Circuit Networks (lecture Notes)

CHAPTER 8 NETWORKS 1: 0909201-03/04 10 December 2003 – Lecture 8b

ROWAN UNIVERSITY College of Engineering Dr Peter Mark Jansson, PP PE DEPARTMENT OF ELECTRICAL & COMPUTER ENGINEERING

Autumn Semester 2003

admin hw 7 due today, hw 8 due at final test review 5.15pm thurs. at end of lab last lab 6 due by end of next week’s normal lab day (no later than 5 PM) final exam: Next Mon 15 Dec 2:45pm „

Rowan Hall Auditorium

take – home portion Š Assignment 8 (15%) Š Tool Kit (10%)

networks I Today’s learning objectives – „ „ „

„

master first order circuits build knowledge of the complete response use Thevenin and Norton equivalents to simplify analysis of first order circuits calculate the natural (transient) response and forced (steady-state) response

new concepts from ch. 8 response of first-order circuits „

to a constant input

the complete response stability of first order circuits response of first-order circuits „

to a nonconstant (sinusoidal) source

What does First Order mean? circuits that contain capacitors and inductors can be defined by differential equations circuits with ONLY ONE capacitor OR ONLY ONE inductor can be defined by a first order differential equation such circuits are called First Order

Circuits

what’s the complete response (CR)? Complete response = transient response + steady state response OR…. Complete response = natural response + forced response

finding the CR of

st 1

Ord. Cir

1) Find the forced response before the

disturbance. Evaluate at t = t(0-) to determine initial conditions [v(0) or i(0)] 2) Find forced response (steady state) after the disturbance t= t(∞) [Voc or Isc ] 3) Add the natural response (Ke-t/τ) to the new forced response. Use initial conditions to calculate K

Figure 8.0-1 (p. 290)

A plan for analyzing first-order circuits. (a) First, separate the energy storage element from the rest of the circuit. (b) Next, replace the circuit connected to a capacitor by its Thévenin equivalent circuit, or replace the circuit connected to an inductor by its Norton equivalent circuit.

RC and RL circuits RC circuit complete response: RL circuit complete response:

v (t ) = VOC + (v (0) − VOC )e

i (t ) = I SC + (i (0) − I SC )e

− t /( Rt C )

− ( Rt / L ) t

simplifying for analysis Using Thevenin and Norton Equivalent circuits can greatly simplify the analysis of first order circuits „ „

We use a Thevenin with a Capacitor and a Norton with an Inductor

Thevenin Equivalent at t=0+ i(t)

+

Rt

Voc +–

C

+ v(t) -

Norton equivalent at Isc

Rt

+ v(t) -

L

i(t)

+ t=0

1st ORDER CIRCUITS WITH CONSTANT INPUT t=0

R1 vs

R2

+ –

R3

( )

v0

−

R3 = vs R1 + R2 + R3

C

+ v(t) -

Example (before switch closes) If vs = 4V, R1 = 20kohms, R2 = 20 kohms R3 = 40 kohms What is v(0-) ?

( )

v0

−

R3 = vs R1 + R2 + R3

as the switch closes… THREE PERIODS emerge….. 1. system change (switch closure) 2. (immediately after) capacitor or inductor in system will store / release energy (adjust and/or oscillate) as system moves its new level of steady state (a.k.a. transient or natural response) …. WHY??? 3. new steady state is then achieved (a.k.a. the forced response)

Thevenin Equivalent at t=0+ i(t)

+

Rt

Voc +–

C

R2 R3 Rt = R2 + R3 KVL

+ v(t) -

R3 Voc = vs R2 + R3

+ Voc − i ( t )Rt − v ( t ) = 0

dv ( t ) + Voc − Rt C − v( t ) = 0 dt

dv ( t ) v ( t ) Voc + = dt Rt C Rt C

SOLUTION OF 1st ORDER EQUATION

dv ( t ) v ( t ) Voc + = dt Rt C Rt C dv ( t ) Voc v ( t ) = − dt Rt C Rt C

dv ( t ) 1 = dt Voc − v ( t ) Rt C

Voc − v ( t ) dv ( t ) = dt Rt C

dv ( t ) 1 =− dt v ( t ) − Voc Rt C

dv ( t ) 1 =− ∫ ∫ dt + D v ( t ) − Voc Rt C

SOLUTION CONTINUED dv ( t ) 1 =− ∫ ∫ dt + D v ( t ) − Voc Rt C ln(v ( t ) − Voc ) = −

t Rt C

+D

⎛ t ⎞ ⎟⎟ v ( t ) − Voc = exp(D ) exp⎜⎜ − ⎝ Rt C ⎠ ⎛ 0 ⎞ ⎟⎟ + Voc v ( 0 ) = exp(D ) exp⎜⎜ − ⎝ Rt C ⎠

⎛ ⎞ t v ( t ) − Voc = exp⎜⎜ − + D ⎟⎟ ⎝ Rt C ⎠ ⎛ t ⎞ ⎟⎟ + Voc v ( t ) = exp(D ) exp⎜⎜ − ⎝ Rt C ⎠

exp(D ) = v ( 0 ) − Voc

SOLUTION CONTINUED ⎛ t ⎞ ⎟⎟ + Voc v ( t ) = (v ( 0 ) − Voc ) exp⎜⎜ − ⎝ Rt C ⎠

⎛ t ⎞ ⎟⎟ v ( t ) = Voc + (v ( 0 ) − Voc ) exp⎜⎜ − ⎝ Rt C ⎠

so complete response is… complete response = v(t) = forced response (steady state) = Voc + natural response (transient) = (v(0-) –Voc) * e -t/RtC) NOTE: τ =RtC ⎛ t ⎞ ⎟⎟ v ( t ) = Voc + (v ( 0 ) − Voc ) exp⎜⎜ − ⎝ Rt C ⎠

Figure 8.3-4 (p. 301)

(a) A first-order circuit and (b) an equivalent circuit that is valid after the switch opens. (c) A plot of the complete response, v(t), given in Eq. 8.3-8.

WITH AN INDUCTOR t=0

R1 vs

R2

+ –

R3

( )

i0

−

vs = R1 + R2

L

Why ?

i(t)

Norton equivalent at Isc

KCL

Rt

+ v(t) -

+ t=0

R2 R3 Rt = R2 + R3 i(t)

L

vs I sc = R2

Why ?

v( t ) + I sc − − i( t ) = 0 Rt

1 di ( t ) L + I sc − − i( t ) = 0 Rt dt

Rt di ( t ) Rt + i( t ) = + I sc dt L L

SOLUTION Rt di ( t ) Rt + i( t ) = + I sc dt L L

dv ( t ) v ( t ) Voc + = dt Rt C Rt C

Rt 1 ⇔ L Rt C

⎛ t ⎞ ⎟⎟ v ( t ) = Voc + (v ( 0 ) − Voc ) exp⎜⎜ − ⎝ Rt C ⎠

⎛ Rt ⎞ i ( t ) = I sc + (i ( 0 ) − I sc ) exp⎜ − t⎟ ⎝ L ⎠

so complete response is… complete response = i(t) = forced response (steady state) = Isc + natural response (transient) = (i(0-) –isc) * e –t(Rt/L)) NOTE: τ =L/Rt

⎛ Rt ⎞ i ( t ) = I sc + (i ( 0 ) − I sc ) exp⎜ − t⎟ ⎝ L ⎠

Figure 8.3-5 (p. 302)

(a) A first-order circuit and (b) an equivalent circuit that is valid after the switch closes. (c) A plot of the complete response, i(t), given by Eq. 8.3-9.

Figure E8.3-1 (p. 308)

Figure E8.3-2 (p. 309)

Figure E8.3-3 (p. 309)

Figure E8.3-4 (p. 309)

Figure E8.3-5 (p. 309)

Stability of

st 1

order circuits

when τ>0 the natural response vanishes as t Æ∞ „

THIS IS A STABLE CIRCUIT

when τ<0 the natural response grows without bound as tÆ∞ „

THIS IS AN UNSTABLE CIRCUIT

forced response summary Forcing function y(t)

Forced response xf(t)

(steady-state before)

(steady-state after)

Constant y(t) = M

Constant: xf(t) = N

Exponential y(t) = Me-bt Sinusoid y(t) = M sin (ωt +

Exponential xf(t) = Ne-bt Sinusoid xf(t) = Asin (ωt+ ) + Bcos(ωt+ )

)

Unit step or pulse signal vo(t) = A + Be-at for t > 0

Example 8.6-2, p. 321-323

Figure 8.6-12 (p. 322)

The circuit considered in Example 8.6-2

Figure 8.6-13 (p. 322)

Circuits used to calculate the steady-state response (a) before t = 0 and (b) after t = 0.

HANDY CHART ELEMENT

R

C L

CURRENT

VOLTAGE

V I= R

V = I∗R

dvc ic = C dt t 1

1 t vc = ∫ ic dt C −∞

di L iL = ∫ v Ldt v L = L L −∞ dt

IMPORTANT CONCEPTS FROM CHAPTER 8 determining Initial Conditions determining T or N equivalent to simplify setting up differential equations solving for v(t) or i(t)

Don’t forget HW 8 (test review) Thursday 5.15 pm 11 Dec after lab