seen at the terminals, it follows that

-4

(10.58)

Z ^ . = Z, + Z + • • • + Z

10.8

2

KIRCHHOFF'S LAWS A N D IMPEDANCE COMBINATIONS

as in the case of series resistors.

Kirchhoff's laws hold for phasors as well as for their corresponding time-domain voltages or currents. We may see this by observing that if a complex excitation, SM V e "'*"\s applied to a circuit, then complex voltages, such as V c' """i\ V e''""* \, appear across the elements in the circuit. Since Kirchhoff's laws hold in the time domain, KVL applied around a typical loop results in an equation such as m 1

A

N

]

FIGURE 10.17

(

I m p e d a n c e s c o n n e c t e d in s e r i e s

l>1

Vie'*"'** ' + Vze*" *' ' + • • • + Vse*"**"^ = 0 1

1

2

—I

V

Dividing out the common factor e ", we have V, + v + • • • + v „ = 0 where V„ = V„/ft,, n = 1, 2, . . . , N Ja

2

332

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

S e c t i o n 10.8

Kirchhoff's Laws and Impedance Combinations

333

An alternative method of solution is to observe that the impedance Z seen at KM source terminals is the impedance of the inductor, jwL, and the res.stor, R, connected in series. Therefore, Z = joL + R and v_ y»ZQ! Z R + je>L as obtained earlier.

Similarly, as was the case for parallel conductances in Chapter 2, the eqiiivn lent admittance Y«, of N parallel admittances is (10.59)

Y„, = Y, + Y + • • • + Y/v 2

In the case of two parallel elements (N = 2), we have 1 . 1 Y„ Y, + Y

Z,Z Z, + Z, 2

2

(10.60)

IXIK(

In like manner, voltage and current division rules hold for phasor circuiti, with impedances and frequency-domain quantities, in exactly the same way that they held for resistive circuits, with resistances and time-domain quantities. The reader i i asked to establish these rules in Exercise 10.8.2.

ISES

III M.I 10.8.2

Derive (10.59). Show in (a) that the voltage division rule, z, + z

2

and in (b) that the current division rule,

EXAMPLE 10.13 Let us return to the RL circuit considered in Sec. 10.2. The circuit and its phasor counterpart are shown in Fig. 10.18(a) and (b), respectively. By KVL in the phasor circuit we have

I

Z J + RI = V„/0°

Y 1« — r, . rw Y, + Y Z, + Z 2

*<

2 8

are valid, where Zi = 1/Yi and Z = 1/Y . 2

2

or (jtoL + R)l = V /0° m

from which the phasor current is R + ja>L =

, VR

2

+ a> L / 2

/ -tan

- 1

2

— *

Therefore, in the time domain we have, as before,

VR FIGURE 10.18

10.8.3

= = = = = cos wt - tan + a> L \ 2

(b)

(a)

— I

EXERCISE 10.8.2

Find the steady-statt current i using phasors. Answer A cos (4f - 36.9°) A 2fl

—wv

2

(a) T i m e - d o m a i n c i r c u i t ;

(b)

equivalent phasor circuit

10 cos

TW^-

41 V I

EXERCISE 10.8.3

10.8.4

Find the steady-state voltage v in Exercise 10.8.3 using phasors and voltage diviAnswer 2 cos (4t - 126.9°) V

(a)

334

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

(b) S e c t i o n 10.8

Kirchhoff's Laws and Impedance Combinations

335

10.9

Therefore, we have

PHASOR CIRCUITS

Z

= i | ( + 3

m-jD

3 + 73 - ;3

5/-36.9°

1

and by current division, j

=

( J ) I, = V 2 781.9° A \ + ji - jv — 3

+

3

In the time domain, the answer is i = V2cos(3r + 81.9°) A In the case of a dependent source, such as a source kv volts controlled by a voltage v, it will appear in the phasor circuit as a source * V „ where V, is the phasor representation of »„ because t>, = V„ cos (tot + hen a complex excitation is applied. Then dividing e " out of the equations leaves v, represented by its phasor V„e*. In the same way, kv. Jtv„ cos (wt +
me

EXAMPLE 10.14 Let us find the steady-state current i in Fig. 10.19(a). The phasor circuit, shown in Fig. 10.19(b), is obtained by replacing the voltage source and the currents by their phasors and labeling the elements with their impedances. In the phasor circuit the impedance seen from the source terminals is

4 - ;3

1

As the discussion in the preceding section suggests, we may omit the steps of finduiji the describing equation in the time domain, replacing the excitations and response* by their complex forcing functions and then dividing the equation through by <•"•' M obtain the phasor equation. We may simply start with the phasor circuit, which we will now formally define as the time-domain circuit with the voltages and currentl replaced by their phasors and the elements identified by their impedances, as illus trated previously in Fig. 10.18(b). The describing equation obtained from this circuit is then the phasor equation. Solving this equation yields the phasor of the answci which then may be converted to the time-domain answer. The procedure from starting with the phasor circuit to obtaining the phasor answer is identical to that used earlier in resistive circuits. The only difference is that the numbers are complex.

w

1

m

I I 10 15 As an example of a circuit containing a dependent source, let us consider Fig. 10 20(a), in which it is required to find the steady-state value of i. The corresponding phasor circuit is shown in Fig. 10.20(b). Since phasor circuits are analyzed exactly like resistive circuits, we may apply KCL at node a in Fig. 10.20(b), resulting

= 4 - j3 ft FIGURE 10.T9 RLC

t i m e - d o m a i n a n d phasor circuits

I +

v,-iv, •_* = 3/0!

(10.61)

By Ohm's law we have V , = 41, which substituted into (10.61) yields -j2l

5 cos

I =

+ -(41) =

-J6

-J6 2 - j2

2V2/-45

V r

(a) FIGURE 10.20

5/_0° V

(a) C i r c u i t c o n t a i n i n g a d e p e n d e n t s o u r c e ; (b) c o r r e s p o n d i n g p h a s o r c i r c u i t

~ -13 n 4r A

(b)

336

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

S e c t i o n 10.9

Phasor C i r c u i t s

337

Therefore, we have i I

=

V2

1

nnnp

cos (4/ - 45°) A 2 11

EXAMPLE 10.16 Let us find the steady-state current i, in Example 10.14 if the source voltuga is v, = 5 sin 3f V. Since v, = 5 cos (3» - 90°) V, the phasor voltage is V, • 5 / - 9 0 ° V, and, as before, the impedance seen by the source is 5 / - 3 6 . 9 ° ft Thui we have 1/-53.T A

'

4 11

i

EXERCISE 10.9.2

10 t l

and

Find the steady-state voltage v in Exercise 10.4.1(b) using the phasor circuit. Find the steady-state voltage t; using the phasor circuit, given that v, = 4 cos 10» V. Answer V2COS (10/ + 135°) V

i'i = cos (3f - 53.1°) = sin (3r + 36.9°) A 0 05 F

If we had based the phasors on the sine instead of the cosine, we would save the steps of first converting the sine to the cosine and then converting the cosine back in the sine, or equivalently, of first subtracting 90° and then adding 90° to the phase Based on the sine, V , = 5/_0° V, and

I l

=

V ^ 9 °

=

1

^

-H(2

n

A

or t, = sin (3f + 36.9°) A

EXERCISE 10.9.4

In the case of an op amp, the phasor circuit is the same as the time-domain circuit. That is, an ideal op amp in the time-domain circuit appears as an ideal op amp in the phasor circuit, because the time-domain equations i = 0,

v =0

which characterize the current into and the voltage across the input terminals retain the identical form,

Solve Exercise 10.2.2 by means of the phasor circuit. Find the steady-state voltage v using the phasor circuit. Answer 8 cos (8t - 53.1°) V

C h a p t e r 10

S e c t i o n 10.10

V = 0

in the phasor equations. As a final note we observe that the phasor method of finding, say i , by first finding I = V / Z and then converting I to i, fails to work if Z( ju>) = 0. This is the case when the circuit is excited at a natural frequency ja>, and we must then use the method of Sec. 9.6. An example of this case is considered in Prob. 10.38.

EXERCISES

338

SUMMARY

In this chapter we have considered in some detail the sinusoidal function, the dominant signal in the electrical power industry. We have defined its amplitude, phase, and frequency, and have considered its phasor representation. We have seen that steady-state sinusoidal responses may be obtained using phasors, a shortcut method based on complex excitations. The phasor voltage-current relations for resistors, inductors, and capacitors are identical in form to Ohm's law, with impedance, the ratio of the element's phasor voltage to its phasor current, playing the role of resistance. The reciprocal of impedance is admittance, the analogy to conductance. A phasor circuit may be obtained by replacing the voltages and currents in the given time-domain circuit by their phasors and labeling each passive element with its impedance. Since KVL and KCL hold for phasors, and impedances behave like resistances, the phasor circuit is solved exactly like a resistive circuit. The phasor solutions are then converted to their time-domain counterparts, completing the analysis.

1 = 0,

10.9.1 10.9.2

10.10

S i n u s o i d a l Excitation a n d P h a s o r s

Summary

339

JO milt, find the time-domain tunc waaanlrrl by the phasors (a) -5 + y'5, A . ( c ) 5 - ;12, (d) 10. and (e)

PROBLEMS 10.1

Given the voltage v = 50 cos (200irt + 60°) V, find (a) its amplitude, (b) its phase angle in degrees, (c) its phase angle in radians, (d) its period in ms, (e) its frequency in rad/s, (f) its frequency in Hz, and (g) by how many degrees it leads or lags the current i = 2 cos (20On-/ - 17°) A.

Convert the following functions to cosine functions with positive amplitudes: (a) 6 sin (2t + 15°), (b) - 2 cos (4t + 10°), (c) 8 cos St - 15 sin 5t. 10.3 Determine if v, leads or lags v and by what amount: (a) v, = 5 cos (4/ - 60°), v = 5 sin 4/, (b) d, = 10 cos 4/, Vz = 5 cos 4/ + 12 sin 4t, (c) P, = 10 (cos 4/ + V3 sin 4/), V2 = 4 cos 4/ + 3 sin 4/. 10.4 Find z' , using only the properties of sinusoids, if (a) ii = 6 cos 3/ A, i = 4 cos (3/ - 30°) A, and i, = - 4 V 3 cos (3r + 60°) A, (b) ii = 5 cos (3/ + 30°) A, i = 5 sin 3/ A, and h = 5 cos (it + 150°) A, and (c) 1, = 25 cos (3t - 53.1°) A, i , = 2 sin 3/ A, and h = 13 cos (3/ - 22.6°) A. (Hint: cos 22.6°

A voltage V„ sin cut V, a resistor R . and an 1 ductor L are all in scries. Show that the for response < is identical to (10.12) except t the cosine is replaced by the sine in the folio ing two ways, (a) Noting that V, sin cor V cos (tot — 90°). use the complex exci tion V e ~'~ > and take the real part of current response, (b) Noting that V„ sin wt lm(V e'™), use the complex excitation V„ and take the imaginary part of the current sponse.

10.8

m

m

10.2

2

10.9

Find t, from the differential equation and the result to find the forced response v to input voltage of (a) 34 cos 4/ V, and 17 sin 4/ V.

1 i) r

4

2

2

n )

9r

m

2

-

l<

20

z

10

0.02 F ;

PROBLEM 10.9

10.10 Find the response v, to the source 4e^ A and use the result to find the response 0 to (a) 4 cos 8/ A, and (b) 4 sin 8/ A.

10.6

10.7

340

PROBLEM 10.10

10.11 A complex voltage input 10>' ' ' V produces a current output of Se '~' A. Find the output current if the input voltage is (a)

In the figure of Exercise 10.2.2, if the source is 4 cos 4000/ mA and the output is t> = 24 cos (4000/ - 53.1°) V, find R and C.

10.12 Find the phasor representations of the timedomain functions (a) 10 cos (5/ + 18°), (b) - 8 cos 5/ + 6 sin 5/, (c) 18 sin 5/, and (d| - 2 sin (5/ - 10°).

C h a p t e r 10

_ 5(1 4 » ( 3 + j*>) ja>(2 + jo>)

m

ja,+«r)

40e

y

(

b

)

20 cos

2/

V,

and

S i n u s o i d a l Excitation a n d P h a s o r s

10.23 Find the steady-state values of i and c.

2

2

Find the resistance and reactance at at = 1,2, and 3 rad/s. If the time-domain voltage applied to the circuit is 64 cos cur V,findthe Meady-state current in each case. It In Prob. 10.9, use phasors, impedance, and milage division tofindv. Find the steady-state value of i if (a) cu = Irad/sand (b) cu = 2 rad/s. Note that in the latter case the impedance seen at the terminals <>l the source is purely resistive.

xn

(c)

4 sin (It - 15°) V.

PROBLEM 10.22

16(2 + » ( 8 - w - j2u>) oi* - \5a> + 64

<2 +25 )

l0

10.22 Find the reactance X so that the impedance seen by the source is real. For this case, find the steady-state current i(t) corresponding to I if cu = 10 rad/s.

a

I MNI the resistance, reactance, conductance, •ml susceptance at o> = 1 rad/s. If the iinir domain voltage applied to the circuit is 30 cot / V, find the steady-state current. A I ircuit has an impedance

4-

(1 <*

PROBLEM 10.21

0

m

a

Find the forced response 1 in Fig. 10.4 if L = 4 mH, R = 6 kfl, V = 5 V, and cu = 2 x 10* rad/s. Find R and L in Fig. 10.4 if the source is 13 cos 60O0t V and the response is i = 2 cos (6000/ - 67.4°) mA. (Take tan 67.4° = m

1

V\

PROBLEM 10.4

10.5

i tolvr I'n* 10 4 using phasors I IM. 1 I I K impedance of the circuit of Fig 10.15 If the time-domain functions repreMMed by the phasors V and I are -30 cos 2/ + 16 sin It V. ~ ' i - 1.7 cot (2/ + 20°) A. (fc) V - Re| je* ] V, I - Re|0 + ^V ***"] mA. H 1 1 •• aV cos (wt + B) V, I - V. cos («/ + 8 - a) A. • M l that if Re Z = R is positive, then Re Y - Re|l/Z] = G is also positive. A circuit has an impedance

10.21 For the phasor circuit shown, find Z^ and use the result to find the phasor current I. If cu = 7 rad/s, find the forced response i corresponding to I.

Mi

K » 1*

V

PROBLEM 10.20

Problems

2uF;±; v

4 cos 2500r V

PROBLEM 10.23

10.24 Find C so that the impedance seen by the source is real. Find the power absorbed by the 12-11 resistor in this case. 10.25 Find the steady-state current i. 10.26 Find the steady-state voltage v. 10.27 Find the steady-state value of I. 10.28 Find the steady-state voltage e. 10.29 Find the steady-state currents i and Ii using phasors. 10.30 Find the steady-state voltage t> if v, = 2.5 cos 8/ V. 10.31 Find the steady-state value of v.

341

16cos8; V I

1 hr

so;

PROBLEM 10.24

PROBLEM 10.29

PROBLEM 10.30

> +

10 cos 10/ V

cos 30,000r mA ^

0.05 F


Ml

;0.01/iF

< I kn

3l00mH

PROBLEM 10.25 PROBLEM 10.31

Find the steady-state values of t> and ci. 11 Find the steady-state value of i when (a) at = I rad/s, (b) at = 2 rad/s, and (c) at = 4 rad/s. | Note that (b) is the resonant case] 19.M Find the steady-state voltage v.

IIM2

•

PROBLEM 10.26

5

K

"

IS

Ml

10.35 Find the steady-state current i. 10.36 Find the steady-state voltage v if 10 cos lOOOf V. 10.37'Find the steady-state current i if 2 cos 2000 V.

-vw-

I

1 il 10 cos 40,000r V

3 11

PROBLEM 10.27

PROBLEM 10.32

5 cos 2 0 0 0 ; V

PROBLEM 10.28

342

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

PROBLEM 10.33 Problems

11

AC Steady-State Analysis

PROBLEM 10.34

14 cos 3/

v

( ^ )

PROBLEM 10.35

Samuel f. a. Morse 1791-1S7!

4 kil

PROBLEM 10.36 10.38

Find the forced response i . [Suggestion: Note that I = V / Z fails to work since Z ( j l ) = 0. Solve the describing equation by the method of Sec. 9.6.]

PROBLEM 10.37 10.39

I H

20 cos 3/ V (

-^rJ0T>2 cos / V

( _

*

PROBLEM 10.39 10.40

PROBLEM 10.38

344

Find the complete response i if ;(0) = 2 A and v (0) = 6 V. (Suggestion: Use phasors to get i) and the differential equation to get i„.)

C h a p t e r 10

S i n u s o i d a l Excitation a n d P h a s o r s

Design. But the previous year his I im lirst practical application of elec- What hath God wrought! wife had died, in 1826 his father died, ti II ity is said by many to be the (The famous message and in 1828 his mother died. The foltelegraph, developed by Samuel F. tapped out on the first lowing year the distressed Morse • Morse, an American portrait telegraph] painter and inventor. Morse built on Samuel F. B. Morse went to Europe to recover and study further. In 1832, while returning home the ideas of the famous American on board the passenger ship Sully, physicist Joseph Henry, using the he met an eccentric inventor and beopening and closing of relays to produce the dots and dashes (or Morse code) that repre- came intrigued with developing a telegraph, the principle of which had already been considered by Henry. nent letters and numbers. Morse was born in Charlestown, Massachu- By 1836 Morse had a working model, and in 1837 he setts, the son of a minister and author. He studied to acquired a partner, Alfred Vail, who financed the probe an artist at Yale and the Royal Academy of Arts in ject. Their efforts were rewarded with a patent and the i "iidon, and by 1815 he was considered to be moder- financing by Congress of a telegraph in 1844, over ately successful. In 1826 he helped found and be- which Morse—on May 24, 1844—sent his nowcame the first president of the National Academy of famous message, "What hath God wrought!" •

Determine i(0) and t>(0) in Prob. 10.39 so that the natural component vanishes and i is simply the forced component.

345

' ODD

FIGURE 11.1

Circuit to be analyzed by the phasor method

it a

I n Chapter 10 we have seen that in the case of circuits with sinusoidal inputs we may obtain the ac steady-state response by analyzing the corresponding phasor circuits. The circuits encountered in most cases were relatively simple ones that could be analyzed by the use of voltage-current relations and current and voltage division rules. It should be clear, because of the close kinship between phasor circuits and resistive circuits, that we may extend the methods of Chapter 10 to more general circuits using nodal analysis, loop analysis, Thevenin's and Norton's theorems, superposition, and so on. In this chapter we formally consider these more general analysis procedures, limiting ourselves to obtaining the forced, or ac steady-state response.

§

1

15^0° A

n<

11.1 NODAL

A N ALYSIS

As we have seen, the voltage-current relation

-/1 n

V = ZI

for passive elements is identical in form to Ohm's law, and KVL and KCL hold in phasor circuits exactly as they did in resistive circuits. Therefore, the only difference in analyzing phasor circuits and resistive circuits is that the excitations and responses are complex quantities in the former case and real quantities in the latter case. Thus we may analyze phasor circuits in exactly the same manner in which we analyzed resistive circuits. Specifically, nodal and mesh, or loop, analysis methods apply. We illustrate nodal analysis in this section and loop analysis in the following section. EX A M P LE 11.1

To illustrate the nodal method, let us find the ac steady-state voltages t>i and 02 of Fig. 11.1. First we obtain the phasor circuit by replacing the element values by their impedances for a> = 2 rad/s and the sources and node voltages by their phasors. This results in the circuit of Fig. 11.2(a). Since we are interested in finding V, and V 2 , the node voltage phasors, we may replace the two sets of parallel impedances by their equivalent impedances, resulting in the simpler, equivalent circuit of Fig. 11.2(b).

The nodal equations, from Fig. 11.2(b), are 2(V, - 5/0°) +

V -/I

y i ^ L + - j \+/2)/5

346

Chapter 11

AC Steady-State Analysis

V — V

+ —

~

- / l

v*

= 0 =

ll f i

-/1 n

5/0° V

a

05/0° A

(b)

FIGURE 11.2

Two versions of the phasor circuit corresponding to Fig.

11.1

which in simplified form are (2 + /2)V, - /'IV, = 10 - / I V , + (1 - /'DV2 = 5 Solving these equations by determinants, we have 5 2 +10 j2 - / ' I / 1 - 5/ I 1 -1 --jl / i 2 + j2 10

-n

5/0°

^ Section 11.1

Nodal Analysis

5

5

10 + /'20 5

347

Eliminating

In polar form these quantities are V, = V 5 / - 2 6 . 6 0 V V = 2V5/63.4" V Therefore, the time-domain solutions are c, = V 5 cos (2t - 26.6°) V v2 = 2 V I cos (2/ + 63.4°) V 2

V

between these two equations and solving lor I , wc h»vc I = 24 x 10 3/53.1° A = 24/53.1° mA

Therefore, in the time domain, we have

i = 24 cos (5000t + 53.1°) mA | u | » | I I 1.3 Let us tind the forced response v in Fig. 11.5 if v, = Vm cos u>t V We note first that the op amp and the two 2-kO resistors conrtitotejVCVSwhfc gain J + = 2 (see Sec. 3.4). Therefore, v = 2t>2, or v2 - v/2, as indicated by the phasor V / 2 in the phasor circuit of Fig. 11.6. AM

EX A M P LE

11. 2

*

As an example involving a dependent source, let us consider Fig. 11.3, in which it is required to find the forced response /'. Taking the ground node as shown, we have the two unknown node voltages v and v + 3000i, as indicated. The phasor circuit in its simplest form is shown in Fig. 11.4, from which we may observe that only one nodal equation is needed. Writing KCL at the generalized node, shown dashed, we have V - 4 V V + 30001 i(10 3 ) | ( 1 - >2)(10 3 ) (2 - 7 1)(10 3 ) Also, from the phasor circuit we have 4 ^

FIGURE 11.5

Hio )

o—AV

3

FIGURE 11.3

•

Circuit containing an op amp

AAA—

Circuit containing a dependent source

2

1

•A / vV

4 cos 5000r V

-jiF

-r-

3000/ V

FIGURE 11.6

Phasor circuit of Fig.

11.5

-/IOOO/w kfi

FIGURE 11.4

Phasor circuit of Fig.

11.3 l/VIkfi

0 0 2

/

-30001

v

s/l

kn

v/2

^

-o

-AAA-

4/iO

348

V

30001 / looo / u k n - p

|(2- / l)kJi

Section 11.1

Nodal Analysis

v

2kS2 W \ AA-

349

Writing nodal equations at the nodes labeled V, and V/2, we have v - y„Zo:

v, - (v/2)

M

• V i + wiooo)

where

6 = -tan

It V

v, - v

= 0 V2(10 3 ) V/2 (V/2) - V, V2(10 3 ) -/10 6 /a> = 0 Eliminating V, and solving for V results in 2V m [1 - (&»2/106)] + 7(V2tt)/10 3 ) In polar form this is y 2 V ^ (1/V2K10- 1 )

10 cos

Mil .'' A

EXERCISE 11.1.2 11.1.3

11.1.4

(11.1)

Find the amplitude of v in (11.3) if Vm = 10 V for (a) *» r 0, (b> a> - 1000 rad/s, (c) a) = 10,000 rad/s, and (d) to = 100,000 rad/s. Answer (a) 20; (b) 14.14; (c) 0.2; (d) 0.002 V Find the steady-state value of 0 using nodal analysis. Answer \SV2 cos (At - 135°) V

4

V2o)/1000 Wiooo)

2

p

(11.2)

In the time domain we have 2V„

V i + wiooo) cos (cot 4

+ 0)

(11.3)

5 •

11

W v

;

If 60

ion • >

EXERCISE 11.1.4

We might note in Example 11.3 that for low frequencies, say 0 < to < 1000, the amplitude of the output voltage v is relatively large, and for higher frequencies, its amplitude is relatively small. Thus the circuit of Fig. 11.5 filters out higher frequencies and allows lower frequencies to "pass." Such a circuit is called a filter and is considered in more detail in Chapter 15.

f i

| J 2 sin 41 f

11.2 M ESH

EXERCISES

W

A N ALYSIS

I X A M P L E 11.4

11.1.1 Find the forced response v using nodal analysis. Answer 10 sin 3f V

To illustrate mesh analysis of an ac steady-state circuit, let us find v, in Fig. 11.1, which was obtained, using nodal analysis, in the preceding section. We shall use the phasor circuit of Fig. 11.2(b), which is redrawn in Fig. 11.7, with mesh currents I and I , as indicated. Evidently, the phasor voltage V may be obtained as 2

V, = 5 - | 10 cos

3i V

The two mesh equations are 2

EXERCISE 11.1.1

11.1.2 Find the steady-state value of v using nodal analysis. Answer 2 5 V l cos (2r - 81.9°) V 350

Chapter 11

AC Steady-State Analysis

(11.4)

_

J Section 11.2

y l ( l 2

Mesh Analysis

-

I l )

-

; i l 2 +

Ii "

j

(i^)

HI, "

( l 2 +

h) = 5 5) =

o 351

sin It

A

I

FIGURE 11.7

(a)

Solving these equations for I i , we have I, =

6 + j2

A

1 / -90" = - ; l A

s

which substituted into (11.4) yields V, « 2 -

/IV

This is the same result that was obtained in the preceding section and may be used to obtain the time-domain voltage t>i. The same shortcut procedures for writing loop and nodal equations, discussed in Sees. 4.1 and 4.5 for resistive circuits, apply to phasor circuits. For example, in Fig. 11.7, if I = —5 is the mesh current in the right mesh in the clockwise direction, the two mesh equations are written down by inspection as

— I t — 4/0°

3

A

( i - ; i ) i , - (-7-1)1,-5

_ _, , ( . (

llI

+

J 1

.

J 1 + H

i?) ,_(Lii2) , 1

1

352

AC Steady-State Analysis

V

,

g / i -n/ i

a /- -v, 1 (b)

(a) Time- domain circuit; (b) phasor counterpart

= 0

Let us consider the circuit of Fig. 11.8(a), where the response is the steady-state value of t>i. The phasor circuit is shown in Fig. 11.8(b), with the loop currents as indicated. Applying KVL around the loop labeled I, we have - V , - 7K-7I + ! ) + ( ! + 72KI + 2V,) = 0

Chapter 11

^)

FIGURE 11.8

Also, from the figure we see that

These are equivalent to (11.5) and are formed as in the resistive circuit case. That is, in the first equation the coefficient of the first variable is the sum of the impedances around the first mesh. The other coefficients are the negatives of the impedances common to the first mesh and the meshes whose numbers correspond to the currents. The right member is the sum of the voltage sources in the mesh with polarities consistent with the direction of the mesh current. Replacing "first" by "second" applies to the next equation, and so on. The dual development, as described in Sec. 4.1, holds for nodal equations. EX A M P LE 1 1 . 5

II

Circuit of Fig. 11.2 redrawn for mesh analysis

V, = 7 l ( 4 - D

Eliminating I from these equations and solving for V,, we have - 4 + 7'3 V, = 1/143.1° V Therefore, in the time domain, the voltage is p, = cos {It + 143.1°) V EXERCISES

Find the forced response i in Fig. 11.3 using mesh analysis. 11.2.2 Solve Exercise 11.1.4 using mesh analysis.

11.2.1

Section*11.2

Mesh Analysis

353

.2.3 Find the steady-state current i using loop analysis. Answer V 2 cos (2l - 45°) A : ii -Tinr 1

41'

I"

* T

FIGURE 11.9

EXERCISE 11.2.3

( ireuil with an ac and a dc source

11.3 NETWORK

THEOREMS

Because the phasor circuits are exactly like the resistive circuits except for the nature of the currents, voltages, and impedances, all the network theorems discussed in Chapter 5 for resistive circuits apply to phasor circuits. In this section we shall illustrate superposition, Thevenin's theorem, Norton's theorem, and the proportionality principle, as applied to linear phasor circuits. In the case of superposition, if a phasor circuit has two or more inputs, we may find the phasor currents or voltages due to each input acting alone (i.e., with the others dead) and add the individual corresponding time-domain responses to obtain the total. In the case of a circuit like Fig. I l l , we may solve the corresponding phasor circuit of Fig. 11.2 by mesh or nodal analysis or by superposition because both sources are operating at the same frequency, namely ) allows us to use only one frequency at a time, and thus we cannot even construct a phasor circuit. EXAMPLE

11. 6

To illustrate superposition, let us find the forced response i in Fig. 11.9. There are two sources, one an ac source with ID = 2 rad/s and one a dc source with to = 0. Therefore, i = ii + i2

where i, is due to the voltage source acting alone and i2 is due to the current source acting alone. Using phasors, we may find ii and k by finding their respective phasor representations I and I , which are the phasor currents shown in Fig. 11.10(a) and (b), respectively. Figure 11.10(a) is a phasor circuit representing the time-domain circuit with the current source killed and to = 2 rad/s. Figure 11.10(b) is also a phasor circuit, with the voltage source killed and to = 0. In the latter case,
354

Chapter 11

AC Steady-State Analysis

4/0° A

FIGURE 11.10

Phasor circuits representing Fig.

11.9

its phasor representation is I , = 4/0° A In short, this is just the dc case considered earlier, as is evident from Fig. 11.10(b). ' From Fig. 11.10(a) we have 5/0° I 3 + j2 + |(1 + j2)(-j\)/(\ j2 = V2/-8.1° from which i, = V 2 c o s (2t - 8.1°) A

Section 11.3

Network Theorems

355

•

From Fig. 11.10(b) we have, by current division,

Mil

3

* - - ( T 7 l ) w I from which, since a> = 0, 12 = - 1 A Therefore, the total forced response is i = ii + h = V2 cos (2/ - 8.1°) - 1 A For an example of a circuit with two sinusoidal sources with nonzero frequen< cies, the reader is referred to Exercise 11.3.1. The procedure is, of course, exactlji the same as in Example 11.6. In the case of Thevenin's and Norton's theorems the procedure is identical to that for resistive circuits. The only change is that vt<- and iK, the time-domain opencircuit voltage and short-circuit current, are replaced by their phasor representations, Voc and In, and Ra,, the Thevenin resistance, is replaced by Zu,, the Thevenin impedance (of the dead circuit). There must be only a single frequency present, of] course; otherwise we must use superposition to break the problem up into single-fre-l quency problems, in each of which Thevenin's or Norton's theorem may be applied,] In general, the Thevenin and Norton equivalent circuits in the frequency domain are shown in Fig. 11.11. There is, of course, a close similarity with the resistive case.

FIGURE 11.11 EXAMPLE

11.7

FIGURE 11.13

C i r c u i t w i t h a d e p e n d e n t c u rr e n t s o u rc e

Ph as o r c i r c u i t s f or use i n T h e v e n i n 's t h e o r e m

(a) T h e v e n i n a n d (b) N o rt o n e q u i v a l e n t p h as o r c irc u i ts

Let us use Thevenin's theorem to find the forced response v of Fig. 11.12. We shall find its phasor representation V using the Thevenin equivalent of the phasor circuit to the left of terminals a-b. The open-circuit phasor voltage V* is found from Fig. 11.13(a), and the short-circuit phasor current is found from Fig. 11.13(b). Then the Thevenin impedance, as for resistive circuits, is Z* = ^

356

FIGURE 11.12

Chapter 11

AC Steady-State Analysis

(11.6)

Section 11.3

Network Theorems

357

In Fig. 11.13(a), since terminals a-b are open, the current 2/0" flows in the sistor and the current 2V, flows in the capacitor. Therefore, by KVL we have

v. - v, - (-;i)(2v,)

where

V, - 2(1) = 2 V

H Ut us consider the ladder network of Fig. 11.15 and use the proportionality princi pie to obtain the steady-state response V. This requires, as in the resistive case, that we assume V to have some convenient value like V = 1 and work backward to find the corresponding V,. Then the correct value of V is found by multiplying the assumed value by an appropriate constant.

Thus we have V„ = 2 + jA V In Fig. 11.13(b), the two nodal equations needed are

li»

in

Au. /ia

Wv

'TOP

+

+

-iia ; ^ v ,

and

-fiat

1 n•; v

(•

he = - V , + 2

FIGURE 11.15

From these we find

he = 3 + jl A The Thevenin impedance, by (11.6), is therefore

Phasor ladder network

Let us begin by assuming V =

1V

Then from the circuit we have The Thevenin equivalent circuit is shown in Fig. 11.14, with the i -F capacitor, corresponding to —jl A, connected to terminals a-b. It is a simple matter now to see, by voltage division, that

v=

,,=

V

V

T +

— = 1

+

, 1 A

Continuing, we have

[(TT7irrr7T)](2 + ^

= / I I , + V = 71(1 + jl) + 1 - jl V I2 = — r r + Ii = - - 1 + (1 + j l ) - j \

V,

= 4 - jl = 2V5/-26.6" V Therefore, in the time domain, we have o = 2V5 cos (3/ - 26.6°) V

-/I V , = 1I2 + V, = jl + jl = J2W

Therefore, V = 1 is the response to V, = jl. If we multiply this value of V, by 6/ jl to get the correct value of V s , then by the proportionality principle, we must multiply the assumed response of 1 by the same factor, 6/72, to get the correct value of V. Therefore, we have

FIGURE 11.14 Thevenin equivalent phasor circuit of Fig. 11.12

V

=(A)(1)=-,-3V

EXERCISES 11.3.1

358

Chapter 11 AC Steady-State Analysis

Find the steady-state current i. Answer 1 cos (4r - 36.9°) + 3 cos (2r + 73.7°) A Section 11.3 Network Theorems

359

JH '"mnr-

8n -AAA/

The voltage phasors of the circuit are V, = Rl = R\l\ \ = jwLl = a , L \ l \ / 9 0 °

vc= - AIOiC- ^ HQ)CI Z z W EXERCISE 11.3.1

11.3.2 For the phasor circuit corresponding to Exercise 11.2.3, replace the part to of terminals a-b by its Thevenin equivalent and find the steady-state current Answer V« = °(2 - j\) V, Zu, = ^(18 + jl) ft, r, = cos 2t A 11.3.3 Find V,, I , , and h in Fig. 11.15. Answer 3 V, 3 - _/3 A, 3 A

and

v

V* + V, + c These are shown in the phasor diagram of Fig. 11.17(a), where it is assumed that |vJ>|V |. The cases, \ \ \ |V -| and |V | = |V |, are shown in Fig. 11.17(b) and (c), respectively. In all cases the lengths representing the units of current and voltage are not necessarily the same, so that for clarity I is shown longer than V». c

C

t

C

11.4 PHASOR

DIAGRAMS

Since phasors are complex numbers, they may be represented by vectors in a planej where operations, such as addition of phasors, may be carried out geometrically, Such a sketch is called a phasor diagram and may be quite helpful in analyzing ac steady-state circuits. EXAMPLE

11.9 Let us consider the phasor circuit of Fig. 11.16, for which we shall draw a l l t voltages and currents on a phasor diagram. To begin with, let us observe that t current I is common to all elements and take it as our reference phasor, denoting il by i-|i|fflr We have taken the angle of I arbitrarily to be zero, since we want I to be our reference. We may always adjust this assumed value to the true value by the proportionality principle discussed in the preceding section. FIGURE 11.16

RL C s e ri es p h asor c i r c u i t

v, =v.

(a)

(c)

(b)

FIGURE 11.17

Ph as o r d i a g r a m s f or F i g . 1 1 . 1 6

In case (a) the net reactance is inductive, and the current lags the source voltage" by the angle 6 that can be measured. In (b) the circuit has a net capacitive reactance, and the current leads the voltage. Finally, in (c) the current and voltage are in phase, since the inductive and capacitive reactance components exactly cancel each other. These conclusions follow also from the equation I =

Z

R + j[a>L - (1/wC)]

(11.7)

Case (c) is characterized by

360

Chapter 11

A C Steady-State Analysis

Section 11.4

Phasor Diagrams

361

or

vie

If the current in Fig. 11.16 is fixed, then the real component of the voltage is fixed, since it is R\1. In this case the locus of the phasor V, (its possible local' on the phasor diagram) is the dashed line of Fig. 11.18. The voltage phasor ' up and down this line as to varies between zero and infinity. The minimum a m tude of the voltage occurs when to — 1/VJLC, as seen from the figure. For a other frequency, a larger amplitude of voltage is required for the same current.

FIGURE 11.19

RL p h as o r c i r c u i t

Substituting this value of R into (11.11) we have, after some simplification, This result may be rewritten as

*2 + {y + £L] - &

JTC

FIGURE 11.18 Locus of the voltage E X A M P L E 11.10

(1112)

which is the equation of a circle with center at [0, -(Vm/2o)L)] and radius Vm/2wL. The circle (11.12) appears to be the locus, as R varies, of the phasor I = x + jy. However, by (11.10), x a 0; thus the locus is actually the semicircle shown dashed on the phasor diagram of Fig. 11.20. The voltage Vm/(f, taken as reference, is also shown, along with the phasor I. If R = 0, we have, from (11.10) and (11.11), x = 0 and y = -VjwL. If R -» °°, then x -» 0 and y - * 0. Thus as./? varies from 0 to a, the current phasor moves counterclockwise along the circle.

phasor for a fixed current

FIGURE 11.20

Locus of the phasor

I

Let us find the locus of 1 as R varies in Fig. 11.19. The current is given by Therefore, if

R + jtoL

we have

R2 + w2L2

I = x + jy

(11.9):

RVm (11.10) x = Re I = R2 + to2L2 -a>LVm y = Im I = R2 + to2L2 (11.11) The equation of the locus is the equation satisfied by x and y as R varies; thus we need to eliminate R between these last two equations. If we divide the first of these two equations by the second, we have R_ x toL from which y otLx R = -: 362

Chapter 11 AC Steady-State Analysis

Section 11.4

Phasor Diagrams

363

• SOLUTION CONTROL STATEMENT FOR AC ANALYSIS

If I is as shown in Fig. 11.20, the current phasor may be resolved into two components, one having amplitude lm cos 0 in phase with the voltage and one wi amplitude /„ sin 0, which is 90° out of phase with the voltage. This construction indicated by the dashed and dotted vertical line. As we shall see in Chapter 12, t in-phase component of the current is important in calculating the average power do* livered by the source. Thus the phasor diagram gives us a method of seeing at glance the maximum in-phase component of current. Evidently this occurs at po : a, which corresponds to 0 = 45°. This is the case x = -y, or R = a>L. EXERCISES 11.4.1

Eliminate

locus of t

•OUTPUT CONTROL STATEMENT FOR V l l ) & I ( R 2 I . PRINT, AC VM (1) V P d l

+

2

FREQ 3.183E - 01

AC STEADY-STATE SOLUTION FOR C IRC UIT OF F I G . 1 1 . 1 . » DATA STATEMENTS VI Rl CI C2 LI L2 R2 II

364

100 100 1 0 1 2 12 2 0 2 0 0 2

0 AC 5 0 1 .5 5 1 .5 25 1 AC 5 0

A C Steady-State Analysis

IRIR2) 2.000E + 00

I I (R2) 4 0O0E+OO

* DATA STATEMENTS V 1 0 AC 4 0 Rl 1 2 0.5K R2 2 0 2K C I 2 0 0 2UF H 3 2 V - 3000 R3 3 4 2K C2 4 0 0. 2UF » SOLUTION CONTROL STATEMENT FOR f = 5 0 0 0 / ( 2 * 3 . 1 4 1 6 ) AC L I N 1 795.77

795.77

PRINT AC I M (Rl )

IP(Rl)

. END

The solution is FREQ 7.958E + 02

AMPLE 11.13

I M (Rl ) 2. 400E - 02

IP ( R1) 5.313E + 01

Let us find the phasor output voltage of the op amp circuit of Fig. 11.5 if the input voltage is ve = 10 cos (lOOOr + 30°) V. A circuit file for the nodes of the op amp inverting input, op amp output, and input source, assigned as 3, 4, and 10, respectively, with nodes 1 and 2 as shown, is AC STEADY-STATE SOLUTION OF F I G . 1 1 . 5 . * DATA STATEMENTS USING OPAMP.CKT OF CHAPTER 4 . L I B OPAMP. CKT VG 10 0 AC 10 30 R l 10 1 0.707K R2 1 2 1.414K C I 1 4 1UF C2 2 0 1UF R3 3 0 2K R4 3 4 2K XOPAMP 3 2 4 OPAMP

Section 11.5 Chapter 11

VP(l) - 2 657E+01

AC STEADY-STATE SOLUTION FOR F I G 1 1 . 3 .

SPICE F O R A C STEA D Y - ST A TE C I R C U I T S

Consider the circuit of Fig. 11.1. Let us find the voltage of node 1 in polar form and the current of the 1-fl resistor in rectangular form. A circuit file for calculating these values is

V M UI 2.236E + 00

LE 11.12 Consider finding the phasor current I in the circuit of Fig. 11.3. A circuit file for nodes being numbered sequentially clockwise, beginning with 1 at the positive terminal of the sinusoidal source, is

11.5

E X A M P L E 11.11

IKR2)

The solution printed in this case is

0

SPICE is a powerful tool in the analysis of ac steady-state circuits and is particularly helpful in performing the many tedious operations of complex arithmetic associated with obtaining these solutions. The procedure is very similar to that described for the dc case using the .AC solution control statement. SPICE, in this instance, analyzes the phasor circuit for the phasor currents and voltages resulting from both independent and dependent sources, which are also expressed as phasors. Since SPICE uses the phasor circuit in the frequency domain, all ac sources must have the same frequency. The reader should review the .AC statement of Appendix E before proceeding with this section.

IR( R2)

END

Answer 1 x - S ) H S ) ^ Find coL in Exercise 11.4.1 so that Im I has its largest negative value. Also find I for this case. Answer R, (VjV2 R)/-45° 2

11.4.2

the

(f = 2 / 2»PI Hzl

AC L I N 1 . 3183 . 3183

SPICE for AC Steady-State Circuits

365

» SOLUTION CONTROL STATEMENT [f

2H

= 1000 / (2*3.1410) Hz]

i

AC L I N 1 159.15 159.15

runr*-

12(1 AAV

. PRINT AC VM(4) VP(4) . END

This circuit file gives a solution FREQ 1 592E+ 02

VM(4) 1.414E + 01

VP (4) - 5.999E + 01

Our discussion has been limited to solutions for circuits at a single frequency Plots of responses for varying frequency inputs are discussed in Chapter 15.

PROBLEM 11.1

EXERCISES 11.5.1 11.5.2

Use SPICE to find the phasor representation of t>i and v2 in Fig. 11.1 f o r / = Answer 3.706/-35.83" V, 4.677/-34.40" V Use SPICE to find the phasor current for i in Fig. 11.3 f o r / = 1 kHz. Answer 25.79 + y'3.414 mA

1

)

11.6

11.1 11.2

366

PROBLEMS Find the steady-state voltage v using nodal analysis. Use nodal analysis to find the steady-state value of v if vg = V„ cos tot and find the amplitude of v for the cases to = (a) 0, (b) 1000, and (c) 106 rad/s. Compare the results for this circuit with those for Fig. 11.6. Chapter 11 AC Steady-State Analysis

11.3 11.4 11.5

Find the steady-state current i using nodal analysis. Find the steady-state voltage t; using nodal analysis. Find the steady-state voltage v using nodal analysis.

M / *

0.1 H

0.1 H

Tf flP

—nSoV

+

SU M M A RY

Since phasor currents and voltages satisfy KCL and KVL, and the voltage-current relationships for passive elements are identical to Ohm's law, with impedance playing the role of resistance, all the methods of resistive circuit analysis apply to phasor circuits. Thus nodal analysis and loop analysis techniques are applicable, and Thevenin's and Norton's theorems apply, as well. The open-circuit voltages and shortcircuit currents are phasors and the Thevenin impedance replaces the Thevenin resistance. Phasor diagrams, which are sketches displaying operations such as addition of phasors, are useful in analyzing ac circuits. They show at a glance when magnitudes and phase angles of phasor quantities reach their maximum or minimum values, and allow us to find the locus of a given phasor as a circuit element value or a frequency changes. Finally, SPICE is also a powerful tool for analyzing ac circuits in the steady state. It is particularly useful in this case because it relieves us of the tedious task of dealing with complex numbers by hand.

toon

20

MF

-

100fi < 1

V

PROBLEM 11.2

14 cos 2r V

PROBLEM 11.3

2 cos

It PROBLEM 11.4

Chapter 11

Problems

367

6 cos 8f V

8 cos

PROBLEM 11.10 PROBLEM 11.5 11.6 11.7 11.8 11.9 11.10 11.11 11.12 11.13 11.14 11. IS 11.16 11.17 11.18

Find the steady-state current i, using nodal analysis. Solve Exercise 11.1.1 using loop analysis. Solve Exercise 11.1.2 using mesh analysis. Find the steady-state voltage v using loop analysis if I,I = 6 cos 4/ A and i,2 = 2 cos 4t A. Find the steady-state voltage v. Find the steady-state current i. Find the steady-state current i. Find the steady-state voltage v. Find the steady-state voltage v. Find the forced response i if u , = 8 cos 2000/ V. Find the steady-state voltage o. Find the steady-state value of v. Show that if Z, Z 4 = Z 2 Z, in the "bridge" circuit shown, then I = V = 0 and therefore all the other currents and voltages remain unchanged for any value of Z 5 . Thus it may be replaced by an open circuit, a short circuit, etc. In this case, the circuit is said to be a bal-

11.19 Show that the circuit of Prob. 11.13 is a balanced bridge, with the series combination of 3 11 and 5 H constituting Z 5 in Prob. 11.18. Replace Z 5 by a short, and show that the same v is obtained as before. 11.20 Note that the corresponding phasor circuit is a balanced bridge, and use the method of Prob. 11.18 with Z5 replaced by an open circuit to find the forced response v. 11.21 Find the steady-state current i if vs = 4 cos lOOOr V. 11.22 Find the steady-state voltage v if u, = 2 cos 2l V. 11.23 Find the forced response c if u, = 2 cos t V. 11.24 Find the steady-state voltage if 5 cos 3» V. 11.25 Find the steady-state voltage 11.26 2 cos 5/ V. 11.27 Find the steady-state voltage v. Find the steady-state value of v.

1 It

-/TTinpio n . 15 cos

5 cos 5r V

» < * )

7k

5 n:

0.08 F

PROBLEM 11.11

anced bridge.

40

2 cos

5 cos 30001 V

PROBLEM 11.6 368

Chapter 11

A C Steady-State Analysis

PROBLEM 11.12

PROBLEM 11.9 Chapter 11

Problems

369

(^)

8

cos 6r V

PROBLEM 11.13

jn

PROBLEM 11.14

PROBLEM 11.15 370

Chapter 11

AC Steady-State Analysis

Chapter 11

Problems

371

os2r V 2 O < "

PROBLEM 11.27 Find the steady-state current i if i, = 9 20 cos t - 39 cos 2/ + 18 cos 3» A.

11.34 In the phasor circuit corresponding to Prob. 11.24, replace everything except the 2-fl resistor between terminals a-b by its Thevenin equivalent and use the result to find the steadystate current i. 11.35 Solve Prob. 11.2 for the steady-state voltage v by applying the proportionality principle to the corresponding phasor circuit. Assume that v, = 2 cos lOOOf V. (Suggestion: Assume that the phasor of v is 100 V.) 11.36 Find the steady-state current i using die principle of proportionality. 11.37 Find the steady-state current i in Prob. 11.24 using the proportionality principle. 11.38 Find I, the phasor representation of i, using a phasor diagram. Show the phasors of i,, ic, and i t , with the phasor of the source voltage as reference.

11.29 Find the steady-state voltage v. II..HI For the phasor circuit corresponding to Exercise 11.2.3, replace everything except the 2-11 resistor between terminals c-d by its Thevenin equivalent and find the steady-state current i. •1.31 For the phasor circuit corresponding to Exercise 11.2.3, replace the circuit to the left of terminals a-b by its Norton equivalent and find the steady-state current i,. 11.32 Replace the circuit to the left of terminals a-b by its Norton equivalent, and find V. 11.33 In the phasor circuit corresponding to Prob. 11.10, replace everything except the 1-fl resistor by its Thevenin equivalent and use the result to find the steady-state voltage v.

8!)

i

>

:• in

Vvv

1

^

PROBLEM 11.28

1H

i

F

2

n

PROBLEM 11.29 Chapter 11

Problems

373

J6CI

-i

-n

A

o

CO M PUTER APPLICATIO N PROBLEMS 11.41 11.42 11.43

12 0 -AAA-

12 0 >

y4(l

11.44 v


Using,SPICE, find r in Prob. 11-5. UsingSPICE, find the Thevenin equivalent circuit in Prob. 11.17 for the network to the Ml of the 1-fl resistor. (Hint: Find V« and U.) Using SPICE, find t in Prob 11.25 if output e of the circuit of Prob. 11.24 supplies o, to the 6-fl resistor of this network. In the circuit shown, find v if vt = 15 sin (10*r + 75°) V.

r;-/4n

PROBLEM 11.32

PROBLEM 11.36

PROBLEM 11.38 11.39 Show that the locus of V c in Fig. 11.16 is a circle if R = 4 O, C = J F, V, = V, w = 2 rad/s, and L varies from 0 to °°. Select the value of L that gives the maximum amplitude of the time-domain voltage t>c, and find vc in this case. 374

Chapter 11 AC Steady-State Analysis

11.40 Find the locus of V c in Prob. 11.39 if L = 1 H, C = { F, V, = 2/f/ V, a> = 2 rad/s, and /? varies from 0 to °°. Select the value of R for which 1m V c has its largest negative value, and find vc for this case. Chapter 11

Computer Application Problems

375

12 AC Steady-State Power

Jam— Pntcott Jout«

rsr8-1889

The m a n to w h o m w e are indebted for the familiar expression i R for the power dissipated in a resistor is the English physicist J a m e s Prescott Joule, w h o published the result as Joule's law in 1 8 4 1 . He also shared 2

The heating of a conductor depends upon its resistance and the square of the current passing through it. James P. Joule

in the famous discovery of the conservation of energy. Joule was born in Salford, England, the second of five children of a wealthy brewer. He taught himself electricity and magnetism at h o m e as a young boy and obtained his formal education at nearby Manchester University. His experiments on heat were

c o n d u c t e d in his h o m e laboratory, and to maintain the •accuracy of his m e a s u r e m e n t s he w a s forced to develop his o w n system of units. His chief claim to fame is that he did more tha n any other person to establish the idea that heat is a form of energy. Throughout most of his life Joule was an isolated amateur scientist, but toward the e n d of his years his work w a s recognized by honorary doctorates from Dublin a n d Oxford. In his honor the unit of energy w a s n a m e d the joule, m

I n this chapter we consider power relationships for networks that are excited by periodic currents and voltages. We concern ourselves primarily with sinusoidal currents and voltages since nearly all electrical power is generated in this form. Instantaneous power, as we now well know, is the rate at which energy is absorbed by an element, and it varies as a function of time. The instantaneous power is an important quantity in engineering applications because its maximum value must be limited for all physical devices. For this reason, the maximum instantaneous power, or peak power, is a commonly used specification for characterizing electrical devices. In an electronic amplifier, for instance, i f the specified peak power at the input is exceeded, the output signal will be distorted. Greatly exceeding this input rating may even permanently damage the amplifier. A more important measure of power, particularly for periodic currents and voltages, is that of average power. The average power is equal to the average rate at which energy is absorbed by an element, and it is independent of time. This power, for example, is what is monitored by the electric company in determining monthly electricity bills. Average powers are encountered which range from a few picowatts, in applications such as satellite communications, to millions of watts, in applications such as supplying the electrical needs of a large city. Our discussion begins with a study of average power. We then consider superposition once again and introduce a mathematical measure for characterizing periodic currents or voltages, known as effective or rms values. Next, we consider the power factor associated with a load and present a complex power. Finally, we describe the measurement of power.

AVERAGE POWER In linear networks which have inputs that are periodic functions of time, the steadystate currents and voltages produced are periodic, each having identical periods. Consider an instantaneous power p = vi where v and /' are periodic of period T. i(l + T) = i(t). I n this case

(12.1) That is, v(t + T) = v{t),

and

p{t + T) = vit + D l ( l + T) = vit)iit) =

376

Section 12.1

Average Power

(12.2)

Pit)

377

Therefore, the instantaneous power is also periodic of period T. That is, p repeats n self every T seconds. The fundamental period 7"i of p (the minimum time in which p repeats itself) m not necessarily equal to T, however, but T must contain an integral number of periods T]. In other words, T = ri7",

i(/),p(r)U

(12.3)

where n is a positive integer. EXAMPLE 12.1

Suppose that a resistor R carries a current i = l Then

m

cos ait with period T =

lit/to.

p = Ri

1

= Rli cos (at 2

= ^ = ( 1 + cos 2u>t) Evidently, 7i = TT/O>, and, therefore, T - 2T\. Thus, for this case, n = 2 in (12.3). This is illustrated by the graph of p and i shown in Fig. 12.1(a). I f we now take i = /„(1 + cos tot), then p = RP (\ cos
In this case, T, = T = Itr/m, and n = 1 in (12.3). This may be seen also from the graph of the function in Fig. 12.1(b). Mathematically, the average value of a periodic function is defined as the time integral of the function over a complete period, divided by the period. Therefore, the average power P for a periodic instantaneous power p is given by (12.4) (b)

where t\s arbitrary. A periodic instantaneous power p is shown in Fig. 12.2. It is clear that i f we integrate over an integral number of periods, say mTi (where m is a positive integer), then the total area is simply m times that of the integral in (12.4). Thus we may write P = - \

p d t

FIGURE 12.1

FIGURE 12.2

(12.5)

Graphs of p and i

Periodic instantaneous power

P
I f we select m such that T = mT (the period of v or i ) , then t

1

pdt J

(12.6)

'i i

Therefore, we may obtain the average power by integrating over the period of t rel="nofollow"> or i.

378

Chapter 12

AC Steady-State Power

Section 12.1

Average Power

V

' V

Let us now consider several examples of the average power associated with s,

Referring to Table 12.1, entry 5, we find, for m = n « 1. o • ^ , i n d

« - f).

nusoidal currents and voltages. A number o f very important integrals which often g

T

a

b

I

e

1

2

1

V

« o n

e

of these integrals is left as an exercise

TABLE 12.1 Integrals for Sinusoidal Functions and Their Products

(12.9)

cos 8

P =

Thus the average power absorbed by a two-terminal device is determined by the M l plitudes V and l„ and the angle 0 by which the voltage v leads the current i . In terms of the phasors of v and i , m

/

1. 2. 3. 4. 5. .

(

,

I

)

sin (ait + a), cos (at + a) sin (mot + a), cos (mot + a)" sin (at + a), cos (lot + a) sin (mot + a)cos (not + a)' cos (mat + a)cos (mot + $y 2

/ ( / ) dt, a » 0

V = V [±

0 0 „ / 0 0 m *n

i = u/


a

!

,

= |V|&

m

-

g

we have, from (12.9),

v cos (a - 0)/
f

'm and n arc integers.

First, let us consider the general two-terminal device o f Fig. 12.3, which is assumed to be in ac steady state. I f , in the frequency domain, z =

= ^ | V || 11 cos (ang V - ang I)

(12.10)

where ang V = d>,

ang I = d> - 8

are the angles of the phasors V and I . I f the two-terminal device is a resistor R, then 0 = 0, and V = RI ,

\z\(e

m

is the input impedance of the device, then for

m

so that

(12.9) becomes

v = V cos (tot + d>) m

(12.7) P„ = | « i

we have / = I„ cos (tot + - 0) where

(12.8)

It is worth noting at this point that i f = / , a constant (dc) current, then a> = 4> = 0 = 0, and / „ = in (12.8). In this special case Table 12.1 does not apply, but by (12.6) we have d c

P„ = The average power delivered to the device, taking U = 0 in (12.6), is _ r

0>V .

m m

2ir~ J

C O S

*


+

^

c

o

s

(<°t + ~ 6) dt

FIGURE 12.3 General two-terminal device I

Ml

In the case of an inductor, 0 = 90°, and in the case of a capacitor, 0 = - 9 0 ° , and thus for either one, by (12.9), P = 0. Therefore, an inductor or a capacitor, or for that matter any network composed entirely of ideal inductors and capacitors, in any combination, dissipates zero average power. For this reason, ideal inductors and capacitors are sometimes called lossless elements. Physically, lossless elements store energy during part of the period and release it during the other part, so that the average delivered power is zero. A very useful alternative form of (12.9) may be obtained by recalling that

— »

Z = Re Z + j I m Z = | Z \/0 V

Twoterminal device

and therefore ReZ cos6 =

380

Chapter 12

AC Steady-State Power

Section 12.1

Average Power

w 381

Alternatively, from (12.11),

Substituting this value into (12.9) and noting that V = | Z | / „ , we have m

P = ^llReZ

25 W

(12.11)

We may also note that the power absorbed by the 100-ft resistor is Let us now consider this result i f the device is a passive load. We know from the definition of passivity in (1.7) that the net energy delivered to a passive load I i nonnegative. Since the average power is the average rate at which energy is delivered to a load, it follows that the average power is nonnegative. That is, P a 0. This requires, by (12.11), that

^

(100)(1/V2>

Vm I m

Z = 100 + 7100

Verify the integrals of Table 12.1. For a capacitor o f C farads carrying a current i = L cos ait, verify that the average power is zero from (12.6). Repeat this for an inductor of L henrys.

12.1.3

Find the average power delivered to a 10-ft resistor carrying a current o f (a) 1 = 5| sin 10/1 m A , (b) 1 = 10 sin 10* m A , 0 s i < TT/10 S = C'I w / 1 0 s t < v/S s; T = i r / 5 s (c) i = 5 m A , 0 s t < 10 ms = - 5 m A , 10 £ t < 20 ms; T = 20 ms (d) i = 2t, 0 s f < 2 s; T = 2 s Answer (a) 125 >iW; (b) 0.25 mW; (c) 0.25 mW; (d) *f W Find the average power absorbed by the capacitor, the two resistors, and the source. Answer 0 , f J , - 4 W

The maximum current is = A = _ L |Z| V2

Therefore, from (12.9), the power delivered to Z

A

is

100 P = —-p cos 45° = 25 W 2V2

12.1.4

EXERCISE 12.1.4

FIGURE 12.4 RL circuit in the ac steady state

3G A^A

382

Chapter 12

AC Steady-State Power

cos 0 = - 2 5 W

12.1.1 12.1.2

0

"

W

EXERCISES

= 100V2/45 n

/

5

where the minus sign is used because the current is flowing out of the positive terminal of the source. The source therefore is delivering 25 W to Z . We note that the power flowing from the source is equal to that absorbed by the load, which illustrates the principle of conservation of power.

I f 0 = 0, the device is equivalent to a resistance, and i f 0 = ir/2 (or -ir/2), the device is equivalent to an inductance (or capacitance). For —w/2 <0 < 0, the device is equivalent to an RC combination, whereas for 0 < 8 < ir/2, it is equivalent to an RL combination.

Let us find the power delivered by the source of Fig. 12.4. The impedance across the source is

2

power. The power absorbed by the source is

or, equivalently,

Finally, i f |01 > 7r/2, then P < 0, and the device is active rather than passive. In this case the device is delivering power from its terminals and, o f course, acts like a source.

=

This power, of course, is equal to that delivered to Z since the inductor absorbs no

Re Z ( » a 0

EXAMPLE 12.2

=

Section 12.1

Average Power

!

383

12.1.5

I f / i W is periodic of period 7", and f (t) is periodic of period T , show t M') + fi(t) is periodic of period T i f positive integers m and n exist such that 2

2

T = mT, = nTi Extend this result to the function (1 + cos u>t)'\d in this section, to ft its period, T = 2tr/a>.

where P and P are the average powers from v , and v, , respectively, acting alone. (We are assuming, of course, that each component of p is periodic of period / ) Superposition for average power applies when t

2

g

P = />, + P

2

(12.12)

2

Clearly, this condition holds i f

12.2

(12.13)

ii i dt = 0 2

SUPERPOSITION AND POWER In this section we consider the power in networks containing two or more sourcei, such as the simple circuit of Fig. 12.5. By superposition, we know that i = h + h

The most important case in which this equation is satisfied is when i(t) is composed of sinusoidal components of different frequencies. Suppose that it

=

/„i

cos (
and

where i, and i are the currents in R due to v., and v , neous power is 2

g2

respectively. The instanu]

ii = Imt cos (w t +


2

Since we are assuming that i = i, + h is periodic of period T, we must have

p = R(i, + i f 2

= Ri , + Ri 2

= Pi + p

2

+

22

+

2Ri,i 2Ri,i

/ „ , cos [u>,{t + T) + ,] + I 2

2

cos [a> (t + T) + d> ] = / „ , cos (a>,t + d>i) + l 2

2

2

gl

a>] T = 2irm,

2

rT

2

2

a> T = 2trn 2

r ir/w I cos (mtot + <) cos (mat + ) dt 2

i,i

2

dt = I„\I

m2

2

o

fll Imdmllt

COS

(d>|

-

) 2

m # n

= 0, Simple circuit w i t h t w o sources

cos (to t + 4> )

where m and n are positive integers. Therefore, i f , then a>i = men and a>2 = nto. In this case the integral in (12.13) becomes, using Table 12.1,

J FIGURE 12.5

m2

which requires that

where p, and p are the instantaneous powers, respectively, due to v acting alone and to v,i acting alone. In general, 2Ri,i * 0; thusp * p , + p , and superposition does not apply for instantaneous power. 2

m 2

Thus i f m = n(o>, = io ), superposition does not apply. However, i f m ^ n, superposition does apply. We may generalize this result to the case of a periodic sinusoid with any number of sinusoidal components of different frequencies. The average power due to the sum of the components is the sum of the average powers due to each component acting alone. It can be shown that superposition of average power holds for sinusoids whose frequencies are not integral multiples of some frequency u>, provided that we generalize the definition of average power to 2

In the case of p periodic with period T, the average power is

i r = f \ + Pi + 2RUi ) 2

P, + P +

2R

dt

1

dt

2

lim -

'o

384

Chapter 12

AC Steady-State Power

Section 12.2

Superposition and Power

r

p dt

385

Extending the foregoing procedure to a periodic current which il the sum ol N + 1 sinusoids of different frequencies,

This generalization applies to the periodic case just considered as well as to the c i = ii + i , where 2

11

= cos t

i =

12

= cos irt

In this case i is not even periodic (the ratio wjuhi m/n), but

+ /»i cos (o),» + «pi) + / . i cos (a> / + < M + • • • + /«« cos (a>,vf + d>„)

(12.14)

2

we find that the average power delivered to a resistance R is

= 1/TT is not a rational num

1 f lim - I hit dt = 0 Jo

• +

+ | ( f t i + fla +

P = Rll

(12.15)

lis)

T

EXAMPLE 12.3

The first term, in which the factor \s missing, is the special case of zero frequency and must be considered separately, as in Sec. 12.1. That is, by (12.6),

Suppose in Fig. 12.5 that t>„ = 100 cos (377* + 60°) V , i> = 50 cos 377. V , a R = 100 ft. Since w, =
2

cos ( w f + 4>,)dt = 0,

11 = 1/60! A

N

i = 1, 2, 3,

and

1 = -0.5 A 2

i f

Therefore, I = I, + l

2

2

P = Hl00)(0.866) = 37.5 W 2

Let us now repeat Example 12.3 with v = 50 V dc. Since 0 , 1 and v are sinusoid! with ail = 377 and a> = 0 rad/s, respectively, superposition for the average power is applicable. Proceeding as before, we find s2

s2

I , = 1/60°

for

I

for o) = 0

2

= -0.5

01

EXERCISES 12.2.1

= 377

2

2

P = RIi 2

= 50

= *!k

2

w

Find the average power delivered to the resistor in Fig. 12.5 i f R = 10 f l and (a) 0(i = 20 cos lOOt and v, = 10 cos (100/ + 60°) V. (b) o,, = 20 cos (r + 60°) and t>, - 100 sin (2. - 30°) V. (c) « „ = 50 cos (t + 30°) and v,i = 100 sin (t + 60°) V. (d) o , i = 20 cos (t + 25°) and t),2 = 30 sin (5t - 35°) V. Answer (a) 15 W; (b) 625 W; (c) 375 W; (d) 65 W Find the average power absorbed by each resistor and each source. Answer 8 W , 24 W , - 8 W , - 2 4 W 2

where I is now a dc current. Therefore, P l

12.2.2

= 25 W

and the average power is

i—Tinp—' i

P = P, + P = 75 W 2

Example 12.4 illustrates a very important case for electronic amplifiers with si nusoidal inputs. These amplifiers contain dc power supplies that produce dc currents which provide the energy for the amplified ac signals. Thus superposition is very useful i n finding the average power associated with each frequency, including w = 0.

Chapter 12

AC Steady-State Power

2

P = Pic + Pl + Pl + • • • + P"

2

386

Rll

This last expression is P*, the average power delivered by Thus, i f we denote the other terms in (12.15) by />, (due to i , ) , P (due to i ) , and so on, we have superposition of power in the form

= 7O.866 A

so t h a t / „ = 0.866 A . From (12.11), we find

EXAMPLE 12.4

RIl dt = Jo

; i n

W v

;

—ripr—

C

^ ) 8 cos 4r V

EXERCISE 12.2.2

Section 12.2

Superposition and Power

387

12.2.3

Suppose that w e now oinsulci ;I stniisonl.il c i i i i r n l /

l - i n d Ihc average p o w e r absorbe d by the resistor and each source

Answer X W .

4 W.

4 W

/~ CO* (toll * ) Then,

from (12.16) and Tabic 12.1, we lind

f

.

-

^2

Thus a sinusoidal current having an amplitude /„ delivers the same average power to a resistance R as does a dc current which is equal to I . / V 2 . We also see that the rms current is independent of the frequency o> or the phase d> of the current t. Similarly, in the case of a sinusoidal voltage, wc find that

EXERCISE 12.2.3

v„ RMS VALUES We have seen in the previous sections that periodic currents and voltages deliver average power to resistive loads. The amount of power that is delivered depends < the characteristics of the particular waveform. A method of comparing the pow delivered by different waveforms is therefore very useful. One such method is f use o f rms or effective values for periodic currents or voltages. The rms value of a periodic current (voltage) is a constant that is equal to t dc current (voltage) that would deliver the same average power to a resistance K Thus i f /m,, is the rms value o f i , we may write

Substituting these values into the important power relations of (12 9) and (12.11) for the two-terminal network, wc have (12.17)

P = Vtm.lnm COS 6

1

p = M L

=

f j

**

d

t

from which the rms current is

(12.16)

In a similar manner, it is easily shown that the rms voltage is

and P =

f L Re

(12.18)

Z

In practice, rms values are usually used in the fields of power generation and distribution For instance, the nominal 115-V ac power which is commonly used for household appliances is an rms value. Thus the power supplied to our homes is provided by a 60-Hz voltage having a maximum value of 115V2 « 163 V. On the other hand, maximum values are more commonly used in electronics and communications. Finally, let us consider the rms value of the current in (12.14), which is made up of sinusoids of different frequencies. In terms of rms currents, (12.15) becomes P = RUi Since P =

+ /i™. + IU,

+ • • • + fL.1

(12.19)

we see that the rms value of a sinusoidal current consisting of dif-

ferent frequencies is v ~ = \/?

f

<"

v2

The term rms is an abbreviation for root mean square. Inspecting (12.16), w« see that we are indeed taking the square root of the average, or mean, value of the square o f the current.

388

Similarly,

.

—,—

= V v i + v?™ + vL» + • • • + vi™

From our definition, the rms value of a constant (dc) is simply the constant it self. The dc case is a special case (o> = 0) of the most important type of waveform, the sinusoidal current or voltage.

These results are particularly important in the study of noise in electrical networks,

Chapter 12

Section 12.3

AC Steady-State Power

a subject of later courses.

rms Values

389

EXERCISES 12.3.1

Find the rms value of a periodic current for which one period is defined by (a) i = / , 0 < / < 2 s = -/, 2< / <4s (b) i = It, Os l
12.3.2

Find the rms values of (a) i = 10 cos tot + 20 sin (tot - 30°), (b) 8 cos (
12.3.3

Find V™,. Answer 4 V

In the case of purely resistive loads, the voltage and current are in phase Therefore, 8 = 0, pf = l , and the average and apparent powers are equal. A unity power factor (pf = 1) can also exist for loads which contain inductors and capacitors if the reactances of these elements are such that they cancel one another. Adjusting the reactance of loads so as to approximate this condition is very important in electrical power systems, as we shall see shortly. In a purely reactive load, 0 = ± 9 0 ° , pf = 0, and the average power is zero. In this case, the equivalent load is an inductance (0 = 90°) or a capacitance (8 = - 9 0 ° ) , and the current and voltage differ in phase by 90°. A load for which - 9 0 ° < 0 < 0 is equivalent to an RC combination, whereas one having 0 < 0 < 90° is an equivalent RL combination. Since cos 0 = cos (-0), it is evident that the pf for an RC load having f? = -0,, where 0 < 8, < 90°, is equal to that of an RL load with 0 = 0,. To avoid this difficulty in identifying such loads, the pf is characterized as leading or lagging by the phase of the current with respect to that of the voltage. Therefore, an RC load has a leading pf and an RL load has a lagging pf. For example, the series connection of a 100-ft resistor and a 0.1-H inductor at 60 Hz has Z = 100 + ;37.7 = 106.9/20.66° ft and has a pf of cos 20.66° = 0.936 lagging. In practice, the power factor of a load is very important. In industrial applications, for instance, loads may require thousands of watts to operate, and the power factor greatly affects the electric b i l l . Suppose that a mill consumes 100 kW from a 220-V rms line. At a pf of 0.85 lagging, we see that the rms current into the mill is

EXERCISE 12.3.3

12.4 /

P O W E R F A C TO R

~

V~~pf

=

=

(220X0.85)

=

5

3

4

8

A

which means that the apparent power supplied is

The average power delivered to a load in the ac steady state, repeating (12.17), P =

cos 0

Now suppose that the pf by some means is increased to 0.95 lagging. Then

The power is thus equal to the product of the rms voltage, the rms current, and cosine of the angle between the voltage and current phasors. In practice, the current and voltage are easily measured and their product, Km,/™, is called the parent power. The apparent power is usually referred to in terms of its Ui voltamperes (VA) or kilovoltamperes (kVA), in order to avoid confusing it with unit o f average power, the watt. I t is clear that the average power can never greater than the apparent power. The ratio of the average power to the apparent power is defined as the po factor. Thus i f we denote the power factor by pf, then in the sinusoidal case

pf =

= cos 9

VU/™, = (220)(534.8) VA = 117.66 kVA

/ n M

=

(220)(0.95)

=

4

7

8

5

A

and the, apparent power is reduced to V™/™ = 1 0 5 . 3 kVA Comparing the latter case with the former, we see that 7™, was reduced by 56.3 A (10.5%). Therefore, the generating station must generate a larger current in the case of the lower pf. Since the transmission lines supplying the power have resistance, the generator must produce a larger average power to supply the 100 kW to the load. I f the resistance is 0.1 ft, for instance, the power generated by the source must be

(12.21

P„ = 10 + 0 . 1 / L . 3

Therefore, we find which, o f course, is dimensionless. The angle 8, in this case, is often referred in the PF angle. We also recognize it as the angle of the impedance Z of the load.

390

Chapter 12

P

AC Steady-State Power Section 12.4

Power Factor

g

= 128.6 kW,

pf = 0.85

= 122.9 kW,

pf = 0.95

391

which requires that the power station produce 5.7 kW (4.64%) more power to su the lower pf load. It is for this reason that power companies encourage a pl exC ing say 0.9 and impose a penalty on large industrial users who do not comply

I

12.6

As an example of the application of (12.22), let us change the power factor lor the circuit of Fig. 12 4 to 0.95 lagging We have already found that Z = 100 + 7 100 = 141.4/45°

Let us now consider a method of correcting the power factor of a load hav

Therefore, before a parallel reactance is added across Z , the power factor is

an impedance

pf = cos 0 = cos 45° = 0.707 lagging

Z = R + jX

Since we desire a power factor of 0.95 lagging, tan (cos ' PF) is positive, so that by (12.22) we have

We may alter the power factor by connecting an impedance Z, in parallel with S shown in Fig. 12.6. For this connection, it is clear that the load voltage dins change. Since Z is fixed. I does not change, and the power delivered to the lof not affected. The current I , supplied by the generator, however, does change

X, =

I00 + 100 2

100 tan (cos

1

2

0.95) -

100

= -297.92 f l

Since X, < 0, the reactance is a capacitance C = - I / a ) * , = 33.6 u.F. The load impedance now becomes Z,

_ (100 + yTOOX- /297.92) 100 + y 100 -

>297.92

190.0/18.2°

Therefore, the power to the corrected load is P = FIGURE 12.6

Circuit for correcting the power factor

Z

Z

2

cos (18.2°) = 25 W

which is the same as that delivered to Z in Fig. 12.4. The current, however, is U

Let us denote the impedance of the parallel combination by T

100

2(190.0)

'

=

I90V2

= 0.372 A

as compared to that of Fig. 12.4, given by

In general, we select Z , so that ( I ) Z, absorbs zero average power, and (2) Z, lies the desired power factor pf = PF. The first condition requires that Z, be reactive. That is,

A™ = - 4 = = 0.5 A

VI

We see, therefore, that the current has been reduced by 0.128 A , or 25.6%.

Z , = jXi The second condition requires that [ COS

Substituting Z

T

[

EXERCISES

,/Im ZA TAN

= PF

IrTz tV

12.4.1

Find the apparent power for (a) a load that requires 20 A rms from a 115-V rms line and (b) a load consisting of a 100-11 resistor in parallel with a 25-/xF capacitor connected to a 120-V rms 60-Hz source. Answer (a) 2.3 kVA; (b) 197.9 VA

12.4.2

Find the power factor for (a) a load consisting o f a series connection of a 10-ft resistor and a 10-mH inductor operating at 60 Hz, (b) a capacitive load requiring 25 A rms and 5 kW at 230 V rms, and (c) a load that is a parallel connection of a 5-kW load with a power factor of 0.9 leading and a 10-kW load with a 0.95 lagging power factor. Answer (a) 0.936 lagging; (b) 0.87 leading; (c) 0.998 lagging

in terms of R, X, and X , into this equation, we find that (see

12.25) R

2

+ X

2

R tan (cos' PF) - X

(121

1

where we note that tan "(cos ' PF) is positive i f PF is lagging and negative i l leading.

392

Chapter 12

AC Steady-State Power

Section 12.4

Power Factor

393

12.4.3

Use (12.22) to correct the power factor seen by the source of Fig. 12.4 to (a) leading and (b) 0.9 lagging. Answer (a) C = 74.22 /uF;«(b) C = 25.78 p-f

From (12.25), wc see that Q = Im S = V J m

mn

sin 8

(12.26)

For an-impedance Z , we know that sin 6 = (lm Z ) / | Z | . so that

12.5 COMPLEX

POWER We shall now introduce a complex power in the ac steady state, which is very u for determining and correcting power factors associated with interconnected lo Let us begin by defining rms phasors for general sinusoidal voltages and curre The phasor representations for (12.7) and (12.8) are V =

l e**-» m

The rms phasors for these quantities are defined as V = —

V5

Inns

Q = lL>lm Z 2

-

v (12.

ImZ

< - > 1 2

2 7

Izp

\e»

I =

V

Therefore, since Vrrm/|Z| = A™,, we see that

A phasor diagram of V ™ and I™ is shown in Fig. 12.7. We see that the phasor current can be resolved into the two components 1^ cos 8 and / » , sin 9. The component /„„> cos 8 is in phase with V™., and it produces the real power P. In contrast, Am, sin 8 is 90° out of phase with V ™ , and it causes the reactive power Q. Since /„„> sin 8 is 90° out of phase with V™,, it is called the quadrature component of I,ms. As a consequence, the reactive power is sometimes referred to as the quadrature power.

^ ~ lm\

V2

Let us now consider the average power given in (12.17). Using Euler's mula, we may write P = V ™ / ™ cos 8 =

ReiV^J^e'")

Next, inspecting (12.23), we see that V

I * =

V

I

e'*

where I™» is the complex conjugate of I™,. Thus P = Re(VrmsI^)

(12.24)

and the product Vrmslrt™ is a complex power whose real part is the average po Denoting this complex power by S, we have S = V „ . I * , = P + jQ

(12.2S)

where Q is the reactive power. Dimensionally, P and Q have the same units; he ever, the unit of Q is defined as. the var (voltampere reactive) to distinguish it In the watt. The magnitude of the complex power is

It is often convenient to view the complex power in terms of a diagram such as that of Fig. 12.8. It is evident that for an inductive load (lagging p f ) , 0 < 0 £ 90°, Q is positive, and S lies in the first quadrant. For a capacitive load (leading p f ) , - 9 0 ° £ 8 < 0, Q is negative, and S lies in the fourth quadrant. A load having a unity power factor requires that Q = 0 since 8 = 0. In general, we see that

|S| = | V „ „ I * , | = IVrmsHl^sl = VrmsAms

(12.28)

which, of course, is equal to the apparent power.

394

Chapter 12

AC Steady-State Power

Section 12.5

Complex Power

395

Mil

It is evident that the addition of Z , does not affect the average power P delivered to the load. It does, however, affect the net reactive power. We can therefore select to obtain our desired power factor, which, of course, is usually raised. This causes ;i reduction j n the current required to produce P, as discussed previously. I 12.7 f

Re

Let us again consider the circuit of Fig. 12.4 and change the power factor to PF = 0.95 lagging. The complex power of the uncorrected load is

FIGURE 12.8 Diagram lor the complex power

S = VmJSm = P + jQ = 25 + 725 since

Let us now consider the complex power associated with a load consisting two impedances Z , and Z , as shown in Fig. 12.9. The complex power deliver*! the combined impedances is

V™, = 70.7 V

2

S = -

= V

I *

+ V

Vm,

I

+ Iw)* If

From (12.28), we see that Q

T

Therefore, the complex power delivered by the source to the interconnected l o t J the sum of that delivered to each individual load, and the complex power is I ' conserved. This statement is true no matter how many individual loads there are how they are interconnected, because it depends only on Kirchhoff's laws and definition of complex power. This principle is known as conservation of co power.

= 0.3535(1 - j ' 1) A

= Q + Q, must satisfy

Therefore. cos 8 = PF = cos

tan

and Q

T

= P tan (cos ' PF) = 25 tan 18.2° = 8.22 var

The required Q, is C?i = Qi " Q = 8.22 - 25 = - 1 6 . 7 8 var Since Q, = V L O m Z , ) / | Z , p and Z , = y X , , we may write FIGURE 12.9 Load consisting of Z, and Zi in parallel V

Conservation of complex power may be used in a straightforward manner correct the power factor. As an example, let us consider the circuit of Fig. 12.6 n again. The complex power to the uncorrected load Z is

2

Solving for X, in the case under consideration, we have

S = P + jQ

16.78

Connecting a pure reactance Z , in parallel with Z results in a complex power to

which represents a capacitance C previous result in Sec. 12.4.

of

= -297.9 f l

1/oiXi = 33.6 ^iF. This is identical to our

S, = JQ, Therefore, from conservation of complex power, for the composite load we have ST

= S + S,

PLE 12.8

Let us find the power factor of two loads connected in parallel, as shown in Fig. 12.9. Suppose that Z , represents a 10-kW load with a power factor pf, = 0.9 lagging and Z a 5-kW load with pf = 0.95 leading. For Z , we have 2

2

= P + HQ + G i )

396

Chapter 12

AC Steady-State Power

S, = />, + Section 12.5

Complex Power

jQ,

397

where P, = 10 W , 0, = c o s 4

1

pf, = 25.84°, and from (12.28), Current coil

(2, = />, tan 0, = 4843 var Similarly, for Z we have

-AAJkJ

2

• Y Y Y V

S = Pi + jQi

Voltage coil

2

where P = 5 x 10 W , 0 = -cos 2

3

2

Q

1

pf = - 1 8 . 2 ° , and 2

= 5 x 10 tan 0 = -1643 var

2

3

2

FIGURE 12.10 Typical connection of a wattmeter

The total complex power is

respect to its other terminal, then the meter gives a positive, or upscale, reading. In Fig. 12.10 this corresponds to the load absorbing power. I f the terminal connections of either the current coil or voltage coil (but not both) are reversed, a negative, or downscale, reading is indicated. Most meters cannot read downscale—the pointer simply rests on the downscale stop. Thus such a reading requires reversing the connections of one of the coils, usually the voltage coil. Reversing the connections of both coils does not affect the reading. The wattmeter of Fig. 12.10, represented by the rectangle and the two coils, is connected to read

S, = P, + Pi + J ' ( C + Q ) 2

= 1.5 x I 0 + >3200 4

Therefore, for the combined loads, = 12.04° and pf = 0.978 lagging

P = IVI

EXERCISES 12.5.1

Find the complex power delivered to a load which has a 0.85 leading power and (a) absorbs 10 kW, (b) 10 kvar, and (c) 1 kVA. Answer (a) 1 1 . 7 6 / - 3 1 . 8 ° ; (b) 1 8 . 9 8 / - 3 1 . 8 ° ; (c) 1 / - 3 1 . 8 ' kVA

12.5.2

Find the impedance of the loads in Exercise 12.5.1 i f Km* = 240 V. Answer (a) 4 . 9 0 / - 3 1 . 8 ° ; (b) 3 . 0 3 / - 3 1 . 8 ° ; (c) 5 7 . 6 / - 3 1 . 8 ° ft

12.5.3

Repeat Exercise 12.4.2 using the concept of complex power.

12.5.4

Repeat Exercise 12.4.3 using the concept of complex power.

P O W E R MEASUREMENT A wattmeter is a device which measures the average power that is delivered to load. I t contains a rotating high-resistance voltage, or potential, coil, connected parallel with the load and a fixed low-resistance current coil which is connected series with the load. The device has four terminals, a pair to accommodate «' coil. A typical connection is shown in Fig. 12.10. We see that the current coil sponds to the load current, whereas the voltage coil responds to the load volr For frequencies above a few hertz, the meter movement responds to the avc power. Ideally, the voltage across the current coil and the current in the voltage o: are both zero, so that the presence of the meter does not influence the power it measuring.

EXERCISES Determine the power reading of each wattmeter after assigning the terminal markings required for a positive reading. Answer 100 W , 75 W , 0 W 50 n

VsA

200

rms

V

Mi. TOT

HOT ,4 1.;

One terminal on each of the coils is marked ± so that i f the current enters t ± terminal of the current coil and the ± terminal of the voltage coil is positive wl

398

Chapter 12

AC Steady-State Power

cos e

where V and I are as indicated and 0 is the angle between V and I (or, equivalently, the angle of the load impedance). This, of course, is the power delivered to the load. Other types of meters are available for measuring the apparent power and the reactive power. A n apparent power or VA meter simply measures the product of the rms current and rms voltage. The varmeter, on the other hand, measures the reactive power.

12.6.1

12.6

111

:50 n

AM.

A rms

j4

6

EXERCISE 12.6.1 Section 12.6

Power Measurement

399

12.6.2

Repeat Exercise 12.6.1 for the circuit shown. Answer

10, 30

• \ lc ,il a periodic current is as shown. I f cuircnl flows in a 2 0 - ( l resistor, find the . I P power

mW 5001! V

12.6

Find the average power absorbed by each N sistor and the power delivered by the SOUTOC ; H

2 n -WA

I H

TOT—i

2il 5 cos 3000r V PROBLEM 12.6 12.7

PROBLEM 12.2

EXERCISE 12.6.2

I inil the average power dissipated in a resistor ill icsistance R i f the current is i = [ / . ( I + cos wt) A .

12.7

Find the average power absorbed by the resistors, the inductor, and the source.

SUMMARY In this chapter we have defined the average power Pdelivered to an element by ,i |Kriodic current as the average of the instantaneous power delivered over one period The average power in the ac case may be expressed quite simply in terms of the maM nitudes of the current and voltage phasors of the element and the cosine of the anglf between them. Alternatively, P may be expressed in terms of the current magnitude and the real part of the impedance of the element. I f two or more sinusoidal sourcili are present and they each have different frequencies, the average power is the sum of the average powers delivered by each source. In other words, superposition holds tut average powers in this case. The computation of average power in the ac case is simplified by the use uf rms

or effective

value

of a current or voltage phasor, defined as the square root

•

the average over a period of the square of the current or voltage. Another usclnl term in average power computation is the power factor, which is the cosine of lit* power

factor

angle

whose imaginary part is the reactive power. The magnitude of the complex power I the apparent power and its phase is the power factor angle. A wattmeter is used to

0.5 H

10 cos 800/ A

PROBLEM 12.7 12.8

Find the average power absorbed by the 3 - k i l resistor and the dependent source.

12.9

Find the average power R = 0.4 il.

12.10

Find the average power delivered to the 4-11 resistor.

12.11

Find the average power delivered to R.

PROBLEM 12.4 A voltage source v = 4 cos 4r V is applied to i n RL series circuit w i t h R = 4 f l and / = 0.5 H . Find the power supplied by the source

delivered

to R i f

i kn

(the angle by which the voltage leads the current). The powa

factor may be corrected (made larger) by adding parallel elements to diminish the power factor angle. Complex power is a complex number whose real part is the average power ant)

Find the average power absorbed by the resistors and the source.

0.4 H

20 cos 10,000(

measure the average power and a varmeter measures the reactive power. Average power is the power measured by the power company to determine our utility bills, and thus is a very important quantity which we consider in the later chapters.

PROBLEM 12.8

PROBLEMS 12.1

One cycle o f a periodic current is given by i = 10 A , = 0,

400

0 < / < 1 ms I

Chapter 12

I f the current flows in a 2 0 - f i resistor, find If* average power.

6 cos 4r V

3 cos 4r A

< K 4 m s AC Steady-State Power

PROBLEM 12.9 Chapter 12

Problems

401

1.5

H

PROBLEM 12.10

{ ^ ^ 9 cos 4l V

PROBLEM 12.11 12.12

L = 1 8 - 1 0 cos t -

PROBLEM 12.18 that

Find the power delivered to the 8 - f l resistor i f

P = P, + P

39 cos 2r + 9 cos 3/ A

2

where P, and P are the average powers as. ciated w i t h i, and i , respectively, acti alone. Note that this includes the case "I /'II being nonperiodic. 2

(a) Find R so that it absorbs the maximum power and find the maximum power, (b) Find the power absorbed by R if R = 2 O. Find the rms value of the voltage

12.24 Find the rms value of the steady-state current through the ; - f l resistor.

mi i = 5 +

12.25 Derive (12.22).

2

12.15

PROBLEM 12.12 12.13

L

We have defined the average power for an i n stantaneous power p(l), which is not necessarily periodic, by P = lim

circuits. (I n both cases, V, and Z , are fixed,, and the load is variable.)

definition yields the same result as (12.6). G i v e n: i = i , + i , where 2

/ , „ cos 01,t 2

is the current flowing in a resistor R . Using the definition o f Prob. 12.13, show for
402

Chapter 12

AC Steady-State Power

L

L

12.16

!z„ cos ai t

L

the maximum power transfer theorem for af

Show for c and i o f ( 1 2 . 7 ) - ( 1 2 . 8 ) that this 12.14

For a Thevenin equivalent circuit consistm of a voltage source V„ and an impedan Z , = R , + jXg, (a) show that the circuit
12.17

Using the result o f Prob. 12.15, replace K in Prob. 12.9 by (a) a resistive load that will' draw the m a x i m u m power and (b) a general load (resistive and reactive elements) that will draw the m a x i m u m power. In both cases, find the resulting power. (a) Find the power absorbed by R i f R = 111 (b) Find the m a x i m u m power that can be dc livered to R and the value o f R for maximum power.

12.23 Find the rms value of the steady-state current through the 2-fl resistor if t>, = 3 cos / V.

12V2 cos t V.

(b) v = 12 cos (2t - 30°) + 4 cos 5f + V 2 cos (8/ + 60°) V. (C) v = 4 cos t - 6 V 2 cos (2/ + 45°) + VTO cos (5r - 10°) V. find the rms value of a periodic current for which one cycle is given by = 4 A, 0 < t < 1s (•) 1 < t < 2 s. = 0, (b) = 3r A , 0 < t < 2 s = -3(« = Im sin

(O

4) A ,

2nt . -T A

T

0 < t < j s.

T

= 0, I'll

0 < t < 4 s.

= / „ sin —

A

< t


0 < t <

Ts.

Find the rms value o f the steady-state volt-

age c. Find the rms value o f the steady-state current

in

R

Chapter 12

Problems

12.26 Find the rms value of the steady-state current 1 and the power factor seen from the source terminals. What element connected in parallel with the source will correct the power lactor to 0.8 lagging? 12.27 Find the power factor seen from the terminals of the source and the reactance necessary to connect in parallel with the source to change the power factor to unity. 12.28 Find the power factor seen from the terminals of the independent source and the reactance which must be connected in parallel with the independent source to change the power factor to 0.8 lagging. 12.29 Find the real power, the reactive power, and the complex power delivered by the independent source. Find also the reactive element to be placed in parallel with the independent source to correct the power factor seen by the source to 0.8 leading.

403

1 i

H

- nn r

*

1(

H

2n vVr 28 cos 16l V

12

n

PROBLEM 12.26 PROBLEM 12.21 211

10 cos

"6

6 0

V

PROBLEM 12.27

10/0° Vrms

PROBLEM 12.28 t,/2V

10 cos

PROBLEM 12.23

PROBLEM 12.3Q

PROBLEM 12.29

Find the real power, the reactive power, and the complex power delivered by the source, and the reactive element which must be connected in parallel with the source to make the power factor seen by the source (a) 0.9 leading and (b) 0.8 lagging.

S cos (2» - 4 5 » ) V ©

PROBLEM 12.24

404

Chapter 12

AC Steady-State Power

Two loads in parallel draw respectively 420 W at a power factor of 0.6 lagging and 80 W at a power factor of 0.8 leading. If the voltage source across the parallel combination is Chapter 12

Problems

V , = 100/15° V rms, find the current I delivered by the source. 12.32 Three parallel passive loads, Z , , Z , and Z , are receiving complex power values of 6 - 75, 8 + j 10, and 2 + jl V A , respectively. I f a voltage source of 50/0° V rms is connected across these loads, find the rms value of the current that flows from the source and the power factor seen by the source. 2

3

405

12.33

12.34

rect the power factor to 0.95 lagging l
A load is supplied the complex power S = 60 + / 80 V A by a voltage source o f 50 cos 377/ V. F i n d the capacitance that should be connected in parallel w i t h the load so that the power factor seen by the source is (a) unity and (b) 0.8 lagging.

12.35

T w o loads i n parallel draw a total o f 3 k W at a 0.9 lagging power factor f r o m a 115-V rms 60-Hz line. One load is k n o w n to absorb 1000 W at a 0.8 lagging power factor. Find (a) the power factor o f the second load and (b) the parallel reactive element necessary to cor-

i

Find the real power, the reactive power, the complex power delivered by the sourea

12.36

F i nd the wattmeter reading.

12.37

F i n d the wattmeter reading.

12.38

Find the wattmeter 4 cos 1000/ V .

reading

if

12.39

I f Z = 1 0 / 3 0 ° 17, find the wattmeter rea

12.40

F i nd the wattmeter reading.

•

PROBLEM 12.38

1 H

6 0

TSTiP

w v -

-cJtM^

-nsw^-

300/rr ( + V nil:

PROBLEM 12.35

300/-120° V rms

PROBLEM 12.39

—'VAA*

( ^ ) -

>2 n PROBLEM 12.36

PROBLEM 12.40 4 H 4

4

H

i — T S W — i — n n n r -

^ 0 0 0 / -

—TSoTT) cos

16(

V

(^M

PROBLEM 12.37

406

Chapter 12

AC Steady-State Power

Chapter 12

Problems

407

13

Three-Phase Circuits

Ttwrrms Aiva Edison 1847-1931

T h e greatest A m e r i c a n inventor and to get along with the schoolmastM Genius is one per cent perhaps the greatest inventor in hisHe w a s e x e m pt from military servi inspiration and ninety-nine tory w a s T h o m a s Alva Edison, w h o per cent perspiration. because of deafness, and during c h a n g e d the lives of people everyThomas A. Edison Civil War he r o a m e d from city to w h e r e with s u c h inventions as the as a telegraph operator. Duhng electric light and the phonograph. He time h e patented improvements patented over 1100 inventions of his the stock ticker a n d sold the pate o w n a n d improved many other persons' inventions, for the t h e n astounding price of $40,000. In 1876 such as the telephone, the typewriter, the electric gen- moved t o Menlo Park, N e w Jersey, a n d from there h erator, and the motion picture. Perhaps most impor- steady s t r e a m of inventions m a d e him world famou tant of all, he w a s one of the first to organize re- The electric light w a s his greatest invention, but search, at one time employing some 3 0 0 0 helpers. supply it to the world he also designed the first elect Edison w a s born in Milan, Ohio, the youngest of seven children. He had only three months of formal education because his mother took him out of school and taught him herself. He asked too many questions

408

power station. His discovery of the Edison effect,' m o v e m e n t of electrons in the v a c u u m of his light bulb, also m a r k e d the beginning of the age of electronics,

. ^ ^ . s we have already noted, one very important use of ac steady-state analysis is its application to power systems, most of which are alternating current systems. One principal reason for this is that it is economically feasible to transmit power over long distances only i f the voltages involved are very high, and it is easier to raise and lower voltages in ac systems than in dc systems. Alternating voltage can be stepped up for transmission and stepped down for distribution with transformers, as we shall see in Chapter 16. Transformers have no moving parts and are relatively simple to construct, whereas with the present technology, rotating machines are generally needed to raise and lower dc voltages. Also, for reasons of economics and performance, almost all electric power is produced by polyphase sources (those generating voltages w i t h more than one phase). In a single-phase circuit, the instantaneous power delivered to a load is pulsating, even i f the current and voltage are in phase. A polyphase system, on the other hand, is somewhat like a multicylinder automobile engine i n that the power delivered is steadier. Consequently, there is less vibration in the rotating machinery, which, in turn, performs more efficiently. A n economic advantage is that the weight of the conductors and associated components required in a polyphase system is appreciably less than that required in a single-phase system that delivers the same power. Virtually all the power produced in the world is polyphase power at SO or 60 Hz. In the United States 60 Hz is the standard frequency. In this chapter we begin with single-phase, three-wire systems, but we concentrate on three-phase circuits, which are by far the most common of the polyphase systems. In the latter case the sources are three-phase generators that produce a balanced set of voltages, by which we mean three sinusoidal voltages having the same amplitude and frequency but displaced in phase by 120°. Thus the three-phase source is equivalent to three interconnected single-phase sources, each generating a voltage w i t h a different phase. I f the three currents drawn f r o m the sources also constitute a balanced set, the system is said to be a balanced three-phase system. This is the case to which we shall restrict ourselves, for the most part.

SI N GLE - P H A SE, T H REE - W IRE SYSTE M S Before proceeding to the three-phase case, let us digress i n this section to establish our notation and consider an example of a single-phase system that is in common household use. This example w i l l also serve to give us some practice with a singlephase system, with which we are already familiar and which w i l l serve as an introduction to the three-phase systems to be considered next. Section 13.1

Single-Phase, Three-Wire Systems

409

In this chapter we shall find extremely useful the double-subscript notation traduced in Chapter 1 for voltages. In the case of phasors, the notation is V.,,, Im voltage of point a with respect to point b. We shall also use a double-subscript m tion for current, taking, for example, L , as the current flowing in the direi i \ from point a to point b. These quantities are illustrated in Fig. 13.1, w h i n l direct path from a to b is distinguished from the alternative path from a • through c.

This is evident without referring to a circuit since by K V L the voltage between two points a and b is the same regardless of the path, which in this case is the path a, n, b. Also, since V^, = - V ( „ , we have

= 100 - 1007-120°

c FIGURE 13.1 Illustration of double-subscript notation

which, after simplification, is V * = 100V3 /_30! V rms

Because of the simpler expressions for average power that result, we use f values of voltage and current throughout this chapter. (These are also the val read by most meters.) That is, i f V = |V I

=

|/0°

11 |/_HJ

These steps are shown graphically in Fig. 13.2(b). A single-phase, three-wire source, as shown in Fig. 13.3, is one that has three output terminals a, b, and a neutral terminal n, for which the terminal voltages are equal. That is,

V rms

(13.

V J > - V r f * V,

A rms

are the phasors associated with an element having impedance,

z = |z|/en

(13

the average power delivered to the element is

FIGURE 13.3 Single-phase, three-wire source a

P = I V I • I I I cos 0 • =

|I|

2

(13.4)

This is a common arrangement in a normal house supplied with both 115 V and 230 V rms, since i f |V„| = |V,| = 115 V, then |V^| = | 2 V, | = 230 V.

(I3,

1

Re Z W

In the time domain the voltage and current are c =

V2|

V | cos wl V

EXAMPLE 13.1

i = V2\1 cos (tot - 0) A The use of double subscripts makes it easier to handle phasors both analytically a geometrically. For example, in Fig. 13.2(a), the voltage is

410

Chapter 13

Three-Phase Circuits

Section 13.1

Single-Phase, Three-Wire Systems

411

I

Let us now consider the source of Fig. 13.3 loaded with two identical lo«4| both having an impedance Z i , as shown in Fig. 13.4. The currents in the linei and bB are

A

Therefore, the current in the neutral wire, nN, by K C L is Lv = ~(LA

+ IM) =

0

FIGURE 13.5

Thus the neutral could be removed without changing any current or voltage in iht system.

Symmetrical single-phase, three-wire system

If the symmetry of Fig. 13.5 is destroyed by having unequal loads at terminals A-N and N-B or unequal line impedances in lines aA and bB, then there w i l l be a neutral current. Let us consider the situation in Fig. 13.6, which has two loads operating at approximately 115 V and one at approximately 2 3 0 V. The mesh equations are 431, -21, -401, -

60I

2

2I 2

+ 63I

2

-

40I

3

= 115

60I

3

= 115

+ (110 + j\0)h

= 0

Solving for the currents, we have FIGURE 13.4

Single-phase, three-wire system w i t h t w o identical loads

I , = 16.327 - 33.7° A rms

I f the lines aA and bB are not perfect conductors but have equal impedance! Z j , then Uv is still zero because we may simply add the series impedances Z , and Z i and have essentially the same situation as in Fig. 13.4. Indeed, in the more general case shown in Fig. 13.5, the neutral current l is still zero. This may be seen by writing the two mesh equations

1

2

= 15.737-35.4° A rms

1

3

= 14.467-39.9° A rms

M

(Z, + Z Z LA 3

2

+ Z )L,., + Z I M 3

3

+ (Z, + Z

2

z,i

3

+ Z )LM + Z , I 3

3

=

v,

=

-V,

FIGURE 13.6

1

•4on

• ion

-V\ A /-

(13.5)

115/0° V rms (

Since by K C L the left member of the last equation is - I ^ v , the neutral current it zero. This is, o f course, a consequence of the symmetry o f Fig. 1 3 . 5 . Three-Phase Circuits

A -t

2S1

(Z, + Z j + Z,)(L^ + I M ) + ZjfJU + I M ) = 0

Chapter 13

in Wv

115/0° V r m s (

and adding the result, which yields

LA + I M = 0

Unsymmetrical single-phase, three-wire system

3

I /ion •6on

1 n -Af\A— Section 13.1

Single-Phase, Three-Wire Systems

413

The voltages V„„, V,„,, l l l d V,„ between the line terminals and the neutral terminal are called phase voltagtl u d in most cases we shall consider are given by

Therefore, the neutral current is I * = I - I , = 0.76/184.3° A rms 2

and of course, is not zero.

V™ = v«/o° V

EXER C I SES

ta

= VV-120°

(13.6)

V,„ = Vp/120

0

13.1.1

Derive (13.5) by superposition applied to Fig. 13.5.

13.1.2

Find the power PAO-, Pea, and P\o+jio delivered to the loads, 40 ft, 60 ft, t 10 + j 10 ft, respectively, of Fig. 13.6. Answer 249, 181, 2091 W

13.1.3

Find the power P A, PbB> and P„N lost in the lines in Fig. 13.6. Answer 266.3, 247.4, 1.2 W

13.1.4

Find the power P and P*, delivered by the two sources in Fig. 13.6. Check the r l suits in Exercises 13.1.2 and 13.1.3 for conservation of power. Answer 1561.4, 1474.5 W

or

v „ = vvoj; V * = VV120°

A

(13.7)

V „ = V„/-120° r

m

13.2

In both cases, each phase voltage has the same rms magnitude V , and the phases are displaced 120°, with V „ arbitrarily selected as the reference phasor. Such a set of voltages is called a balanced set and is characterized by p

V„ + V

THREE- P H ASE Y-Y

SYSTE M S

Let us consider the three-phase source of Fig. 13.7(a), which has line terminals
t a

+ V„ = 0

(13.8)

e

as may be seen from (13.6) or (13.7). The sequence of voltages in (13.6) is called the positive sequence, or abc sequence, while that of (13.7) is called the negative, or acb, sequence. Phasor diagrams of the two sequences are shown in Fig. 13.8, where we may see by inspection that (13.8) holds. Evidently, the only difference between the positive and negative sequence is the arbitrary choice of the terminal labels, a, b, and c. Thus without loss in generality we shall consider only the positive sequence.

FIGURE 13.7 T w o representations of a Y-connected source FIGURE 13.8 (a) Positive and (b) negative phase sequence

M

414

Chapter 13

Three-Phase Circuits

(a)

(b)

Section 13.2

Three-Phase Y-Y Systems

(b)

415

Let us now consider the system of Fig. 13.10, which is a balanced Y-Y, threephase, four-wire system, if the source voltages are given by (13.6). The term Y-Y applies since both the source and the load are Y-connected. The system is said.to be balanced since the source voltages constitute a balanced set and the load is balanced (each phase impedance is equal—in this case—to Z„). The fourth wire is the neutral line nN, which may be omitted to form a three-phase, three-wire system.

By (13.6), the voltages in the abc sequence may each be related to V „ lationships, which w i l l be useful later, are

v „ = v.y-i2o

c

V „ = V^120° by which we mean the indicated angle is added to that o f V „ . The line-to-line voltages, or simply line voltages, in Fig. 13.7 are V,,,, and V „ , which may be found from the phase voltages. For example, v„ = V. + =

fl

V* -

yj-120°

= V J V,/30° In like manner, V* =

K/-90°

v„ = V 5 ty-2io° I f we denote the magnitude of the line voltages by V , we have L

|

VL

= V3

(13.B

vT

FIGURE 13.10 Balanced Y-Y system The line currents of Fig. 13.10 are evidently

and thus V„» = V^30°,

= VJ-W.

V„

= V J - 2 1 V

I I I

LA

These results also may be obtained graphically from the phasor diagram s h o w f l Fig. 13.9.

=

W-120

IM

FIGURE 13.9 Phasor diagram showing phase and line voltages

=

he =

0

(13.12)

LA/120°

The last two results are a consequence of (13.9) and show that the line currents also form a balanced set. Therefore, their sum is

0

- K v = L A + I M + Ice =

Thus the neutral carries no current in a balanced Y-Y four-wire system. In the case of Y-connected loads, the currents in the lines aA, bB, and cC are also the phase currents (the currents carried by the phase impedances). I f the magnitudes of the phase and line currents are /, and l , respectively, then h = 1 , and (13.12) becomes L

\

416

Chapter 13

Three-Phase Circuits

L A = htli

=

I M = h/-e

-

120°

he = hl-e

+

120° =

where 0 is the angle of Section 13.2

P

l
+

120°

(13.13)

120°

Z. p

Three-Phase Y-Y Systems

417

/

M the effective phase load. Since by the foregoing discussion there is no neutral current, we have

The average power P delivered to each phase o f Fig. 13.10 is p

P. = V.I. cos 0 (II,

= 1} Re Z ,

U a

and the total power delivered to the load is

5 / S

=

=

2 0 /

-

3 6

- ' 9

A

r

m

s

The currents form a balanced, positive sequence set, so that we also have

P = 3P ,

I m = 20/-156.9° A rms,

The angle 0 of the phase impedance is thus the power factor angle of the threeload as well as that of a single phase. Suppose now that an impedance Z is inserted in each of the lines a A, bB, cC and that an impedance Z not necessarily equal to Z , is inserted i n lint ( In other words, the lines are not to be perfect conductors but are to c o r f impedances. Evidently, the line impedances, except for Z
N

L

p

L

N

N

he = 20/-276.9° A rms

Example 13.3 was solved on a "per-phase" basis. Since the impedance in the neutral is immaterial in a balanced Y - Y system, we may imagine the neutral line to be a short circuit. We may do this i f it contains an impedance or even i f the neutral wire is not present (a three-wire system). We may then look at only one phase, say "phase A , " consisting o f the source V„, i n series with Z and Z , as shown i n F i g . 13.12. (The line nN is replaced by a short circuit.) The line current L A , the phase voltage L A Z , , and the voltage drop i n the line L A Z J . may all be found from this single-phase analysis. The other voltages and currents i n the system may be found similarly, or from the previous results, since the system is balanced. L

p

FIGURE 13.12 Single phase for a per-phase analysis EX A M P LE 13.3

Let us find the line currents in Fig. 13.11. We may combine the 1 - f t line i m and (3 + y'3)-ft phase impedance to obtain Z„ = 4 + y3 = 5/36.9° ft

Suppose that we have a balanced Y-connected source, having line voltage V = 200 V rms, which is supplying a balanced Y-connected load with P = 900 W at a power factor of 0.9 lagging. Let us find the line current I and the phase impedance Z . Since the power supplied to the load is 900 W , the power supplied to each phase is P = ?f = 300 W , and from L

L

p

FIGURE 13.11 Balanced system w i t h line impedances

Al

p

we have 300 Therefore since for a Y-connected load the phase current is also the line current, we have 3V3

h = l

p

= 2(b~9)

=

2

8

9

A

r

m

s

4

Q

The magnitude of Z is given by p

]

z

l

=

K

=

h

418

Chapter 13

Three-Phase Circuits

Section 13.2

Three-Phase Y-Y Systems

200/V3

=

f

l

3V3/(2)(0.9)

419

and since 8 = cos;

1

0.9 = 25.84° is the angle of Z,,. we have Z

p

= 40/25.84° ft

If the load is unbalanced but there is a neutral wire which is a perfect cod tor, we may still use the per-phase method of solution for each phase Howcvd this is not the case, this shortcut method does not apply. There is a very method employing so-called symmetrical components which is applicable to anced systems and which the reader may encounter in a course on power syste is important to note, in any case, that a three-phase circuit is still a circuit and, anced or unbalanced, may always be solved by general analysis procedures.

EXERC ISES 13.2.1

Given: = 230/0° V rms is a line voltage of a balanced Y-connected three-p source. I f the phase sequence is abc, find the phase voltages. Answer 1 3 2 . 8 / - 3 0 , 132.8/-150°, 132.8/-270" V rms

13.2.2

In Fig. 13.10 the source voltages are determined by Exercise 13.2.1, and the lot each phase is a series combination of a 30-ft resistor, a 1-mF capacitor, and 0,. inductor. The frequency is to = 100 rad/s. Find the line currents and the power livered to the load. Answer 2 . 6 6 / - 8 3 . T , 2.66/-203.1°, 2.66/36.9° A rms, 634.8 W

13.2.3

Show that i f a balanced three-phase, three-wire system has two balanced th phase loads connected in parallel, as shown, the load is equivalent to that of t 13.10 with

I.2.4

I f in Exercise 13.2.3 Z, = 4 + j 3 ft, Z , = 4 - >3 ft, and the line voltage is V, = 2 0 0 V 3 V rms, find the current h in each line. Answer 64 A rms

I I >l I T A

CON NECTION

Another method of connecting a three-phase load to a line is the delta connection, or A connection. A balanced A-connected load (with equal phase impedances) is shown in Fig. 13.13(a), in a way that resembles a A, and in an equivalent way in Fig. 13.13(b). If the source is Y-or A-connected, the system is a Y-A or a A-A system.

0

bo-

FIGURE 13.13 T w o versions of a A-connected load

EXERCISE 13.2.3 420

Chapter 13

Three-Phase Circuits

A n advantage of a A-connected load over a Y-connected load is that loads may be added or removed more readily on a single phase of a A, since the loads are connected directly across the lines. This may not be possible in the Y connection, since the neutral may not be accessible. Also, for a given power delivered to the load the phase currents in a A are smaller than those in a Y. O n the other hand, the A phase voltages are higher than those of the Y connection. Sources are rarely A-connected, because i f the voltages are not perfectly balanced, there w i l l be a net voltage, and consequently a circulating current, around the delta. This, o f course, causes undesirable heating effects in the generating machinery. Also, the phase voltages are lower in the Y-connected generator, and thus less insulation is required. Obviously, systems with A-connected loads are three-wire systems, since there is no neutral connection. Section 13.3

The Delta Connection

421

From Fig. 13.13 we see that in the case of a A-connected load the voltages are the same as the phase voltages. Therefore, if the line voltages are by (13.11), as before, the phase voltages are v «

= vjw,

y

= v j - w ,

K

V„ c

= VJ-2W

I11

where Vi = V, If Z

p

= \ \[0, the phase currents are l.

=

i *

=

= uyr

~

=

- o

U-w

-

e

i11

Lp = ^

= ij-ivr

- o

where FIGURE 13.14 Phasor diagram for a A-connected load

(13, The current in line aA is

13.5 LA = L »

LA

—

As an example of a three-phase circuit w i t h a A-connected load, let us find the line current h in Fig. 13.13 i f the line voltage is 250 V rms and the load draws 1.5 kW at a lagging power factor o f 0.8. For one phase, P = Mp = 500 W, and thus p

which after some simplification is

9

500 = 250/„(0.8) LA =

V3IJ-0

or

The other line currents, obtained similarly, are

/, = 2.5 A rms

IM = V3L/-120" - 0

Therefore, we have

he = V 3 7^-240° - 6

h = V 3 /„ = 4.33 A rms

Evidently, the relation between the line and phase current magnitudes in the A c a i i h = Vil

(13.l|

P

Finally, in this section, let us derive a formula for the power delivered to a balanced three-phase load with a power factor angle 6. Whether the load is Y-connected or A-connected, we have P = 3P„ = 3V I

P P

and the line currents are thus

cos 0

I n the Y-connected case, V = V t / V J and /, = h, and in the A-connected case, P

LA = IJ-0,

h* = hi-120°

- 8,

Lc = / /-240° - 9 t

(13.

V, = V and L

= / t / V ^ . I n either case, then,

Thus the currents and voltages are balanced sets, as expected. The relations betw line and phase currents for the A-connected load are summed up in the phasor gram of F i g . 13.14.

422

/> =

Section 13.3 Chapter 13

Three-Phase Circuits

The Delta Connection

3 ^ c o s 0

423

or

P = V3 VI

L L

COS 6

As a check on Example 13.5, (13.21) yields 1500 = V3(250)/ (0.8) L

or, as before, l

L

= 4.33 A rms

EXERC ISES 13.3.1

FIGURE 13.15 Y-connected and A-connected loads in parallel

Solve Exercise 13.2.2 i f the source and load are unchanged except that the 1 A-connected. [Suggestion: Note that in (13.15), (13.17), and (13.20) 30° nr subtracted from every angle.] Answer 4.6V3/-83.1°, 4.6V3/-203.1°, 4.6V3/36.9 A rms, 1904.4 W A balanced A-connected load has Z = 4 + 7 3 ft, and the line voltage is 200 V rms at the load terminals. Find the total power delivered to the load. Answer 19.2 kW A balanced A-connected load has a line voltage of VL = 100 V rms at the load minals and absorbs a total power of 4.8 kW. I f the power factor of the load it leading, find the phase impedance. Answer 4 - j 3 ft I f in Fig. 13.13, = 100/0° V rms, the voltages are a balanced positive s set, ZAB = 20 ft, ZBC = 10 ft, and ZCA = 710 ft, find the line currents. Answer 6.20/233.8°, 13.23/-139.1°, 19.32/45° A rms 0

13.3.2

13.3.3

13.3.4

P

FIGURE 13.16 (a) Y, (bj a connection, ana (c) tne two superimposed I,

'1

13.4 Y-A

TRA NSFORM ATIO NS In many power systems applications it is important to be able to convert fr< Y-connected load to an equivalent A-connected load, and vice versa. For exam suppose that we have a Y-connected load in parallel with a A-connected load, shown in Fig. 13.15, and wish to replace the combination by an equivalent thr phase load. I f both loads were A-connected, this would be relatively easy since responding phase impedances would be in parallel. Also, as we saw in Exe 13.2.3, i f both loads are Y-connected and balanced, the phase impedances may I be combined as parallel impedances. To obtain Y-to-A or A-to-Y conversion formulas, let us consider the Y and connections of Fig. 13.16. To effect a Y-A transformation we need expressions Yd,, Y*f, and Y-„ of the A in terms of Y„, Y,, and Y, of the Y so that the A con tion is equivalent to the Y connection at the terminals A, B, and C. That is, i f the is replaced by the A, the same node voltages V A , V « , and V w i l l appear, and same currents Ii and I2 will flow. Conversely, a A-Y transformation is an express of the Y parameters in terms of the A parameters.

00

(b)

r

C

424

Chapter 13

Three-Phase Circuits

Section 13.4

Y-A Transformations

425

Let us begin by writing nodal equations for both circuits. I f node C it I reference, in the case of the Y network we have

Y V o — Ii

YoVt

D

Y V, + (Y„ + Y„ + Y,)V = T

D

(13.24)

z„.=

d

Y»V. - Y„V = I -Y.V,

z. =

2

z

0

Solving for V in the third equation and substituting its value into the first two tions, we have, after simplification,

c

=

B

/ Y.V» + Y.V,\ /

Y.Y,

VY. + Y . + Y . r

\Y. + Y» + Y.)

\

+

\Y + Y„ a

+

where the Z's are the reciprocals of the Y s of Fig. 13.16. The rule is as follows:

YJ

\Y. + Y. + v J

" ~

V

s

The impedance of an arm of the Y is equal to the product of the impedances of the adjacent arms of the A divided by the sum of the A impedances.

1 1

(By adjacent here we mean "on each side of and terminating on the same node as." For example, in the superimposed drawing of the Y and A, Z . lies between Z * and Z „ and all three have a common terminal A. Thus Z * and Z*. are adjacent arms of Z..)

1 2

The nodal equations for the A circuit are

(Y* + Y„)V - Y^Va = I, A

-Y^V^ + (Y* + YtcJV, = I

2

Equating coefficients o f like terms in these equations and (13.22) and solving admittances o f the A circuit, we have the Y-A transformation

Y.Y, Y, + Y„ + Y Y*, =

Y„Y Y_

c

c

Yob ~

C

(13.

+ Y» + Y

c

. Y Y„ Y„ = Y, + Y» + Y

Let us find the input impedance Z of Fig. 13.17(a). This is a problem that would have required us to write loop or nodal equations in the past, because we cannot simplify the circuit by combining series and/or parallel impedances. Replacing the 6 , 3 - , and 2-il resistors, which constitute a Y , by their equivalent A, as shown i n Fig. 13.17(b), however, enables us to solve the problem readily. Comparing Figs. 13.17(a) and 13.16(a), we see that Y„ = i , Y„ = i , and Y = { S. Therefore, from (13.23), we have

|

FIGURE 13.17 T w o equivalent circuits

C

C

5 n W v -

a

I f we imagine the Y and A circuits superimposed on a single diagram as in 13.16(c), then Y„ and Y» are adjacent to Y„», Y, and Y are adjacent to Y * , a~ on. Thus we may state (13.23) in words, as follows:

36 n

36 n

f

Z—*•

The admittance of an arm of the A is equal to the product of the admittances o f the jacent arms of the Y divided by the sum of the Y admittances. To obtain the A-Y transformation we may solve (13.23) for the Y admittana difficult task, or we may write two sets of loop equations for the Y and A circul. In the latter case we shall have the dual of the procedure which led to (13.23),] either case, as the reader is asked to snow in FYob. 13.28, the A-Y transformation

426

Chapter 13

Three-Phase Circuits

24 SI

0

(b)

(a) Section 13.4

'—

Y-A Transformations

427

.

id)

i

_

meed A-connected load with phase impedance Z2 = 12 + j9ft.Find the power delivered by the source.

\mwer 2.4 kW £+ 5+ 1~ 12

\w that the Y-A transformation of (13.23) is equivalent to S

Therefore, in Fig. 13.17(b) we have = 18 11,

Z . Z + Zc.Zc + ZcZ„ Zc t

Z*

= 6

Zc

ft,

a

= 12 ft

Thus R g . 13.17(b) may be simplified by combining parallel and series resin

Z EX A M P L E 13. 7

Z* =

= 12 ft

Zee =

Suppose that we have a balanced Y-connected load with phase impedance ZJ wish to convert it to an equivalent A-connected load. By (13.23), since Y„ Y,., Y are all equal to Y,, the reciprocal of Z , the equivalent A-connected load is balanced because c

=

V I be = V X co =

Y

2

y

Y

Z , = 3Z,

( l l

t

t

c

C

z„

The impedance of an arm of the A is equal to the sum of the products of the impedances of the Y, taken two at a time, divided by the impedance of the opposite arm of the Y.

= ll ^

Thus i f Z j is the phase impedance of the equivalent balanced A-connected load,

a

or in words:

v

Yab

Z Z + Z Z + Z Z„

1 • I f the lines in Exercise 13.3.2 each have a resistance of 0.1 ft, find the power lost in the lines. Answer 1.44 kW

which may be used to convert from Y to A, and vice versa.

EXERC ISES

™

~

—

~

™

Find the input impedance seen by the source using a Y-A or A-Y transformatiod

13.4.1

5»s-

,ow

t h i s

resultfindtheaverage

^-Xrr

|

vw , 1

T

5 ii

H

1 H

-wv—'TSOIP—f

F

1

5n

25

T*

25

EXERCISE 13.4.1

1

3

'

4

'

2

> W !R

MEASUREMENT It appears to be a simple matter to measure the power delivered to a three-phase load by using one wattmeter for each of the three phases. This is illustrated for the threewire Y-connected load in Fig. 13.18. Each wattmeter has its current coil i n series with one phase of the load and its potential coil across one phase of the load. The connections are theoretically correct but may be useless i n practice because the neutral point N may not be accessible (as, for example, i n the case of a A-connected load). It would be better, i n general, to be able to make the measurements using only the lines a, b, and c. In this section we shall show that this is, i n fact, possible and that, moreover, only two wattmeters are required, instead of three. The method is general and is applicable to unbalanced as well as balanced systems. Let us consider the three-wire Y-connected load of Fig. 13.19, which has three wattmeters connected so that each has its current c o i l in one line and its potential coil between that line and a common point x. I f T is the period of the source voltages and i , ib, and i are the time-domain line currents, directed into the loads, then the total power P, indicated by the three meters is a

1 r .Pi = (tw'a + v Jb + v„ic) dt

„~H v " , Vi = 100 V rms is delivering power to a anced Y-connected load with phase impedance Z , = 8 + , 6 ft in parallel wi.hl

a A

l a

e d

t h r e e

p h

a s e

s o u r c e

c

w i t h

b

(13.26)

0

428

Chapter 13

Three-Phase Circuits

Section 13.5

Power Measurement

429

Regardless o f the point Jt, which is completely arbitrary, we have Vox = OoN + VN* ,

Ob =

Vm + VN,

Vc, = VcN

+

VHI

J

Substituting these results into (13.26) and rearranging, we obtain

f By K CL

( t W „ + Omit + Vcnic) dt +

we have i

0

OAfcO. + J*

+ «r)

dt

+ ib + ic

a

so that

If

(VONL + vwib + Vcnic) dt

(13.27)

Thus the sum of the three wattmeter readings is precisely the total average power delivered to the three-phase load, since the three terms in the integrand of (13.27) are the instantaneous phase powers. Since the point x in Fig. 13.19 is arbitrary, we may place it on one of the lines. Then the meter whose current coil is in that line w i l l read zero because the voltage across its potential coil is zero. Therefore, the total power delivered to the load is measured by the other two meters, and the meter reading zero is unnecessary. For example, the point x is placed on line b in Fig. 13.20, and the total power delivered to the load is P

=

PA +

Pc

FIGURE 13.20 T w o wattmeters reading the total load power

A

430

Chapter 13

Section 13.5

Power Measurement

431

where PA and P are the readings o f meters A and C It is important to note t or the other o f the two wattmeters may indicate a negative reading, and thus t of the two readings is the algebraic sum. The proof of the two-wattmeter method of measuring the total three power has been carried out for a Y-connected load. However, it holds also A-connected load, as the reader is asked to show in Prob, 13.30. C

t X I K ( ISES U.S.I

11.5.2 EX A M PLE 13.8

In Fig. 13.20 let the line voltages be a balanced abc sequence with V * = 100V3/iT V rms and phase impedances given by

IJ.5.3

z, = z = z = 2

3

10 + / i o n

In Fig. 13.18, let Z, = Z = Z = 20/30" ft, and let the line voltages be a balanced abc sequence set, wit h V * = 100/0° V rms. Find the reading o f each meter. Answer 2 5 0 / V 3 W I f the power delivered to the load o f Exercise 13.5.1 is measured by the two wattmeters A and C connected as shown in Fig. 13.20, find the wattmeter readings. Check for consistency with the answer o f Exercise 13.5.1. Answer 2 5 0 / V 3 , 5O0/V5 W Find the wattmeter readings PA and P and the total power P in Fig. 13.20 i f the line voltages are as given in Exercise 13.5.1 and Z i = Z = Z3 = 20/75" ft. Check the answer by using P = iP . Answer - 7 4 . 7 , 204.1, 129.4 2

3

c

2

p

Then we have Vrf, = -V

= - 1 0 0 V 3 / - 1 2 0 = 100V3/60 V rms

k

o

0

. \AN 100V3/V3/-30° 1*A = — = ' " = 5V2/-75 A rms 0

z

i

10V2/45

Ice = 5V2/-315° A rms The meter readings are thus PA =

IV* || 1,A

0

I cos (ang

|3.t> UM< I FOR THREE-PHASE CIRCUIT ANALYSIS The analysis o f three-phase networks presented previously has been restricted to balanced systems whose solutions can be expressed in terms o f a single phase. The application o f SPICE is easily used for both balanced and unbalanced systems when applied to the entire network. A l l principles necessary for using SPICE have been presented in previous chapters.

L

- ang I*,)

= (10OV3)(5V2) cos (0° + 75°)

= 317 W

IE 13.9

and Pc = I Veil LcI cos (ang \ t - ang he) = ( 1 0 0 V 3 ) ( 5 V 2 ) cos (60° + 315°)

= 1183 W with a total o f

Consider finding the line voltage and phase current at the load for phase A of the balanced Y - Y system o f Fig. 13.21. The transmission line for interconnecting the generator and load has losses that are represented by 2-ft resistors. A circuit file for this network is 3-PHASE «

DATA

VAN

1 0 AC 120 0

VBN

2 0 AC 1 2 0 - 1 2 0

VCN 3 0 A C 1 2 0 RLOSSA

P = 1500 W As a check, the power delivered to phase A is

Pp = lYwIIUl cos (ang Yw - ang \) = (^)(5V2)cos(-30°

+

75')

2 5 2

RLOSSC

3 6 2

RLOSSN

10 0 2

RA

7 10 0 . 1

RB

5 8 10

LB

8 10 0 . 1

Since the system is balanced, the total power is P = 3P = 1500 W P

which agrees with the previous result.

*

I N RMS

120

4 7 10

LA

IX

LINE LOSSES

1 4 2

RLOSSB

RC

= 500 W

432

V - Y SYSTEM WITH TRANSMISSION STATEMENTS VOLTAGES E X P R E S S E D

6 9 10

9

10 0 . 1

SOLUTION

CONTROL

S T A T E M E N T F O R f = 6 0 Hz

AC L I N 1 6 0 6 0 •

OUTPUT

.PRINT

CONTROL

AC V M ( 4 , 5 )

STATEMENT VP(4,5I

IM(VAN)

IP(VAN)

END

Chapter 13

Three-Phase Circuits Section 13.6

SPICE for Three-Phase Circuit Analysis

433

/=60Hz

/=

120 ziQ!

Vnns

120 / - 1 2 0 ° V r

©

0

2 n

'©

10(1

2fl

^ 0

0.1 H

ion

0.1 H

•AVv-

- W v

©

fjq)

®

iy—:—^—r iy—t—^—r

120/^120-Vrrni' ©

ion

FIGURE 13.22

FREQ

VM(4,

5)

V P ( 4 , 5)

6.0OOE+01

IM(VAN)

2.049E+02

3.280E+01

3.033E+00

Y-D

UNBALANCED

STATEMENTS WITH

VAN

1 0 AC

120

0

VBN

2

0 AC

120

-120

VCN

3 0 AC

120

120

RLOSSA

1 4

1

RLOSSB

2 5

2

RLOSSC

3 5

RBC 5

7 8

CBC

7

1000UF

RAC 4

8

10

»

6 8

.PRINT

CONTROL 1 400

STATEMENT FOR f

=

400

Hz

400

AC V M ( 4 , 5 )

VP(4,5)

IM(VAN)

IP(VAN)

. END

The computer output contains FREQ 4.0O0E+02

434

Chapter 13

VM(4.

5)

V P ( 4 , 5)

IM(VAN)

1. 6 0 1 E + 0 2

2.219E+01

1.255E+01

Three-Phase Circuits

ion

•8tl

®

0

T

1000 j i F

u

Unbalanced Y-A system

13.6.1

Find the line voltage and phase current for the load of phase A of the system of Fig. 13.21 i f the load of phase C is short-circuited (called a phase fault). Answer 204.9/32.8° V, 4.07/90.54° A

13.6.2

Repeat Exercise 13.6.1 i f the 2-fJ neutral line between nodes 0 and 10 is removed. Answer 204.9/32.8° V, 5.33/82.41" A

MMARY

VALUES

0.05

OUTPUT

.AC L I N

T - L I N E LOSSES RMS

IN

12

6

LAC

S Y S T E M WITH VOLTAGES

6 3

RAB 4

n

IXERCISES

Let us find the line voltage and line current for the unbalanced Y-A sys Fig. 13.22. The transmission line losses are denoted by the 1-, 2-, and 3-i tors, respectively. A circuit file is DATA

: 12

©

Ky—?—™—?—

0.1 H

The computer output yields

•

n

3«

©

2u

©

3-PHASE

2

0

120 / 120° V rms.

FIGURE 13.21 Balanced V-V system for SPICE analysis

EX A M P LE 13.10

m

0.05 H

20 120 / 120° V rms

400 Hz 120 / 0 ° V r r n i

IP

(VAN)

-1.627E+02

In this chapter we have considered three-phase circuits, which are excited by a balanced three-phase sequence of voltages. This is the common system used in the power industry to supply electricity to homes and businesses. The sources are generally Y-connected, consisting of three generators connected in a Y w i t h phases displaced 120°. A system may be Y-Y, in which case the load consists of three impedances also Y-connected. There may be three wires connecting the three-phase source to the load, or there may be a fourth wire or neutral line. I f the load impedances are identical, the system is a balanced Y-Y system. We may also have a Y-A system in which the three-phase load is connected in a delta. A n analysis of a three-phase circuit often entails finding phase currents, phase voltages, line currents, line-to-line voltages, and three-phase power. We may convert a Y to an equivalent A, or vice versa, by means of a Y-A transformation. This transformation may also be used in ordinary circuits, enabling us to find input impedances without writing Kirchhoff's laws. Three-phase power measurements may be made with three wattmeters, one for each phase of the load. As we have seen, however, it is also possible to find the total power with only two wattmeters. Finally, SPICE is also very useful in three-phase circuit analysis because the numbers involved are often quite complicated.

Section 13.7

Summary

435

V A irystem o f Prob

PROBLEMS 13.1

I f in F i g . 13.4 V „ = V the

impedance

between

= 100/0° V r m s ,

r t

terminals A-N

13.11

is

10/60° fl, and that between terminals N-B is 10/—60° 11, find the neutral current L * 13.2

I n F i g . 13.5, let V, = 100/0° V r m s , Z , = 5 + y 5 n , Z = 0.5 a , Z , « 1 f l , and Z. = 10 - j5 fl. Find the average power absorbed by the loads, lost i n the lines, and delivered by the sources.

A balanced Y-connected source, 200/0° V rms, positive sequence, is by four perfect conductors to an unhal Y-connected load, Z A « = 8 + j 6 11 >20 fl, and Z,.« = 10 fl. Find the neuli rent.

2

13.3

13.4

A balanced Y - Y three-wire, positive-sequence system has V„j, = 200/0° V rms and frequency 200 rad/s. I f the load i n each phase is a series connection o f a 4 0 - A resistor, a 0 . 1 - H inductor, and a 100-/U.F capacitor, find the line currents and the power delivered to the load.

13.12

I f the neutral wire is removed in the sy4 Prob. 13.11 , find the line currents

13.13

I f i n Exercise 13.2.3, Z , = 3 - j4 11, 3 + j4 ft, and the line voltage is 1 0 0 V 3 V r m s , find the current l in etch I t

13.14

I n F i g . 13.10 a balanced, positive-sequence source has V„t = 120/0° V rms and L A = 10/-60° A rms. Find Z and the power delivered to the three-phase load. P

13.5

13.6

A balanced three-phase Y-connected load draws 1.2 k W at a power factor o f 0.6 leading. I f the line voltages are a balanced 200-V rms set, find the line current hA balanced Y-Y system with Z = 3V3 /30° fl delivers 9.6 kW to the load. Find the line voltage V and the line current h •

13.15

P

I n F i g . 13.10, the source is balanced, w i t h positive phase sequence, and V«, = 100/0° V rms. Find Z i f the source delivers 3.6 k W at a power factor o f 0 . 6 leading. P

13.8

A balanced Y - Y three-wire, positive-sequence system has V „ = 200/0° V rms and Z , = 3 + j i 11. The lines each have a resistance o f 1 fl. Find the line current l , the power delivered to the load, and the power dissipated i n the lines. L

13.9

I n F i g . 13.13 the source is balanced with pdH live phase sequence and V „ = 100/0° V nH I f the phase impedance is 3 V 3 / 3 0 I I , • the line current and the power delivered t
13.16

L

13.7

A balanced three-phase Y-connected draws 3 k W at a power factor o f 0.8 l a g g l M ^ balanced Y o f capacitors is to be placed mf^l allel w i t h the load so that the power facltj^H the combination is 0.85 lagging I f the quency is 6 0 H z and the line voltages i^H balanced 200-V rms set, find the capacita^H required.

A balanced Y-connected source, V„, = 200/0° V rms, positive sequence, is connected by four perfect conductors (having zero impedance) to an unbalanced Y-connected load, Z = 10 fl, Z,s = 10/-30° fl, and ZCN = 2 0 V 2 /75° fl. Find the neutral current and the power absorbed by the load.

13.17

I n F i g . 13.13 the positive sequence s y s t e ^ H V „ = 200/0° V rms. Find Z„ i f the sourcl M livers 2.4 k W at a power factor of 0.8 lag||H I n the Y - A system shown, the source is p f l tive sequence w i t h V™ = 100/0° V r m s ^ H the phase impedance is Z = 3 - j4 11. H f l the line voltage V , the line current /,., umiW power delivered to the load P

436

I f the neutral w i r e is removed in the system o f Prob. 13.9, find the line currents.

Three-Phase Circuits

A

13.25

balanced

three-phase

positive-sequence

source w i t h V „ = 240/0° V r m s is supplying a parallel combination o f a Y-connected load

|.Vs

and a A-connected load. If the Y and A loads

1 ind the line current /, and the

.1,.,.., voltages i f V.,„ is the reference,

are

balanced

balanced Y - A system, the load draws

8 -

jt fl and 24 + ; 2 4 fl, respectively, find

I a power factor o f 0.8 lagging. L i k e

the line current /, and the power supplied by

us are connected across each

the

phase

HHSC ihe load power factor to 0.9 lagin,l |hc value o f capacitance required i f juency is 6 0 Hz and the line voltages

phase

assuming

balanced A 'source'wUh

13.26

impedances

perfectly

of

conducting

three phase

positive-sequence

V * - 240/0°. V rms supplies

43 2 kW at unity power factor to a parallel

Y A system o f Prob. 13.17 the source

1

source,

with

lines.

I V rm§. V „ = 100/y V

rms and Z

combination o f a balanced Y-connected load

=

w i t h phase impedances 4 + ) 4 fl and a bal-

I I . Find the line voltage, the line cur-

anced A-connected load. Determine the phase

P

he load current magnitudes, and the delivered to the load

impedances o f the A-connected load. 13.27

i» line current h ir power delivered to the load in Prob. If the magnitude o f the source voltages

V rins and Z = P

~

Find the line currents L * , L>». Lc 100/0° Vrms

"

•need three-phase, positive-sequence w i t h V * = 200/0° V r m s is supplying onnected load, ZAB = 20 fl, Z « c = • fl, and Z o , = 50/30° fl. Find the line currents and the power delivered ! load (Assume perfectly conducting

/ion 100 / -120 Vrms b

-/ion

:

20 n lanced

three-phase,

positive-sequence

100/120° Vrms

: w i t h V * = 200/0° V rms is supplying onnected

load,

Z « = 50 fl, Z » c =

, 2 0 fl, and ZCA = 30 -

y 4 0 fl. Find

PROBLEM 13.27

•it currents and the power absorbed by id.

-o-

120/0° Vrms

1 n ^vVv9-/12n

.._e_s.

120 / -120° V rms

1 n

9-ma

-Wv-

9

-e-

120/120° Vrms

-;T2n

1 n

-WvPROBLEM 13.21

PROBLEM 13.17

Chapter 13 Chapter 13

=

r

L

W

13.10

13 17, Z

11 ,,,,1 (he power delivered to the load

Problems

437

13.28

Derive (13.24).

13.29

For a balanced three-phase phase voltages are

13.34 system, i f the

««(»)

= V cos tut

Mt)

= V cos (cur -

v (t)

= V„ cos (ait - 240°)

m

120°)

m

c

Find the readings l\d P of the wallir and the total power delivered to the In the source is a balanced Y-connected sequence source w i t h V „ = 100/0° V and the phase impedances are each 30/30° fl. B

i ilns ii-sult show that tan e =

V3

Pc + PA

Find the readings o f the wattmeters connected as in Prob. 13.34 for the system o f Prob. 13.23.

13.39

Repeat Prob.

mi from the t w o wattmeter readings,

13.38 for the system of Prob.

13.27.

we may find the power factor, cos 9, o f h*.

13.40

Find the wattmeter readings.

13.36, find the pf of the load i f

L - Pc, (b) PA = -Pc,

then the phase currents are

13.38

• 0, and (e) PA =

(C) PA = 0 , (d)

2P C

««(') = L cos (cur - 6) i„(t) = /„ cos (a>t - 6 ic(t) = /„ cos (at - 6 -

120°) 240°)

Show that the total instantaneous power, p(t)

= t)„i„ + vtit, +

vi

c c

is a constant given by

PROBLEM 13.34

pit) = iv i cot e

13.35

m m

which is also P, the total average power. [Suggestion: Recall that cos a + cos (a - 120°) + cos (o - 240°) = 0.) 13.30

Show that P given by (13.26) is equal to the total average power delivered to a A-connected load.

13.31

I n the system o f F i g . 13.20, the line voltages are a balanced, positive-sequence set, w i t h V * = 200/0° V rms, and Z, = Z = Z = 10/30° fl. Find the power delivered to the load (a) by finding the readings o f the t w o wattmeters and (b) by P iP .

For the system shown, the line voltages I N balanced positive sequence set w i t h V«t 300/0° V rms. F i n d the meter readings P„» P and the total power delivered to the lo.nl Z , = 10/30° fl. B

x

2

COMPUTER APPLICATION PROBLEMS 13.41

Use SPICE to solve the system o f Fig. 13.21 i f the 0.1-H inductor o f phase A is replaced by

13.42

Use SPICE to find the current o f the phase A source i f phase voltage C is zero (phase C gen-

a 1000-ftF capacitor.

3

erator fault) in the Y - Y system o f Fig. 13.21. 13.43

Use SPICE to find the current in the 1 2 - f l load o f Fig. 13.22 i f the load between phases B

13.44

Repeat Prob. 13.42 for the system o f Fig. 13.22.

and C (8 fl and 1000 p.F) is open-circuited.

p

13.32

Repeat Prob. 13.31 i f Z , = Z = Z = 10/60° fl. Note that in this case wattmeter C reads the total power since wattmeter A reads zero.

13.33

Show that in Fig. 13.20 i f V,* = V [a, V = Vja - 120°, V „ = V /a - 240° V rms, and Z i = Z = Z = | Z1/60°, then wattmeter A reads zero and wattmeter C reads the total average power delivered to the load,

2

L

3

k

PROBLEM 13.35

L

2

3

P =

VI 2|Z|

13.36

Show that i f a balanced, positive-sequeixi source is connected by three perfectly conJ ducting wires to a balanced load having phaal impedance Z , = | Z , the t w o wattmeteri] connected as i n Fig. 13.20, read, respectively] PA and P , where C

This is a generalization o f the result o f Prob. 13.32.

438

Chapter 13

Three-Phase Circuits

cos (30° + 8) cos (30° -

0)

Chapter 13

Computer Application Problems

439

1

4

Complex Frequency and Network Functions Ln the previous chapters we have considered resistive circuit analysis, natural and forced responses of circuits containing storage elements, and, in particular, ac steady-state analysis. The excitations we have considered were, for the most part, constants, exponentials and sinusoids. In this chapter we consider an excitation, the damped sinusoid, which includes all these excitations as special cases. From this function we develop generalized phasors and general network functions which include, as special cases, the phasors and impedances of Chapters 10-13. The network functions are expressed in terms of a complex frequency that includes the frequency jco as a special case. The concepts of complex frequency and general network functions enable us to combine all of our earlier results into one common procedure. Both the natural and forced responses of a circuit may be found from its excitation and its network function, as we shall see. In addition, the network function may be used to determine the frequency-domain properties of the circuit, which are the subject of Chapter 15.

14.1 I HI DAMPED SINUSOID If Thomas A. Edison has a rival for the title of the world's greatest inven- He [Tesla] was the greatest ability to recite by heart entire bocj tor, it is certainly the Croatian-Ameri- inventor in the realm of and poems. He spent two years I can engineer, Nikola Tesla. When the electrical engineering. the Polytechnic Institute of Gr« tall, lanky Tesla arrived in the United W. H. Eccles Austria, where he conceived the id« States in 1884, the country was in the of the rotating magnetic field that WM middle of the "battle of the currents" the later basis for his induction motfll between Thomas A. Edison, promoting dc, and At this point in Tesla s life his fatha George Westinghouse, leading the ac forces. Tesla died, and he decided to leave school, taking a job • quickly settled the argument in favor of ac with his Paris with the Continental Edison Company I wo] marvelous inventions, such as the polyphase ac years later he came to America, where he remain*! power system, the induction motor, the Tesla coil, and until his death. During his remarkable lifetime he h M fluorescent lights. over 700 patents, settled the ac versus dc dispuHJ Tesla was born in Smiljan, Austria-Hungary and was primarily responsible for the selection of 69 (now Yugoslavia), the son of a clergyman of the Hz as the standard ac frequency in the United Stattl Greek Orthodox Church. As a boy Tesla had a talent and throughout much of the world. After his death hfl for mathematics and an incredible memory, with the was honored by the choice of tesla as the unit of magi netic flux density. •

In this section we consider the damped sinusoid. v = V e"' cos (wt + 0) m

which is a sinusoid, like those of the previous chapters, multiplied by a damping factor e . The constant a (the Greek lowercase letter sigma) is real and is usually negative or zero, which accounts for the term damping factor. The damped sinusoid contains, as special cases, most of the functions we have considered thus far. For example, if a = 0, we have the pure sinusoid m

v=

V„ cos (tat + <£)

(14.2)

of the previous chapters. If w = 0, we have the exponential function v = Voe"'

(14.3)

and if a = m = 0, we have the constant (dc) case v = V

0

(14.4)

where, in both (14.3) and (14.4), V = V cos . These functions are sketched in Fig. 14.1 for the various cases of a and to. 0

440

(14.1)

Section 14.1

The Damped Sinusoid

m

441

dimensionless unit, neper (Np), was used for at Thus < ^ t h e _ ^ j N f f g ^ nepers per second (Np/s). The neper was chosen in honor of the Scottish mathematician John Napier (1550-1617), who invented logarithms. 14 1

As an example of a circuit having a damped sinusoidal excitation, let us find the forced response i in Fig. 14.2. The loop equation is 2* + - = 25c dt 5l

lie

1

cos

1

cos It

(145)

It V ( *

F I G U R E 14.2 Circuit excited by a damped sinusoid

Trying as a forced response I

=

e~'(A cos 2l + fl sin It)

which is the excitation and all its possible derivatives, we have le~'{-2A

sin It - A cos It + 2B cos It - B sin It) + 5e~'{A cos 2r + B sin It) = 25e"' cos It

Since this must be an identity, we have 3A + 4fl = 25 - 4 A + 35 = 0 or A — 3 and B = 4. Therefore, the forced response is i = e~"'(3 cos 2l + 4 sin 2r) A <

c ) o

=

or equivalently,

(f)»-w-0 F I G U RE 14.1 Various cases of (14.1) 0

i = 5e ' cos (2» - 53.1°) A

(14.6)

As expected, the forced response is, like the excitation, a damped sinusoid. As we see from Fig. 14.1. 0 represents growing oscillations (b) or growing exponential (e), and a < 0 represents decaying oscillations (a) or a dec k i n g exponential (d). Finally, cr = 0 represents ac steady state (c) or dc steady state (t). t

The units of to are radians per second and of d> are radians or degrees, as betore. Since at is dimensionless, cris in „ f 1 per second (1/s). This unit was encountered earlier, in connection with natural frequencies, in Chapter 9 where the u

442

Chapter 14

l t s

EXERCISES 14.1.1

Excite the circuit of Fig. 14.2 by the complex function t>, = 25e"e*'

0

Complex Frequency and Network Functions

= 25e ,

Section 14.1

The Damped Sinusoid

,+j1)

*

443

and show that the forced response is i, = ( 5 / - 5 3 . 1 V - " * 14.1.2

nusoids? It seems plausible that we can, because the properties of sinusoids that made the phasors possible are shared by damped sinusoids. That is, the sum or difference of two damped sinusoids is a damped sinusoid, and the derivative, indefinite integral, or constant multiple of a damped sinusoid is a damped sinusoid In all these operations only V and d> may change. This is exactly the case with undamped sinusoids

1

Show that if i, is the response to Ci in

m

2— + 5ii = ui at

then Re i, is the response to Re v,. Apply this result to the functions of Excrcl 14.1.1 to obtain i given by (14.6). 14.1.3

Find the forced response t using the methods of Exercises 14.1.1 and 14.1.2 if dv 2

-TT +

dt

2

v = V e" cos (tot + d>) m

dv

= RelV^e"^-**']

2— + v = Se 'cos 2t dt

= Re[V„,

Answer —2e~' cos 2/ 14.1.4

Let us see what follows if we write (14.10) in the form analogous to (14.9). That is,

If we define the quantity s = a+ja>

Use the method of Exercise 14.1.3 to find the forced response v if

then we have

v = Re(Ve")

where i = 4
/ I n s n w 2 e c o s (« + 30°)

14.2 COMPLEX FREQUENCY A N D GENERALIZED PHASORS In Chapter 10 we considered circuits with sinusoidal excitations such as v, = V cos (tot + d>) m

(14.71

(14.11)

(14.12)

where V is the phasor of (14.8). Evidently, since V is the same for both, the undamped sinusoid given by (14.9) is identical in form to the damped sinusoid of (14.12). The only difference is that the number jto is used in one case and the number s is used in the other. Obviously, then, we may do anything with the damped sinusoids that we did with undamped sinusoids as long as we use s instead of jto. We may define the phasor V of (14.8) to be the phasor representation of v in (14.10) and use it for damped sinusoidal circuit problems in exactly the same way that we used it for the sinusoidal problems. In the damped sinusoidal case we may wish to write the phasor as \{s) to distinguish it from the undamped sinusoidal case, \(j
which we also wrote in the equivalent form The damped sinusoid

v, = Re(V„,«>*e*")

t; = 25e-' cos It V

(14.13)

Using the phasor representation of t>,, of Fig. 14.2 has the phasor representation V = V„e*

= V [£ m

(14.8)1

we may also write

V = V ( i ) = 25/0!

t>, = Re(Ve**) (14.9) The phasor representation was extremely useful in solving ac steady-state circuits in which the voltages and currents were sinusoids of the form of (14.7). In this chapter we wish to consider circuits in which the excitations and forced responses are damped sinusoids, such as v = V . ^ ' c o s (tot +
(14.10)

v is given by (14.13). Some authors prefer to call \(s), corresponding to v (t) in (14.10), a generalized phasor, even though it is identical to the phasors of sinusoidal functions. It is, however, a function of a generalized frequency, namely s, given by (14.11). Since s is a complex number, it is more often called a complex frequency. Its components are

A good question at this point is: Can we define phasor representations of damped si-

a = R e s Np/s

nusoids that will work as well for us as the phasors, such as (14 8). of undamped si-

444 Chapter 14

(14.14)

where s = - 1 + y'2. Conversely, if V is given by (14.14) and s is as specified, then

Complex Frequency and Network Functions

to = I m s rad/s Section 14.2

Complex Frequency and Generalized Phasors

445

which have units of frequency, and, indeed, s is the coefficient of t in an exponetH function, as was the case in Chapter 9 for the natural frequencies of a circuit f l units of s are I /second and are sometimes called complex nepers per second or i ufl plex radians per second.

IMPI DANCE AN D ADMITTANCE Because of the identical form of the phasor representations for sinusoids and damped sinusoids, we may find forced responses to damped sinusoidal excitations using phasors precisely as we did in the previous chapters. All the concepts and rules, such as impedance, admittance, K C L , K V L , Thevenin's theorem, Norton's theorem, and superposition, carry over to the damped sinusoidal case exactly. We need only to use s = a + jto rather than jto. It follows that in the J domain, the phasor current \(s) and voltage \(s), associated with a two-terminal device, are related by

It might be worth noting at this point that a function which can be written! the form / ( f ) = K,t' ' 1

+ Kit'*' + • • • + K.e'"'

where the K, and s, are independent of r, may be said to be characterized by complex frequencies l\, si,.

. . , s„. For example, writing (14.10) in the form

2

)

results in

\(S)

where Z(s) is the generalized impedance, or simply impedance, of the device. We may obtain Z(s) from Z(jto), the impedance in the ac steady-state case, by simply replacing jto by s. For a resistance R, the impedance is, therefore,

where K, = V e*/2 and K = V e~»/2 = Kf (the complex conjugate). Thus possesses not one but two complex frequencies, namely *i = a + jto • s = s* = a - jto. This concept of complex frequency is consistent with that dj the natural frequencies considered in Chapter 9. m

2

= Z(5)I(J)

m

2

R

EXERCISES 14.2.1

14.2.2

In the case of an inductance L , the impedance is

Find the complex frequencies associated with (a) 5 + 3e *', (b) cos tot, (c) sin (oil + fl), (d) 6e " sin (4t + 10°), and (e) e~'(l + cos 2t). Answer (a) 0, - 4 ; (b) ± jto; (c) ± jto; (d) - 3 ± j4; (e) - 1 , - 1 ± j2 Show that if i is a damped sinusoid,

Ms)

= sL

and for a capacitance C , it is

i = I me" cos (tot + d>) then r, defined by

Zeis) =

1 sC

v = L ~ + Ri dt In a similar manner the admittances are, respectively,

is also a damped sinusoid of the same complex frequency. Answer v = I V(R m

14.2.3

+ trLf

+ to L 2

2

e" cos (tot +


-1

„

m

L

Find s and \(s) if v(t) is given by (a) 6, (b) 6
Find c(<) if (a) V = 8/0!, * = - 3 ; (b) V = 5 / 1 5 ° , s = y4; and (c) V = 6/30°, s = - 3 + ;'2. Answer (a) 8
3

VL(S)

= 4 . sL

Y c W = sC

3

Answer (a) 0, 6 / 0 ° ; (b) - 2 , 6/CT; (c) - 3 + J 4 , 6/10°; (d) j2, 6 / 1 0 ° 14.2.4

Y * « = ^ = G,

. |

M PLE 14.3

Impedances in series or parallel are combined in exactly the same way as in the ac steady-state case, since we merely replace jto by *. Let us reconsider the circuit of Fig. 14.2, whose phasor circuit is shown in Fig. 14.3, where J = — 1 + j2. The impedance seen from the source terminals consists of the impedances 2s and 5 in series and thus is given by Z(s) = 2s + 5 n

446

Chapter 14

Complex Frequency and Network Functions Section 14.3

Impedance and Admittance

447

F I G U R E 14.3 Phasor circuit of Fig. 14.2

Therefore, since the input voltage phasor is V,(j) = 25/0° V we have V,W

IW

2s + 5

_ _ 25/0° = 5/—53,1° A 2 f - i + jl) + 5

Thus in the time domain the forced response, as before, is «(») = 5 e - ' c o s ( 2 / - 53.1°) A EX A M P LE 14.4

Let us consider the time-domain circuit of Fig. 14.4(a), where it is required to fi the forced response t>„(r) for a given damped sinusoidal input v,(t). The phasor cuit is shown in Fig. 14.4(b), from which we have the nodal equations,

Q i i,)v,-|v,-v,-i,v.-o +

+

V, = 0

(b)

We note that V - V„/2 since the op amp and its two connected resistors constitul a VCVS with a gain of 2. Therefore, eliminating V, and V , we have 2

F I G U R E 14.4 (a) Time- domain circuit; (b) its phasor circuit

2

V.W =

16 jV (*) s* + 2s + 8 f

(14.1S1

Let us find the forced response i in Fig. 14.5 if v i = &e~' cos l V g

If we have v,(t) = e~* cos 4/ V

and 1,2 =

then s = -2

+ j4 and \,(s)

= 1/0°.. Thus from (14.15) we have

2e ' A 5

Since the complex frequencies of u,i and i , are different, we must use superposi2

\(s)

= V2/135° V

tion. That is, i =

and therefore C ( r ) = V2 e-- cos (4i + 135°) V

I'I +

k

where i, is due to v , acting alone and l'j is due to i,2 acting alone. g

21

448

Chapter 14

Complex Frequency and Network Functions

Section 14.3

Impedance and Admittance

449

[XEROSES 14 1.1

Find the forced response 1 if v, - 4*"' cos » V. Answer 2V2 e-' cos (t - 45°) A

F I G U R E 14.5 Circuit with two sources

The phasor circuits for i„ = 0 and u„ = 0 are shown in Fig. 14.6(a) and ( respectively. The phasor currents are I , and I , as shown. In Fig. 14.6(a), using c rent division, we have 2

I, =

8/0°

which, since s = - 1 + jl

EXERCISE 14.3.1

4/s

1 + {[(2* + 4)(4/*)]/(2J + 4 + 4/s)}

2s + 4 + 4/s

14.3.2

for this case, becomes

Find the forced response t; if o „ = 4e- " cos (t - 45°) V and 1,2 = 2e~' A . Answer 2\Pi e ' cos (t + 90°) + 4e'' V 1

I, = 2 V 2 / - 4 5 A 0

Therefore, the forced component due to v

alone is

tl

I'I = 2V2V' cos (t - 45°) A In Fig. 14.6(b), again by current division, we obtain \{[1(4A)]/(1 which, since * = - 5 , is

1 1 / ( 1

+

4

A

k m

)

+ 4/*)} + 2s + 4 /

( 2 Z 2

-'

EXERCISE 14.3.2

14.3.3

I2 = 0.8/0° A Therefore, the forced component due to i, alone is 2

14.3.4

k = 0.8*-" A The complete forced response of Fig. 14.5 is therefore i = 2V2

cos (t - 45°) + O.&e* A

For the circuit of Exercise 14.3.2, take 1,2 = e~* cos (1 + 45°) A , leaving o as it is, and find c by (a) using superposition, (b) writing a single node equation, and (c) using source transformations. Answer 4e " cos (t + 135°) V In the phasor circuit for Exercise 14.3.2 with 1,2 = 0 and v„ as it is, replace everything except the capacitor by its Thevenin equivalent and use the result to find \(s), the phasor of v. Note that the result leads to the first component of v given in Exercise 14.3.2. Answer fl

:V,(j), Z*(j) = + 4

F I G U R E 14.6 Phasor circuits associated with Fig 14 5 4

2) n, + 4

\(s)

s +2 v,(*) + 3s + 4

-k

^ vVv

14.4

8/0°

NETWORK FUNCTIONS

(a)

450

Chapter 14

(b)

Complex Frequency and Network Functions

A generalization of impedance and admittance is the so-called network function, which, in the case of a single excitation and response, is defined as the ratio of the response phasor to the excitation phasor. For example, if \(s) and l(s) are the

Section 14.4

Network Functions

451

from which

voltage and current phasors associated with a two-lerminal network, the impedance

„.

V.W = W»)V W (

( 1 4 1 8 )

In general, stnee , - complex, H « is complex. Therefore, we may write, in is the network function if Us) is the excitation and V(.v) is the response. On the hand, if \(s) is the input and l(s) is the output, the input admittance

P

°

l a r f 0 r m

'

HWHBWIZ!

where | H(s) | is the amplitude a « l . the phase of

Therefore, if

I W

VW

y.(s) = v4±

is the network function. The foregoing examples are special cases in that both the input and output measured at the same pair of terminals. In general, for a given input current voltage at a specified pair of terminals, the output may be a current or voltage al place in the circuit. Thus the network function, which we designate in general) H(s), may be a ratio of a voltage to a current (in which case its units are ohimi, current to a voltage (with units of Siemens), a voltage to a voltage, or a current Id] current. In the last two cases, H(s) is a dimensionless quantity. If the input and c put are measured at different terminals, the network function is also called a tranr function. EX A M P L E 1 4 . 6

Consider the circuit of Fig. 14.4(b), which was analyzed in the preceding section, \„(s) is the output phasor and V„(5) is the input phasor, then, by (14.15), the work function is

v.(5) = v J H « | Z £ + J 1

114.8

'

2 0 )

( 1 4 2 1 )

ouJuTis that of the input plus that of the network function.

!,(*) in (14.16) is given by Us) where V,(s) = 8 / 0 ° V a n d , -

= H(*)V,(s)

- 1 + ; 1 , as given in the preceding section. There-

fore,

2 H(-l

+ ji)

16

H(*) =

( 1 4

= piTTlF+'oX-i

+

>

l )

, ^ / - 4 5 ! 4 '

+ Zs + 8 and

EX A M P L E 14 . 7

If the source V , = 8/0° V is the input and I> is the output of Fig. 14.6(a), we m show that H(5)

Us) V,(*)

s

2

+ 6s + 10

I, = ( ^ ) ( 8 ) / 0 _ J l « !

= 2V2 / - 4 5 ° A

(14.1

In the former example H(.v) is dimensionless and in the latter case its units Siemens. The network function H(s) is independent of the input, being a function only of the network elements and their interconnections. Of course, when the input It specified this determines the value of s to be used in a given application. From 4 knowledge of the network function and the input function, we may then find the output phasor and subsequently the time-domain output. To be specific, suppose thai \,(s) is the input and \„(s) is the output. Then

as was obtained earlier.

the network function

^

H,t» = w

h

e

r

e

=7-

, = - 1 + j 1 and % = 8/0! V. Then from Fig. 14.6(b) (with V „ - 0), we

have the network function

lUW

452

Chapter 14

Complex Frequency and Network Functions

Section 14.4

Network Functions

=

r

453

= 2/V

where s = - 5 and l

g2

A

.

W

e

m

t h e n

find

,

I S A N D ZEROS In general, the network function is a ratio of polynomials in s with real coefficients that are independent of the excitation. To illustrate this, let us consider the example of Fig. 14.2, described by

EXERCISES

2 ^ + 5i = v 14.4.1

Given the network function

HO)

find ^ f o r c e d A » „ 14.4.2

(a)

2 4 e

J

V,0)

and the input

+

2

4s + 5

= 2/0°

response M r ) if ( ) , = - , (b) , . a

- ' ; (b) - 2 * - * i „ , ( 2

S

;

dt

4Q + 5)

2

3V2

c )

-

4

+ yi, a„

d

( C )

I

cos (3r - 135°)

S

+

3

s

+

1

3

3D

w

1 u o + 2 ) 0 + 3)

;

( c )

(2* + 5)Ie" = V
Find H(s) if the response is (a) 1,0), (b) ! , ( , ) , and (c) V . ( i ) . Answer (a) .,M 0 + IX* + 2 ) 0 + ) • o +

where i is the output and v is the input. Using the same technique with complex forcing functions that we used in Chapter 10, we note that if v = \e", the output must have the same form, namely 1 = Ie", where, of course, V = \(s) and I = l(s) are the phasor representations of v and 1. Substituting these values into the differential equation, we have

1 (7+1x7+"

1^ 2s + 5

V

In the general case, if the input and output of the circuit are e,-(/) and v„(t), respectively, the describing equation is

1 H

1 Hf

EXERCISE 14.4.2

14.4.3

Find HO) if the response is v. Use the result to find the forced response if v, 1 5 cos t V. Anww

s

.

~f , 8 cos (/ + 4s + 4 S

143.1°) V

The a's and b's of course, are real constants and are independent of o . As before, if o, = V,e", the output must have the form v„ = \e", where \i(s) and V „ 0 ) are the phasor representations of t>, and v„. Substituting these values into (14.22), we have (

(a„s

n

+ a„-,s"~'

+ •••+«!* +

ao)V e"

= (i.s" + b

EXERCISE 14.4.3

0

m

,s"

1

+ • • • + b,s + baHVit*

(14.23)

From this we obtain the network function V„(J)

v,0)

8fi

"If

_

= HO) =

b„s" + *»-,*""•' a.s" + a„.

b (s - zi)Q - z ) . . . Q -

z)

a„(s - p,)(s - p ) . . . 0 -

P.)

m

2

2

Chapter 14

Complex Frequency and Network Functions

0

(14.24)

which is a ratio of polynomials in s. We may also write the network function (14.24) in the factored form

HO)

454

+ • • • + b,s + b + • • • + a\s + ao

Section 14.5

Poles and Zeros

m

(14.25)

455

In this case the numbers z„z „ are called zeros of the network f u m h . J cause they are values of s for which the function becomes zero The number* Pz,, . . . , p„ are values of * for which the function becomes infinite and ared poles of the network function. The values of the poles and zeros, along with (hit ues of the factors a. and b , uniquely determine the network function 2

z

rXIK < ISES 14 V I

m

II the / cms of H U ) are s = - 1 , - 1 ± y l , the poles are s = - 2 , - 1 ± y'2, and H(0) = 4, find H ( 5 ) . 20(5 + 1 )(5 + 2s + 2) Answer — , rr ~ 2

EX A M P L E 14.10

(s + 2 ) ( 5 + Is + 5) 2

The network function 6 ( 1 + 1)Q + 25 + 2)

14.5.2

Draw the pole-zero plot of the network function of Exercise 14.5.1.

14.5.3

If the describing equation of a circuit is d Vo dv , _ dv>i _ + 4 — + 13*, = 2 - + 4 c

2

s(s + 2){s

W

=

2

+ 45+13)

2

6 ( 5 + \)(s + 1 + j\)(s

+ 1 -

jl)

s(s + 2)(5 + 2 + ; 3 ) ( 5 + 2 -

y3)

0

find H(.v) and the poles and zeros. 2J + 4 Answer H(s) = -z : — : poles: - 2 ± y'3; zeros: —2, infinity 5 + 4 5 + 1 3

has zeros at - 1 , - 1 + Jl, and - 1 - y l and poles at 0, - 2 , - 2 + y'3,1 —2 — y'3. As is generally the case, because the as and b's of (14.24) are rj complex poles or zeros always occur in conjugate pairs. Since the network funcf is a ratio of a third-degree to a fourth-degree polynomial, it approaches zero as .T comes infinite. Thus we also have a zero at 5 = » . If the numerator were higher degree than the denominator, s = would be a pole. 0 0

The poles and zeros of a network function may be sketched as a pole -id plot, which is simply the s-plane, consisting of a- and y'oi-axes, with the poles | zeros located on it. Zeros are represented oy a small circle and poles by a sir cross. As an example, the pole-zero plot of (14.26) is shown in Fig. 14.7. The ues of a are on the horizontal (a) axis and those of to are on the vertical (so-caL jto) axis. A s we shall see in Chapter I S , pole-zero plots are useful in consider) frequency-domain properties of circuits.

F I G U R E 14.7 Pole-zero plot -2+/3X

-3 •2

o — i

9

a -2

I n

- 1

0

O

-1

-1 - ; 1

14.6 T H E NATURAL RESPONSE FROM THE NETWORK FUNCTIO N As we know from our previous experience, an output of a circuit consists of the sum of a natural and a forced response. In the last few chapters we have been concerned entirely with finding the forced response, and we have seen that the phasor technique enables us to do this in a very easy, straightforward way in the cases of sinusoidal, damped sinusoidal, and exponential excitations. In power systems studies the forced response is, of course, an ac steady-state response and is always present. Therefore, the forced response is usually of more interest than the natural response, which is transient and gone after a very short time. With damped sinusoidal excitations, on the other hand, both the forced and natural responses are transients. (In an actual circuit the natural response must be a transient, for otherwise we would have either a sustained or a growing response without an external excitation.) Therefore, the natural response assumes more importance, relative to the forced response, than it does in ac steady-state cases. In finding natural responses in Chapters 8 and 9 we considered only first- and second-order circuits, for the simple reason that our methods were applied to the describing differential equations and became more difficult to use as the order of the circuit increased. However, as we shall see in this section, the natural response may be obtained relatively easily from the phasor representation. The results of the preceding section illustrate that we may obtain the network function quite readily from the describing equation. For example, if (14.22) is the describing equation, the network function (14.24) is

•-2 -2-/3

x

|_

„ / ,! _ V.(.) _ N(s) - W) ~ o w

3

H ( , )

456

Chapter 14

Complex Frequency and Network Functions

Section 14.6

The Natural Response from the Network Function

( 1 4

2 7 )

457

where V„ and d> are constants, then, by (14.27), with the numerator

v„(5) = (v„Z4>)H(5)

N(s)

= b s"

+

m

+ • • • + b,s

+ b

Q

Therefore, smce V , , ) has no

and the denominator

l^rbetrandTe

o

D(s)

= OnS" + a -is"~'

+ • • • + as

m

t

+ ao

(l
The last two expressions are merely the right member and left member, respccttof, of (14.22) with the derivatives replaced by powers of 5. EXAMPLE 14.11

2

dt

MPI1

14.12

the natural response may be

frTwhich

are the p o * o,

constructed.

Let us find the complete response i of the citcuit of Fig. 14.8. Using current division, the phasor 1(5) of t is given by

If the describing equation is d v, ,

Zsf,

n a L , frequences

mtjt i. + 4— + dt

2

iv = c

dt

then the network function is

H(*) =

v.(*) _ V,(5)

2* + 1

YM

_

5

"

(5 + 1)(5 + 12)

2

3s + 25 + 6

since V , = 2 / 0 ° . Since the poles of I are 5 = - 1 , - 1 2 , the natural response is

+ 45 + 3

s

35

x

12 + {[35(25 + 6)]/(35 + 25 + 6)}

W

2 — + u,

i„(f) = A,e-' + A e2

A

,2>

The process is reversible if there has been no cancellation of common termt the network function. For example, it is a simple matter to write down (14.' given (14.31). In general, from (14.28) and (14.29) we may reconstruct the describing tion (14.22). In the special case of (14.29) equated to zero D(s)

= 0

(14.

replacing powers of 5 by corresponding derivatives of v„ results in the homogen equation of the system d"Vo " dt"

+

a.

+ fl| —

dt"

+ OoVo =

F I G U R E 14.8 Second-order circuit

For $ •

0

I = 0.16/12.7"

as discussed in Chapter 9. Therefore, (14.32) is the characteristic equation, and i' roots are the natural frequencies of the circuit. Since these roots, by (14.29), f also the poles of the network function, we see that the natural response of the circ' is o. = A,e'i'

- 2 + j 1, the phasor representation is

dt

+ Aie^'

+

+ A„e"-'

, so that the forced response is i,(t)

2 1

cos (r + 12.7°) A

The complete response is therefore

(14.33

where the natural frequencies pi, pi, . . . . p„ are the poles of the network functi and A,, Ai, . . . , A. are arbitrary constants. Modifications, of course, must made, as described in Chapter 9, if the natural frequencies are not distinct. We now have a very simple method, based on phasors, for finding the com" plete response of a circuit. A l l we need to find is the network function from which by (14.27) we may obtain the output phasor. The forced response is found from the phasor response in the usual way, and the natural response is given by (14.33), where the natural frequencies are the poles of the network function. If the input phasor is of the form

= 0. l k r

i(t)

= Aie~' + Aie"'

21

+ 0.16f- ' cos (t + 12.7°) A 2

The arbitrary constants may be evaluated, as in Chapter 9, if we know the initial energy conditions.

EXERCISES 14.6.1

( h a t t h e r e

,

sn o

c a n c e l l a t i o n

Find the natural response in Exercise 1 4 . * . , in the network function. Answer e'^A, cos f + A sin f) 2

V,(5) =

458

V l± m

Section 14.6 Chapter 14

Complex Frequency and Network Functions

The Natural Response from the Network Function

459

14.6.2

lulu the circuit is sometimes called the pliers entry since we cut a wire and insert a voltage source I he second entry is a soldering entry since we solder the source across two nodes Any other method of entry would be improper because killing the uiseiicd sources would result in a different dead circuit. In the pliers entry, if V,(s) is the voltage source inserted and \,(s), the current inlo the circuit at the source, is the response, then the network function is

Find the complete response in Exercise 14.4 1(c) if the natural response is as g i in Exercise 14.6.1 and at Answer e~ '[3 cos t - 9 sin t + 3 V 2 cos (3f 2

14.6.3

135°)]

Find the complete response i if i, = 10 cos 2f A , j(0) = 0, and 8 A/s. Answer 2e~' - le'V + sin 2f A

di(0*)/dU

'

}

\,(s)

Thus the natural frequencies are the poles of Y,(s) or the zeros of its reciprocal, Z.(.v), the input impedance at x-x'. In this case we have \ . - . 12(3*) Z.(s) = Is + 6 + — m

(

£

2(s + l)(s + 12)

=

s + 4 Therefore, the natural frequencies are - 1 , —12, as before. In the case of Fig. 14.9(b), if a current source l (s) is placed across y-y' and the voltage \,(s) across the source is the output, then the network function is

EXERCISE 14.6.3

y

14.7

l,(s)

N A TUR A L F RE Q U E N C I ES

Therefore, the natural frequencies are the poles of Z,(s) or the zeros of its recipro-

As we have seen, the natural frequencies are the poles of the network function there is no cancellation of a common pole and zero. Also, as discussed in Chapter the natural frequencies are the same for any response in a given circuit unless portion of the circuit is physically separated from the rest. Thus if we are looki only for the natural frequencies, we may consider any response, and it is obvi better to choose one that is easier to find.

cal, YAs), given by 12 =

35

25 + 6

(5 + l)(s + 12)

125(5 + 3) Again, the natural frequencies are —1, - 1 2 .

E X A M P L E 14.13

Let us consider the circuit of Fig. 14.8 with the source killed (since the natural rti sponse corresponds to a zero source). The natural frequencies are the poles of an network function. Therefore, we may excite the circuit in some proper manner and for some chosen output, determine the network function. Figure 14.9 illustrates I In two proper means of applying an excitation to the dead circuit. We may insert I voltage spurce in series with an element as in x-x' of Fig. 14.9(a), or we may place a current source across an element, as across y-y' of Fig. 14.9(b). The first entry

EXERCISES 14.7.1

ing (a) a pliers entry in series with the capacitor and (b) a soldering entry across the

14.7.2

F I G U R E 14.9 Circuit with two possible entries

Find the natural frequencies of the circuit of Fig. 14.5 by killing the sources and us-

14.7.3

capacitor. Answer — 3 ± jl Find the natural frequencies of the circuit of Exercise 14.4.2 by killing the source and using (a) a pliers entry in series with the capacitor and (b) a soldering entry across the capacitor. (Note: Do not cancel the common pole and zero, 5 = - 3 . ) Answer — 1, —2, —3 Find the network function Z(s) with a soldering entry across the 6-ft resistor of Fig. 14.9, and use the result to find the natural frequencies. 6 5 ( 5 + 10)

Answer

(b>

(a)

460

-77 (5 + l)(s

Section 14.7 Chapter 14

Complex Frequency and Network Functions

. — — 1, —\Z +12)

Natural Frequencies

461

14.8 TW O-PORT

NETW ORKS

One of the most important applications of the network function concept is to M works for which the input and output signals are measured at different parrs of lertfl nals. The simplest and most-often-encountered circuit for which this is possibH the two-port network, a port being defined as a pair of terminals at which a sigi may enter or leave. A general two-port network is symbolized by Fig 14.10(b) contrasted with the one-port network of F i g . 14.10(a).

F I G U R E 14.12 General two-port network

and I ) . For example, if port 2 is open-circuited ( I = 0), then from (14.34) we have 2

2

V, III

(a)

T II

=

v

(b)

1*21

« If

—

F I G U R E 14.10 (a) One- port and (b) two-port networks

In general, as seen in Fig. 14.10(b), a two-port network has four terminals. It is possible, of course, for two of the terminals to be the same, in which case we havd a three-terminal, or grounded, network. A general example of this case is shown in Fig. 14.11.

1

o F I G U R E 14.11

l -0 2

Similarly, if port 1 is open-circuited (I, = 0), we have

o

Three-terminal two-port network

We may associate two pairs of currents and voltages with a general two poii network, as shown in the frequency-domain case of Fig. 14.12, with variables \,{s), I i M , V (s), and l (s) as indicated. In case the network is linear, these variables may be related in a number of ways. For example, if I, and I are inputs and Vi and Vi are outputs, by superposition V, and V each have components proportional to I and I . That is, 2

(14.35)

2

2

Accordingly the z's are called open-circuit impedances, or open-circuit parameters, or simply z-parameters. In any case, they are examples of network functions. By (14 35) and (14.36), we see that z „ is the impedance seen looking in the primary port (port 1) when the secondary port (port 2) is open, and z is that seen at the secondary when the primary is open. The parameters z and z , are transfer impedances, which are ratios of a voltage at one port to a current at another. 2 2

12

2

2

2

2

V, = z , , I . + z , I 2

V = z ,I, + z I 2

2

2 2

2

(14.34) 2

where the z's are proportionality factors, which in general are functions of s. Since the z's multiply currents to yield voltages, they must be measured in ohms. Therefore, they are impedance functions. We may find the z's from the network by open-circuiting either port 1 (with variables V, and I , ) or port 2 (with V,

462

Chapter 14

Complex Frequency and Network Functions

Let us consider the three-terminal network of Fig. 14.13Jfecause of its shape, it a sometimes called a T network. Evidently, it is simply a Y network, as discussed in Chapter 13. To find • „ and we open port 2 and excite port 1 with a current source I , . Since I = 0, we have 2

(Z, + Z ) I , 3

and V

Section 14.8

Two-Port Networks

2

= Zjl,

463

V,

•f

v

=

2

( 4 ^ ) l ,

= ^ i ,

(2

+

f)l

+

2

( 6 ^ ) i 2

+

Comparing these results with the definition (14.34), the z-parameters, the coefficients of the currents in the last two equations, are given by F I G U R E 14.1 J

T nelwork Zn

' 6 = 4 + s

1,2

= 2 + s

Therefore, *n - "J

Z i + Z3

ll

2 1 z i - . - £ . Ifi li 2

The other two parameters are found in a similar manner with the primary opt (Ii = 0) and the secondary excited with a source I . The result is

Z22

= 6 + s

2

Zn

= Zi +

We may find another set of parameters considering Vi and V as inputs and I , and I as outputs in Fig. 14.12. In this case we have, by superposition, 2

Z3

2

t\i — Z22

Z21

=

— Z + 2

I 14 l / |

Z3 Z3

I, = y , V , + y , V 2

I2 = y 2 , V , + y V 2 2

(.4.38)

2

2

EXAMPLE 14.15 Let us find the z-parameters for the phasor circuit of Fig. 14.14. We will illust the procedure in this case by simply writing the two loop equations, V, = 2 I +

(4

2

V = ^6 2

+

2

) l

+

2

+

2

(I,

+

where, evidently, the proportionality factors are admittances given by

i ) 2

I. v -o 2

+ * d i + It)

h V12 =

77

V

2 v

F I G U R E 14.14 Two-port network

'-°

(14.39)

I2

v -o 2

4/s

-K-

2/i

h

H f —Wr

v,-o

Accordingly, the y's are short-circuit admittances (V, = 0 or V = 0), or shortcircuit parameters, or simply y-parameters. 2

464

Chapter 14

Complex Frequency and Network Functions

Section 14.8

Two-Port Networks

465

The parameters y,i and ya are admittances seen looking in one port will other port short circuited, and yn and y i are transfer admittances, or m current at one port to the voltage at another, under the appropriate short-cm ml ditions.

Vi -

2

h n l

1

N

h

(14.41)

I , - k u l i + iiuVj

and E X A M P L E

1 4 . 1 6

Let us find the y-parameters of the three-terminal circuit of Fig. 1 4 . 1 5 . Becagfl its shape, it is sometimes called a n network. Evidently, it is simply a A m considered in Chapter 1 3 . If the secondary is short circuited ( V = 0 ) , Y , is \ out, so that Y „ and Y * are in parallel. Therefore,

= gi.V, +

I,

%uh

(14.42)

2

y„ = Also in this case, - I

2

Y„ +

2

=

g22l2

2

Y»

flows to the right through Y I

V = gjlV, +

Thev are called hybrid parameters because they relate a mixture of a current B

so that

and

voltage, rather than two currents to two voltages,

or vice versa. From

Y»V,

(14.41)

and

(14.42),

we see that

or h y^i = y

. =

h nn — ,

v

- Y »

The other two y's may be found in a similar way with the primary short circui

h

,

2

=

/ -o 2

v i

,-0

(14.43)

hi = . 2

2-0

D 2 2 =

\

1,-0

and F I G U R E 14.15 TT network I,

The result is

B n = % |j-0 yu =

Y« +

yn =

y2i =

yn =

Y» +

Y„ I,

Y

V,-0

(14.

- Y » C

R» -

y;

(14.44)

I =0 2

If Z12 = z i (or, equivalently, y i = y i ) , the network is a reciprocal netw This is always the case when the elements inside the network box of Fig. 1 4 . 1 2 resistors, inductors, and capacitors. Examples, as we note from ( 1 4 . 3 7 ) and ( 1 4 . are the circuits of Figs. 1 4 . 1 3 and 1 4 . 1 5 . The circuit of Fig. 1 4 . 1 4 contains a pendent source, and is nonreciprocal, since I12 # Z21. The z- and y-parameters are but two sets of parameters associated with a two port network. Two other sets are the hybrid parameters, h n , h u , h i , ha and g g i 2 , gzi, g22, which are also very important, especially in electronic circuits, are defined, respectively, by 2

2

2

466

Chapter 14

v

2

Complex Frequency and Network Functions

g 2 2 =

2

I

v,-o

_

ll(

analyze the g's in the same manner, of course.

Section 14.8

Two-Port Networks

467

E X A M P L E 14.17

Let us find the /(-parameters for the circuit of Fig 14.13, if Z , = 6 ft, Zj • and Z = 10 ft. From the I ' S found in (14.37). we know that 3

Therefore, A is a voltage open, and B is a transfer •hort circuited. A second and I in terms of V, and 2

ratio and C is a transfer admittance, with the secondary impedance and D is a current ratio, with the secondary set of transmission parameters may be defined giving V - I , . They are illustrated in Exercise 14.8.5.

2

V , = 161, + 10I 2

The derivation of the A-parameters of (14.46) suggests a general method of ob-

I8I2

taining one set of parameters from another. For example, if we solve (14.38) for the

V

2

=

101, +

If we substitute the value of I from the second of these into the first, we hav 2

161, + 10

fV

2

V ' s , we have

101,"

-

(14.49)

18

which simplifies to V, = ? I , + | V

2

where A r is the determinant,

If we solve the second of (14.45) for I , we have 2

l

2

= - H i + rsV

Ay = JTiiJta - Jrt* ' ° have the z-parameters in terms of the Comparing this result with (14.34), we ( 1 4

2

Comparing these last two results with the definition (14.41) of the h's, we have

5

)

h„ h, = I h

2 l

"22

-y

y22

2

(14 4h |

= - §

,

H

"

Ar

1 , 2

""

2 2

Ar

(14.51)

yn

= A

* ' 2

-

~ A 7

1 2 2

~ z

Still another set of two-port parameters is the transmission parameters, A. II. C , and D, defined for the general two-port network by V, = A V - B I 2

I, = C V - D I 2

parameters to the other sets ror me 10

2

parameters in terms

(14.47) :

These are important to transmission engineers because they express the primary (sending end) variables V, and I , in terms of the secondary (receiving end) variables V and - I . (The negative of I is used because that is the current entering the re ceiving end load.) Evidently, from (14.47), we may write 2

r

2

TABLE 14.1

Two-Port Parameter Conversion Formulas ' AM ho -fcii

2

a,* | ho _L ha.

ho

-h, '

1

2

En h , h„

k.,

-A» h ,

-h„ h

A„ h" .

2

2

-1

h„

h , 2

ft" h , 2

468

Chapter 14

Complex Frequency and Network Functions

Section 14.8

Two-Port Networks

bo J

469

first row. Comparing corresponding enteries in the first matrix with the sec trix of that row yields (14.51). The determinants A , Ac, A«, and A r are the t, y, h, and transmission matrices, respectively. z

A I T l ICAHO NS OF TWO-PORT PARAMETERS I he two-port parameters of the preceding section are useful in many ways, as we will see in this^ection. Our first use of them will be to obtatn various network tunctions.

EXERCISES 14.8.1

Find the z-parameters and the ABCD parameters of the circuit shown. B Answer Z n = 6, Zn — Z21 • 4, Z22 = 10 ft; A 11 ft, C = J S , D

I l f l I 14.18 If port 2 is open ( I = 0), then from (14.34) we may find the voltage ratio function 2

(14.52) V,

6S2

Zn

Also, for a short-circuited secondary (V = 0), we may write, using (14.38), 2

h

(14.53)

= ?il .... yn

I

I.

4 n .

MIM 14 19 Let find the ratio^ function for ^ the ^two-port loaded with ^1ft,^ as shown _m M P I II 14.19 Let us us^nnd t ^ current ^ . ^ y

EXERCISE 14.8.1

14.8.2

Show that the y-parameters may be obtained from the z-parameters by Z22

yn

y

2 ]

y i 2

Zn

5l A '

=

-AT

=

y 2

Az

2

z

I,

i_

14.8.4

2

— 2

v,

+

y

2 2

2

6.6.

2

2

— h i , (b) Ar = 1 2

A second set of transmission parameters may be defined, expressing the output variables in terms of the input variables, by V = aV,

bl,

I = cV,

dl,

2

2

F I G U R E 14.16 Two-port loaded with a 1-11 resistor

r X A M P L E 14 20 Let us find the transfer function fe/I, for the circuit of Fig. 14.14 if the secondary is terminated in a 1 -ft resistor. We have previously found

Find the parameters a, b, c, and d in terms of the other transmission parameters A, B, C , and D. D . B C A Answer a = — , D = - r - , C = — , d = — Ar Ar Ar Ar

470

1

Using Table 14.1, find the conditions on (a) the A-parameters and (b) the transmit' sion parameters that a circuit be reciprocal (z, = z i ) . Answer (a) b.12 =

14.8.5

1 + Zl2

If we make the same substitution ( - I 2 for V ) in (14.38), we have the voltage ratio . ' 1 (14.55)

where A z = Z i i Z - Z12Z21. (This is given in the first entry of the second row d Table 14.1.) Use this result to find the y-parameters of the circuit of Exercise 14.8.1, Answer y „ = ^ , y , = y i = - A, y * = 22 S Find the /i-parameters and the ^-parameters of the network of Exercise 14.8.1. .* i Answer h's: g s: 2

1 (14.54)

-Z21

2 2

14.8.3

e q u a

tion of (14.34) yields, after some rearrangement, -——

Z12

Az'

=

.2 Z 2 , -

z Section 14.9

Chapter 14

Complex Frequency and Network Functions

M

7

4 = 6 + -

Applications of Two-Port Parameters

471

so that by (14.54) the transfer function is h Ii

which substituted into (14.34) yields

-2/s

-2

1 + [6 + (4/5) |

75 + 4

_

Z|2 ,

V, - Z . I , = z „ I ,

(14.56) E X A M P L E 14.21

For the TT network of Fig. 14.17 we may write down by

inspection the parameter v

V

= Z |Ii - — V

2

2

2

Z.1

25 yv = 5 +

If we solve each of these for I , . we have -

, j

25

V, + ( z , / Z ) V

=

2

Zn

so that by (14.55) the voltage ratio is V

2

V,

_

t

Z

+

2

V + (z /ZJV

=

2

2 2

2

Z i

G

2

From the last two members, we may solve for the voltage ratio function.

1/25 1 + 5 + (1/25)]

25 + 25 + 1 2

Z21Z/. V,

(z,, + Z , ) ( z

Z, Z ,

+ Zi.) -

22

2

(14.57)

2

•—nsTSTy\ v, MI

-

i

•

If we divide both sides of (14.57) by — Z i , we have the transfer admittance function.

M

i .a

7'

h

. 1

m

V,

(Zn+ ZJ(z

2 2

Z12Z21

+ ZL) -

(14.58)

F I C U R E 14.17 LC two-port terminated in 1 J l

Next we will consider a doubly terminated two-port network as shown in Fig 14.18. The load impedance at the output port is Z , and Z , at the input port is the internal impedance of the source V , . If Z = 0 and Z = 1, we have the case of Fig. 14.16.

Others that may be readily derived in terms of the z-parameters are the current ratio function,

L

8

L

-Z21

(14.59)

Z

Z22 +

L

the transfer impedance function, ZL V,

Vj._

Ii F I C U R E 14.18 Doubly terminated two-port network

Z21Z1.

(14.60)

ZL

Z22 +

and the input impedance,

From the circuit we may write

_ V Z„ = - = z„ + Z s

G

-

^

/

/ 1 T

Y

L

(14.61)

-V2 ZL

and Vi - V , - Z I , 8

472

Chapter 14

Complex Frequency and Network Functions

These same functions, along with a number of others, may also be derived in terms of the other two-port parameters. We consider some of these in the exercises and problems. Section 14.9

Applications of Two-Port Parameters

473

E X A M P L E 14.22

currents are the same, then the port voltages are the same, and vice versa, for both

To illustrate the use of (14.57), let us find the voltage ratio function for the circuil t Fig. 14.19. By inspection we have Zn

Zl2

Z22

=

—

networks If the circuit of Fig. 14.12 is a reciprocal network (Z12 = Z21), a simpler, passive equivalent circuit can be found by solving (14.37) for Z i , Z j , Z j . The result,

2 s + — is

Zj =

_ 2

Z21 —

—

Z12

Zi = Zn - Z = Zn —

*12

Zl =

I|2

3

is

= 2s +

Z22

— Z3

=

Z22

—

substituted into Fig. 14.13 yields the equivalent passive circuit shown in Fig. 14.21. It is an equivalent circuit for a general reciprocal three-terminal network.

3s Z, = 1 ZL

= 2 D

which substituted into (14.57) gives \

V,

c

2/3

=

s

3

+ 2s

2

+ 2s + 1

F I G U R E 14.21 Equivalent circuit of a general reciprocal three-terminal network

Many other circuits equivalent to the general two-port network may be obtained from the defining equations of the two-port parameters. For example, the circuit of Fig. 14.22 is another such equivalent circuit using the fc-pararneters, as the reader is asked to verify in Exercise 14.9.4. This is a popular equivalent circuit, often used to represent a transistor.

Another use for the two-port parameters is in the construction of equivalent circuits. For example, in the first equation of (14.34), Vi is a sum of two terms, z I i and Z 1 2 I 2 . The first may be obtained by an impedance Z u carrying a current I , , and the second may be obtained by a dependent voltage source controlled by I . In 1 similar way we may interpret the second equation of (14.34) and draw the circuit representing these equations. The result, as may be verified by inspection, is shown in Fig. 14.20. We say that this circuit is equivalent (i.e., at the terminals) to that of Fig. 14.12, because they both have the same two-port parameters. Thus if the port M

2

F I G U R E 14.22 Equivalent circuit of the general three-terminal two-port, using h-parameters

F I C U R E 14.20 Equivalent circuit for Fig. 14.22

EXERCISES 14.9.1

Find the voltage ratio transfer function for the two-port terminated in 1 ft, shown i Fig. 14.16, with z-parameters z,, = 8 ft, l i j = Z21 = 6 ft, and Z22 -

14.9.2 14.9.3

474

Chapter 14

14 ft.

Answer 74 Show that the given circuit is equivalent to the general reciprocal three-terminal nei work by showing that (14.38) holds. Show that the given circuit is equivalent to the general two-port network.

47

Complex Frequency and Network Functions Section 14.9

Applications of Two-Port Parameters

~ ": <

-

l2«

-—a •

lit

-

—-

EXERCISE 14.9.2

—

z

ll

Z

t

I2

z

J 3

+

J

b

1 2

- o

M 21 Z

-^12)

I K . I IRE 14.23 Parallel connection of (a) two general two-ports and (b) two grounded two-ports

and In. = yn*Vift + y m V ^ fe* = y2i*Vi» + y22»V2i,

EXERCISE 14.9.3

14.9.4

where V, = V,„ = Vu,

Verify that the circuit of Fig. 14.22 is equivalent to the general three-terminal new work by showing that (14.41) holds.

V

2

= V*. = Va,

I | = Ilo + Il k I2

14.10 I N T E R C O N N E C T I O N S O F T W O - P O R T N ET W O RKS The two-port networks of the previous sections may be used as building blocks In design more complicated circuits. That is, subsections may be designed as two-potl networks and then interconnected to form the overall circuit. In this section we will consider some of these interconnections and see how the overall circuit may be analyzed by analyzing its component two-port parts. The first interconnection we consider is that of Fig. 14.23(a), which is called n parallel connection for reasons that will be clear later. The parallel connection of two grounded two-port networks is shown in Fig. 14.23(b). In the work to follow, we want each subnetwork to retain its integrity as a two-port network, which will be true in Fig. 14.23(a) if the currents into the top leads of each network come out the bottom leads. To ensure this in some cases, it may be necessary to place an ideal transformer (discussed in Chapter 16) at one of the four ports. In any case, the integrity of the two-port networks in the grounded case of Fig. 14.23(b) is always maintained, because the lower (or grounded) terminals are common to the subnetworks and the overall network. By Fig. 14.23(a), we may write Ii„ = y u V o

l 0

+

=

1.2a + I2*

Combining these results, we have 11 = ( y n . + y t u O V i + ( y i i . +

yi2»)V

2

12 = (y2i« + y i » ) V , + (y22« + y i , ) V 2 2

2 2

Therefore, we see that the y-parameters of the interconnection are the sums of the y-parameters of each subnetwork. That is, yn =

yn. + y n * ,

yu = y i ^ + yn*

7n =

yu + y»».

V22 = y22«. + Jin

2

(14.62)

The fact that the admittances add, as they do for parallel circuit elements, is motivation for the name parallel connection. I X A M P L E 14.23 Let us find the transfer function V / V , in the circuit of Fig. 14.24, which is a parallel connection of two two-ports terminated in a 1-ft resistor. By inspection we have 2

y . 2

y2ia + y2i* =

-

-

yi^Va, y

I i , = y 2 1 „ V l „ + y ;u.V2<,

2 2

= y

2 2 o

+ y22» = s + - + 1 + * = 2 J + S

1 +

7 •

2

476

Chapter 14

Complex Frequency and Network Functions

Section 14.10

Interconnections of Two-Port Networks

477

II

V

1 + [2s + 1 + (l/s)]

8

2s

2

I HI

14.25 Series connection of (a) two general two-ports and (b) two grounded two-ports

The last interconnection we will consider is the cascade connection of Fig. 14.26, in which the output port of network a is the input port of network b. From the figure and the definition of the transmission parameters, we see that

+ 2s + 1

The connection of Fig. 14.25(a) is a series connection of the two networks a and b, so called because, as we will see, the z-parameters add, as the y-parameter» do in the parallel connection. For the two networks a and b we have Via — \la

Ziialia

l2\a\\a

=

2\2 \

+

+

V,

a

TlZtXla

= V,. = A.V2. -

Bali,

=

AaVli, +

Balu,

=

A.(A»Va, -

=

(A«Aft + B . C » ) V

B»I») + B . ( C » V

- D»Iza)

a

a

-

(A„B

t

+

B D*)I»

2

-

(A„B

fc

+

B„D»)I

0

and V, -

Z2l(,Ilfc +

= (A.A

f c

+ B.C*)V

In a similar manner, using the transmission parameter equation for I , , we may ob-

lllhllb

tain

Also, from the figure we see that

I , = ( C A * + D „ C » ) V 2 - (C„B* + D „ D ) I t

1 = I , , = Ii»

and

—•

V , = V , „ + V i » = ( z „ + z » ) I , + (zii, + z,2*)I n

V

2

= \la

( Z j l a + *21»)Il + (122. +

=

+ \

M

1

-

"

T

"

~ A

=

A o A j , + B,/Cft

2

B = A„B,, + B„D»

*22l,)h

D = C B a

Z\\b,

Zl2 — Z,2o + Z,2»

Z21 — Z l

Z lft,

Z22 — Z22« + Z22/,

2

a

+

2

(14.63)

6

+

D Db

Chapter 14

Complex Frequency and Network Functions

a

A reader familiar with matrix multiplication will recognize this result to be a statement that the transmission matrix of the overall network is the product of the transmission matrices of networks a and b. That is,

Again, we are assuming that the two-ports of Fig. 14.25(a) maintain their integrity as two-port networks. This is always the case in Fig. 14.25(b), where the subnetworks are grounded two-ports.

478

(14.64)

C = C . A * + D„C»

Therefore, the z-parameters of the overall network are given by Zll — Zlla +

2

Comparing these last two equations with (14.47) for the overall network, we have for the cascade connection.

— 12j,

I:

2

B

a Section 14.10

Interconnections of Two-Port Networks

f

(14.65)

479

F I C U R E 14.26 Casca d e connection of two two-port networks

F I C U R E 14.28 Two-port network with a 1-fl termination

E X A M P L E 14.24 We may show that the transmission parameters of the simple two-ports of F l l 14.27(a) and (b) are, respectively, A„ = C

so that V2

B„ = Z„

= 0

V,

(14.661

D„ =

B„ = 0

C

D„ = 1

= Y„

(14.671

and therefore those of their cascade connection (c) are, by (14.64), A = 1 + Z Y»,

B = Z„

C = Y„,

D = 1

0

2

There are two other interconnections of two-port networks that we will mention but will not consider further. The first is the series-parallel connection in which the primary is connected like the primary of the series connection of Fig. 14.25 and the secondary is connected like the secondary of the parallel connection of Fig. 14.23. The second is the parallel-series connection, in which the primary is parallel connected and the secondary is series connected. If the integrity of each two-port network is maintained with the interconnections, the /i-parameters add to give the overall /i-parameters in the series-parallel connection, and the g-parameters add to give the overall g-parameters in the parallel-series connection.

and A* = 1

1 2 s + Is + 1

(14.68)

EXERCISES 14.10.1

Find V / V , for the circuit shown. 4s + 6s Answer 2s + I5s + 35s + 2$s + 3 2

2

t

(a)

(b)

(c)

}

2

2s

F I C U R E 14.27 T w o simple two - port networks, and their cascade connection

E X A M P L E 14.25 We may use the result in (14.68) to find the voltage ratio transfer function V 2 / V 1 for the circuit of Fig. 14.28. As the reader is asked to show in Prob. 14.33 for the general two-port network terminated in 1 CI, the voltage transfer function is V2

V,

^

1 A + B

(14.69)

The two port in this case is like that of Fig. 14.27(c) with Z„ = Is and Y = s. Therefore, we have b

A = 1 + 2s

2

B = Is

480

Chapter 14

Complex Frequency and Network Functions

EXERCISE 14.10.1 Section 14.10

Interconnections of Two-Port Networks

481

14.10.2

14.9

Use the ABCD parameters to find V / V , for the two-port network terminal l - f l resistor if the two-port is the T network of Fig. 14.13 with Z i = Z • Z = l/s. (Suggestion: Note that the two port is that of Fig. 14.28 with impedance added at the output.) 1 Answer . .• —r—;—;—:——4 i + 2s + 4s + 1 2

2

2

2

3

3

Use the principle of proportionality to lind the network function V / V i , and use the rcsull to find the forced response c (/) when Ui(f) 4 cos t V.

1

VVV

2H

1

i—nrvT>—!

2

PR O BLEM 14.4 ul ihr forced component o f v in Prob. 14.4

J"

14.11

I c V " cos 2t V.

itw that

SU M M A RY In this chapter we have considered the damped sinusoid and its generalized phatm which is a function of a complex frequency s. This leads to a generalization impedance, admittance, and phasor circuits based on s. A further generalization impedance results in the network function or transfer function, defined as the ratio the generalized phasor output to the input. Poles and zeros of the transfer fundi are the values of s for which the function is infinite and zero, respectively. T knowledge of the poles and zeros allows us to construct the transfer function ,i predict the behavior of the circuit. In particular, the natural frequencies of the m cuit are the poles of the transfer function. Two-port networks were defined and a number of two-port parameters wo used to describe the networks. These included the open-circuit impedances I short-circuit admittances, the hybrid parameters, and the transmission paramcui \ These parameters are also useful in obtaining general transfer functions of arbitral y circuits, and in obtaining general equivalent circuits. Finally, interconnections 01 two-port networks enables us to combine the two-port parameters of simple circuits to obtain those of more complicated circuits. Some typical interconnections are the series, parallel, and cascade connections, which are very important in analysis a* well as design of two-port networks.

1

Y = Yi +

r -

f ^IF

ins

PR O BLEM 14.9

14.10 If the voltage source in Prob. 9.23 is v,(t) = 6 e c o s 2/ V, find the network function l(s)/\,(s) and the forced response i in each case.

;

Z + 2

Y, +

14.11 \SkKKestion: Note

that

Y = Y , + Y„ =

Vt + 1 / Z . = Y , + l / I Z , + ! > ) « • • • . ] liar Ihe method of Prob.

14.6 to find the

Impedance seen by the source, and use the reMill to find the forced response v. In Prob.

14.7 find the network

function

M(.t) = I0) / I „ (s ) by using the proportionality principle. Use the result to find the forced response i

Find H(s) = \(s)/\,(s) and use the result to find the forced response c if c, = e' ' cos 2t V. 14.12 Find the forced response «J using the transfer function if v, = 5e cos 20r V. 2

14.13 Find the network function \{s)/V,(s) and use the result to find the forced response v if o, = 6e-" cos 8r V . 14.14 Find the network function and the forced response v(t) if Vg(t) = 6e V. 21

PROBLE M S 14.1 Find the phasor representation \(s) and the frequency J of (a) v(t) = 5e' ' cos (lOr — 30°), (b) v(t) = e" sin (5/ + 45°), (c) v(t) = 2e"(3 cos 3» + 4 sin 3r), and (d) v(t) = 5*~". 2

14.2 Find v(t) if (a) V(J) = 6/10° with s = -2 + >8, (b) \(s) = 5/0° with s = - 1 0 , (c) V(j) = 4 + j3 with s = -1 + j2, and (d) V 0 ) = -j6 with s = j4. 14.3 Find the admittance Y(s) seen by the source and locate its poles and zeros. If the source is v = 4e~' cos ( V, find the forced response jf. g

482

Chapter 14

PR O BLE M 14.6 4ft

PR O BLE M 14.7

PR O BLE M 14.3 14.4 Find the impedance Z(s) seen by the sourc and locate its poles and zeros. If the source i v = lOe' cos 4( V, find the forced compo nent of current it delivers. g

Complex Frequency and Network Functions

- 5e

-Jr

cos r A

21

Chapter 14

Problems

483

)M network function V ( i ) / V , ( j ) and the

14.

For the corresponding phasoi m u m irpliuc everything to the left of terminals a h by its Thevenin equivalent and use the irsull In hnil

14.

Gi ve n the network function

rMponM v if v, = 2e " " cos 5t V. he network funiHion V ( J ) / V , ( J ) and the

20 W v

",o

8ft

rttponse v if (a) «, = 6e

AAV

41

the forced response i.

cos 2l V

) v, - 6 cos 2t V V network function \(s)/\,(s)

4fi

icsptinse « if v, = 6e

61

and the

VjU)

cos 3t V.

H

r in the corresponding phasor circuit ev-

U

'

V,U)

=

4s(s + 2) (s + \)(s + 3)

If no cancellation has occurred, find the com plete response f ( ( ) , for ( > 0, if «<(') = 6e~' cos 2t V and v„(Q*) = <Ji>„(0 )/dr = 0.

I I I except the inductor by its Thevenin lent circuit and use the result to find the

0

+

response i if c , = 4 e " cos 3» V. PR O BLE M 14.11 4(1 —WVo

4J2 WV-

8n -AAV-

PR O BLE M 14.12 1 H PR O BLE M 14.16

ii

F

PR O BLE M 14.17 I PR O BLE M 14.13

40 WV-

P R O BLE M 14.14 P R O BLE M 14.15 1 r

2

n

6fi

-wv-

V

484

Chapter 14

Complex Frequency and Network Functions

Chapter 14

Problems

485

'p

8 e ' cos

It

A ^~

PROBLEM 14.18 14.21

14.22

PROBLEM 14.32

PROBLEM 14.19

Repeat Prob. 14.20 if v,{t) = 6e ' V. (Sugges-

14.27

Show

that

the

»'on: Observe the describing differential equa-

in

(14.42)

tion.)

z-parameters by

Find the complete response v for t > 0 using the network function, if u ( 0 ) and +

be

obtained

dv(Q*)/dt

gn

2

14.28

5

14.29

In

2

'

14.38

A + B

Show that the given circuit is equivalent to the

hi, = 1 M l , h

1 2

two circuits differ. 14.39

= 10 \ > = 100, and

a

and cross arms both equal to

[III -

10 * S in Fig. 14.22 and find the net-

Z * . It is called a symmetrical lattice because

w.,ik

Iunction V / V , if port 2 is open cir-

the series arms are equal and the cross arms

Hlg.

Show that the transmission parameters, A . I

The circuit shown is a lattice with series arms both equal to Z

2

2

are equal. Show that the z- and y-parameters are given by

Show for the doubly terminated network of

14.25.

Show that the given circuits are equivalent to the general two-port network. Note how the

, iniiil

Find the h- and ^-parameters of the Iwo pu network of Prob

14.37

general two-port network. Note how it differs

i , rIIiui.iied network of Prob. 14 32.

z, z i. 2

in terms of the

Check the result, using this formula, for the

Zl t

-

2 2

V,

A,

Z |

1 5 c - ' cos 1 0 / A

14.16,

from the equivalent circuit of Exercise 14.9.3.

Zi i

g2. = — , III 2

that tor Fig

mission parameters we have

Zll

gl2

where A = Z n Z

I

from

Yi = 1

are both zero. •

• M

hybrid parameters

may

14.18 that the voltage ratio transfer funcZu

tion is given by the three equations

= z

2 2

—

z i

= ^(Z + Z J t

C , and D , defined for the two-port network i PROBLEM 14.22

Fig.

y iZ

14.12 by

2

V, 14.23

Find the complete response v if v(0)

V, = A V

= 0. Use

network functions and superposition.

1, =

cv

2

2

BI

2

DI

2

V

2

y, y iZ Z,. 2

2

f

t

2 2

2

21

cos 4 / V I

)

c

in v in Prob.

by killing the source and using (a) the

14.30

14.25

14.26 4 8 6

= - L , *2I

I) 2

14.31

= 1 kfl,

(This is the first entry in the third row of Tahlo

h,

S, Z , =

14.1.)

360 11, and Z

= 1 0 ' , h , = 100, h22 = 10 2

L

2 2

= - ( Y » + Y„) 2

4

fc

The symmetrical lattice of Prob. 14.39 is terminated in 1 il and Z„ and Z

b

are as shown.

Find the voltage ratio transfer function. PROBLEM 14.37

Derive the Y - A transformation of Chapter 11 the terminals (that is, equal two-port parame

Y.)

2

where Y„ = 1 / Z . and Y» = 1 / Z . 14.40

= 1 kfl.

Find the transmission parameters of the two-

by making Figs. 14.13 and 14.15 equivalent at

Chapter 14

y,

Use the second of the formulas of Prob. 14.35 doubly terminated network with h,

the resistor.

Find the y-parameters in Prob. 14.16.

+ B + DZ*

to tind the voltage ratio transfer function for a

Z I

soldering entry with a current source across Considering the figure for Prob. 14.22 as a two-port network with terminals as shown, find the z- and y-parameters as functions of s.

t

port network of Prob. 14.25.

pliers entry with a voltage source inserted between the capacitor and the resistor and (b) the

g

Wm

PROBLEM 14.23

14.22

— — (Zfr Z„)

t

y,, = y

(A + C Z ) Z

in

tti

(

Find the natural frequencies

2

A,

n

14.24

2

und

i — E ^~v

2

and h, h2,Z

are given by

*Vv\

r "- 5 ^

4e

;

~h iZi

_

VJ " (hi, + Z , ) ( l + h z j Z J 1

Zi

(1 + y „ Z , ) ( l + y Z . )

*ll ~ * J I

z

22

*2,

ters). 14.32

Find the y-parameters of the network shown Terminate the output port with a 1-11 resistor, and find the resulting network function V / V , .

Complex Frequency and Network Functions

2

Chapter 14

Problems

4 8 7

'2

15

Frequency Response

Alexander Graham Bell

PROBLEM 14.39

488

Chapter 14

Complex Frequency and Network Functions

PROBLEM 14.40

ing the University of E dinburgh and Ubtodly, the most c o m m o n a n d Mr. Watson, come here. I the University of London, also bewidely used electrical instruwant you. Ihe telephone, inv ented by Alexander Graham Bell c a m e a sp e ech teacher. In 1866 Bell b e c a m e interested in trying to transtish-American scientist Alexmit sp e ech electrically after reading a G r a h a m Bell. T h e d ate w a s book d escribing how vowel sounds I 8, 1875, wh e n Bell a n d his asni i h o m a s W a ts o n transmitted a musical note. could b e m a d e with tuning forks. Shortly afterward, Hrtl intelligible telephon e sentences, "Mr. W a t- Bell's two broth ers died of tuberculosis a n d Melville COme here. I want y o u , " we r e spoken inadver- Bell m o v e d his family to C a n a d a for health reasons. In hy Bell himself o n M a rch 10, 1876, wh e n he 1873 yo u n g G r a h a m b e c a m e a professor at Boston W a tso n to co m e to a n adjoining room to help University and b e g a n his electrical exp eriments in his sp are time. It w a s there that h e f o rm e d his partnership ne spilled acid. Bell w a s born in E dinburgh, S cotland. His fa- with W a tso n a n d went o n to his great invention. Bell's Alex ander Melville Bell, w a s a well-known tel ephon e patent w a s t h e most valuable o n e ever isteacher a n d his grandfather, Alex and er Bell, su e d , a n d the t e l e p h o n e op en ed a n e w a g e in the d ei a sp e e ch teacher. Y o u n g Bell, after attend- v e lopm e nt of civiliz ation. •

489

and Im H Q )

d> (cu) = tan

(15.4)

Re H ( »

The amplitude and phase responses are, o f course, special cases of frequency re sponses. Wit A requency-domain functions are very useful, as we have seen, for finding con sponding time-domain functions. The frequency response of a circuit, however, extremely useful in its own right, as we shall see in this chapter. For example, if are interested in which frequencies are dominant in an output signal, say V( then we need only consider the amplitude | \(jat) |. The dominant frequencies ct spond to relatively large amplitudes, and frequencies that are virtually suppre correspond to relatively small amplitudes. There are many applications in which frequency responses are important. ( very common application is in the design of electric filters, which are networks ill pass signals of certain frequencies and block signals of other frequencies. That is. the output signal of the filter has amplitude \ then oi, passes i f | V O i ) | I relatively large and is blocked i f | V(j'a>i) | is relatively small. There are many exalt pies of electric filters in our modern society, some of the more common being thai in our television sets, which allow us to tune in a certain channel by passing ii of frequencies while filtering out those of the other channels. In this chapter we consider frequency responses, both amplitude and phase. \Ml also define resonance and quality factor and show how they are related to the f M quency responses. Finally, we consider methods of scaling networks to yield a given frequency response with practical circuit element values.

15.1

Suppose that the network function of the RLC parallel circuit o f Fig. 15.1 is the input impedance HI \ ' m

S

_ -ii \ h(s)

} (\/R)

C K S }

+ sC +

(UC)s

B(s)

s + {l/RC)s

+

2

(15.6)

(\/LC)

HO)

=

1 (\/R)

+

(15.7)

j[a>C-(\/(oL)]

so that the amplitude and phase responses are IHO) I =

1 V ( l / / ? ) + [o>C 2

(\/
(15.8)

2

and ) = - t a n "

(15.9)

Since R, L , and C are constants, the maximum amplitude occurs at the frequency u> = o) for which the denominator in (15.8) is a minimum. Evidently, this occurs when

A network function H(jco), as well as any phasor quantity, is in general a ci function, having a real and an imaginary part. That is, H O ) = Re H O ) + j I m H O )

(15.5)

For s = jtov/e have, from (15.5),

15.1 AMPLITUDE A N D PHASE RESPONSES

(i/sL)

0

(15.1)

wC - \ 0

As we know, we may also write the network function in the polar form FIGURE 15.1 RLC H O ) = |HO)k*<"'

parallel c i r c u i t

(15.2)

where | H(yo>) | is the amplitude, or magnitude, response and ) is the phase r* sponse, given, respectively, by | H O ) | = VRe H ( » + Im H O ) 2

490

C h a p t e r 15

Frequency Response

2

(15,1

S e c t i o n 15.1

A m p l i t u d e a n d Phase R e s p o n s e s

491

EXERCISES 15.1.1

Let 7? * 4 A , L = H , and C = ^ F i n Fig. 15.1, and find the maximum umpli tude and the point where it occurs. A l s o , sketch the amplitude and phase rtH\x Answer |H|mu 4, too = 20

15.1.2

For the RLC series circuit with a voltage source v , let the network function be H = I / V , , where I is the phasor current. Show that the amplitude and phase responses are similar to those of Fig. 15.2 with | H | „ « = \/R and o>o = 1/VEc.

15.1.3

Let the network function o f Fig. 15.1 be H = I t / I , , where L is the inductor phasor current directed downward. Show that

Thus

=

| H(»U

= | H(/wo)|

=R

Also, it is clear that | H(jto) | - » 0 as to - » 0 and to - » =°. Therefore, the amplnuut response has the form shown in Fig. 15.2(a). In a like manner we may sketch ilw phase response, shown in Fig. 15.2(b), since tb(a>a) = 0, tf>() —* -ir/2 as to - » <*>. I f the input of Fig. 15.1 is the time-domain functlofl

t

i'i(f) = /m cos tot

v

the input phasor is I , = /™/0j\d the output phasor is V = I Z = I H. Thuii tht amplitude of the output is simply that of the network function multiplied by a OB stant. Therefore, we may obtain as much information from the network function t». sponse as from the output response. For this reason and the reason that the netwoHl function depends only on the network, not on how it is excited, we shall usually consider the frequency response of the network function. 2

FIGURE 15.2

(a) A m p l i t u d e a n d

(b) p h a s e

m

r e s p o n s e of

* + (\/RC)s 2

+ (1/LC)

m

m

(15.7)

'

and that | H | „ = 1, occurring at too = 0, provided that 2RC

2

£ L

15.2 FILTERS W i t h reference to Fig. 15.2(a) we see that frequencies clustered around to = 1/VZc rad/s, o r / = l / ( 2 7 r V Z c ) Hz, correspond to relatively large amplitudes, while those near zero and larger than too correspond to relatively small amplitudes. Thus Fig. 15.1 is an example of a bandpass filter, which passes the band of frequencies centered around to . In the general amplitude case, shown in Fig. 15.3, we say that to , the frequency at which the maximum amplitude occurs, is the center frequency. The band of frequencies that passes, or the passband, is defined to be 0

0

lH(/w)l

0

0

to

C[

s a> s to

C2

(b)

492

C h a p t e r 15

Frequency Response S e c t i o n 15.2

Filters

493

where to and to are called the cutoff points and are denned as the frequem i<which the amplitude is 1 / V 2 = 0.707 times the maximum amplitude. The width the passband, given by C[

n

(IVI

B = o» , - o»c, c

is called the bandwidth. As we shall see in Sec. 15.4, the bandwidth in the case of Fig. 15.1 it i

(a)

E X A M P L E 15.2

Let us consider the circuit of Fig. 15.4. Analyzing the circuit, we may readily O ^ H the voltage-ratio function, HM M

W

= M )

= _

V,(s)

s

2

1

m

_

(b)

FIGURE 15.5

l/RC.

i

Low-pass frequency responses

There is only one cutoff point, as indicated in the figure. This follows from the definition |H(M)I

l

= ^ | H ( » U

+ 2s + 2

1 V2

Letting s = jto and calculating the amplitude response, we have |H(>w)| =

(1)

2

V(2 - m ) 2

2

+ 4a.

V l + (mil A)

2

or, after simplification,

whose only real positive answer is to — V 2 . Thus the band of frequencies which passes is the low-frequency band c

| H ( » |

=

Vl

0 < to < V 2

+ (a. /4) 4

The phase response for Fig. 15.4 may be easily shown from H(jto) to be . i

.

tb(to) = - t a n

_,

2to _ ^

2

which is sketched in Fig. 15.5(b).

FIGURE 15.4

L o w - p a s s filter

A n example of a passive, low-pass filter is that o f Fig. 15.1, where the network function is as defined in Exercise 15.1.3. In fact, i f R = 1 Cl, L — 1 H , and C = j F, the network function is the same as that of Fig. 15.4. There are many types of filters other than low-pass and bandpass. Two of the more common are high-pass filters, which pass high frequencies and reject low frequencies, and band-reject filters, which pass all frequencies except a single band. Typical amplitude responses are shown in Fig. 15.6, and examples are considered in Exercises 15.2.2 and 15.2.3. In Fig. 15.6(a), w is the cutoff point, and the passband is to > to . I n Fig. 15.6(b), to is the center frequency o f the rejected band of bandwidth B = to — to • In general, the order of a filter is the degree of the denominator polynomial of its network function. Thus the filters of Figs. 15.1 and 15.4 are second-order filters, as seen from (15.6) and (15.12). Higher-order filters are more expensive to construct but they have better frequency responses than lower-order filters. This w i l l be seen for a special type o f filter in Prob. 15.18. c

c

The amplitude function continuously decreases as to increases, because n merator is constant and its denominator continuously increases with freqo Therefore, the amplitude response attains its maximum of | H |„«x = 1 at m - ' thus has the shape of Fig. 15.5(a). From this we see that the circuit of Fig I ! low-pass filter. That is, it passes low frequencies (relatively large amplitude, rejects high frequencies (relatively small amplitudes).

494

C2

S e c t i o n 15.2 C h a p t e r 15

Frequency Response

0

Filters

cl

495

|H(/u)l IHI „ m

K I M »NAN< I A physical system that has a sinusoidal type of natural response reacts vigorously, and sometimes violently, when it is excited at. or near, one of its natural frequenThis effect may have been noticed by the reader in Sec. 9.6, particularly in the case of Exercise 9.6.3. The system in this respect is somewhat like all of us. When urged to do what it naturally wants to do, it responds with enthusiasm. This phenomenon is known as resonance, and its side effects may be good or they may be bad. As an example, a singer may break a crystal goblet with his voice alone by properly producing a note at precisely the right frequency. Also, a bridge may be destroyed i f it is subjected to a periodic force with the same frequency as one of its natural frequencies. This is why no thoughtful troop commander w i l l march his men in step across a bridge. On the other hand, without resonance there could be no electric filters. We define a sinusoidally excited network to be in resonance when the amplitude of the network function attains a pronounced maximum or minimum value. The frequency at which this occurs is called the resonant frequency. As an example, the RLC parallel circuit of Fig. 15.1 is in resonance when the frequency of the driving function is too = 1/VZc. This was shown in Sec. 15.1 where it was demonstrated that the maximum network function amplitude occurred at to . The amplitude response of Fig. 15.2(a) is typical, with its relatively high peak at the resonant frequency. The parallel RLC circuit is so important that the term parallel resonance is reserved for its resonant condition.

(a)

FIGURE 15.6

(a) H i g h - p a s s a n d (b) b a n d - r e j e c t a m p l i t u d e r e s p o n s e

EXERCISES 15.2.1

Show that

2s s

2

+ OAs + 4

is the network function of a bandpass filter and find to , o i , to , and B. 0

Answer 2; + 0 . 2 + V4M 15.2.2

q

C2

= 1.81, 2.21; 0.4

Show that

2s

2

W

• ^T17T8

is the network function of a high-pass filter, and find | H ( » |„»» and to .

a

c

Answer |H|m« = 2, to = 2 V 2 c

15.2.3

Show that

3(iS

H ( , )

=

2

+

2 5

)

+ , + 25

is the network function of a band-reject filter, and find | H(ja>) I ™ , m

and

IO

c

Answer | H | = 3. « o = 5, to „ = ( + 1 + VToT)/2 = 4.525, 5.525 A n all-pass filter is one whose amplitude response is constant. (Thus, it passes all frequencies equally well.) I t can be cascaded with another filter to keep a desired amplitude response but shift the phase. Show by finding the network function V / V i , the amplitude response, and the phase response that the given circuit is a first-order all-pass filter. M

15.2.4

c

C2

2

The reader may recall encountering the term resonant frequency earlier in Sec. 9.8 in connection w i t h the underdamped case of the parallel RLC circuit. The two frequencies, there and here, are exactly the same. The natural frequencies of the parallel RLC circuit are the poles of the network function, given from (15.6) by S\,2 = -a

:

L

(15.13)

where a =

Answer ^ •4.+ 4 ' 1. 180° - 2 arctan to/4 EXERCISE 15.2.4

± jtoj

iic

(15.14)

and to = Vtol - a d

(15.15)

2

From the pole-zero plot shown in Fig. 15.7 we see that the resonant frequency too, or s = jtoo, is very near the natural frequency s = —a + jto , since, by (15.15), too is the radius of the dashed semicircle. I f R is made larger, then a is made smaller and the resonant frequency is even closer to the natural frequency. I n this case, as is clear in Fig. 15.2(a), the peak is more pronounced. O f course, i f R were infinite (open-circuited), then the resonant frequency would coincide with the natural frequency and the amplitude would be infinite. Before leaving the parallel RLC circuit, let us note that the network function is actually the input impedance, as stated in (15.5). The resonant frequency d

496

S e c t i o n 15.3 C h a p t e r 15

Frequency Response

Resonance

497

f<4

,3

In I xcrcise 15.3.2(a), find the amplitude of the voltage across the combination il ihr current source is i , = cos tot mA and to is (a) 10", (b) 5 x 10 , and (c) 25 • I 0 rad/s. Answer (a) 0.0417; (b) 2; (c) 0.0417 V 4

4

|\

/

| M BANDPASS FUNCTIONS A N D QUALITY FACTOR

1 1

N 1'

-«

\

0

\

The network function (15.6) is a special case of the general second-order bandpass function Ks (15.16) s + as + b 2

where K, a > 0, and b > 0 are real constants. To see that the function is o f the bandpass type, let us consider its amplitude

\

I HOW) | =

—

V(b - to ) + a to 2

FIGURE 15.7

P o l e - z e r o plot o f t h e p a r a l l e l

RLC

Va

2

+ [(b - to )/to]

2

0

2

1*1

circuit

o> = 1/VZc is the frequency for which the input impedance is purely real, as seM in (15.7). Indeed, many authors define the resonant frequency precisely this way ! the case of a two-terminal network. In the general case, maximum amplitude iliwi not always occur exactly at the frequency of real impedance, but usually there is vera little difference. In the case of the RLC series circuit excited by a voltage source \ , il tb» phasor current I is the output, then

2

2

2

Evidently, the maximum value is | H ( » U

= L!M

a

occurring at the center, or resonant, frequency m, satisfying b = too

(15.17)

At a cutoff frequency to , we must have c

»<»

= fgj

-

Y

W

= « +

* * - ( l / . C ) ]

where Y is the input admittance seen at the source terminals. Evidently, series reio nance occurs at to = 1/VZc, yielding a maximum amplitude of l/R. As in I N parallel resonance case, the resonant frequency is also the frequency of real input impedance or admittance. A t resonance the effect of the storage elements exactly cancels, and the source sees only the resistance.

|H(M)I =

^ | H ( » U

0

Va

2

+ [(b -

toD/tOc?)

Via

which evidently holds i f b - to

2

EXERCISES 15.3.1 15.3.2

498

Show that the resonant frequency in the case of the function of Exercise 15.2.1 co« incides with the frequency for which the function is real. Find the resonant frequency for the parallel RLC circuit described by (a) R = 2 k l l L = 4 m H , and C = 0.1 / i F ; (b) to* = 5 rad/s and a = 12 Np/s; and (c) a 1 Np/s, R = 4 H, and L = 2 H . Answer (a) 50,000; (b) 13; (c) 2 rad/s C h a p t e r 15

Frequency Response

Thus we have to ± aw, — b = 0 2

(15.18)

which, because of the double-sign possibility, has four solutions. Using the positive sign, we have, by the quadratic formula, *c, * S e c t i o n 15.4

-a + Va ~

2

B a n d p a s s F u n c t i o n s a n d Q u a l i t y Factor

+ 4b

(15.19)

499

We have discarded the negative sign on the radical because this gives a ncgalW off frequency. Using the negative sign in (15.18) and again suppressing the n t root, we have the other cutoff frequency,

The filler described in Exercise 15.2.1 is a bandpass filter with o> - 1 rad/s, B = 0.2 rad/s, and Q - 5. Another example is the parallel KLCcircuit with the network function given in (15 6) In that case, o» = 1 / V l C , B = \/RC, and Q = wo/B, which has the equivalent values 0

0

a + Va

+ 4b

2

I

11

Q = Evidently, from (15.19) and (15.20) we have the bandwidth, r2

f|

R

Thus in view o f (15.17) and (15.21) we may write the network function as Ki s

2

(00 L

(15 22)

+ Bs + wh

We should mention that the quantity we have called Q is defined as selectivity by some authors, who reserve for Q the definition _

which is the general network function of a second-order bandpass filter having c frequency m and bandwidth B. The amplitude response is shown, o f course, ii 15.3. Another result worth noting from (15.17), (15.19), and (15.20) is (15.23)

O>o = 0),,(O.

(15.26)

•4l

(13 i | |

B = a> -
mRC

which demonstrates that the center frequency o> is the geometric mean V w , I the cutoff points. A good measure of selectivity or sharpness of peak in a resonant circuit is ihc so-called quality factor Q, which is defined as the ratio of the resonant frequency to the bandwidth. That is,

^

_

total energy stored at resonance n

^

27)

energy dissipated per cycle at resonance

In the examples we have considered, the two definitions are the same, as the reader is asked to show in Exercises 15.4.1 and 15.4.2. However, in general, there is a slight difference. Finally, let us consider the effect of Q on resonance. Incorporating (15.19) and (15.20) into one equation and replacing a and b by their values, we have

W

^

2

=

+

2 Q

+

0

Evidently, i f Q is high, we may neglect (\/2Q) _

v

in comparison with 1, and write

2

too

,

(15.24)

B

or, approximately, (The letter Q is also our symbol for reactive power, as the reader w i l l recall. How ever, the two quantities will never be used in the same context, so there should lxno confusion.) Evidently, since B = m/Q, a low Q corresponds to a relatively large bandwidth, and a high Q (sometimes arbitrarily taken as 5 or more) indicates a small bandwidth, or a more selective circuit. W i t h this definition of Q we may write (15.22) in the form Ks

0*0 - T (15.29) B

Thus as Q increases, the amplitude response approaches arithmetic symmetry. That is, the cutoff points are half the bandwidth above and below the center frequency. In the example of Exercise 15.2.1, we have o> = 1 and Q = 5, which we consider as high. Thus by (15.29), the approximate cutoff frequencies are a> , = 0.9 and a>„ = 1.1 rad/s. The exact values are 0

H

to

=

i • /

, . n

•

2

(15.25)

so that we see at a glance the center frequency, the quality factor, and the bandwidth.

c

a)<-|.c = 0.905,

1.105

2

500

C h a p t e r 15

Frequency Response

S e c t i o n 15.4

B a n d p a s s F u n c t i o n s a n d Q u a l i t y Factor

501

EXERCISES 15.4.1

For the RLC parallel circuit in resonance, show that the total energy stored >» w(t)

cos

= \r CH 2

2

2

=

o> t + a

2a>oL

sin

2

mt

\r>c,i

where the excitation is i = I cos u> t. m

15.4.2

0

Show that for the circuit of Exercise 15.4.1 the energy dissipated per cycle is [

Rli cos cool dt = 2

FIGURE 15.8

l r R I m

and thus by definition (15.27) and the result of Exercise 15.4.1 that

s

V e c t o r r e p r e s e n t a t i o n of

- s.

The typical vector s - s,, drawn from s to s, may be written in polar form where its magnitude is its length and its phase is the angle it makes with the positive real axis. I f s = j
Q = a> RC 0

15.4.3

For the RLC series circuit with excitation v = V cos oL/R. The response is the loop current.

15.4.4

Show that the circuit is a bandpass filter, and find a> and Q (H = V2/V1). Answer 8 rad/s, 0.4

s

m

N,e"',

i =1,2,.

M,e "*,

* = 1, 2, .

joj - z , =

0

jw

-

p

=

k

J

Therefore, the network function is IK>) =

-Wv2n 2

IHO^K*"

where, for AT > 0, the amplitude is

il

IHO) | =

KN,N M,M

(15.31)

2

2

and the phase is

EXERCISE 15.4.4

) = (a, + a

15.5 The pole-zero plot of a network function may often be used to readily sketch the frequency responses. To see how this may be done, let us write the network function in the form K(s - Z , ) ( J - zi) • • • ( ? - z ) (15.30) (S - p,){s T pi) •:••... ( j - p„) m

t

502

C h a p t e r 15

Frequency Response

+ • • • + « „ ) - (/3, + B + • • • + B„) 2

(15.32)

both of which may be measured directly from the pole-zero plot. I f K < 0, then K = | K | / 1 8 0 ° , which must be accounted for in the amplitude and phase. Thus for any point ja) we simply draw vectors from all the poles and zeros to j
USE O F P O L E - Z E R O PLOTS

where, as before, the z's and p's are the zeros and poles. Each of the factors is I complex number of the form s - s, and may be represented in the s-plane by a vector, drawn from si to s, as shown in Fig. 15.8. This is true since by vector addition the vector s is clearly the sum of the vectors s and s — s .

2

PLE 15.4

Suppose that we have

H(*) =

t

S e c t i o n 15.5

U s e of P o l e - Z e r o Plots

4s s* +• 2s + 401

503

- ] +/20

X—,

—

;20

For to = 20 [Fig. 15.9(c)] we have

«\

|H|-

\

4(20)

_

:

(l)Vl60T 90° - (0 + tan" 40) « 0 1

This point is in the region where the amplitude changes the fastest, since M, = 1 is a minimum and is changing percentage-wise much faster than N or M . The amplitude therefore will reach its peak near this point. Actually, we know from our previous work that o> = V 4 0 1 yields the peak amplitude of 2. If co = o>», a very high value, say, such as 10 , then all three vectors w i l l be essentially vertical, so that 2

7

k

0

6

.-/20

N

*

a "

M, «• M - to 2

fit"

fi,

*

h

W

Therefore, we have | H | = 4 / w » , a very small value, and tb " - 9 0 ° . Sketching the functions, we have the forms of Fig. 15.2, as expected, with to — 20 and | H |m„ ~ 2. We may get a rough idea of the cutoff points from Fig. 15.9(d). In this figure A/i = V 2 , which is V 2 times its value at the approximate peak, represented by Fig. 15.9(c). Since M and N have changed by much smaller percentages, the amplitude is then approximately 1 / V 2 times its peak, so that to — 19. By a similar argument at to = 2 1 , we have to •» 2 1 . The exact values, by (15.19) and (15.20), are to , = 19.05, 21.05. In this example, by comparing the network function w i t h (15.25), we see that Q = 10.01. Thus we have a high Q so that the poles are very near the jto-axis. Thus to is very near m, and the approximations we have made are very near the exact values. 0

2

C)

C2

cl C7

d

EXERCISES (c)

FIGURE 15.9

(d)

S t e p s in t h e c o n s t r u c t i o n o f the f r e q u e n c y r e s p o n s e s

15.5.1

Use the method of this section to sketch the amplitude and phase responses for

(figures not d r a w n to s c a l e )

H(*) = which has a zero at 0 and poles at - 1 ± y'20. These are shown on the pole-zen plots of Fig. 15.9. By (15.31) and (15.32) we have |H ( » | =

X

2

whose components are identified in Fig. 15.9(a). First we note that i f to varies from 0 to + ° ° , then a = 90°. For to = 0 we have, from Fig. 15.9(b), +


504

C h a p t e r 15

4(0) /401V

0

401

= 90° - ( - t a n

1

+ 4s + 2504

Answer to = 50, Q = 12.5, B = 4, to . - 48, 52 Find the exact values for the answers given in Exercise 15.5.1. Answer 50.04, 12.51, 4, 48.08, 52.08 cl C2

2

tb(a>) = a - Q3, + Q )

HI

2

0

15.5.2

AN MM

16s s

20 + t a n

1

20) = 90°

15.6 SCALING THE NETWORK

FUNCTION

The reader may have noticed that in many of our examples we have used network elements such as 1 ft, 1 F, and 2 H , which are extremely nice numbers to have when we are analyzing a network. For example, in F i g . 15.4 we had elements of 2 ft, j F, and j F, and the network was a low-pass filter with a cutoff frequency of V 2 rad/s,

Frequency Response S e c t i o n 15.6

Scaling the Network Function

505

or 0.23 Hz. However, such element values as these are not very practical, to say the least, and there is very little demand for filters that pats frequencies between 0 and 0.23 Hz. (As an illustration, a l - F capacitor, constructed of two parallel plates 1 CM apart with air as the dielectric, would require a face area of 1.13 X 10' m . i It would be an ideal situation i f the circuits we analyze or design contain sun pie element values such as 1 F, while the circuits we build have practical element values such as 0.047 /*F, and useful characteristics such as cutoff points of 100 Hz, As we shall see in this section, network scaling allows us to have it both ways. We shall consider two types of network scaling, impedance scaling and frequency scaling. To illustrate the former, let us suppose that the network function i l an impedance given by :

Z'(s)

= sL' + R

+

To frequency-scale a network function by a frequency scale factor k , we sun ply replace s by s/kf. That is, i f the unsealed network has the network function H'(S), then the scaled network function H(i) is obtained by letting S = s/kf, resulting in f

Ms)

Thus i f the unsealed network had a property, such as center frequency, when S = j \ then the scaled network has this property at s = jk . This is clear from (15.35), which gives

Hfjfc) = H ' O D Another way to consider frequency scaling is to note that s = k S, so that i f i = ja> corresponds to S = jil, then to = fc/ft. Thus the values on the frequency axis have been multiplied by the scale factor k , without affecting values on the vertical axis of a frequency response. f

(15.33) +

l

+

s(C

'

The scaling of the network to effect the transformation of (15.35) is quite simple I f Z ' ( 5 ) , given by

/k.) Z'(5) = SL' + R' + ^ 7

If the impedance of the scaled network is

is any impedance in the unsealed network, then the corresponding impedance in the

Z(s) = sL + R + ^-

scaled network is

then we see by comparison with (15.33) that L =

C

_

Z(s) = sL + R +

k,L'

R =

k,R' c

(15.35)

f

+ R' +

= s(k,L )

= H ' (j - )

f

— sC

The network has been impedance-scaled by an impedance scale factor fc, i f the impedance Z(s) of the scaled network is Z(s) = kZ'is). In other words, we must have Z(s) = k^sL'

mams an infinite gain device, and thus is unchanged. The network lun, n V (v)/Vi(s). being dimensionless, is also unchanged

-

(15.34)

SL

*(|)

' k, Comparing these results, we have

In summary, to impedance-scale a network by the factor k,, we multiply the L's and R's byfc,and divide the C's by fc,. We have illustrated this for a special case, but it may be shown to hold in general. Also, i f there are dependent sources, the scaling is accomplished by multiplying gain constants having units of ohms by it, dividing those with units of Siemens by k,, ar.d leaving unchanged those that are di mensionless. The scaling multiplies by fc,, divides by fc,, or leaves unchanged network functions that have units of ohms or Siemens or are dimensionless. EXAMPLE 15.5

506

To illustrate, let us impedance-scale the network of Fig. 15.4 by fc, • 5. The 2-H resistors become 2 x 5 = 10 ft, the j - F capacitor becomes | + 5 « 0.1 F , and the J-F capacitor becomes a + 5 = 0.05 F. The op amp, an infinite gain device, re C h a p t e r 15

Frequency Response

L =

V k,

R = R' C =

(15.36)

C kt

Therefore, to frequency-scale a network by a factor fc/, we divide the L's and C's by k and leave the R's unchanged. I f there are dependent sources w i t h constant gains, these also are left unchanged. f

S e c t i o n 15.6

Scaling the Network Function

507

EXAMPLE 15.6

ill I I t . Suppose that we amplitude- and frequency-scale the network to obtain a resmi.mi frequency of 10* rad/s using a capacitor of 1 nF. Then we have k » 10*. and the new capacitance is

The circuit of Fig. 15.4 is a low-pass filter with a cutoff frequency of at V 2 rad/s, as we have seen previously. Suppose that we want to frequency-scale t network so that the cutoff frequency is 2 rad/s. Then the scale factor k is given by

t

r

f

V 2 kf=

c=.o->-

f

f

1

Ikt x 10"

2k L = — = 10 H = 1 mH kf R = Ik, = 500 ft 1

O f course, we may perform both impedance and frequency scaling on a net work. To obtain a network with a practical property such as center or cutoff fro* quency, we may first apply frequency scaling with the proper factor kf. Then tn make the resulting element values more practical we may impedance-scale by factor k<. I f the parallel RLC network of Fig. 15.1 contains a 1 - f t resistor, a 2-H inductor, and a j - F capacitor, then for the network function of (15.5) we have an amplitude ra« sponse as in Fig. 15.2(a), with a resonant frequency of 1 rad/s and a peak amplitude

-

Therefore, k, = 500, and the new inductance and resistance are

or */ = V 2 . Dividing the C's in Fig. 15.4 by k we have, in the scaled network, pacitances of 1 / 2 V 2 F and 1/4 V 2 F. The rest o f the circuit remains unchanged there being no inductors to scale.

EXAMPLE 15.7

1 / 2

kik

2

3

The scaled network is shown in Fig. 15.10(a), and its amplitude response, a scaled version of Fig. 15.2(a), is shown in Fig. 15.10(b).

EXERCISE 15.6.1

Frequency- and impedance-scale the circuit of Fig. 15.4 to obtain tu = 200O7T rad/s (f = 1000 Hz), using capacitors of 0.01 and 0.005 pF. c

c

Answer k = (1000V2)tt, *, = 1 0 / 2 V 2 i r , R's = 22.5 k f t each f

FIGURE 15.10 I

(a) N e t w o r k ; (b) its a m p l i t u d e r e s p o n s e

jl mH

15.6.2

5

Show that the circuit is a bandpass filter with tu = 1 rad/s, and frequency- and impedance-scale it to obtain a center frequency of 1000 rad/s using capacitors o f 0.1 and 0.4 ^iF. Answer Both R's = 5 k f t 0

:t nF

-o>

15.7 THE DECIBEL I f the output of a network is used to provide some quantity that is to be sensed by a human being, the function is sensed in a noncontinuous manner. For example, i f the network provides sound, as in the case o f the telephone, the listener cannot detect continuous changes in intensity. Moreover, i f the sound intensity is 1 (on some arbitrary scale) and it must be increased to 1.1 before the listener detects any change, S e c t i o n 15.7

T h e Decibel

509

then the same listener can detect no change, i f the original level is 2, until II in in creased to 2.2. In other words, the ear i:. not a linear device but more like I logfl< rithmic device, since the differences.

I wen more often in practice i t is of interest to consider the attenuation or loss. defined by a (to) = - 2 0 1 o g , | H p ) | 0

log 1.1 - log 1 = log 1.1 and log 2.2 - log 2 = log

= 20 logi,

(¥)

=

Suppose that

Vi(»

0

s + Vis + 2

(15.37)

'

p number of dB = 10 logm — P\ If the two average powers P, and P are referred to equal impedances, the last expression may be given in terms of the corresponding voltages by 2

number of dB = 10 logio

'

J-

Vl +

;

to*

which is that of a low-pass filter with m, = 1 rad/s. The linear sketch looks somewhat like that of Fig. 15.5(a). The attenuation is given in decibels by

p number of bels = logio — Pi However, this proved to be a large unit, so that the unit decibel (is be') became com mon. That is,

i

As the reader may verify, the amplitude response is IH P ) | =

Historically, the logarithmic unit, now known as the bel, was defined originally by Alexander Graham Bell (1847-1922), who, of course, invented the telephone, ai the power unit,

a (to) = 20 log.o V l

+ to'

(15.39)

= 10 log,o (1 + <•>') and is shown in Fig. 15.11 for 0 £ to £ 5 . FIGURE 15.11

A t t e n u a t i o n o f a l o w - p a s s filter

dB 30

-

Vi(» V,(jw)

which, of course, is equivalent to (15.37). In any case, (15.37) is taken as the standard definition. In practice, frequencies are not ideally blocked or attenuated in the filtering process, as may be seen in the low-pass response of Fig. 15.5(a). A zero amplitude would correspond to absolute, or infinite, attenuation, and any approach to such an ideal situation would be difficult to appreciate on a linear sketch. For example, if | H p ) | = 0.001 in Fig. 15.5(a) ( ^ of 1% of its peak value of 1), this would correspond to —60 dB (or 60 dB below its peak value of 0 dB). The latter figure means much more to a telephone engineer than the linear figure.

Frequency Response

dB

V p)

V»P)

number of dB = 20 l o g , | H p ) |

C h a p t e r 15

v , p )

W

In this case, a frequency w, passes if aM is relatively small and is attenuated i f a (to,) is relatively large. The decibel units enable us to tell with some standard precision the degree to which the frequency is attenuated.

is more commonly expressed in decibels, abbreviated dB and defined by

510

Vi(jw)

2

in the two cases are the same. For this reason, among others, the amplitude response, |Hp)|

lis

V2P)

= - 2 0 log,

10

S u

S e c t i o n 15.7

The Decibel

511

EXERCISES 15.7.1

4i A

Let the amplitude response |H(ja>)| be such that iHO'tuJlmax = | H ( y O ) | A, that at cutoff | H ( y o ) ) | = K/Vl. Find the loss, given by (15.38), at to = 0 and to = to . Note that to corresponds to the "3-dB point," meaning that at to «• loss is approximately 3 dB more than at the point of minimum loss, to = 0 in case. c

c

c

Answer a(0) = - 2 0 log K, a(to ) c

15.7.2

(I)

Ikft

©

w

= - 2 0 log K + 3

Given the bandpass filter function „, .

0-25 s

+ 0.2s + 1

find the loss in decibels at to = 0.0001, 0.5, 0.905, 1.0, 1.105, 10, and 100 rail/ Answer 94, 18, 3, 0, 3, 34, 54

15.8

FIGURE 15.12 RLC

circuit with

CCCS

F R E Q U E N C Y RESPONSE W I T H SPICE The frequency response of a network is easily obtained using SPICE by sim| specifying a frequency range and the number of points desired within the range the .AC solution control statement. The generation of a circuit file for a given i quency response is almost identical to procedures described in previous chapters I ac analysis. The .PRINT or .PLOT output control statements can be used for out to the console or the printer. EXAMPLE 15.9

Let us plot the frequency response for a linear frequency sweep in the network • Fig. 15.12 from 10 to 500 kHz with v having an amplitude of 1 V. The Vd = 0 (shown dashed) connected from node 5 to the reference node has been added for specifying the current i of the CCCS. A circuit file is g

FREQUENCY RESPONSE FOR FIGURE 15.12 •DATA STATEMENTS VG 0 1 AC 1 0 Rl 1 2 IK L I 2 0 1UH C I 2 3 0. 1UF F 2 3 VD 4 L2 3 0 1UH C2 3 4 0.1UF R2 4 5 I K VD 5 0 AC 0 0 •SOLUTION CONTROL STATEMENT AC L I N 25 10K 500K •OUTPUT CONTROL STATEMENT

FIGURE 15.13

•! +i

F r e q u e n c y r e s p o n s e for F i g . 1 5 . 1 2

1M(R2) IP(R2) FREQ

(•)( + )1. 00OE+04 3 . 042E+04 5. 083E+04 7. 1 2 5 E + 0 4 9. 1 6 7 E + 0 4 1. 1 2 1 E + 0 5 1. 3 2 5 E * 0 5 1. 529E- 05 1. 7 3 3 E + 0 5 1..938E+05 2..142E+05 2.,346^+05 2..550E+05 2 .754E+05 2 .958E+05 3 .163E+05 3 .367E+05 3 .571E+05 3 .775E+05 3 .979E+05 4 .183E+05 4 .388E+05 4 .592E+05 4 .796E+05 5 .00OE+05

IM(R2) 1.0000E-11 -2.0000E+02 * 2.452E-11 . 7.022E-10 . 3.324E-09 . 9.344E-09 . 2.046E-08 . 3.877E-08 . 6.698E-08 . 1.088E-07 . 1.693E-07 . 2.584E-07 . 3.820E-07 . 5.662E-07 . 8.452E-07 . 1.292E-06 . 2.078E-06 . 3.728E-06 . 8.989E-06 . 1.188E-04 . 1.065E-05 6.245E-06 4.756E-06 4.029E-06 3.613E-06 3.353E-06 3.184E-06

1.0000E-09 -1.0000E+02

1.0000E-07 O.OOOOE+00

1.0000E-05 1.0000E+02

1.0000E-03 2.0000E+02

*

* *

+

+ +

.+ .+ .+ .+

PLOT AC IM(R2) I P ( R 2 ) .END

The resulting plot for this example is shown in Fig. 15.13.

512

C h a p t e r 15

Frequency Response

S e c t i o n 15.8

Frequency Response with SPICE

513

EXAMPLE 1 5 . 1 0 Consider the network o f Fig. 15.4 having node assignments o f 1 at v,. 2 i t the c mon node o f the 2 - 0 resistors, 3 at the op amp noninverting input, and 4 at IM put. Let us determine the frequency response for a linear sweep from OtMit i<> I for v having a 1-V amplitude. A circuit file using the subcircuit file OPAMP CK Chapter 4 is

In this chapter we have considered amplitude and phase responses of circuits. These arc special cases of frequency responses and enable us to consider which frequencies in a signal are dominant (correspond to large amplitudes) and which are suppressed (correspond to small amplitudes). This allows us to consider electric filters, which arc circuits that pass certain bands of frequencies and suppress or filter out others. The band of frequencies that pass has a bandwidth centered about a center frequency. Popular filter types are bandpass (those that pass a band between two positive frequencies called cutoff points), low-pass (those passing low frequencies), high-pass (those passing high frequencies), and band-reject filters (those passing all but a finite band, which is rejected).

FREQUENCY RESPONSE FOR LOW-PASS F I L T E R OF F I G . 1 5 . 4 •DATA STATEMENTS VI 1 0 AC 1 0 HI 1 2 2 R2 2 3 2 C I 2 4 0.5 C2 3 0 0.25 XOPAMP 4

3

4

OPAMP

Other terms of interest are resonance (the phenomenon associated with vigorous behavior such as pronounced peaks or dips in amplitude of the output signal), resonant frequency, and quality factor. Scaling the network moves the resonant frequency to a desired value and usually results in more practical circuit elements. The decibel is a convenient unit for measuring the amplitude of a transfer function, and is used extensively in filter theory. Finally, SPICE is extremely useful in plotting frequency responses, particularly in practical circuits where the numbers involved are necessarily complicated.

•DEFINE SUBCIRCUIT F I L E .LIB

OPAMP.CKT

•SOLUTION CONTROL STATEMENT .AC L I N 25 0.001 1 •OUTPUT CONTROL STATEMENT PLOT AC VM(4) V P ( 4 ) . END

The plot for this response is similar to that o f Fig. 15.5.

PROBLEMS

EXERCISES 15.8.1

15.8.2

Using SPICE, find the frequency response from 10 Hz to 400 k H z for V in Fig 15.10(a) i f the capacitor and inductor values are 10 p,¥ and 0.1 p.H, respectively,] Use a source current o f 1 m A . Using SPICE, find the frequency response in the interval from 0.001 to 1 Hz. U t t u, = 1 V.

For the circuit shown, R = 4 CI, L = 0.4 H, and C = 1 mF. If the input and output are V, and V , respectively, find the network function and sketch the amplitude and phase responses. Show that the peak amplitude and zero phase occur at o> — 50 rad/s. 2

EXERCISE 15.8.2

v

-Tnnr>|

15.3

1

L

Find H(s) = Vi(s)/\,(s) and sketch the amplitude and phase responses. Shnw that the peak amplitude and zero phase cccur at OJ = 4 rad/s.

3

-If-

•0

PROBLEM 15.3 15.4

Show that the transfer function V2/V1 is given by

PROBLEM 15.1 HW = If H{s) = V / V , in the circuit of Prob. 15.1, and L = 1 H, find R and C so that the peak amplitude occurs at u> = 10 rad/s and | H ( y 6 ) | = 0.707. 2

514

C h a p t e r 15

Frequency Response

C h a p t e r 15

Problems

V»d) V,(i)'

H

1 R-,

1

s + —

R3

Find Ri, Ri, and R, so that the transfer function is identical to that of Prob. 15.1.

515

PROBLEM 15.12

Show that the network function Vi/V, ifure for Prob. 15.1 is given by

•o v.

in the

15.16 Show that the circuit is a low-pass filter by finding the transfer function H(s) = V (J)/V,(S), the amplitude and phase responses, and to,. (Note that this circuit and that of Prob. 15.14 yield the same amplitude response except for a constant multiplier.) 2

VJ _ 1/LC V, s + (R/L)s + (1/LC )

PROBLEM 15.4 15.5

15.6

If R = R = 0.5 11 and R, = 0.01 n in Prob. 15.4, find His), the frequency aio at which the peak amplitude and zero phase occur, and the magnitude of H at io«. t

2

Show that in the general case the network function in the figure for Prob. 15.1 is given by m

A

VfW ; \,(s)

w

W

15.7

(R/L)s s + (R/L)s + 2

Select C = 0.1 F and find R and / Prob. 15.6 so that o) = 10 rad/s and f 10 rad/s. M

0

Show that in Prob. 15.6 if L = Q, C and R = 1, then H<5)

(i/LC)

2

PROBLEM 15.7

1/

Let R = I n, L = 1/V2 H, and C = V 2 F, and show that the result is a low-pass filter and " 1 rad/s by finding the amplitude renpon.se. Show that the circuit is a low-pass filter by finding the transfer function H(s) = V (J)/VI(J), the amplitude and phase responses, and to,.

(1/0)* s

+ (\/Q)s

+ 1

Frequency Response


+

15.10 Show that in the general case the circuit Prob. 15.4 is a bandpass filter with center Itf quency 1/Vfl^ rad/s, bandwidth i/R nA% gain R /R\, and quality factor R /VRi. I Ri = Q/G, R = Q, and R, = 1 to obtain I network function

1

PROBLEM 15.16

2

PROBLEM 15.14

2

2

2

2

H(5)

=

(G/Q)s \,(s)

' s + (l/Q)s 2

+1

Using the results of Prob. 15.10, obtain i bandpass filter having the form of the iinur for Prob. 15.4 with
For the circuit shown, find the network function, H(J) = V (s)/V,U), and sketch the amplitude and phase responses. Show that the peak amplitude and zero phase occur at to = 0. 2

PROBLEM 15.15 I n

1

y/2_H

nsw^-

PROBLEM 15.17 15.18 One type of low-pass filter is the nth-order Butter worth filter, whose amplitude response is

IH(» I =

= , n = 1, 2, 3,

where i t is a constant. Ideally, a filter would pass all frequencies in its passband equally

2

C h a p t e r 15

1 H

v w

15.17 Show that the given circuit is a low-pass filter with to, = 1 rad/s by finding H(s) = V ( J ) / Vi(s) and the amplitude response.

15.12 Show by finding V / V , that the circuit is a bandpass filter, and find the gain, the hanil width, and the center frequency.

516

211

1

2

which is a bandpass network function wll quality factor Q, center frequency 1 rad/s, bandwidth B = l/Q rad/s, and y» G = 1.

15.11

il

4

4

15.9

2

Thus by comparison with (15.22) and (15.25), show that the circuit is a bandpass filter with resonant, or center, frequency too = 1/VLC, bandwidth B = R/L, and quality factor Q = a>o/B = (l/R)VL/C. Show also that the gain G of the filter, defined as its peak amplitude value, is given in this case by G = I . Show by finding the network function H(s) = V ( i ) / V , ( j ) that the circuit is a bandpass filter with tol = \/(LC), B = 1/C, and gain = 1. Select L and C so that B = 0.5 and tuo = 1 rad/s.

1

15.8

2

C h a p t e r 15

Problems

517

IH(/u)l

15.25

. by hndinK the transfer function / V , ( s ) . the amplitude response, the gain, „ that thr CUM* >« • third-order high-

Use the results of Prob. 15.24 to dcMgn • band-reject filter with ui = I rad/s, Q - 2, and a gain of ; . 0

15.26

Apply impedance and frequency scaling to lh* circuit of Prob. 15.1 with R = 1 CI, L 0.5 H. and C 0.02 F to obtain a bandpass filter with a center frequency f = 10* 11/, with a quality factor (2 = 5, and using a capacitor of 10 pF. (Suggestion: See Prob 15.9.) 15.27 Obtain an active bandpass filter using the configuration of Prob. 15.4 with / = 2000 Hz, Q - 10, and G = 2, using capacitors of 5 nF. (Suggestion: See Prob. 15.10.) 0

0

PROBLEM 15.23

15.28

Show thai the network function of the given circuit is V

H(J)

PROBLEM 15.18 well (| H | > 0 would be constant) and perfectly block all other frequencies (| H | would be zero). As the accompanying figure shows, for n - 2, 3, and 8, the Butterworth filter improves (approaches ideal) as the order n increases. Show that to, = 1 rad/s for any n and that the filters of Probs. 15.13 and 15.17 are Butterworth filters of second and third orders, respectively. Finally, sketch a fourth-order Butterworth response and compare it with the second, third, and eighth orders. 15.19 Show that for the figure for Prob. 15.4 Vitf s

2

l/R, + (\/R )s + ( l / « ) 2

3

Choose values of the resistances so that the circuit is a low-pass Butterworth filter with to, - 1 rad/s and H(0) = - 2 . [Suggestion By Prob. 15.13, the denominator of H(.v) is required to be s + Vis + 1.] 15.20 If the gain of a low-pass filter with the network function H(s) is defined to be K = | H(0) |, compare the gains in the general cases of Probs. 15.13 and 15.19, which are, respectively, passive and active circuits that perform low-pass filtering. Note that gains higher than 1 are possible with active elements present. This feature and the absence of in2

518

ductors, which are undesirable at lower inquencies, are advantages of active filters ovtf passive filters.

V,

K ****

2

s + (R/L)s + (1/LC)

Frequency Response

+

15.29

1

f

0 ( 1 - AT) I

2

•6

Determine R, and R so that ihe circuit is a first-order low-pass filter ( H = V / V i ) with to, = I rad/s and a gain of 4. Scale the result to obtain to, = 10* rad/s using a It)- n l capacitor. 2

2

|_

|

i n>v

2

0(1-

PROBLEM 15.29

PROBLEM 15.24

15.22 Show by finding the transfer function V u ) / V i ( j ) , the amplitude response, the gain (see Prob. 15.21), and the cutoff point to, thai the circuit is a second-order high-pass filter ;

PROBLEM 15.28 0.05 F

PROBLEM 15.22

519 C h a p t e r 15

C h a p t e r 15

»

2

i -it

s

Let R = 1 ft, L = 1/V2 H, and C = V2 »J and show that the result is a high-pass tilin with cutoff to, = 1 rad/s. The gain of a high' pass filter is defined as the limit of its transfer function as s —» » . Note that in this case tht gain is 1.

+

2

Thus the circuit is a band-reject filter with center frequency (rejected) o>„ = 1 rad/s. Also, as in the bandpass case, Q is the quality factor, and B = too/Q = \/Q is the bandwidth. Note that the gain is H(0) = K. where • < K < I.

15.21 Show that in the figure for Prob. 15.1

Vi _

_

2

W, " s + (1/0*

Show by finding H = V / V , that the given circuit is a bandpass filter, and find the center frequency and the bandwidth. Scale the circuit so that the center frequerfcy^js 20,000 rad/s using 0.01 nF capacitors.

Problems

l F 15.30 Show that the given circuit is a third-order low-pass Butterworth filter witH to, = 1 rad/s and a gain of I . Scale the circuit so that the capacitances are 0.01 jxF each and f, = 1 0 0 0 Hz. 15.31

Scale the circuit of Prob. 15.23 to obtain a high-pass filter with to, = 10 rad/s using 20-nF capacitors. (Hint: to, = 1 rad/s in Prob. 4

15.23.)

15.32 Show that the circuit, with H(s) = V2(s)/V,(j), is a second-order high-pass filter with the same transfer function, except for the gain, as that of Prob. 15.22. 15.33 Scale the circuit of Prob. 15.32 to obtain a high-pass filter with to, = 10,000 rad/s using 0.02-^iF capacitors. 15.34 Show that if R, = R R*,

transfer function is

y (s) 2

2

15.35 Scale the network in the figure for Prob so that too = 10° rad/s, Q = 5, If - 0. the capacitance is 50 pF. 15.36 Show that the given circuit is a band filter with a gain of 1, Q — 1, and a frequency to = 1 rad/s. Scale the net» obtain a center frequency / = 60 Hz, capacitances of 1 and 2 nF. 0

0

2

1 F

in

-AAA/—

«1

i rt

-WV-

15.37 Show that the transfer function is V (J)

s - Is + 2

V,(s)

s + 2s + 2 2

PROBLEM 15.34

and find the amplitude and phase responi Note from Exercise 15.2.4 that the circuit an all-pass filter. Note also that the phaw ±180° when to = to = V5 rad/s. By scalu move this point to to = 20,000 rad/s 0.01- and 0.005-/U.F capacitors.

~(VR<)s + (\/R )s + ( l / t f j ) 2

- W r

I rs

0

and thus the circuit is a second-order high-pass filter. Find R , R , / f , and R, so that the t

2

0

3

1 F

1 F v,

o!

T "

in -vw-

VW-

-ov,

2

2

2

s

t n -JyVv-

2

s + 2s + 2

the transfer function

2

-3J

_

-A/$/—

2Si

PROBLEM 15.36 PROBLEM 15.37 PROBLEM 15.30

1 H

PROBLEM 15.32

in

vw— 1 F

1 F IB

520

C h a p t e r 15 C h a p t e r 15

Frequency Response

Problems


521

15.38

Show by finding H = V / V , and the amplitude and phase responses that the circuit is a second-order all-pass filter. (Suggestion: See Exercise 15.2.4.)

satisfy the appropriate L-qu.iiiiui H 0. its maximum occurs when 11. mum of | H | occurs Also | H mil It functions of «i ; thus the maximum m l when

2

2

J

;

a

(

W

. ^Vi V,

dto

C2

0

2

= 0 2

:

2

0

|H(»|

Since (d/dw)\H\ = (d/cuo )\H\U 2wid ,/o) HH,' = 0, we must check
i£ P + is> + %s + is + 4

and so forth. [Suggestion: Find the amplitude response and verify that o> = V2 yields the peak point and that to,,, o> = o> + (a)o/10)

Ploi the frequency response of the network shown and determine the tiller iy|>r ( I a . band puss etc.) and the characteristics (passband, cutoff frequency, etc.) using SPICE.

:

15.39 The given circuit is a fourth order bandpass filter (the denominator of the network function is fourth degree), with o> = V2 rad/s and (2 = 5. Verify this by finding H

15.44

15.40 Scale the circuit of Prob. 15.39 so I wo = V2 x 10* rad/s using capacilorn 0.01 and 0.25 fif.

PROBLEM 15.38

PROBLEM 15.39

COMPUTER APPLICATION PROBLEMS 15.41 15.42 15.43

522

Using SPICE, plot the Using SPICE, plot 0.001 < / < 0.5 Hz. Using SPICE, plot the pacitor values being 1 < / < 1200 Hz.

frequency response of V in Prob. 15.16 for 0.001 < / < 2 Hz. the frequency response of the network of Prob. 15.23 2

frequency response of the network of Prob. 15.32 for resistor ami replaced by k f l and fiF values, respectively, in the interVQ

C h a p t e r 15 C h a p t e r 15

Frequency Response

C o m p u t e r Application Problems

523

16 Transformers I n our study of inductance in Chapter 7, we found that a changing current produces a changing magnetic flux which induces a voltage in a coil. In this chapter we consider the effect of a changing magnetic flux that is common to two or more distinct coils. Neighboring coils which share a common magnetic flux are said to be mutually coupled. In mutually coupled circuits, a changing current in one coil winding produces an induced voltage in the remaining mutually coupled windings. The i n duced voltage is characterized by a mutual inductance which exists between the neighboring coils. A system of mutually coupled coils which are wound on a composite form, or core, is commonly called a transformer. Transformers are available in a wide variety of sizes and shapes that are designed for numerous applications. Devices as small as an aspirin tablet, for instance, are common in radios, television sets, and stereos for connecting various amplifier stages of the system. On the other hand, transformers are designed for 60-Hz power applications which range in size from that of a pingpong ball to those which are larger than an automobile. We shall restrict ourselves to linear transformers (those whose constituent coils are linear) and begin by considering the properties of self- and mutual inductances. Energy-storage and impedance properties are then analyzed, and an important special case of the linear transformer, the ideal transformer, is introduced. We conclude the chapter with a discussion of equivalent circuits that are useful for representing linear transformers. The circuit analyses presented in the chapter include both time-domain and frequency-domain cases. Particular emphasis is given to the frequency-domain case because the most important transformer applications occur in the ac steady state.

George Westinghouse

In trie battle of the currents of the 1880s, ac won over dc because of the fabulous inventions of Nikola Tesla, the availability of transformers to step up and step down the ac voltages, and the genius of George Westinghouse. Westinghouse had already made his fortune in 1869 with the invention of the air brake for railroad trains. He was shrewd enough to use his wealth to hire Tesla and to buy the patent from Lucien Gaulard and John D. Gibbs for their newly developed practical transformer. Westinghouse was born in Central Bridge, New York, the son of a prosperous machine factory owner.

nized the Westinghouse Electric Company in 1888, which he used as a base to advocate successfully tht ac system. Westinghouse was one of America'! greatest inventors and one of the true giants of US, industry. •

16.1 MUTUAL INDUCTANCE In Sec. 7.4 we found that the inductance L o f a linear inductor, as shown in Fig. 16.1, is related to the flux linkage A by the expression A = N


524 Section 16.1

Mutual Inductance

525

F I G U R E 16.1 Simple inductor

It is evident from Faraday's law that a voltage is induced in a coil wh tains a time-varying magnetic flux, regardless of the source o f the flux. Let u i fore consider a second coil, having N turns, positioned in the neighborhood first coil, having N = N turns, as shown in Fig. 16.2(a). I n this case, formed a simple transformer having two pairs of terminals in which coil 1 is to as the primary winding and coil 2 as the secondary winding. To introduce several important inductive quantities, let us begin by c x a f l ^ H the open-circuited secondary case of Fig. 16.2(a). The current i'i produce netic flux n given by 2

t

u = u + 2l where <j> \s the flux of U that links coil 1 and not coil 2, called the leakage Jin * nml 2i is the flux o f i, that links coil 2 and coil 1, called the mutual flux. We shall assume, as in the case o f the linear inductor, that the flux in each tffl links all turns o f the coil. Since the secondary is an open circuit, no curent Howl [ coil 2, and the flux linkage of this coil is L

Therefore, the voltage v is given by 2

d\ d(j> v = — = Nr . dt dt

2

2

(We shall justify later that the polarity is as shown.) In a linear transformer, the flux ,s proportional to i\. Thus we may wnir 2

N <j>2 2

M

2]

i,

(Id I

where M \s a mutual inductance in henrys (H). In terms o f this mutual inductunvti the open-circuit secondary voltage becomes 2

V2

526

Chapter 16

Transformers

dU '~dt

Let us now find the primary voltage v i. We know that dX, ~r dt where A i , the flux linkage o f coil 1, is given by V i

Where the flux linkage is A

Si

=

2

= A/2022

We now define the relation Z.2I2 •

A, = A/,0„

A/2022

(16.5)

where L is the self-inductance o f coil 2 in henrys (H). Therefore, 2

As in the case o f a single linear inductor, the inductance L , of the pri winding, sometimes called the self-inductance of the primary to distinguish it the mutual inductance, is defined by Lti,

= N,

, di L 2

v = 2

2

in the open-circuit primary case. Let us now consider the general case o f Fig. 16.2(c) in which both i, and i are nonzero. The fluxes in coil 1 and coil 2, as shown in the figure, are

u

2

Hence we see that D| = Li

di,

in the case under consideration. Next let us consider Fig. 16.2(b) in which the primary is an open circuit and current i flows in the secondary. Proceeding as before, we have

respectively. Therefore, the flux linkages o f the primary and secondary coils are Ai = A/,0,1 + A/,012

0 2 = 4>L2 + l2

A

2

where 0 is the leakage flux of i that links coil 2 but not coil 1, and 0 i is the m tual flux o f i j that links coil 1 and coil 2. The flux linkage of coil 1 is 2

2

+ 012

02 — L2 + 012 + 021 = 021 + 022

2

t 2

0..

01 = 011 + 021 + 012 =

dt

= A/ 0

2

2

2 1

+ A' 022 2

Substituting from (16.1)—(16.5), we find upon differentiation that the primary and secondary voltages are

Ai = A/1012

Therefore, the primary voltage is given by (16.6)

dX, d, t>, = — = N,—— dt dt 2

-

2

AA0, i2

2

= Afl l 2

l i t . II

2

is a mutual inductance, the open-circuit primary voltage is Vi

= Mr.

di dt 2

2

(16.4)

528

Chapter 16

Transformers

dt

2

2

It is clear that the voltages consist o f self-induced voltages due to the inductances L , and L and mutual voltages due to the mutual inductance M. In the preceding discussion, the coil windings in Fig. 16.2 are such that the algebraic sign o f the mutual voltage terms M dUjdt and M di /dt are positive for the terminal voltage and current assignments as shown. In practice, it is, of course, undesirable to show a detailed sketch o f the windings. This is avoided by the use o f a dot convention which designates the polarity of the mutual voltage. Equivalent circuit symbols for the transformer o f Fig. 16.2 are shown i n Fig. 16.3. The polarity markings are assigned so that a positively increasing current flowing into a dotted (undotted) terminal in one winding induces a positive voltage at the dotted (undotted) terminal o f the other winding. For the purpose of writing the describing equations, we may state the following rule: 2

l2

«2

'

2

In the next section we show that the mutual inductances A / and Af . are equal) therefore, we shall write

and refer to M as the mutual inductance. The secondary voltage is given by

D

v = M— + L — dt dt

If we define

where M

M I

A current i entering a dotted (undotted) terminal in one winding induces a voltage M di/dt with positive polarity at the dotted (undotted) terminal of the other winding. Section 16.1

Mutual Inductance

529

note that a current flowing into b in Fig. 16.4(a) produces a flux in the same direcreceives the polarity dot. The circuit symbol l.n IIH>. 16.4(b).

tion at 0. Thus terminal b transformer is shown in Fig

Evidently, the polarity markings on a transformer are independent of the terminal voltage and current assignments. Consider, for instance, the assignment of Fig. 16.5(a). Applying K V L around the primary and secondary loops, we find 9 (a)

0>)

_ , dii

di

dt

dt

F I C U R E 16.3 Circuit symbols for the transformer of Fig. 16.2: (a) dots on upper tcrmW (b) dots on lower terminals

,,d'< dt

, ,

2

di dt

2

We note that with this rule it is unimportant whether the current i is incrcasIMB not, since the sign of the induced voltage is accounted for by di/dt. That is, t f l increasing, the induced voltage M di/dt is positive, and i f i is decreasing, (lie duced voltage is negative. I f i is a dc current, then, of course, the induce.! zero. EXAMPLE 16.1

Let us write the loop equations for Fig. 16.3(a). We see that i enters a dotted tarf nal. Thus the mutual voltage M dh/dt has positive polarity at the dotted terniinat the primary. Similarly, since t'i enters a dotted terminal, the mutual voltage M di, has positive polarity at the dotted terminal of the secondary. Application ol K around the primary and secondary circuits gives (16.6). In a like manner, identical statements apply to the undotted terminals of 16.3(b), and K V L once again yields (16.6). 2

Let us now establish a method for placing polarity markings on a transfer We begin by arbitrarily assigning a dot to a terminal, such as terminal a of 16.4(a). A current into this terminal produces a flux <j> as shown. (The direction is determined by the right-hand rule, which states that i f the fingers of the ri hand encircle a coil in the direction of the current, the thumb indicates the direct of the flux.) The dotted terminal in the remaining coil is the one that a current ters to produce a flux in the same direction as (b. (This statement, which may used to obtain the polarity of the induced voltage, is a consequence of a rule knas Lenz's law. The reader may study this law, formulated in 1834 by the Ger scientist Heinrich F. E. Lenz, in a later course in electromagnetic field theory.)

In Fig. 16.5(b), the voltage and current assignments have been changed. Application of K V L for these assignments yields

' dt ,.di,

, ,

dt

" dt di

2

dt

The results above can also be obtained by an alternative method for selecting the sign of the mutual voltages, as follows: If both currents enter (or leave) the dotted terminals of the coils, the mutual and self-inductance terms for each terminal pair have the same sign; otherwise, they have opposite signs.

F I G U R E 16.4 (a) Model of a transformer to determine polarity markings; (b) circuit sym

The student can easily verify the use of this method for the cases of Figs. 16.3 and

5

• (

\

2

1<

3

530

Let us find the open-circuit voltage v in the circuit of Fig. 16.6, given that «'i(0 ) = 0. For t > 0, since no current flows in the secondary, applying K V L around the primary yields

•

di\ — + 107, = 20 at

(a) Chapter 16

Transformers

Section 16.1

Mutual Inductance

531

H-l

<<•••'

I-el us find the phasor voltage V..(.sl in th* MtWOfti ..I I \y l ( . | dm |o the complex forcing function V,(s). The loop equations in ilu |IMIIIUIV mid secondary arc V, = (« + 2)1, + ft] 0 = sh + (2s + 3>I

a

Therefore, s + 2

FIGURE 16.6 Circuit with an open-circuit secondary

This is a first-order differential equation with a general solution of i, = 2 +

V,

*

0

s + 2

5

i

2s + 3

Ae'

a

-a-V,

Since t'(0 ) = i ( 0 " ) = 0, A = - 2 and

j

+

2

+ Is + 6

i , = 2(1 - e- '*) A 1

Therefore, in the secondary, v, =

-0.25^ at

=

-5e- V ,a

Let us now consider the transformer of Fig. 16.3(a) in the case of an excitation having a complex frequency s. In this case, since differentiation in the time d o m i H is equivalent to multiplication by s in the frequency domain, the phasor equation! corresponding to (16.6) are

V , ( i ) = sL,l,(s)

+

sMli(s)

Y (s)

+

sL h(s)

2

= sMh(s)

F I G U R E 16.8 Circuit with a complex excitation

and i f H{s) is the voltage ratio function, then H

( )

M

W

= ^

S

V,

= 2k =

V,

s

~

2

3

'

+ 7i + 6

-3s

2

(s + DCs + 6) A pole-zero plot and a sketch of | H ( » | are shown in Fig. 16.9. We may show that

The phasor circuit for this network is shown in Fig. 16.7. In the case of a purely sinusoidal excitation, we simply replace s by jio.

the maximum ac steady-state response occurs for co = Vf5 rad/s. Suppose, as an example, that v\ 100 cos lOr V. Then V, = 100 V ,

F I G U R E 16.7 Phasor circuit for a transformer

and V

2

= HV,

i = 710 rad/s

(-J30)(100) -100 + ; 7 0 . + 6

which simplifies to V = 2 5 . 6 / 1 2 6 . 7° V 2

Thus the ac steady-state response is u (r) = 25.6 cos (10/ + 126.7°) V 2

532

Chapter 16

Transformers

Section 16.1

Mutual Inductance

533

those of Figs. 16.2-16.4. In so doing, we shall show that M„ - M„ » M and establish limits for the magnitude of M. m Fig. 16.3, the stored energy is the sum of the energies supplied 10 the primary and secondary terminals. The instantaneous powers delivered to these tcrmi nals, from (16.6), are 0

2

4

(•)

6

8

p , = ck, =

u

(16.7)

FIGURE 16.9 (a) Pole-zero and (b) |H(;u)| plot for the network of Fig. 16.8

Pi = V I

=

2 2

EXERCISES 16.1.1

+

2

^,/'

2

where M has been replaced by Ma and M , in the appropriate terms. Let us now perform a simple experiment. Suppose that we start at time to with t'i(»o) = i (fo) = 0. Since the magnetic flux is zero, no energy is stored in the magnetic field; that is, w(h) = 0. Next, assume, beginning at time to, that we maintain i = 0 and increase i i until, at time r,, i , ( t , ) = h and i ( r , ) = 0. During this interval, h = 0 and di /dt = 0. Thus the energy accumulated during this time is 2

In the circuit of Fig. 16.3(a), L , = 4 H , L = 6 H and M = 3 H . Find t), and i , the time when the rates of change of the currents are 2

di, dt

= -2

A/s

di* dt

2

2

2 K/s

2

2

f'

=

Answer —2, 6 V 16.1.2

^ + * » f

(b)

In the circuit of Fig. 16.3(a), L , = L = 0.1 H and M = 10 m H . Find c, and | (a) i i = 0.4 sin / A and i .= 0.2 cos t A and (b) i'i = 0, h = 10 cos lOOt m A . Answer (a) 40 cos t - 2 sin t, 4 cos t — 20 sin t mV; (b) - 1 0 sin 100/, - 1 0 0 sin lOOf mV

('

1

(p, + Pi) dt =

di W\-rd%

1

2

2

16.1.3

Verify that the coil windings of F i g . 16.2 are consistent with the polarity mark in of Fig. 16.3.

16.1.4

Find the phasor currents I , and I . Answer 2 — j2, j2 A

Lii,

- f

di, =

\L,I]

•'n i , = I, and increase i until, at time t , i' (» ) = / . As a final step, let us maintain During this time interval, di,/dt = 0, and the energy accumulated, using (16.7), is 2

2

iv

2

2

2

2

2

f'2 (M, I,

= M I,l i2

6/0° V

+ L ii)

2

- f

+

2

2

\L l\ 2

Thus the energy stored in the transformer at time r is 2

-19 CI

w(r ) 2

= IV (to) + w, + w

2

= \Lil\ M, I,I

EXERCISE 16.1.4

2

+

2

\L ll 2

Let us now repeat our experiment but reverse the order in which we increase i, and i . That is, in the interval from to to h, we increase i so that i' (fi) = I while holding I'I = 0. Finally, we maintain i = I while increasing t'i so that i i ( / ) = l . Using the same steps as before, we find in this case that

16.2

2

2

2

ENERGY STORAGE

w(t )

We have shown previously that the energy stored in an inductor at time t is w(t)

=

\Li\t)

Evidently, for a given inductance L , the energy is completely specified in terms of i(t). Let us now determine the energy stored in a pair o f mutually coupled inductors, 534

di

2

Chapter 16

Transformers

2

2

2

= \L,l\ M ,I,I 2

2

2

2

+

t

\Uli

Since i i ( / j ) = I, and i ( r ) • h in both experiments, then w(t ) should be the same in both cases. Comparing our results, we see that this requires 2

2

2

M,

2

Section 16.2

Energy Storage

= M, = M 2

535

I f we repeat our experiment with the polarity markings of Fig. 16.4 il« of the mutual voltage is negative, which causes the mutual term in the energy to be negative. Thus a general expression for the energy at any time t is given w(t) = {L,ii

± Mi i

+

{ 2

fail

where the sign in the mutual term is positive if both currents enter dotted (or U ^ H ted) terminals; otherwise, it is negative. The coefficient of coupling between the inductors indicates the amount of o f l pling and is defined by M

M

\/M, M l 2

VL,L

VLlL

2

since 0 i / 0 u 2

1 and 0 i / 0 2

_

2

16.2.1 16.2.2

Find the coefficient of coupling i f L , = 4 m H , L = 16 m H , and M = 2 m H . Answer 0.25 Find M i f L , = 0.2 H , L = 0.8 H , and (a) k = 1, (b) k = 0.5, and (c) k = 0.02. Answer (a) 0.4 H ; (b) 0.2 H ; (c) 8 m H Determine the energy stored at t = 0 in each case of Exercise 16.1.2. Answer (a) 2 m j ; (b) 5 u.J 2

2

16.3 CIRCUITS WITH LINEAR TRANSFORMERS


A two-winding transformer is in general a four-terminal device in which the reference potential in the primary can be different from that of the secondary without altering the values of v , v , i,, or it.

' 011 ' 022

2

2 2

EXERCISES

16.2.3

Clearly, k = 0 i f no coupling exists between the coils, since M = 0. We can cstu lish the upper limit of k by substituting from (16.1)—(16.5), which yields k =

he sign of the mutual term is negative i f both currents enter dotted (oi Ulldol led) terminals; otherwise, it is positive. The second equation of (16.10) shows ili.u «•(/)'can be zero in a unity-coupled (A = 1) transformer even though i, and r H I nonzero.

£ 1. Thus we must have

t

2

0 < t < l or, equivalently,

I» AMPLE 16.4

Consider the circuit of Fig. 16.10. K V L around the primary and secondary circuits gives

0 £ M If k = 1, all of the flux links all of the turns of both windings, which is a unity* coupled transformer. The value of k (and hence M) depends on the physical dimensions and numhci of turns of each coil, their relative positions to one another, and the magnetic properties of the core on which they are wound. Coils are said to be loosely coupled lf| k < 0.5, whereas those for which it > 0.5 are tightly coupled. Most air-core tramformers are loosely coupled, in contrast to iron-core devices for which k can approach 1. Let us now examine the values of i, and i in (16.8) for which w(t) is zero. From the quadratic formula, we may write

0 = -M^ Evidently, secondary of course, reference.

dt

+ Ri

+

2 2

L& dt

the voltages and currents are not affected by V . For this reason, the of the transformer is said to have dc isolation from the primary. Point a, is at an absolute potential of V + i R volts with respect to the ground I f we now let V = 0, it is seen that the bottom terminals of the trans0

0

2

2

0

F I C U R E 16.10 Circuit showing different reference potentials in primary and secondary

2

M

Mi

2

Li

LI

For real values of i, and i , we see that for w(t) = 0 we must have 2

ii = i = 0

M < VL,L

2

i, =

536

Chapter 16

Transformers

±

M.

2

(16.10)

—1 , Li 2

Section 16.3

Circuits with Linear Transformers

537

former are connected and that the primary and secondary circuits have a uimihH reference point. The transformer, in this case, is a three-terminal device. EXAMPLE 1 b . 5

Let us find the complete response i lor i - 0 in Fig. 16.11, given Af = l / \ r ^ M and i,(0 ) = / ; ( ( ) ) = 0. Since the forced response for i , is a dc current (by i n i M tion), no steady-state voltage is produced in the secondary; thus iy = 0. To find M natural response, let us first obtain the network function for t > 0. A p p l i c a ^ ^ H K V L for a complex excitation V\(s) yields

- V I / 2 In iln» 0 * M

Suppose that we now repeat Example 16.5 for M = VL,L the network function for / > 0 is

2

2

V , « = (l,

+

2

) l ,-

H

5,

W

4

+

and the complete response is k = Ae-°-" We know that t , ( 0 ) = i ( 0 ) = 0; however, i f we take » (0*) = i ( 0 " ) as before, then the unreasonable solution i = 0 results. Also, applying K V L gives _

± s k

2

2

2

2

0 =

+ (s + 2 ) I

2

V2 from which we find

V372

Multiplying the latter equation by

and adding it to the former, we have

12 = 2r, + V 6 i

(16.11)

2

This result contradicts i , ( 0 ) = i ( 0 ) = 0. Recalling (16.10), we see that the energy is zero in the unity-coupled case +

2

+

when i,(0 ) = ^ < ( 0 ) = +

2

^( (0 )

+

2

+

Combining the last result and (16.11) gives i' (0 ) « 2.94 A , and therefore 2

F I G U R E 16.11 Switching circuit containing a transformer

i

The poles of H(s) are the natural frequencies, - 1 , - 4 , o f the natural response Thus, for / > 0, (2 = 12/ + 12. = 0 + A , c " ' +

Let us find the steady-state response v in Fig. 16.12. Applying K V L for a complex 2

frequency s, we have

, (, +

2

v =

2

3
2 Q

+

V2 Solving, we find di (0*)/dt = 6V2 di (Q )/dt, the solution becomes 2

2

t

Chapter 16

Transformers

2*

+ 1 + - II

for which

41,(0+)

V

dt

A/s. Evaluating A i and A

2

using J ( 0 ) and 2

2

V,

I

s '+ 1

2

2

V,

s

i

+ Is

2

+ 4s + 4

+

Substituting s = j2 rad/s and V , = 16 V , we find V = 2[QT V. Thus

+

2

l)l , -(i+i)l

+ - III + I

dt

*

i = 2V2(e-

538

2

1 dh(0 )

=

3+

1
V2

dt

08

2

+

+

= 2.94c- ' A

Thus the current in a unity-coupled transformer can change instantaneously as a result of the application of a finite forcing function.

A e"

From (16.8), w ( 0 " ) = 0; therefore, since the energy cannot change instantaneously in the absence of infinite forcing functions, t v ( 0 ) = 0. From (16.10), noting thai M < VL,L , we see that t'i(0 ) = i (0*) = 0. To obtain a second initial condition for finding A, and A , we apply K V L around the primary and secondary at t = 0*1 which yields 2

2

+

2

- e")

A

02 = Section 16.3

2 cos

Circuits with Linear Transformers

2t V

539

3a

T

-wv»

.

2H -nnns^-

I H

' t n t s-

EXERCISES

3

v, = 16 cos

16.3.1

Determine i , for t > 0 in the network of Fig. 16.11, given M = 1 / V 2 H that the circuit is in steady state at / = 0 " . Answer 6 - 4e'' - 2e'" A

16.3.2

Find the forced response t> i f v, Answer Vie- ' cos (t + 45°) V

ie-

21

2

Assume

cos t V.

2

F I G U R E 16.12 Circuit containing a linear transformer

EXAMPLE 16.8

Let us find the network function V / V , for the phasor circuit of F i g . 16.13. Let first examine the voltage V , that appears across winding A in loop 1. We see that voltage due to I , is the self-inductance term (2JI,) and the mutual term of wind C(2sl,). The voltage due to I arises from the mutual terms of winding B(-sh) winding C ( - 2 J T ) . Thus 2

2

2

\

=

( 2 s

+

2 s ) h

-

( s

+

EXERCISE 16.3.2

2 s ) h 16.3.3

Similarly, we find V

8

= -(s

V

c

= (4s + 2s)h - (4s + 3s)h

+ 3s)l, + (3s +

3s)l

Determine the network function V / V , for Fig. 16.13 i f the polarity dot of coil B is placed on the other terminal. Answer 6(s + i)/(6s + 45s + 11) 2

2

2

16.4

Applying K V L in loops 1 and 2 gives V, = 31, + V , + V 0 = -21, - V

c

c

+ \

- 2I

REFLECTED IMPEDANCE

2

+ 5I

In this section we develop several important impedance relationships for the ac steady-state case. Let us begin by considering the phasor circuit of Fig. 16.14 having a practical source V , with an impedance Z connected in the secondary. Applying K V L at the primary terminals of the transformer we find

2

or

2

V, = (10s + 3)1, 0 -

(10s + 2 ) I

2

- ( 1 0 s + 2)1, + (13* + 5 ) I

V, ™ KiiLJi -

2

(16.12)

Solving these equations for the ratio I / V , , we find the network function to be 2

H(J)

-

V

2

3I _

3(10i + 2)

2

jatMh

0 = -jotMl,

+ (Z + 2

Eliminating I from these equations we have [ jtoM

ja>L )h 2

2

30s

2

+ 49s + 11

(ja>M)]

F I G U R E 16.13 Circuit containing a three-winding transformer F I G U R E 16.14 Circuit for deriving impedance relationships

540

Chapter 16

Transformers

Section 16.4

Reflected Impedance

541

tor

so that the input impedance seen at the primary terminals of the transformc V,

Z,

wM 2

j
Ii

Z

m

(Do, _

_ 01IM

2

2

+

2

to

^ - ^ R T

JIDLI

and the reflected impedance is purely real. Inspecting (16.13) and (16.14), we see that i f the polarity dot on either winding of Fig. 16.14 occurs on the opposite terminal, the current and voltage ratios require a sign change, whereas the impedance relations are unaffected.

The first part, ju>L,, depends entirely on the reactance of the primary. The tm I part is due to the mutual coupling, and it is called the reflected impedance, given hj co M 2

2

Z + j


2

EXERCISES It may be thought of as the impedance inserted into, or reflected into, the prime, by the secondary. The input impedance as seen by the source V , is evidently

16.4.1

Given: In Fig. 16.14, V , = 100/fT V , Z , = 40 ft, L , = 0.6 H , U = 0.1 H , M = 0.2 H , and o> = 100 rad/s. I f Z = 10 - ( j 1000/w) il, find (a) Z,„, (b) I , , (c) I , (d) V , , a n d ( e ) V . Answer (a) 80 + JG0 ft; (b) 0.8 - J0.6 A ; (c) 1.2 + j 1.6 A ; (d) 68 + j24 V; (e) 28 + ; 4 V Repeat Exercise 16.4.1 i f the polarity dot is on the lower terminal of the secondary. Answer (a) 80 + jGO ft; (b) 0.8 - ; 0 . 6 A ; (c) - 1 . 2 - j 1.6 A ; (d) 68 + j24 V; (e) - 2 8 - j4 V Find the frequency for which the reflected impedance in Fig. 16.14 is real if L = 2 H and Z is a 6-ft resistor in series with a ^ - F capacitor. Answer 4 rad/s 2

2

2

Zj

n

=

Zg + Z |

Also the secondary-to-primary current ratio I2/I1 may be found from the secc equation of (16.12), and the voltage ratio V2/V1 may be found from

MiiHv.)

Yi = z. v, v

16.4.2

16.4.3

using (16.13). The results are I

jtoM

2

Z

I,

v

+

2

ja)L

16.5

2

(16.14)

J10MZ2

2

V,

jcoL,(Z

2

+ JC0L2)

+

o) M 2

2

It is interesting to note that Z is independent of the dot locations on the trans former. I f either dot in Fig. 16.14 is placed on the opposite terminal, the sign of the mutual term in each equation of (16.14) changes, which is equivalent to replacing M by -M. Since Z varies as M , its sign is unchanged. A second important property is illustrated by rationalizing Z , which gives R

R

2

R

Z, =

o) M 2

R\ (X2

2

[R

2

+ (0L2)

2

2

j(X

"

2

2

+ WL2)}

THE IDEAL TRANSFORMER A n ideal transformer is a lossless unity-coupled transformer in which the selfinductances of the primary and secondary are infinite but their ratio is finite. Physical transformers which approximate this ideal case are the previously mentioned iron-core transformers. The primary and secondary coils are wound on a laminated iron-core structure such that nearly all of the flux links all of the turns of both coils. The reactances of the primary and secondary self-inductances are very large compared to moderate load impedances, and the coupling coefficient is nearly unity over the frequency range for which the device is designed. The ideal transformer is thus an approximate model for well-constructed iron-core transformers. A n important parameter that is necessary in describing the characteristics of an ideal transformer is the turns ratio n, defined by

where we have used the relation Z = R + jX for the load impedance. I t is seen that the sign of the imaginary part of Z» is minus. Therefore the reflected reactance is opposite that of the net reactance X + oaL of the secondary. In particular, i f Xi is a capacitive reactance whose magnitude is less than o>L or i f it is an inductive reactance, then the reflected reactance is capacitive. Otherwise, the reflected reactance is either inductive or it is zero. In the latter case, X must be a capacitive reactance -\/wC with a resonant frequency f = OI /2TT = 1/2TTVL C. In this case,

tively.

Chapter 16

Section 16.5

2

2

2

2

(16.15)

2

2

2

a

542

Transformers

0

2

where N, and N are the number of turns on the primary and secondary, respec2

The Ideal Transformer

543

The flux produced in a winding o f a transformer due to a current in th# i ing is proportional to the product of the current and the number of lui winding. Thus, in the primary and secondary windings,

Hi I 'i u ing n by N /N,, 2

we have V2

(A,, = aJVii, 022 =

aN ii

N,l,

2

where a is a constant of proportionality which depends on the physical proix-i n the transformer. (The constant a is the same in each case because we are anMI that there is no leakage flux and thus both flux paths are identical.) Suhflll these relations into (16.2) and (16.5), we see that Therefore,

k , ( 'Y N

_

2

2

V

'

t

JIQZ VL\L

=

2

JtoL,(Z

Nl

2 2

Therefore, the voltages are in the same ratio as the turns, and the ampere turns (Nl) are the same for both primary and secondary. The symbol for an ideal transformer is shown in Fig. 16.15(a) with polarities such that (16.17) holds. The vertical lines are used to symbolize the iron core, and 1 : n denotes the turns ratio. I f one or the other, but not both, of the polarity dots is placed on the opposite terminal, then n is replaced by -n in (16.17). Figure 16.15(b) shows an ideal transformer connected to a secondary load Z and a source V , with internal impedance Z „ . The primary impedance Z, of (16.13), in the case of the ideal transformer, is given by

h

second of (16.14) becon

e

* - Yi - Yiln - W

2

+ ja>L ) +

2

N, =

2

In the case of unity coupling, M = VLTL , Yi

N2

V,

w L,L

2

2

l, ~

2

nl

2

~

n

2

2

4, _ Z

_ Z

' ~ n

2

Z,

2

'

(16.18)

Z,

tend to nfin r r ' t**" P 8' inductances L , and . tend to infinity in such a way that the ratio of (16.16) is the constant n From ,1 first equation of (16.14) we see that in this case we have ' 1 ,

0

C O U

l i n

t

h

e

2

km l i m Ii

Zi + =

lim

j(oL

2

JO) VLJL2

LuL^.jw

+

(Z /L ) 2

2

lim FIGURE 16.15 (a) Ideal transformer symbol; (b) circuit containing an ideal transformer

1 Thus the input impedance viewed from the terminals of the voltage source is , rJ » H , ^ " * « y are related simply to the turns ratio by a r

U

P

r i m a r

a

d

e c o

, a r

v o l t a

v

544

Chapter 16

g « and currents of an ideal transforn irannorn Zin

2

12

1

I,

n

(16 17)

—

Zp "I" Z ] — Z

G

j

(16.19)

The lossless property of an ideal transformer is easily demonstrated using (16.17). Taking the complex conjugate of the cutrent ratio, we have, since n is real.

Transformers Section 16.5

The Ideal Transformer

545

from which

- B Z, J

V . I f _ V2I? 2 2

F I G U R E 16.17 Thevenin equivalent circuit of Fig. 16.15(b)

Thus the complex power applied to the primary is delivered to the load hence the transformer absorbs zero power. In analyzing networks containing ideal transformers, it is often conveniaB replace the transformer by an equivalent circuit before performing the analysis us consider, for example, replacing the transformer and load impedance /. ol 16.15(b). Clearly, the input impedance seen by the generator V , is Z,„ giv (16.19), so that an equivalent circuit insofar as V , is concerned is shown in 16.16. The voltages and currents can now be easily determined from the single circuit.

transformer. It may be shown that this statement holds in general whether the result is the Thevenin circuit or not. On the other hand, i f we replace the secondary circuit and transformer by an equivalent circuit, as in Fig. 16.16, we simply multiply each secondary voltage, current, and impedance by n, and 1/n , respectively. I f either dot on the transformer is reversed, we simply replace n by - « . In applying the above-described procedure, the student is cautioned that the technique is valid i f the transformer divides the circuit into two parts. When external connections exist between the windings, the method in general cannot be used. The equivalent circuits to be discussed in the next section are often useful for networks of this type. 2

Let us find V in the circuit o f Fig. 16.18(a), which contains an ideal transformer and a voltage-controlled current source. In Fig. 16.18(b) an equivalent circuit for the primary circuit and the transformer is shown. I t should be noted that a minus sign has been used on primary voltages and currents to account for the polarity dots 2

-.0 F I G U R E 16.16 Equivalent circuit for Fig. 16.15(b) obtained by replacing the seconda

F I G U R E 16.18 (a) Example circuit; (b) equivalent circuit

Let us next replace the primary circuit and the transformer of Fig. 16.15(b) its Thevenin equivalent. By (16.17) we have I i = nh,

1,

—

ion —VVA

V = nV,

M

O

•»

1

<(T>

a

4

2

so that for V « , I = 0 and thus I , = 0. Therefore 2

8/2° v Q ~ )

0.01 v ,

1 1 E

V,

V„c = V = nV, = n V , 2

For Isc, we have V = 0 and thus V, = 0. Therefore, 2

I,c = I = 2

=

and

I kS2

V« Zth = — = I*

-VW nZ 2

g

T )

The resulting equivalent circuit is shown in Fig. 16.17. Inspecting the results for the Thevenin equivalent circuit, we see that each p r i mary voltage is multiplied by n, each primary current is multiplied by 1/n, and cadi primary impedance is multiplied by n when we replace the primary circuit and ilic

-

]

v

2

< ?

I

60 + ;80 S2

2

546

Chapter 16

Transformers

Section 16.5

The Ideal Transformer

547

shown on the transformer. The nodal equation for V gives 2

4-/3(1

V, + 80/0° 10

from which

+

10: V 3

+

2

= 0

60 + ; 8 0

3

: v

32/0° V

V = 5V2/-135" V So far we have considered only the ideal transformer in the ac stculv case. In the general case, we see from (16.6) that

\Li

dt ( Thus since M

2

= L,L

and

2

M

di,\

L

di)

2

2

EXERCISE 16.5.1

16.5.3

Using Norton's theorem, show that the primary circuit and transformer of Fig. 16.15(b) is equivalent to a constant current source of V , / n Z , A in parallel with an impedance of n Z, ft.

16.5.4

Find V in Fig. 16.18(a) by first replacing the secondary circuit and the transformer

2

16.5.5

t>

2

v, = — n

2

2

A

Mdh L , dt

di,

n

s

w

e

r

2

1F

11

'2 _

di

2

dl

=

Find H(s) = l /h • [Suggestion: Note that the circuit is 2 two-ports connected in series and that the transformer provides isolation so that the z-parameters add. Thus H = z . / ( l + 122), and so on.] s + 1 s + 6s + 3 2

Next, let us rearrange the first equation of (16.6) in the form +

2

by an equivalent circuit.

VL /L, = n, we have

^ _ dh L, dt

64+/480

2

, M

M \di,

2

•

2

+

Tt

n

Taking the limit as L , becomes infinite, we have di

2

dt Integrating both sides, we have -ni

2

+ C,

where C, is a constant of integration. Since dc currents produce no time-varying magnetic flux, they do not contribute to the induced voltages or currents in the idea transformer. Therefore, i f we neglect the constant C,, then I'I = -ma

EXERCISE 16.5.5

where the minus sign arises due to the direction assigned i ' in Fig. 16.3(a). Thus the same current and voltage relationships are valid in the time domain as were found in the frequency domain i f we neglect any dc currents. 2

16.6 EQUIVALENT CIRCUITS Equivalent circuits for linear transformers are easily developed by considering the equations for primary and secondary currents and voltages. In Fig. 16.19(a) we

EXERCISES 16.5.1 16.5.2

Find V , , V , I , , and I . Answer 2 0 / 3 6 . 9 ° V , 8 0 / - 1 4 3 . 1 ° V , 4 / 0 ° A , 1/180" A 2

di,

2

n

a

v

e

In Fig. 16.15(b), V , = 100/0! V , Z = 20 i l , and Z = 2 k f t . Find n such that Z i = Z , , and then find the power delivered to Z . Answer 10, 125 W g

di

2

2

2

548

Chapter 16

Transformers

^

~

~dt

L

l

*

549 Section 16.6

Equivalent Circuits

(•)

(b)

F I G U R E 16.19 (a) Linear transformer; (b) equivalent circuit

It is evident that the circuit of Fig. 16.19(b) satisfies these equations. The depended voltage sources, however, are controlled by the time derivatives of the primary i M secondary currents. In the frequency domain, these sources can be considered • current-controlled voltage sources. Let us now rearrange the equations in the form

„,, ,_ | (5 l) (l

w +M

+

I

F I G U R E 16.21 (a) Ideal transformer; (b) and (c) equivalent circuits

EXERCISES 16.6.1 These equations are satisfied by the T network of Fig. 16.20(b). Since this circuit I a three-terminal network, it is equivalent to the transformer connection of P i l l 16.20(a). I f either polarity dot is changed to another terminal, we must replace M M —M in the equivalent circuits.

Use the T equivalent for the linear transformer to find the steady-state value of v. (Note that an inductor with L = 0 is a short circuit.) Answer 6 sin 8/ V

7 sin it A

EXERCISE 16.6.1

16.6.2 (•)

(b)

Find V in Fig. 16.18(a) using the equivalent circuit i n Fig. 16.21(c). Answer 5 V 2 / - 1 3 5 ° V 2

F I G U R E 16.20 (a) Linear transformer with common terminals; (b) equivalent T network

16.7 In the case o f an ideal transformer as shown in Fig. 16.21(a), the currents and voltages are given by

Clearly, the circuits of Figs. 16.21(b) and (c) satisfy these relations. I f either of the dots are reversed, of course, we must replace n by ~n in the equivalent circuits.

SPICE can be used for analyzing circuits that contain linear transformers in the same manner as performed in previous chapters by employing the K data statement for defining the transformer. A description of the K statement is given in Appendix E. The K statement expresses the mutual coupling of the transformer in terms of the coefficient of coupling given in (16.9).

Chapter 16

Section 16.7

'2 =

,

Vi = nv,

n

550

SPICE ANALYSIS FOR TRANSFORMERS

Transformers

SPICE Analysis for Transformers

551

EXAMPLE 1 6 . 1 0 Consider the circuit o f Fig. 16.11, redrawn in Fig . 16.22 for applying SI'K I should be noted that the primary and secondary have been connected togctli.-i m reference node to avoid having the nodes of the secondary elements dangling ( disconnected) for SPICE analysis. A circuit file for the transient response 0 < r < 1.5 s is SOLUTION O F FIG . 16 . 11 USING E Q » DATA

LL 16.11 l e t us find the solution for the network o f F i g . 16.12, which has a coupling Hicient o f 0.707. A circuit file for nodes being numbered sequentially clockwise, with node 1 being at the top o f t>i, is SPICE SOLUTION F O R FIG . 16 . 12 . • DATA

16 . 9 F O R FINDING k

STATEMENTS

VI 1 0 A C 16V Rl 1 2 3 L l 2 3 1H C 3 0 1 L2 4 3 2 H R2 4 0 1

STATEMENTS

VIN 1 0 D C 12V Rl 1 2 2 L l 2 0 1.5 IC=0 L2 3 0 1 I C = 0 R2 3 0 2

K L l L 2 0. 707

K L l L2 0 . 577

* SOLUTION AND OUTPUT CONTROL

* SOLUTION CONTROL

STATEMENT

• TRAN 0 . 1 1.5 U I C * OUTPUT CONTROL

STATEMENT

. AC L I N 1 0 . 3 1 8 3

0 . 3183

. PRINT AC VM(R2)

VP(R2)

STATEMENTS

. END

. P L O T T R A N I(R2)

which yields the solution

. END

FREQ 3 . 183E-01

A plot o f the output o f this program is shown in Fig. 16.23.

VM(R2 | 2 . 000E-00

VP(R2) -3 1 3 9 E - 0 3

In the case o f an ideal transformer, solutions can be found using the equivalent circuits o f Fig. 16.21(b) and (c) employing the procedures discussed in previous chapters.

EXERCISE 16.7.1

®

Using SPICE, find V in Fig. 16.13 i f V , = 12/f£ V , / = 10 k H z , and the resistor and inductor values are replaced by k f t and m H values, respectively. Answer 6 . 6 7 4 / 1 . 6 4 ° V 2

F I G U R E 16.22 Circuit of Fig. 16.11 redrawn for SPICE

552

TIME )

F I G U R E 16.23 Response for the circuit of Fig. 16.22 I(R2) O . OOOOE+OO 5 . 0000E-01 1 . 0000E+00 1 . 5000E+00

0 . 000E+00 1 . 000E-01 2 . OOOE-01 3 . OOOE-01 4 . OOOE-01 5 . 000E-01 6 . 000F-01 7 . OOOE-01 8 . OOOE-01 9 . OOOE-01 1 . 000E+00 1 . 100E+00 1 . 200E+00 1 . 300E+00 1 . 400E+00 1 . 500E+00

3 6 1 1 1 1 1 1 1 1 9 9 8 7 6 6

Chapter 16

602E-01 044E+00 242E+00 324E+00 333E+00 295E+00 232E+O0 155E+00 072E+00 885E-01 065E-01 284E-01 550E-01 8S7E-01 238E-01

Transformers

. . . . . . . . . . . . . . .

•

•

•

16.8 2 . OOOOE+0

SUMMARY In this chapter we considered the transformer, a system o f coils with mutual flux linkage. The common case is two coils used to step up or step down the voltage on the primary coil to a higher or lower voltage on the secondary coil. The inductances of the two coils are called self-inductances to distinguish them from the mutual inductance between the coils. The polarity o f the secondary voltage for a given p r imary voltage is determined by how the coils are wound on the common core, but a dot convention is used on paper to avoid the need for seeing the actual windings. Energy is stored in the transformer in the two coils. There is also a component due to the mutual coupling o f the coils. The coefficient of coupling is a measure o f how tightly or how loosely coupled the coils are. Circuits with transformers are analyzed as before, with the added complication of a second, mutually coupled coil. The Section 16.8

Summary

553

analysis may be simplified by the use of reflected impedance, which may !«• iht of as an impedance reflected into the primary by the secondary. A n ideal transformer is approximated by a very tightly coupled tninnfc with extremely high self inductances. It is characterized by a turns ratio n. the of the number of turns of the secondary coil to that of the primary coil, whi. h m the ratio of the secondary voltage to the primary voltage. The secondary and mary currents are in the ratio 1/n. SPICE is used for transformer circuits in the same way as in the inductor circuits of the previous chapters.

PROBLEMS 16.1

In Fig. 16.2(a). N = 500 turns and
16.5

16.2

If the inductance measured between terminals a and d is 0 1 H when terminals b and c are connected, and the inductance measured between terminals a and c is 0.9 H when terminals b and d are connected, find the mutual inductance M between the two coils, and the position of the dots.

Find i,

and t : in Fig. 16.5(a) and I'

L, = 2 H, L = 5 H, M = 3 H. and the 2

2

rents ii and i are changing at the rates 10 and - 2 A/s, respectively. 2

16.6

Find v for t > 0 across the open-circ secondary if i = 8u(t) A.

16.7

Repeat Prob. 16.6 i f i , = fe^uW A .

16.8

Find the steady-state currents ii and i .

16.9

Find the steady-state value of v.

g

O.l H

2

16.10 Find the power dissipated in the 1-il reslr in the primary circuit and the power seen by the source.

PROBLEM 16.2 16.3

If U = 4 H , L2 = 3 H , M = 2 H, and the currents i, and i are changing at the rates - 2 A/s and 10 A/s, respectively, find v and 02 in (a) Fig. 16.3(a) and (b) Fig. 16.3(b). 2

16.11

Find the steady-state current i .

16.12

Find the steady-state currents i,(r) and i l l ) to = 2 rad/s.

16.13

Find the energy stored in the transformer Prob. 16.12 at t = 0. (Take the transient CUL, rents to be zero.)

16.14

(a) Find the energy stored in the transformt| of Fig. 16.3(a) at a time when i, = 2 A ! i , = 4 A i f L, = J H, L2 = 2 H , and M I J H . (b) Repeat part (a) if one of the dois ll moved to the other terminal.

t

16.4

(a) Find v, and «2 i f L, = 5 H , L = 3 H , I, = - 3 cos 4t A , and i 2 = 5 sin 2r A . (b) Find v if the secondary is open. 2

M = 2 H,

2

P R O B L E M 16.9 0.5 H

i n

P R O B L E M 16.10

2

P R O B L E M 16.11 PROBLEM 16.4

554

Chapter 16

Transformers

Chapter 16

Problems

555

i n

\

10/0^ V 10 cos 4( V

^f-- no a

I H

(^) 3(1

PROBLEM 16.12

16.15

Repeat Prob. / = TT/8 S.

16.13 if ai = 4 rad/s and

16.21 Show from (16.8) that a real transformer (0 < M < V L , LI) satisfies the passivity urn dition w(t) 3: 0.

16.16 Find the coefficient of coupling k if L, • 0.02 H , Z.2 = 0.125 H, andAf = 0.01 H.

16.22 (a) Find v for t > 0 i f v = 4u(t) V and (b) find the steady-state value of v if u, 4 cos Hi V and the output terminals are loaded with a resistor of 8 fl. s

16.17 Find Ai i f L , = 0.4 H , L = 0.9 H , and (a) k = 1, (b) * = 0.5, and (c) k = 0.01. 2

16.18 Note that for coupled coils we may always choose the current assignments into the dots and the positive voltage polarities at the dots, as in Fig. 16.3(a). (This, of course, may require using —i instead of i , etc.) Show that if we solve the describing equations (16.6) for the current derivatives and integrate the results, we have "

W

l « 2

*

A !

=

~ "

1"' | «, + ^ Jf v

"""f

2

Find the network function, H(J) = V (s)/V,(s). 2

2

-

M

PROBLEM 16.31

PROBLEM 16.29

PROBLEM 16.22

16.23

Find i for t > 0 i f i(0) = 0 and v(0) = 4 V.

dt + 1 (0)

Find the natural frequencies present in i(t). i n

2

16.32 Find the power delivered to the (75 + ; 1 2 5 ) - f l load using reflected impedance. 16.33 Find the average power delivered to the 8-fl resistor. 24 0 ' \AA>

i(0

_

^2 n

90

...

2

w v

2fl

16.19 Solve Prob. 16.6 using nodal analysis as described in Prob. 16.18. 16.20 Find v(t) for t > 0 using (a) loop analysis and (b) nodal analysis.

16.24 16.25

1 H

2(2

10 u(r)

Find the steady-state current i, in Prob. 16.8 using reflected impedance. Find the steady-state current i in Prob. 16.8 by replacing everything in the corresponding phasor circuit to the left of terminals a-b by its Thevenin equivalent circuit.

p

1 8n

i

PROBLEM 16.33

PROBLEM 16.30

2

16.26

Find the steady-state current i.

16.27

Find the steady-state current i , in Prob. 16.12 using reflected impedance.

16.28

Transformers

.

1:2

1

H

(3

1

PROBLEM 16.23

PROBLEM 16.20

Chapter 16

+

P(Of«Il)

This result enables us to solve coupled circuits using nodal analysis.

556

31 Find i (r) for / > 0 i f / ( 0 ) = 0.

1

where A is the determinant A = L,L

16

20 e' cos I V( _

V2< + ,(0) V

PROBLEM 16.26

Find the steady-state current i in Prob. 16.12 by replacing everything to the left of terminals a-b by its Thevenin equivalent circuit.

PROBLEM 16.32 1:5 '

i - 1 2a

• 5oze° v ( i

75 +/125 £2

2

Chapter 16

Problems

557

16.34 Find the power delivered to the 2-fi resistor. 411

PROBLEM

16.35

PROBLEM 16.34

16.35 The stepdown autotransformer has the second ary terminal 2 tapped to the primary winding at node 2, as shown, (a) If the secondary winding has N turns and the primary winding has A/, turns, find the voltage and current ratios, (b) Find I and V if V, = 100 / 0 ° V, I , = 2/60" A, N, = 1000 turns, and N = 200 turns.

16.37 Find the z- and y-parameters of the tn former of Fig. 16 20(a) in terms of s. 16.38

:

2

2

2

16.36 Find the turns ratio n so that the maximum power is delivered to the 8-kfi resistor. Find the power delivered by the source in this case if v, = 40 cos 377t V.

Solve Prob. 16.31 using the equivalent cir of (a) Fig. 16.21(b) and (b) Fig. 16.21(c).

16.39 Solve Prob. 16.32 using the equivalent cir of(a) Fig. 16.21(b) and (b) Fig. 16.21(c). 16.40 Find the steady-state voltage t using the eq alent T circuit for the transformer. 16.41 Find H(s) = V ( s ) / V , ( j ) . (Suggestion: Exercise 16.5.5) 2

C O M P U T E R A P P L I C A T I O N PROBLEM S 16.42 16.43 16.44 16.45

Using SPICE, plot i for 0 < t < 5 s in the circuit of Prob. 16.23. Using SPICE, find the steady-state current i in Prob. 16.26 i f / = 10 Hz. Using the equivalent circuit of Fig. 16.21(b) with SPICE, find the voltage of the 8-fi resistor in Prob. 16.33 i f / = 60 Hz. In the circuit of Prob. 16.40, replace the 10-fi resistor bridging the transformer with a series connection of a 10-fi resistor and a 1000-mF capacitor. Use SPICE to plot the frequency response for the voltage v of the resulting circuit in the interval 1 < / < 50 Hz.

PROBLEM 16.36

PROBLEM 16.40

16 cos 1001

558

Chapter 16

Transformers

Chapter 16

Computer Application Problems

559

DP14 Show thai a solution of the equations of Design Prob. DP13 is 4 *

" I. + V . '

„

'

*'

+

•

~bCR

~

DP15 Design a high-pass Butterworth Hit*

8MG - 1>,C

I

*

J " .

M,. |

Appendix

•

V JT- rts, 7 |

Answers to Selected Odd-Numbered Problems

(10//) F . M

2

< C > /?4

(The last two equations minimize the li of the op amp.)

"

I

G/?2

HAPTER 1 1 J 5 7 9 l 1 13 IS

Styron—20.55 mph. Griffith Joyner— 20.96 mph (a) 29 C; (b) 37.5 C; (c) 1.5, 6, 5 A Prob 1.3: 9. 36. 30, 12 W; Prob. 1.4: 9, 0, -18. 0 W (a) 15, 4, 4 W; (b) 84 mJ 25/4, 8 mW -94.25 mJ 6 cos 2r A, 3 C (a) 48f * W. 12(1 - i ") J; (b) -64e W, me - l ) J ; ( c ) -24<- * W, 6(e - 1)J 16(1 - if" ) J (a) 34,560 J: (b) 2880 C (a) 14,400 J; (b) 1200 C H' = 6(e~* - I ) a 0 41

4

2

CHAPTER 2

2.9

200 mA, 6 V 8 A, 15 n - 1 A, 4 V 2 A, 3 A, 10 V 7 A . 6(1

2.11 2.13 2.17

300 n 400(1 2, 4, 6, 8, It), 12 V

2.1 2.3 2.5 2.7

2.19 8 V

760

Appendix F

Circuit Design Methodology

500 (1, 4 mA 1 A 2A,4() (a) 800. 780 (1; (b) 720, 702 il 1.5 W 2 A, 8 V 4 A, I A, 2 V 37.5/^A 4.5 k i i

CHAPTER 3

41

41

17 14 21 2?

2.27 2.29 2.31 2.33 2.35 2.37 2.39 2.41 2.43

2.21 2.23

l O k l i , 60 mA 1.5, 0.5 A

2.25

^ A, 3 A , 8 V

3.1 - 3 V, 3.6 W 3.3 3 A 3.5 (a) 5 A; (b) -2.5 A 3.7 (a) 3 A, 6 V; (b) - 1 A, - 6 V 3.9 (a) 2 cos 2/ A, 6 il; (b) - 8 cos 3/ A, - 4 (1 3.11 - 1 6 V 3.13 6 V 3.15 (a) 2 0 k ( l , 20 k ( l ; (b) I k ( l 3.19 6 s i n 3 r V 3.21 21 a 3.23 1.5 A 3.25 -0.25 A 3.27 sin 3000r mA 3.29 (b) 17 V 3.31 (b) 0.999990<,. 0.9900990ti,, 0.5t),

CHAPTER 4 4.1 4.3 4.5

64,48 V 2, 4 A 4 A

761

4.7 49 4 11 4.13 4.15 4.17 4.19 4.21 4.23 4.25 4.27 4.29 4.31 4.33 4.35 4.41 4.43

1.5 A 4A 20 V 64 W -2. -4 V 8V 2 cos 3/ V 20kil 8 mA 8 W 8 V -4, 8 A 75 W 8 V 6 cos 1000/ mA -0.25 V 0.4 mA

6.11 (a) 3 A; (b) 6.17 26 V 6 19 8 mA

H 1.25(3* " + e ") A 28 + 8c V 21

10.11

14c " - 7e ' A, 42* - 48e 6 - 2e A 4 - It mA (a) (3e " + 6)u(/) V; (b) + 6]«(/ - I) V O " + 6)«(r) - (3e "' 8.33 (14* " - \2e »)u{t) V 8.35 lOe " - 8e V . 8.37 8(1 - e-") V 8.39 (a* -4(1 - e )u(l) V; (b) - 4 ( 1 - e ')u(/) + 4|1 - e "!«(/ - 1) V 2

CHAPTER 7 7.1

CHAPTER 5

7.7

5.1 5.5 5.9 5.11 5.13 5.15 5.17 5.19 5.21 5.23 5.25 5.27 5.29 5.31 5.35 5.37

3.5 A , 35 V, 8 V -100V 3 A 8V 16 W 3A u * = 14 V, R = 10O; 4 W L = 1.75 A, R = 10 il; 1.25 A = 3 A , R = 8 il; 1A u„ = * f V, R* = » i l ; 8 V P i = - 7 0 V, R„, = 16 il; -3.5 A u« = \, /t, = 400 i l ; 0.5 mA t)« = 14 V, /c = 10 i l ; 4 W - 3 mA 0.9 mW, 400 i I 28 il, 1.75 W

5.39

^ i l , 312.5 W

A

79 7.11 7.13 7.15 7.17 7.19 7.21

a

th

h

lh

rw

20*- " V, 12 - e '• A

<><
7.3 7.5

5.41 5.43

10.5 10.7 10.9

5r * A

1 A

= - 6 V, R* = 4 i l ; 4 i l i v = 12 V, /?„, = 56 i l ; 0.2 A

21

i = 2 fiA, 0 < / < 0 5 = -2 0.5 < / < 1 s i ( l + 1) = Hi) p '= 20/ u.W. 0 < / < 0.5 = -20(r - 1) /iW, 0.5 < I < 1 = 200 - 1) MW, 1 < I < 1.5 = -20(/ - 2)M W, 1.5 < I < 2 = 20(/ - 2) /xW, 2 < t < 2.5 = -20(/ - 3) /iW, 2.5 < / < 3 4 ms, 4 V (a) 0; (b) - 2 0 t i A ; (c) 20c ' t i A ; (d) 3 cos 100/ mA 16c ' W

7.33 7.35 7.37 7.39

2

2

6.1 6.3 6.5 6.9

762

8.1 8.3 8.5 8.7 8.11 8.13

CHAPTER 9 'II 9.3 9 7 99 9.11 9 13 9.15

l.2e ' - 0.2<> A < '(8 cos I - 4 sin /) V 6(1 + 4l)e A 54c - 9e V (4 + 10/)f " A c "(cos 5/ + 0.2 sin 5/) A (a) r '(cos / + 4 sin /) A; ( b i l l » 3/)« A; (c) ie ~ — 2c A 4c ' - c- " + 1 A 3c - 9c ' + 6 mA 3c + 9 V, 6 c + 6 V (a) 3(1 - c '(cos / + sin /)] A; (b) 1.5(1 - 2e ' + e ' ) A; (c) 3(1 - (1 + 2/)c ^ A 6(1 + 5 / ) c + 4 A - 2 c ' - c " + 5 cos 3/ + sin 3/ A (2 + 17/ - 8/ )c A (a) (10 - \2t)e - 10 V; (b) (8 - 6/)c" ' - 8 cos 2/ - 5 sin 2/ V (4 f 12/)c" ' + 2 sin 8/ V 8c sin 4/ V 5|1 - c ^ c o s 500/ + sin 500/)] V

9 17 9.19 9.21 9.23

121 k

l0

w

, , ( x a

, o c 0

21

2

21

21

2

9.33 9.35 9.37

IV 2 A 12 V 4V Appendix C

Answers to Selected Odd-Numbered Problems

10.31 10.33 10.35 10.37 10.39

8

10.3

(a) (e) (a) (c)

11

CHAPTER 11 111 11.3 11.5 11.9 11.11 11.13 11.15 11.17 11.21 11.23 11.25 11.27 11.29 11.31

CHAPTER 10 10.1

12

21

40

2

9.25 9.27 9.29 9.31

10.23 10.25 10.27 10 29

21

21

21

f

2

10e"* V (a) 0.4 M i l , 0.5 /i.F, 16 tiJ; (b) 36.8% 8c " V , e " A -6c * A lOc^V 16c " " A

10.21

2,1

CHAPTER 8 CHAPTER 6

10.17

2,

201

2

10.15

21

4

w^, = 50 J at l = - s; M L * = 2 J al / = 0 2 e A 1.6 J, 1.6 J, - 2 A 3 A, 3 A, - } A, 15 V 10/iF 6, 0, - 2 V (a) - 10 sin 10/ mV; (b) - 5 0 sin 20/ tiW; (c) 5 cos 10/ /tiJ; (d) 50 tiW (a) 3/ - 2/ A , (b) 2/ - 3/ + 2 A 15 mA, - 2 0 A/s - 2 A , 7.5 A/s 15 mH (a) 50, 0.5 mH; (b) one answer: two in series with four parallel sets of two each 5, - 2 0 ; 25, 20 V - 1 6 , - 8 V/s 1 nF, 10' i l 2 or 10 V

10.13

1

2

2

7.23 7.25 7.27 7.29 7.31

21

0.5 cos (2 x 10°r - 53.1°) mA lOkfl, i j i F (a) 10 cos (4/ - 28.1°) V; (b) 5 sin (4/ - 28.1°) V (a) 20c' '*""' A; (b) 10 cos (2/ - 45*) A; (c) 2 sin (2/ - 60°) A (a) 5 V 2 cos (20/ + 135°); (b) 5 cos (20/ + 216.9°); (c) 13 cos (20/ - 67.4°); (d) 10 cos 20(; (e) 5 sin 20/ (a) 2 0 / - 1 7 1 . 9 ° il; (b) 1/V2/15° k i l ; (c) a[a il 6 i l , - 8 il, 0.06 S, 0.08 S, 2 cos (/ + 53.1°) A 4/53.1° n , 2/-53.1° A . 2 cos (7/ - 53.1°) A 40 sin 2500/ mA, 4 cos 2500/ V 4 cos (10/ - 53.1°) A V2 cos (40,000/ - 98.1°) mA 2 cos (5/ - 53.1°) A, 0.5 cos (5/ - 53.1°) A 10 cos 30,000/ V (a) 4 cos (/ + 36.9°) A; (b) 5 cos 2/ A; (c) 4 cos (4/ - 36.9°) A 0.75 cos 3/ A V2 cos (2000/ - 135°) mA -5c ' - e + 4 V 5 cos (3/ - 26.6°) A

12 cos (3/ + 233.1°) V 0.5 cos 2/ A 48 sin 8/ V 16V2 cos (4/ — 45°) V V 2 c o s ( 5 / - 135°) A 4 V 2 cos (6/ - 135°) V 6 V 2 cos (2000/ + 45°) mA 2V5 cos (4/ - 10.3°) V - 2 sin 1000/ mA 8 cos / V 4V2cos (5/ + 135°) V 4 + V 2 cos (2/ - 45°) V 3 cos 2/ + 12 cos (3/ + 7.4°) V I« = ^ ( 7 - 7 ' 4 ) A , Z

50 V; (b) 60°; (c) TT/3 rad; (d) 10 ms; 200ir rad/s; (f) 100 Hz; (g) 77° v, leads by 30°; (b) c, leads by 67.4°; v, lags by 23.1° Appendix C

11.33 11.35

,„ =

1 8

+ J a, ,, = cos 2/ A 1

V„c = f[£ V, z * = -jiO, v = 2 cos (8/ - 53.1°) V

V2

cos (1000/ - 135°) V

Answers to Selected Odd-Numbered Problems

763

11.37 11.39

0.5 cos (3f + 126.9°) A * + (v + i ) = J, 2 H, sin 2t\ 2

13.27 13.31 13.35 13.37

CHAPTER 12 12.1 12.3 12.5 12.7 12.9 12.11 12.17 12.19 12.21 12.23

L„ = 17 32. I*. = -8.66 + ,6.34 I,< = —8.66 — 76.34 A rms (a) 2 / V 3 . 4 / V 3 kW; (b) 2V 3 kW 3 V 3 , I . 5 V 3 , 4.5V3 kW (a) 1; (b) 0; (c) 0.5 lagging; (d) 0 5 k'admit (e) 0.866 leading I 500V3, -633.7V3 W

2

CHAPTER 14

-W

14.1

P, = 4000 W, / W i = 400 W, P, = -4400 W 3.4 W 145 W (a) 2 W; (b) ft W, | n (a) 13 V; (b) 9 V; (c) 7 V 4Vrms

0

2

•

3

4

2

14.5 14.7 14.9 14.11 14.13 14.15

- A rms 4

30 W, - 1 8 var, 30 - y 18 VA, 5 \ / 2 / - 3 0 ° A rms (a) 170 j i F ; (b) 7 4 / i F 14 W, 2 var, 14 + j2 VA 112.5W 3V3kW

21

2

4

l0,

O

55

F

CHAPTER 13 10V3/90° A rms 2.31/6.9°, 2.31/-113 . 1° , 2.31/-233.1 A rms, 640 W 0

13,17 13.19 13.21 13.23 13.25

764

2

15 19 R, = ^ n ,

16.35 1 6 3 5

\

~2s + 2s + 1 /?2

|H(/w)|

V2

' vTT n.R,

=

1 n

f

, (a)

) 2 » ^ ; ( b. >)

V

1

4

1

7

c o s ( 3

,4.19

V..=

+

6

7

4

)

v

' V, .= ' " n ; s s V 5 c 'cos (2» - 26.6°) A 6*"' - 6 c " -1 \2te • V -36c -' + 2c ' sin 4r + 36 V s + 250 / = s. I n = z i = s, z = ,

14.21 14.23 , .„ 14.25

1 6 ( 4

+

3

)

Z

2

+

1

)

"

=

s + 250 ^ 5 b V " 5

5/-53.1 n -25 + y 15 A rms, 7.83 kW 2.66 - j 10.32 A rms 24 A rms

15.1

=

y 2 1 =

_

y

"

250

2

( I / O

15.11

*

H, 2 F + ( I / O s + l/LC R, = 2.5 n , R2 = 10 n , R, = 1 n , o) = 0.95 rad/s, cu , = 1 05 rad/s. B = 0 1 rad/s J

M

Appendix C

10s s + 10s + 2500 6s s + 10s + 16 2s wo = 10 rad/s, | H | = 1 J + 2s + 100 2

15.7

1

|

10/60° A.

N ;

20/tf V I n = i-.s, I11 = l u = Ms, in = L M _ _ t ,

16.37

V l + o> '

.y» = y» =

where A = Li Li - M 150 W

16.39

2

16.41

-

J

as

2

2

3s + 2s + 2 2

2

2

3

CHAPTER 17 (a) oo = 3,'a„ = 0,

17.1

2

"

(b) a. = r r ( n + 1) 2

n = 0, 1, 2, . . . ; b„ = -2na „ , n = 1, 2, 3, . . . ; wo = 1;

- 11 — 3.54 k l l

]

15.35 15.37

1 f l —» 8 k f l , 2.5 H —» 20 mH |H| , 22w -< 1, 4> = -2 tan" • 0.25 H, 1 H • 0.5 H,

mr

2(-D " - sinh 7r,

+ - •e ( - 1 ) " (c) i + rns Z h x " c o s n 3 ^P 1

4

w

'

+

For 17.2(a): a„ = 0, b = — [1 - 3 ( - l ) " ] ; nir

17.5

n

1 11 • 7.07 k f l

17.2(b):fr„= 0, oo = 1, !

0

15.5

>

2

CHAPTER 15

100

)

22

2

20V5 A rms

— = A r m s , 15 kW V3 100V3 V rms, 60 A rms, 10.8 kW 17.62 / i F 15V2 A rms L, = 10 - j4, I = -2 0 , Lc = 10 + j4 A rms, P = 3.69 kW 10V5 A rms, 7.2 kW

N

2

s' + 2s + 2s + I ' vTT «), = 1 rad/s (l/2)(s + 1) ;R • I 11, L = 1 H. 15.25 s + (l/2)i + 1 C = 1 F 15.27 R, = 79.6 k f l , R2 - 159.2 k f l , R - 15.92 k l l , other i n resistors become 15.92-kil resistors. 15.29 R, = 1 11, R: = 3 il, in scaled circuit R, = 10 k l l , R - 30 k f l , 1 il becomes 10 V.H 15.31 L = 0.25 H, R's = 5 k l l 15.33 1 n -»7.07 k n , 2 n - » i 4 . i 4 m,

;

y

15.3

|

1 rad/s

15.23

(a) 5 / - 3 0 ° ; (b) 1/45°; (c) 1Q/-53.1 ; (d) 5/0° (i + l ) Xn —, ~2e ' sin / A 2(s + is + 3) - 8 c * sin 2r V 16* sin f V V 2 c o s (/ - 135°) V 4 V 2 c ' cos (2/ + 45°) V 2V5 c ' cos (8/ + 116.6°) V -5 — , -12 c s i n 5 / V J + 15s + 75 18 12 (27T3)5'T3^' ' - °

mt

12.29 12.31 12.33 12.35 12.37 12.39

13.15

1

2

0.6 lagging, - ^ il

13.5 13.7 13.9 13.11 13.13

i +

4

12.27

13.1 13.3

1

15 I

15.21

IRH

V 2 V ' c o s ( r + 45°) - c ' A 4W

16.31 16.33

7 + V^l+ 1

/

13.39

500 W

15.15

2

f|

Answers to Selected Odd-Numbered Problems

r

CHAPTER 16 16.1 16.3 16.5 16.7 16.9 16.11 16.13 16.15 16.17 16.19 16.23 16.25 16.27 16.29

17.2(d): ft. = 0. oo = 1,

10 cos 10» V (a) 12, 26 V; (b) 12, 26 V 26, - 4 0 ; 14, 20 V

(-1)

y i " (-1)"] - + 2 sin 2( t -

17.9

IT

4(e" - 2c" -) V 2

2.5V2 cos (20r + 135°) V 2.5V2 sin (8r + 135°) A 4J 8J (a) 0.6 H; (b) 0.3 H; (c) 6 mH -4c ' V 4(e"' - c ") A 3 sin 8r A 4 V 2 cos (2r - 45°) A (25

8 i cos ir . - , 4n - 1 2

2 A (-D "

1 1

7

1

1

4

(2n - Dirt

2 .?, ( 2 7 ^ 1 ) ° 2 16ri v (4n/ - t a n ' n)] +

C

c

1 7 1 3

7L2

17.15

2-

+

o

S

s

.?,(T^v^rrJ

S

+ l)(s + 2) Appendix C

( «'-T)

+l

Answers to Selected Odd-Numbered Problems

n i r V 4 + n 7r 4

4

765

17.17

-128

x In 25 - 4n

n cos ^ Ant + 90° - tan

17.39

(4n - 1)V(100 - 16n ) + 64n I , 4V2n \ „ n cos 4n/ - tan —— I 2

2

2

2

2

17.9 ^TT2 , - i 17.21

(4„2 1 - 1 - 2 X 7r|_ . - i 4n

l)Vl

- " 256n 1

+

1 6

,7.27

(

CHAPTER 18

8n fT, _ £ < - ! ) • « » , 4« An 2

18.1

(a)

jut

2

1

7

2

9

7 g , 2 ^ l V 4

+

n

0 = (2n -" l1)71 ) i « + tan

[(>

7T

2

2

(In

-

1) [4 2

1) TT 2

+ (2n -

,

A

2

2

l)if

)

1) TT 2

18.3

a + jut —2ia> 2 oi7r (a) a , + a>3. (*>>-;1 - 01; -2r a,1 + .to (a - b)ut 1 - at', . f,2 (a - b)o> . -(a—- b)ot , tan ' tan (a) 2 -1- sin 0) sin 2 L 2 ; 2

2

18.5

17.31 17.33

2

i

2

+ o. + fc b - a. to )) + 4a 4 a a> o. ' , ut , 2aoi tan ' tan ' 2 . a a + b - at

W

a

2

2

2

22

22 22

22

2

8 ,7r(l - 4n ) 2 (b) |Co| = - sinh 7r,
2

2

2

J

2

18.23 18.25 18.27 18.29 18.31

J

2

2

2

,

18.7

(c) |co| = | , s*o = 0; (c„| = j j j j j ,

19.1

e

Appendix G

18.9

+

y

= ' x

+

i

r

19.29

19.5

(a)

19.7

2

(b) (c) (d) (a) (c) (d)

(2 -

20.1

3C")H(»);

(2 sin 2r + 2t)u(t); (3 cos 3r + 2 sin 3/)«(/); (cos t - t + 2e-')«(») c '(2 cos » - 6 sin r)u(/); (b) 2 f c " « ( r ) ; (2 - r ) c ' « ( r ) ; c " ( 2 cos 3* + 2sin3r)u(») 1 — *-C*+2)

-«-»); (b)

(C)

'

1

;

5 + 2 0 > . S < " ; ( d )'-s + 2 % t i i (1 j j (; - l)«(r - ! ) - ( , - )u(t - 2); (2 cos 2» + 3 sin 2t)u(t - 77); j sin 3r(u(/) - u(t - ir)\;

(r + a2V -2a) a ) '' (b),7(7^ ( S )

77

19 11

« J

19.13

(a) (b) (c)

2

Answers to Selected Odd-Numbered Problems

2

2

20.3 20.5 20.9 20.13 20.17

C

2

18.13

1 + C -

C-")(J

2

+

j>'

. s + 1 1 sin 2/ A (a) 2e"' - e ' ; (b) 3 cos 2r + cos t 5c~' - 2 c ' A -6c" A 4c-' + ^c- ' - § c - ' , - 4 c ' + 7c" ' - c " A 3

7

2

-

1

19.9

2

18.11

'(1

s

(c)

19.31 19.33 19.35 19.37 19.39

;(b):

i ( l + c-)

6

2

CHAPTER 20

2

20

8c^' cos — sin o>

(a)

-2

— s + 1

1

2

(b)

1 - 2c * c(a)-! — ;(b)-

19.3

2

d>„ = 0, n even = 180°, n odd For/4 = 1, (a) / ( / ) = 1.273 sin (2TH) + 0.423 sin (6rr») + 0.253 sin (lOrrr) + 0.179 sin (14 irt) + 0.138 sin (187rt) + • • • (c) / ( / ) = 0.642 + 0.418 sin (2TT/ - 92°) + 0.084 sin (4TT( - 94°)

v

2

2

sin 2o> _. oi -—57, t»n ' 1 - 2 c * cos 2tt> a 1 [ 3 + jut 1 \ / 2 l 5 - oi - 2yo)J (a) + 9 O + 2) + 4 ' ^ ' (1 - - » tan"

2

4

l ± ^ ' 1 + Jlxt J,^J 1 fj (a)JJ;(b)|J ;

19.27 (a) [c~'(l - 3/ + ^ r ) + e > ( / ) ;

CHAPTER 19

2

1

766

(d) 8(«) - (1 - cos / + sin Du(i), (e) (cos I - sin 2/)u(l)

2

2

2

(c) — 2a OJ V (tan"' a + * - a> ) +5;4fl o> a + b - at jl + c"*° - 2c" * cos 2a»

2

; ( a ) (1 „ +, ' ja>) . „ ; ( b ) T1 -+^ — = V V 2 - 1 » 0.644 2H 3w , 90° - tan V 4 + 5a, + a.

2

1*1

2

d>„ = tan n, n even = 180° + t a n ' n, n odd

17.37

(O [ c - * i t - IV " *J»W;

2

/ \ >

right: 128 y 1 _T2 S ^ (2n - 1) [4 + (In - l) 7r ] w 77T 2

(a)r-^; 2 + 70)

c o s

2

2

2

2

2

;

(d)

1) TT

1) TT ]

2;

\)TT,

S i

IT m\ — 1 (a) (sinr + sin 2r)u(»); (b) (c' - c ' + 2 sin r)u(r): H

19.25

2

b

b'

2

S

, (2n -

1)77/ - tan

1 + (2n -

2

2 + (2/t -

(2n - 1)V4 + (2n -

77 „ .

left: 128

(a + jo>) +

(2 -l) . "" (2n - l)ir

18.21

a + jut (a + jo>y +

(b)

")«(» - 1)| V

•921 - 2 - _ L - n ( 2 „ -

2

2

•)]

4n

(1 -

(b) (c ' - e ')«(r) V

4

18.19

[4m + 2 tan

2

(a) - c M - r ) ; (b) - » c ' « ( - / )

1817

(b) 12[(1 - c"')«(/)

7r[(r - 2) + a ]' 2

'18.15

(a) 12(1 - c ')u(l) V;

19.19

(O

V

2

cos

+ 0.037 sin (6rrt - 95°) + 0.021 sin (8771 - 97°) + 0.013 sin (lOrr/ - 99*) + • • •' / ( / ) = 6.379 + 1.986 sin (2m + 10 K ' l + sin (47rt + 8.10°) + 0.670 sin (6m + 8.40°) + 0.504 sin (%m + 9.44°) + 0.404 sin (IOTTT + 10.8°) + •

2

(d) sin 2t[e '«(/) - e ' " u ( » - n/2)] (a) cos I + sin /; (b) 1 - cos t + sin t (

19.17

Appendix C

/ 2 )

20.19 20.21 20.23 20.25

(a) c"'(2 cos 2/ + sin 2r) A; (b) \s t - 5 sin r + c"'(| cos 2t - j sin 2») A 2 + 10c"' - l O c " A (a) 8 - c " ( 6 sin t + 8 cos t) V, (b) 15c-' - e"(5 sin t + 15 cos 1) V 20 - c ' ( 4 cos 2r - 2 sin 2t) V (a) 2 sin 2r V; (b) 2 cos 2, V; (c) 2 cos 2» + 2 sin 2/ V (a) (4 - 5e"' + e~ ')u(t), 5(c-' - c " ) u ( / ) ; (b) [1 - c"(cos3r + f sin 3/)|«(/). ^c"sin3r«(j) 2 + 6e" - 8c" V c " ( l - cos/Mr) s

.

.

(J +

. 3) + 1 2

r

2 c " sin/i<(/)

+ Vis + 1'

Answers to Selected Odd-Numbered Problems

vT 767