Nise_control_engineering_solutions_to_sk.pdf

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Solutions to Skill-Assessment Exercises

CHAPTER 2 2.1 The Laplace transform of t is F ðsÞ ¼

1 ð s þ 5Þ 2

1 using Table 2.1, Item 3. Using Table 2.2, Item 4, s2

.

2.2 Expanding F(s) by partial fractions yields: A B C D þ þ F ðsÞ ¼ þ s s þ 2 ð s þ 3Þ 2 ð s þ 3Þ where,



5
¼ A¼ 2

ðs þ 2Þðs þ 3Þ S!0 9 10

C¼




B¼ 2

s ð s þ 3Þ 10

¼ 5 S!2




10

10 40 2 dF ðsÞ
¼ ¼ ; and D ¼ ð s þ 3 Þ
sðs þ 2Þ S!3 3 ds
s!3 9

Taking the inverse Laplace transform yields, f ðt Þ ¼

5 10 40  5e2t þ te3t þ e3t 9 3 9

2.3 Taking the Laplace transform of the differential equation assuming zero initial conditions yields: s3 CðsÞ þ 3s2 CðsÞ þ 7sCðsÞ þ 5CðsÞ ¼ s2 RðsÞ þ 4sRðsÞ þ 3RðsÞ Collecting terms, 3    s þ 3s2 þ 7s þ 5 CðsÞ ¼ s2 þ 4s þ 3 RðsÞ Thus, C ðsÞ s2 þ 4s þ 3 ¼ 3 RðsÞ s þ 3s2 þ 7s þ 5 1

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2.4 G ð sÞ ¼

C ðsÞ 2s þ 1 ¼ RðsÞ s2 þ 6s þ 2

Cross multiplying yields, d2 c dc dr þ 6 þ 2c ¼ 2 þ r dt2 dt dt 2.5 CðsÞ ¼ RðsÞGðsÞ ¼

1 s 1 A B C ¼ ¼ þ þ  s2 ðs þ 4Þðs þ 8Þ sðs þ 4Þðs þ 8Þ s ð s þ 4 Þ ð s þ 8Þ

where

1 1
¼ A¼ ðs þ 4Þðs þ 8Þ
S!0 32



1

1 1

1 ¼  ; and C ¼ ¼ B¼ sðs þ 8Þ
S!4 16 sðs þ 4Þ
S!8 32

Thus, c ðt Þ ¼

1 1 1  e4t þ e8t 32 16 32

2.6 Mesh Analysis Transforming the network yields, s

1

V(s)

I9(s)

+ _

V1(s)

1

+ s I1(s)

s

V2(s) _

I2(s)

Now, writing the mesh equations, ðs þ 1ÞI 1 ðsÞ  sI 2 ðsÞ  I 3 ðsÞ ¼ V ðsÞ sI 1 ðsÞ þ ð2s þ 1ÞI 2 ðsÞ  I 3 ðsÞ ¼ 0 I 1 ðsÞ  I 2 ðsÞ þ ðs þ 2ÞI 3 ðsÞ ¼ 0 Solving the mesh equations for I2(s),


ð s þ 1Þ V ð s Þ 1



s 0 1


2 
1 s þ 2s þ 1 V ðsÞ 0 ð s þ 2Þ

¼ I 2 ðsÞ ¼

sðs2 þ 5s þ 2Þ s 1


ð s þ 1Þ
s ð2s þ 1Þ 1



1 1 ð s þ 2Þ


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Chapter 2 Solutions to Skill-Assessment Exercises

But, VL ðsÞ ¼ sI 2 ðsÞ Hence,

2  s þ 2s þ 1 V ðsÞ VL ðsÞ ¼ ðs2 þ 5s þ 2Þ

or VL ðsÞ s2 þ 2s þ 1 ¼ 2 V ðs Þ s þ 5s þ 2 Nodal Analysis Writing the nodal equations,   1 þ 2 V 1 ðsÞ  VL ðsÞ ¼ V ðsÞ s   2 1 V 1 ðsÞ þ þ 1 VL ðsÞ ¼ V ðsÞ s s Solving for VL(s),




1

þ2 V ðs Þ

s




1 2 
1 V ðsÞ
s þ 2s þ 1 V ðsÞ s 
¼ VL ðsÞ ¼

1
ðs2 þ 5s þ 2Þ

þ 2 1
s

 


2
1
þ 1

s

or VL ðsÞ s2 þ 2s þ 1 ¼ 2 V ðs Þ s þ 5s þ 2 2.7 Inverting G ð sÞ ¼ 

Z2 ðsÞ 100000  ¼ s ¼  Z 1 ðsÞ 105 =s

Noninverting

G ð sÞ ¼

½ Z 1 ðsÞ þ Z ðsÞ ¼ Z1 ðsÞ

105 þ 105 s ! 105 s

! ¼sþ1

2.8 Writing the equations of motion, 2  s þ 3s þ 1 X1 ðsÞ  ð3s þ 1ÞX2 ðsÞ ¼ F ðsÞ   ð3s þ 1ÞX1 ðsÞ þ s2 þ 4s þ 1 X2 ðsÞ ¼ 0

3

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Solving for X2(s),


 2

s þ 3s þ 1 F ðsÞ



ð3s þ 1Þ ð3s þ 1ÞF ðsÞ 0

¼ X 2 ðsÞ ¼

 
sðs3 þ 7s2 þ 5s þ 1Þ 2 s þ 3s þ 1  ð 3s þ 1 Þ

2 


ð3s þ 1Þ s þ 4s þ 1


Hence, X 2 ðsÞ ð3s þ 1Þ ¼ 3 F ðsÞ sðs þ 7s2 þ 5s þ 1Þ 2.9 Writing the equations of motion, 2  s þ s þ 1 u1 ðsÞ  ðs þ 1Þu2 ðsÞ ¼ T ðsÞ ðs þ 1Þu1 ðsÞ þ ð2s þ 2Þu2 ðsÞ ¼ 0 where u1 ðsÞ is the angular displacement of the inertia. Solving for u2 ðsÞ, 
 2

s þ s þ 1 T ðsÞ



 ð s þ 1Þ ðs þ 1ÞF ðsÞ 0

¼ u 2 ðsÞ ¼
 2
s þ s þ 1 ðs þ 1Þ
2s3 þ 3s2 þ 2s þ 1


ðs þ 1Þ ð2s þ 2Þ
From which, after simplification, u 2 ðsÞ ¼

2s2

1 þsþ1

2.10 Transforming the network to one without gears by reflecting the 4 N-m/rad spring to the left and multiplying by (25/50)2, we obtain,

Writing the equations of motion, 2  s þ s u1 ðsÞ  sua ðsÞ ¼ T ðsÞ su1 ðsÞ þ ðs þ 1Þua ðsÞ ¼ 0 where u1 ðsÞ is the angular displacement of the 1-kg inertia. Solving for ua ðsÞ, 
 2

s þ s T ðsÞ



s sT ðsÞ 0

¼ 3 u a ðsÞ ¼
 2
s þs
þ s2 þ s s s


s
ðs þ 1Þ

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From which, ua ðsÞ 1 ¼ T ðsÞ s2 þ s þ 1 1 But, u2 ðsÞ ¼ ua ðsÞ: 2 Thus, u2 ðsÞ 1=2 ¼ 2 T ðsÞ s þ s þ 1 2.11 First find the mechanical constants.     1 1 2 1 Jm ¼ Ja þ JL ¼ 1 þ 400  ¼2 5 4 400     1 1 2 1  ¼7 ¼ 5 þ 800 Dm ¼ Da þ DL 5 4 400 Now find the electrical constants. From the torque-speed equation, set vm ¼ 0 to find stall torque and set T m ¼ 0 to find no-load speed. Hence, T stall ¼ 200 vnoload ¼ 25 which, Kt T stall 200 ¼ ¼ ¼2 Ra Ea 100 Ea 100 Kb ¼ ¼ ¼4 vnoload 25 Substituting all values into the motor transfer function, KT u m ðsÞ 1 R a J m ¼   ¼  1 KT Kb 15 Ea ðsÞ Dm þ s sþ s sþ Ra Jm 2 where um ðsÞ is the angular displacement of the armature. 1 Now uL ðsÞ ¼ um ðsÞ. Thus, 20 u L ðsÞ 1=20  ¼  15 Ea ðsÞ s sþ 2 2.12 Letting u1 ðsÞ ¼ v1 ðsÞ=s u2 ðsÞ ¼ v2 ðsÞ=s

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in Eqs. 2.127, we obtain   K K J 1 s þ D1 þ v1 ðsÞ  v2 ðsÞ ¼ T ðsÞ s s   K K  v 1 ðsÞ þ J 2 s þ D 2 þ v 2 ðsÞ s s From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively. J1

T(t)

D1

J2 1 K

– +

w1(t)

D2 w2(t)

Series analog 1 K

w1(t)

J1

T(t)

w2(t)

1 D1

J2

1 D2

Parallel analog

2.13 Writing the nodal equation, C

dv þ ir  2 ¼ iðtÞ dt

But, C¼1 v ¼ vo þ dv ir ¼ evr ¼ ev ¼ evo þdv Substituting these relationships into the differential equation, dðvo þ dvÞ þ evo þdv  2 ¼ iðtÞ dt We now linearize ev. The general form is f ð v Þ  f ð vo Þ 


df

dv dv
vo

Substituting the function, f ðvÞ ¼ ev , with v ¼ vo þ dv yields,
dev

dv evo þdv  evo  dv
vo

ð1Þ

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Chapter 3 Solutions to Skill-Assessment Exercises

Solving for evo þdv, e

vo þdv


dev

¼e þ dv ¼ evo þ evo dv dv
vo vo

Substituting into Eq. (1) ddv þ evo þ evo dv  2 ¼ iðtÞ dt

ð2Þ

Setting iðtÞ ¼ 0 and letting the circuit reach steady state, the capacitor acts like an open circuit. Thus, vo ¼ vr with ir ¼ 2. But, ir ¼ evr or vr ¼ lnir . Hence, vo ¼ ln 2 ¼ 0:693. Substituting this value of vo into Eq. (2) yields ddv þ 2dv ¼ iðtÞ dt Taking the Laplace transform, ðs þ 2ÞdvðsÞ ¼ I ðsÞ Solving for the transfer function, we obtain dvðsÞ 1 ¼ I ðsÞ sþ2 or V ðs Þ 1 ¼ about equilibrium: I ðsÞ s þ 2 CHAPTER 3 3.1 Identifying appropriate variables on the circuit yields C1

R iR

iC1 v1(t) +

L

–

iL

+

C2

vo(t) –

iC2

Writing the derivative relations dvC1 ¼ iC1 dt diL ¼ vL L dt dvC2 ¼ iC2 C2 dt C1

ð1Þ

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Using Kirchhoff’s current and voltage laws, iC1 ¼ iL þ iR ¼ iL þ

1 ð vL  vC 2 Þ R

vL ¼ vC1 þ vi 1 iC2 ¼ iR ¼ ðvL  vC2 Þ R

Substituting these relationships into Eqs. (1) and simplifying yields the state equations as dvC1 1 1 1 1 ¼ vC 1 þ iL  vC 2 þ vi dt RC1 C1 RC1 RC1 diL 1 1 ¼  vC 1 þ v i dt L L dvC2 1 1 1 ¼ vC 1  vC 2 vi RC2 RC2 RC2 dt where the output equation is vo ¼ vC 2 Putting the equations in vector-matrix form, 2



1 RC1

6 6 6 1 6 x_ ¼ 6  6 L 6 4 1  RC2 y ¼ ½0

0

1 C1 0 0

3 2 3 1 1 6 RC1 7 RC1 7 6 7 7 7 6 7 7 6 1 7 7 vi ð t Þ 0 7x þ 6 7 6 L 7 7 6 7 4 1 5 1 5  RC2 RC2



1 x

3.2 Writing the equations of motion 2  sX 2 ðsÞ ¼ F ðs Þ s þ s þ 1 X 1 ðsÞ 2  sX 1 ðsÞ þ s þ s þ 1 X 2 ðsÞ X 3 ðsÞ ¼ 0 2  X 2 ðsÞ þ s þ s þ 1 X 3 ðsÞ ¼ 0 Taking the inverse Laplace transform and simplifying, x€1 ¼ x_ 1  x1 þ x_ 2 þ f x€2 ¼ x_ 1  x_ 2  x2 þ x3 x€3 ¼ x_ 3  x3 þ x2 Defining state variables, zi, z1 ¼ x1 ; z2 ¼ x_ 1 ; z3 ¼ x2 ; z4 ¼ x_ 2 ; z5 ¼ x3 ; z6 ¼ x_ 3

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Chapter 3 Solutions to Skill-Assessment Exercises

Writing the state equations using the definition of the state variables and the inverse transform of the differential equation, z_ 1 ¼ z2 z_ 2 ¼ x€1 z_ 3 ¼ x_ 2 z_ 4 ¼ x€2 z_ 5 ¼ x_ 3

¼ x_ 1  x1 þ x_ 2 þ f ¼ z2  z1 þ z4 þ f ¼ z4 ¼ x_ 1  x_ 2  x2 þ x3 ¼ z2  z4  z3 þ z5

¼ z6 z_ 6 ¼ x€3 ¼ x_ 3  x3 þ x2 ¼ z6  z5 þ z3

The output is 2 0 6 1 6 6 6 0 z_ ¼ 6 6 0 6 6 4 0 0

z5. Hence, y ¼ z5 . In vector-matrix form, 3 2 3 1 0 0 0 0 0 7 6 7 1 0 1 0 0 7 617 7 6 7 607 0 0 1 0 0 7 7z þ 6 7 f ðtÞ; y ¼ ½ 0 7 607 1 1 1 1 0 7 6 7 7 6 7 405 0 0 0 0 1 5 0

1

1

0

1

0 0

0

1

0 z

0

3.3 First derive the state equations for the transfer function without zeros. X ðs Þ 1 ¼ 2 RðsÞ s þ 7s þ 9 Cross multiplying yields 

 s2 þ 7s þ 9 X ðsÞ ¼ RðsÞ

Taking the inverse Laplace transform assuming zero initial conditions, we get x€ þ 7x_ þ 9x ¼ r Defining the state variables as, x1 ¼ x x2 ¼ x_ Hence, x_ 1 ¼ x2 x_ 2 ¼ x€ ¼ 7x_  9x þ r ¼ 9x1  7x2 þ r Using the zeros of the transfer function, we find the output equation to be, c ¼ 2x_ þ x ¼ x1 þ 2x2 Putting all equation in vector-matrix form yields, " # " # 0 1 0 x_ ¼ xþ r 9 7 1 c ¼ ½ 1 2 x

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3.4 The state equation is converted to a transfer function using GðsÞ ¼ CðsI  AÞ1 B

ð1Þ

where A¼

4 4



2 ;B ¼ ; and C ¼ ½ 1:5 0 0

1:5

Evaluating ðsI  AÞ yields

ðsI  AÞ ¼

sþ4

1:5

4

s

0:625 :



Taking the inverse we obtain ðsI  AÞ1 ¼



s 1:5 1 s2 þ 4s þ 6 4 s þ 4

Substituting all expressions into Eq. (1) yields GðsÞ ¼

3s þ 5 s2 þ 4s þ 6

3.5 Writing the differential equation we obtain d2 x þ 2x2 ¼ 10 þ df ðtÞ dt2

ð1Þ

Letting x ¼ xo þ dx and substituting into Eq. (1) yields d2 ðxo þ dxÞ þ 2ðxo þ dxÞ2 ¼ 10 þ df ðtÞ dt2

ð2Þ

Now, linearize x2. 2

ðxo þ dxÞ 

x2o

 
d x2

¼ dx ¼ 2xo dx dx
xo

from which ðxo þ dxÞ2 ¼ x2o þ 2xo dx

ð3Þ

Substituting Eq. (3) into Eq. (1) and performing the indicated differentiation gives us the linearized intermediate differential equation, d2 dx þ 4xo dx ¼ 2x2o þ 10 þ df ðtÞ dt2

ð4Þ

The force of the spring at equilibrium is 10 N. Thus, since F ¼ 2x2 ; 10 ¼ 2x2o from which pffiffiffi xo ¼ 5

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Chapter 4 Solutions to Skill-Assessment Exercises

Substituting this value of xo into Eq. (4) gives us the final linearized differential equation. pffiffiffi d2 dx þ 4 5 dx ¼ df ðtÞ dt2 Selecting the state variables, x1 ¼ dx _ x2 ¼ dx Writing the state and output equations x_ 1 ¼ x2 pffiffiffi x_ 2 ¼€dx ¼ 4 5x1 þ df ðtÞ y ¼ x1 Converting to vector-matrix form yields the final result as



0 1 0 pffiffiffi x_ ¼ xþ df ðtÞ 4 5 0 1 y ¼ ½1

0 x CHAPTER 4

4.1 For a step input C ðsÞ ¼

10ðs þ 4Þðs þ 6Þ A B C D E ¼ þ þ þ þ sðs þ 1Þðs þ 7Þðs þ 8Þðs þ 10Þ s s þ 1 s þ 7 s þ 8 s þ 10

Taking the inverse Laplace transform, cðtÞ ¼ A þ Bet þ Ce7t þ De8t þ Ee10t 4.2 Since a ¼ 50; T c ¼ Tr ¼

1 1 4 4 ¼ ¼ 0:02s; T s ¼ ¼ ¼ 0:08 s; and a 50 a 50

2:2 2:2 ¼ ¼ 0:044 s. a 50

4.3 a. b. c. d.

Since Since Since Since

4.4 a. b. c. d.

vn vn vn vn

poles poles poles poles

are are are are

at 6  j 19:08; cðtÞ ¼ A þ Be6t cosð19:08t þ fÞ. at 78:54 and 11:46; cðtÞ ¼ A þ Be78:54t þ Ce11:4t . double on the real axis at 15 cðtÞ ¼ A þ Be15t þ Cte15t : at j 25; cðtÞ ¼ A þ B cosð25t þ fÞ.

pffiffiffiffiffiffiffiffi ¼ 400 ¼ 20 and 2zvn pffiffiffiffiffiffiffiffi ¼ 900 ¼ 30 and 2zvn pffiffiffiffiffiffiffiffi ¼ 225 ¼ 15 and 2zvn pffiffiffiffiffiffiffiffi ¼ 625 ¼ 25 and 2zvn

¼ 12; ¼ 90; ¼ 30; ¼ 0;

;z ¼ 0:3 and system is underdamped. ;z ¼ 1:5 and system is overdamped. ;z ¼ 1 and system is critically damped. ;z ¼ 0 and system is undamped.

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4.5 vn ¼

pffiffiffiffiffiffiffiffi 361 ¼ 19 and 2zvn ¼ 16; ;z ¼ 0:421:

4 p ¼ 0:5 s and T p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 0:182 s. zvn vn 1  z 2 From Figure 4.16, vn T r ¼ 1:4998. Therefore, T r ¼ 0:079 s. pzp ffiffiffi 2 Finally, %os ¼ e 1  z  100 ¼ 23:3% Now, T s ¼

4.6 a. The second-order approximation is valid, since the dominant poles have a real part of 2 and the higher-order pole is at 15, i.e. more than five-times further. b. The second-order approximation is not valid, since the dominant poles have a real part of 1 and the higher-order pole is at 4, i.e. not more than five-times further. 4.7 1 0:8942 1:5918 0:3023 a. Expanding G(s) by partial fractions yields GðsÞ ¼ þ   . s s þ 20 s þ 10 s þ 6:5 But 0:3023 is not an order of magnitude less than residues of second-order terms (term 2 and 3). Therefore, a second-order approximation is not valid. 1 0:9782 1:9078 0:0704 b. Expanding G(s) by partial fractions yields GðsÞ ¼ þ   . s s þ 20 s þ 10 s þ 6:5 But 0.0704 is an order of magnitude less than residues of second-order terms (term 2 and 3). Therefore, a second-order approximation is valid. 4.8 See Figure 4.31 in the textbook for the Simulink block diagram and the output responses. 4.9



sþ5 s 2 1 sI  A ¼ ; ðsI  AÞ1 ¼ 2 s þ 5s þ 6 3 " #3 s þ 5 0 BUðsÞ ¼ . 1=ðs þ 1Þ

a. Since

2 : s

Also,

1  The state vector is XðsÞ ¼ ðsI  AÞ1 ½xð0Þ þ BUðsÞ ¼ ð s þ 1 Þ ð s þ 2 Þ ð s þ 3Þ "  #  2 s2 þ 7s þ 7 5s2 þ 2s  4 . The output is Y ð s Þ ¼ ½ 1 3 X ð s Þ ¼ ¼ ðs þ 1Þðs þ 2Þðs þ 3Þ s2  4s  6 0:5 12 17:5  þ . Taking the inverse Laplace transform yields yðtÞ ¼  sþ1 sþ2 sþ3 t 2t 0:5e  12e þ 17:5e3t . b. The eigenvalues are given by the roots of jsI  Aj ¼ s2 þ 5s þ 6, or 2 and 3. 4.10



sþ5 2 1 . Taking the a. Since ðsI  AÞ ¼ ; ðsI  AÞ ¼ 2 s þ 5s þ 4 2 s 2 sþ5 Laplace transform of each term, the state transition matrix is given by 2 3 4 t 1 4t 2 t 2 4t e  e 6 3e  3e 7 3 3 7: FðtÞ ¼ 6 4 2 2 4t 1 t 4 4t 5 t  e þ e  e þ e 3 3 3 3

s

2



1

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Chapter 5 Solutions to Skill-Assessment Exercises

2

3 4 ðttÞ 1 4ðttÞ 2 ðttÞ 2 4ðttÞ  e  e e 6 3e 7 3 3 3 7 and b. Since Fðt  t Þ ¼ 6 4 2 2 4ðttÞ 1 ðttÞ 4 4ðttÞ 5 ðttÞ  e þ e  e þ e 3 3 2 3 3 3 2 2 t t 2t 4t

6 3e e  3e e 7 0 6 7: BuðtÞ ¼ 2t ; Fðt  t ÞBuðtÞ ¼ 4 5 1 4 e t t 2t 4t  e e þ e e 3 3 Rt Thus, xðtÞ ¼ FðtÞxð0Þ þ 0 Fðt  t Þ 2 3 10 t 4 4t 2t 6 3 e  e  3e 7 7 BUðtÞdt ¼ 6 4 5 5 8 t 2t 4t  e þe þ e 3 3 c. yðtÞ ¼ ½ 2 1 x ¼ 5et  e2t CHAPTER 5 5.1 Combine the parallel blocks in the forward path. Then, push

1 to the left past the s

pickoff point. s

R(s)

+

– s2 +

1 s

1 s

C(s)

–

s

Combine the parallel feedback paths and get 2s. Then, apply the feedback formula, s3 þ 1 . simplify, and get, T ðsÞ ¼ 4 2s þ s2 þ 2s 5.2 G ð sÞ 16 , where ¼ 2 1 þ GðsÞH ðsÞ s þ as þ 16 16 a and H ðsÞ ¼ 1. Thus, vn ¼ 4 and 2zvn ¼ a, from which z ¼ . and GðsÞ ¼ s ð s þ aÞ 8   % ln a 100 But, for 5% overshoot, z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ¼ 0:69. Since, z ¼ 8 ; a ¼ 5:52. % p2 þ ln2 100 Find the closed-loop transfer function, T ðsÞ ¼

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5.3 Label nodes.

R(s) +

– –

+

s

s N1 (s)

N2 (s)

N3 (s)

1 s

N5 (s)

1 s

N4 (s)

+

N6 (s)

s

N7 (s)

Draw nodes. R(s)

N1(s)

N3(s)

N2(s) N5(s)

N4(s)

C (s)

N6(s)

N7(s)

Connect nodes and label subsystems. −1

R(s) 1 N1 (s) s

N2 (s)

s

N3 (s)

1

1 −1

1 N4 (s) s

C(s)

1 N5 (s)

1 s

N7 (s)

N6 (s)

s

Eliminate unnecessary nodes. –1

R(s)

1

s

s

1 s –s

1 s

C(s)

C(s)

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5.4 Forward-path gains are G1G2G3 and G1G3. Loop gains are G1 G2 H 1 ; G2 H 2; and G3 H 3. Nontouching loops are ½G1 G2 H 1 ½G3 H 3  ¼ G1 G2 G3 H 1 H 3 and ½G2 H 2 ½G3 H 3  ¼ G2 G3 H 2 H 3 . Also, D ¼ 1 þ G1 G2 H 1 þ G2 H 2 þ G3 H 3 þ G1 G2 G3 H 1 H 3 þ G2 G3 H 2 H 3 : Finally, D1 ¼ 1 and D2 ¼ 1. P Substituting these values into T ðsÞ ¼ T ðsÞ ¼

CðsÞ ¼ R ðsÞ

k

T k Dk D

yields

G1 ðsÞG3 ðsÞ½1 þ G2 ðsÞ ½1 þ G2 ðsÞH 2 ðsÞ þ G1 ðsÞG2 ðsÞH 1 ðsÞ½1 þ G3 ðsÞH 3 ðsÞ

5.5 The state equations are, x_ 1 ¼ 2x1 þ x2 x_ 2 ¼ 3x2 þ x3 x_ 3 ¼ 3x1  4x2  5x3 þ r y ¼ x2 Drawing the signal-flow diagram from the state equations yields 1

r

1

1 s

x3

1 s

1

x2

1 s

1

–3

–5

x1

y

–2

–4 –3

5.6 100ðs þ 5Þ we draw the signal-flow graph in controller canonical form s2 þ 5s þ 6 and add the feedback.

From GðsÞ ¼

100

r

1

1 s

1 s

x1

x2

500 y

–5 –6

–1

15

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Writing the state equations from the signal-flow diagram, we obtain " x¼

105

506

#

" # 1

xþ r 1 0 0 y ¼ ½ 100 500 x

5.7 From the transformation equations, 1

P



3 2 ¼ 1 4



Taking the inverse, P¼

0:4 0:1



0:2 0:3

Now, " 1

P AP ¼

3

2

#"

1

3

#"

0:4

0:2

#

4 6 0:1 0:3 " #" # " # 3 2 1 3 P1 B ¼ ¼ 1 4 3 11 " # 0:4 0:2 CP ¼ ½ 1 4  ¼ ½ 0:8 1:4  0:1 0:3 1

4

" ¼

6:5

8:5

9:5

11:5

#

Therefore, " z_ ¼

6:5

8:5

#

"

3

#

u zþ 11 9:5 11:5 y ¼ ½ 0:8 1:4 z

5.8 First find the eigenvalues.



l 0 1 jlI  Aj ¼

 0 l 4





l  1
¼
6

4 3


3

¼ l2 þ 5l þ 6 l þ 6


From which the eigenvalues are 2 and 3. Now use Axi ¼ lxi for each eigenvalue, l. Thus,





x1 1 3 x1 ¼l 4 6 x2 x2 For l ¼ 2, 3x1 þ 3x2 ¼ 0 4x1  4x2 ¼ 0

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Chapter 6 Solutions to Skill-Assessment Exercises

Thus x1 ¼ x2 For l ¼ 3 4x1 þ 3x2 ¼ 0 4x1  3x2 ¼ 0 Thus x1 ¼ x2 and x1 ¼ 0:75x2 ; from which we let

0:707 0:6 P¼ 0:707 0:8 Taking the inverse yields



5:6577 ¼ 5

1

P

4:2433 5



Hence, " 1

D ¼ P AP ¼ " P1 B ¼ CP ¼ ½ 1

5:6577

5 #" # 4:2433 1

4

#"

5

5:6577 5 "

4:2433

5 0:707

0:6

0:707

0:8

3 #

¼

1

#"

3

4 6 " # 18:39

0:707

0:6

0:707

0:8

#

" ¼

2

0

0

3

#

20

¼ ½ 2:121

2:6 

Finally, " z_ ¼

2

0

#

"

18:39

zþ 0 3 20 y ¼ ½ 2:121 2:6 z

# u

CHAPTER 6 6.1 Make a Routh table. s7

3

6

7

2

s6

9

4

8

6

5

s

4.666666667

4.333333333

0

0

s4

4.35714286

8

6

0

s3

12.90163934

6.426229508

0

0

s2

10.17026684

6

0

0

s1

1.18515742

0

0

0

s0

6

0

0

0

Since there are four sign changes and no complete row of zeros, there are four right half-plane poles and three left half-plane poles. 6.2 Make a Routh table. We encounter a row of zeros on the s3 row. The even polynomial is contained in the previous row as 6s4 þ 0s2 þ 6. Taking the derivative yields

17

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24s3 þ 0s. Replacing the row of zeros with the coefficients of the derivative yields the s3 row. We also encounter a zero in the first column at the s2 row. We replace the zero with e and continue the table. The final result is shown now as s6

1

6

1

6

s

5

1

0

1

0

s

4

6

0

6

0

s3

24

0

0

0

2

e

6

0

0

s1

144=e

0

0

0

s0

6

0

0

0

s

ROZ

There is one sign change below the even polynomial. Thus the even polynomial (4th order) has one right half-plane pole, one left half-plane pole, and 2 imaginary axis poles. From the top of the table down to the even polynomial yields one sign change. Thus, the rest of the polynomial has one right half-plane root, and one left half-plane root. The total for the system is two right half-plane poles, two left halfplane poles, and 2 imaginary poles. 6.3 Since GðsÞ ¼

Kðs þ 20Þ G ð sÞ K ðs þ 20Þ ; T ðsÞ ¼ ¼ sðs þ 2Þðs þ 3Þ 1 þ GðsÞ s3 þ 5s2 þ ð6 þ KÞs þ 20K Form the Routh table. s3

1

ð6 þ K Þ

s2

5 30  15K 5 20K

20K

s

1

s0

From the s1 row, K < 2. From the s0 row, K > 0. Thus, for stability, 0 < K < 2. 6.4 First find


2
s

6 jsI  Aj ¼
4 0

0

0 s 0

0

3

2

2

7 6 05  4 1 s 3

3



ð s  2Þ

7

7 1 5
¼
1

4 5

3 1

1

1 ðs  7Þ 4


1


1

ð s þ 5Þ


¼ s3  4s2  33s þ 51 Now form the Routh table. s3

1

33

s2

4

51

s

1

20:25

s

0

51

There are two sign changes. Thus, there are two rhp poles and one lhp pole.

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Chapter 7 Solutions to Skill-Assessment Exercises

CHAPTER 7 7.1 a. First check stability. GðsÞ 10s2 þ 500s þ 6000 10ðs þ 30Þðs þ 20Þ ¼ ¼ 3 T ðsÞ ¼ 2 1 þ GðsÞ s þ 70s þ 1375s þ 6000 ðs þ 26:03Þðs þ 37:89Þðs þ 6:085Þ Poles are in the lhp. Therefore, the system is stable. Stability also could be checked via Routh-Hurwitz using the denominator of T(s). Thus, 15 15 ¼ ¼0 15uðtÞ : estep ð1Þ ¼ 1 þ lim GðsÞ 1 þ 1 s!0

15 15 ¼ 2:1875 ¼ lim sGðsÞ 10 20 30 s!0 25 35

30 15 30 15t2 uðtÞ : eparabola ð1Þ ¼ ¼ ¼ 1; since L 15t2 ¼ 3 2 0 s lim s GðsÞ 15tuðtÞ :

eramp ð1Þ ¼

s!0

b. First check stability. G ð sÞ 10s2 þ 500s þ 6000 T ðsÞ ¼ ¼ 5 1 þ GðsÞ s þ 110s4 þ 3875s3 þ 4:37e04s2 þ 500s þ 6000 10ðs þ 30Þðs þ 20Þ ¼ ðs þ 50:01Þðs þ 35Þðs þ 25Þðs2  7:189e  04s þ 0:1372Þ From the second-order term in the denominator, we see that the system is unstable. Instability could also be determined using the Routh-Hurwitz criteria on the denominator of T(s). Since the system is unstable, calculations about steady-state error cannot be made. 7.2 a. The system is stable, since GðsÞ 1000ðs þ 8Þ 1000ðs þ 8Þ ¼ ¼ 2 T ðsÞ ¼ 1 þ GðsÞ ðs þ 9Þðs þ 7Þ þ 1000ðs þ 8Þ s þ 1016s þ 8063 and is of Type 0. Therefore, K p ¼ lim GðsÞ ¼ s!0

1000 8 ¼ 127; K v ¼ lim sGðsÞ ¼ 0; s!0 7 9

and Ka ¼ lim s2 GðsÞ ¼ 0 s!0

b. estep ð1Þ ¼

1 1 ¼ ¼ 7:8e  03 1 þ lim GðsÞ 1 þ 127 s!0

eramp ð1Þ ¼

1 1 ¼ ¼1 lim sGðsÞ 0 s!0

eparabola ð1Þ ¼

1 1 ¼ ¼1 lim s2 GðsÞ 0 s!0

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7.3 System is stable for positive K. System is Type 0. Therefore, for a step input 1 12K estep ð1Þ ¼ ¼ 0:1. Solving for Kp yields K p ¼ 9 ¼ lim GðsÞ ¼  ; from s!0 1 þ Kp 14 18 which we obtain K ¼ 189. 7.4 ð s þ 2Þ , ð s þ 4Þ 1 ¼ ¼ 9:98e  04 2 þ 1000

System is stable. Since G1 ðsÞ ¼ 1000, and G2 ðsÞ ¼ e D ð 1Þ ¼ 

1 1 þ lim G1 ðsÞ lim s!0 G2 ðsÞ s!0

7.5 System is stable. Create a unity-feedback system, where H e ðsÞ ¼ The system is as follows: +

R(s) –

Ea(s)

100 s+4

1 s 1¼ sþ1 sþ1

C(s)

– −s s+1

Thus, 100 GðsÞ 100ðs þ 1Þ ð s þ 4Þ ¼ Ge ðsÞ ¼ ¼ 2 100s 1 þ GðSÞH e ðsÞ S  95s þ 4 1 ð s þ 1Þ ð s þ 4 Þ Hence, the system is Type 0. Evaluating Kp yields 100 ¼ 25 Kp ¼ 4 The steady-state error is given by estep ð1Þ ¼

1 1 ¼ 3:846e  02 ¼ 1 þ K p 1 þ 25

7.6 Since GðsÞ ¼

K ð s þ 7Þ 1 ; eð1Þ ¼ ¼ þ 2s þ 10 1 þ Kp

s2

1 10 . ¼ 7K 10 þ 7K 1þ 10

Calculating the sensitivity, we get Se:K ¼

K @e ¼ e @K

K ð10Þ7 7K  ¼ 2 10 10 þ 7K ð10 þ 7K Þ 10 þ 7K

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Chapter 8 Solutions to Skill-Assessment Exercises

7.7 Given

A¼

0

1

3

6

; B¼



0 1

1 ; C ¼ ½ 1 1 ; RðsÞ ¼ : s

Using the final value theorem,

2 " #1 " #3 h i 0 s 1 5 estep ð1Þ ¼ lim sRðsÞ 1  CðsI  AÞ1 B ¼ lim 41  ½ 1 1  s!0 s!0 1 3 sþ6 " # 2 3 sþ6 1 " #7 6 2 0 7 6 3s s 6 7 ¼ lim s þ 5s þ 2 ¼ 2 ¼ lim 61  ½ 1 1  2 2 s!0 4 s þ 6s þ 3 1 7 5 s!0 s þ 6s þ 3 3

Using input substitution, step ð1Þ

" 1

¼ 1 þ CA B ¼ 1  ½ 1 " ¼ 1 þ ½1 1

6 1 3

0

1

0

1

3

6

#1 " # 0 1

# " # 0

3

1

2

¼ 1 þ ½1

3 1  2 1 4 3 5 ¼ 3 0

CHAPTER 8 8.1 ð5 þ j9Þð3 þ j9Þ a. F ð7 þ j9Þ ¼ ð7 þ j9 þ 2Þð7 þ j9 þ 4Þ0:0339 ¼ ð7 þ j9Þð7 þ j9 þ 3Þð7 þ j9 þ 6Þ ð7 þ j9Þð4 þ j9Þð1 þ j9Þ ð66  j72Þ ¼ 0:0339  j0:0899 ¼ 0:096 < 110:7 ¼ ð944  j378Þ b. The arrangement of vectors is shown as follows: jw (–7+j9)

s-plane

M1

–7

X –6

M2

M3

–5

–4

M4

X –3

M5

–2

–1

X 0

s

21

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Solutions to Skill-Assessment Exercises

From the diagram, F ð7 þ j9Þ ¼

M2 M4 ð3 þ j9Þð5 þ j9Þ ð66  j72Þ ¼ ¼ M1 M3 M5 ð1 þ j9Þð4 þ j9Þð7 þ j9Þ ð944  j378Þ

¼ 0:0339  j0:0899 ¼ 0:096 <; 110:7 8.2 a. First draw the vectors. jw X

j3

j2 s-plane j1

–3

–2

–1

s

0

–j1

–j2

X

–j3

From the diagram, P

angles ¼ 180  tan1

    3 3  tan1 ¼ 180  108:43 þ 108:43 ¼ 180 : 1 1

is on the root locus. b. Since the angle is 180, the point pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 2 1 þ3 12 þ 32 P pole lengths ¼ 10 ¼ c. K ¼ 1 P zero lengths 8.3 First, find the asymptotes. sa ¼

P

P poles  zeros ð2  4  6Þ  ð0Þ ¼ ¼ 4 # poles  # zeros 30

ua ¼

ð2k þ 1Þp p 5p ¼ ; p; 3 3 3

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Chapter 8 Solutions to Skill-Assessment Exercises

Next draw root locus following the rules for sketching. 5 4 3 2 Imag Axis

E1SM

1 0 –1 –2 –3 –4 –5 –8

–7

–6

–5

–4

–3

–2

–1

0

1

2

3

Real Axis

8.4 a. jw

j3

X s-plane s

O –2

0

–j3

2

X

b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function. G ð sÞ K ð s þ 2Þ ¼ T ðsÞ ¼ 1 þ GðsÞ s2 þ ðK  4Þs þ ð2K þ 13Þ Using the denominator of T(s), make a Routh table. s2

1

2K þ 13

s1

K  40

0

0

2K þ 13

0

s

We get a row of zeros for K ¼ 4. From the s2 row with K ¼ 4; s2 þ 21 ¼ 0. From pffiffiffiffiffi which we evaluate the imaginary axis crossing at 21. c. From part (b), K ¼ 4. d. Searching for the minimum gain to the left of 2 on the real axis yields 7 at a gain of 18. Thus the break-in point is at 7.

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Solutions to Skill-Assessment Exercises

e. First, draw vectors to a point e close to the complex pole. jw s-plane j3

X

s –2

2

0

–j3

X

At the point e close to the complex pole, the angles must add up to zero. Hence, th st angle from zero –angle  from pole in 4 quadrant – angle from pole in 1 quadrant ¼ 3 180 , or tan1  90  u ¼ 180. Solving for the angle of departure, 4 u ¼ 233:1. 8.5 a. z = 0.5 X

jw j4 s-plane

–3

X

0

–j4

o

o

2

4

s

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Chapter 8 Solutions to Skill-Assessment Exercises

b. Search along the imaginary axis and find the 180 point at s ¼ j4:06. c. For the result in part (b), K ¼ 1. d. Searching between 2 and 4 on the real axis for the minimum gain yields the break-in at s ¼ 2:89. e. Searching along z ¼ 0:5 for the 180 point we find s ¼ 2:42 þ j4:18. f. For the result in part (e), K ¼ 0:108. g. Using the result from part (c) and the root locus, K < 1. 8.6 a. jw

z = 0.591 s-plane

X –6

X –4

X –2

s 0

b. Searching along the z ¼ 0:591 (10% overshoot) line for the 180 point yields 2:028 þ j2:768 with K ¼ 45:55. 4 4 p p ¼ ¼ 1:97 s; T p ¼ ¼ ¼ 1:13 s; vn T r ¼ 1:8346 from the c. T s ¼ jRej 2:028 jImj 2:768 rise-time chart and graph in Chapter 4. Since vn is the radial distance to the pole, pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi vn ¼ 2:0282 þ 2:7682 ¼ 3:431. Thus, T r ¼ 0:53 s; since the system is Type 0, K 45:55 Kp ¼   ¼ ¼ 0:949. Thus, 2 46 48 estep ð1Þ ¼

1 ¼ 0:51: 1 þ Kp

d. Searching the real axis to the left of 6 for the point whose gain is 45.55, we find 7:94. Comparing this value to the real part of the dominant pole, 2:028, we find that it is not five times further. The second-order approximation is not valid. 8.7 Find the closed-loop transfer function and put it the form that yields pi as the root locus variable. Thus, 100 2 G ð sÞ 100 100 ¼ ¼ s þ 100 ¼ T ðsÞ ¼ 1 þ GðsÞ s2 þ pi s þ 100 ðs2 þ 100Þ þ pi s 1 þ pi s s2 þ 100

25

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Hence, KGðsÞH ðsÞ ¼

pi s . The following shows the root locus. s2 þ 100 jw s-plane

X j10

s

O 0

X –j10

8.8 Following the rules for plotting the root locus of positive-feedback systems, we obtain the following root locus: jw s-plane

o –4

X –3

X –2

X –1

0

s

8.9 The closed-loop transfer function is T ðsÞ ¼

K ð s þ 1Þ . Differentiating the s2 þ ðK þ 2Þs þ K

denominator with respect to K yields 2s

@s @s @s þ ð K þ 2Þ þ ðs þ 1Þ ¼ ð2s þ K þ 2Þ þ ð s þ 1Þ ¼ 0 @K @K @K

@s @s ðs þ 1Þ K @s K ðs þ 1Þ , we get ¼ . Thus, Ss:K ¼ ¼ : @K @K ð2s þ K þ 2Þ s @K sð2s þ K þ 2Þ 10ðs þ 1Þ Substituting K ¼ 20 yields Ss:K ¼ . sðs þ 11Þ Now find the closed-loop poles when K ¼ 20. From the denominator of T ðsÞ; s1;2 ¼ 21:05;  0:95, when K ¼ 20. For the pole at 21:05,   DK 10ð21:05 þ 1Þ Ds ¼ sðSs:K Þ ¼ 12:05 0:05 ¼ 0:9975: K 21:05ð21:05 þ 11Þ Solving for

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Chapter 9 Solutions to Skill-Assessment Exercises

For the pole at 0:95, Ds ¼ sðSs:K Þ

  DK 10ð0:95 þ 1Þ ¼ 0:95 0:05 ¼ 0:0025: K 0:95ð0:95 þ 11Þ CHAPTER 9

9.1 a. Searching along the 15% overshoot line, we find the point on the root locus at 3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system, Kv ¼ lim sGðsÞ ¼ K=7 ¼ 45:84=7 ¼ 6:55. s!0

Hence, eramp uncompensated ð1Þ ¼ 1=Kv ¼ 0:1527. b. Compensator zero should be 20x further to the left than the compensator pole. ðs þ 0:2Þ . Arbitrarily select Gc ðsÞ ¼ ðs þ 0:01Þ c. Insert compensator and search along the 15% overshoot line and find the root locus at 3:4 þ j5:63 with a gain, K ¼ 44:64. Thus, for the compensated system, 44:64ð0:2Þ 1 ¼ 127:5 and eramp compensated ð1Þ ¼ ¼ 0:0078. Kv ¼ ð7Þð0:01Þ Kv eramp uncompensated 0:1527 d. ¼ ¼ 19:58 eramp compensated 0:0078 9.2 a. Searching along the 15% overshoot line, we find the point on the root locus at 3:5 þ j5:8 at a gain of K ¼ 45:84. Thus, for the uncompensated system, Ts ¼

4 4 ¼ ¼ 1:143 s: jRej 3:5

b. The real part of the design point must be three times larger than the uncompensated pole’s real part. Thus the design point is 3ð3:5Þþ j 3ð5:8Þ ¼ 10:5 þ j 17:4. The angular contribution of the plant’s poles and compensator zero at the design point is 130:8. Thus, the compensator pole must contribute 180  130:8 ¼ 49:2 . Using the following diagram, jw

j17.4 s-plane 49.2° –pc

s –10.5

17:4 ¼ tan 49:2 , from which, pc ¼ 25:52. Adding this pole, we find Pc  10:5 the gain at the design point to be K ¼ 476:3. A higher-order closed-loop pole is found to be at 11:54. This pole may not be close enough to the closed-loop zero at 10. Thus, we should simulate the system to be sure the design requirements have been met.

we find

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Solutions to Skill-Assessment Exercises

9.3 a. Searching along the 20% overshoot line, we find the point on the root locus at 3:5 þ 6:83 at a gain of K ¼ 58:9. Thus, for the uncompensated system, Ts ¼

4 4 ¼ ¼ 1:143 s: jRej 3:5

b. For the uncompensated system, K v ¼ lim sGðsÞ ¼ K=7 ¼ 58:9=7 ¼ 8:41. Hence, s!0

eramp uncompensated ð1Þ ¼ 1=Kv ¼ 0:1189. c. In order to decrease the settling time by a factor of 2, the design point is twice the uncompensated value, or 7 þ j 13:66. Adding the angles from the plant’s poles and the compensator’s zero at 3 to the design point, we obtain 100:8 . Thus, the compensator pole must contribute 180  100:8 ¼ 79:2. Using the following diagram, jw j13.66

s-plane 79.2° –pc

s

–7

13:66 ¼ tan79:2 , from which, pc ¼ 9:61. Adding this pole, we find the Pc  7 gain at the design point to be K ¼ 204:9. Evaluating Kv for the lead-compensated system: we find

K v ¼ lim sGðsÞGlead ¼ K ð3Þ=½ð7Þð9:61Þ ¼ ð204:9Þð3Þ=½ð7Þð9:61Þ ¼ 9:138: s!0

K v for the uncompensated system was 8.41. For a 10x improvement in steady-state error, Kv must be ð8:41Þð10Þ ¼ 84:1. Since lead compensation gave us K v ¼ 9:138, we need an improvement of 84:1=9:138 ¼ 9:2. Thus, the lag compensator zero should be 9.2x further to the left than the compensator pole. Arbitrarily select ðs þ 0:092Þ . G c ðsÞ ¼ ðs þ 0:01Þ Using all plant and compensator poles, we find the gain at the design point to be K ¼ 205:4. Summarizing the forward path with plant, compensator, and gain yields G e ðsÞ ¼

205:4ðs þ 3Þðs þ 0:092Þ : sðs þ 7Þð9:61Þðs þ 0:01Þ

Higher-order poles are found at 0:928 and 2:6. It would be advisable to simulate the system to see if there is indeed pole-zero cancellation.

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Chapter 9 Solutions to Skill-Assessment Exercises

9.4 The configuration for the system is shown in the figure below.

R(s)

+

+

K

1 s(s+ 7)(s +10)

–

–

C(s)

Kf s

Minor-Loop Design: Kf . Using the following diagram, we ðs þ 7Þðs þ 10Þ find that the minor-loop root locus intersects the 0.7 damping ratio line at 8:5 þ j 8:67. The imaginary part was found as follows: u ¼ cos1 z ¼ 45:57 . Hence, Im ¼ tan45:57 , from which Im ¼ 8:67. 8:5 For the minor loop, GðsÞH ðsÞ ¼

jw

z = 0.7 (-8.5 + j8.67) Im

X −10

−8.5

s-plane X –7

q

s

The gain, Kf , is found from the vector lengths as Kf ¼

pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffipffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1:52 þ 8:672 1:52 þ 8:672 ¼ 77:42

Major-Loop Design: Using the closed-loop poles of the minor loop, we have an equivalent forward-path transfer function of Ge ðsÞ ¼

K K : ¼ sðs þ 8:5 þ j8:67Þðs þ 8:5  j8:67Þ sðs2 þ 17s þ 147:4Þ

Using the three poles of Ge ðsÞ as open-loop poles to plot a root locus, we search along z ¼ 0:5 and find that the root locus intersects this damping ratio line at 4:34 þ j7:51 at a gain, K ¼ 626:3.

29

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Solutions to Skill-Assessment Exercises

9.5 a. An active PID controller must be used. We use the circuit shown in the following figure: Z2(s)

Vi (s)

I2(s)

Z1(s)

V1(s)

–

Vo(s)

Ia(s)

I1(s)

+

where the impedances are shown below as follows: C1 C2

R2 R1 Z1(s)

Z2(s)

Matching the given transfer function with the transfer function of the PID controller yields ðs þ 0:1Þðs þ 5Þ s2 þ 5:1s þ 0:5 0:5 ¼ ¼ s þ 5:1 þ s s 8 2 3 1   6 R2 C 1 R C 7 6 þ R2 C 1 s þ 1 2 7 þ ¼ 4 R1 C 2 s 5

G c ðsÞ ¼

Equating coefficients 1 ¼ 0:5 R1 C 2

ð1Þ

R2 C 1 ¼ 1   R2 C 1 ¼ 5:1 þ R1 C 2

ð2Þ ð3Þ

In Eq. (2) we arbitrarily let C1 ¼ 105 . Thus, R2 ¼ 105 . Using these values along with Eqs. (1) and (3) we find C2 ¼ 100mF and R1 ¼ 20 kV. b. The lag-lead compensator can be implemented with the following passive network, since the ratio of the lead pole-to-zero is the inverse of the ratio of the lag pole-to-zero: R1 + vi (t)

+ C1

R2 vo (t) C2

–

–

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Chapter 10 Solutions to Skill-Assessment Exercises

Matching the given transfer function with the transfer function of the passive lag-lead compensator yields ðs þ 0:1Þðs þ 2Þ ðs þ 0:1Þðs þ 2Þ ¼ G c ðsÞ ¼ ðs þ 0:01Þðs þ 20Þ s2 þ 20:01s þ 0:2    1 1 sþ sþ R1 C 1 R2 C2   ¼ 1 1 1 1 2 s þ sþ þ þ R1 C 1 R2 C 2 R2 C 1 R1 R2 C1 C2 Equating coefficients



1 ¼ 0:1 R1 C 1 1 ¼ 0:1 R2 C 2

1 1 1 þ þ R1 C1 R2 C2 R2 C1

ð1Þ ð2Þ  ¼ 20:01

ð3Þ

Substituting Eqs. (1) and (2) in Eq. (3) yields 1 ¼ 17:91 R2 C 1

ð4Þ

Arbitrarily letting C1 ¼ 100 mF in Eq. (1) yields R1 ¼ 100 kV. Substituting C1 ¼ 100 mF into Eq. (4) yields R2 ¼ 558 kV. Substituting R2 ¼ 558 kV into Eq. (2) yields C2 ¼ 900 mF. CHAPTER 10 10.1 a.

1 1 ; GðjvÞ ¼ 2 ðs þ 2Þðs þ 4Þ ð8 þ v Þ þ j6v qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 MðvÞ ¼ ð8  v2 Þ þ ð6vÞ2   6v 1 : fðvÞ ¼ tan 2 8v 

 6v : fðvÞ ¼  p þ tan1 8  v2 GðsÞ ¼

For v <

pffiffiffi 8,

For v <

pffiffiffi 8,

b. Bode Diagrams 0 –20 Phase (deg); Magnitude (dB)

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–40 –60 –80 –100 0 –50 –100 –150 –200 10–1

100 101 Frequency (rad/sec)

102

31

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Solutions to Skill-Assessment Exercises

c. Nyquist Diagrams 0.08 0.06 0.04 0.02

Imaginary Axis

32

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0

–0.02 –0.04 –0.06 –0.08

–0.05

0

0.05

0.1

0.15

0.2

Real Axis

10.2 Asymptotic –20 dB/dec –40

–40 dB/dec Actual

–60

–20 dB/dec

–80

–40 dB/dec

–100 –120

1

0.1

1000

100

10 Frequency (rad/s)

–45o/dec Phase (degrees)

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20 log M

E1SM

–50

–90o/dec –45o/dec –90o/dec

–100

–45o/dec –150

–45o/dec

Asymptotic –200

0.1

1

10 Frequency (rad/s)

Actual

100

1000

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Chapter 10 Solutions to Skill-Assessment Exercises

10.3 The frequency response is 1/8 at an angle of zero degrees at v ¼ 0. Each pole rotates 90 in going from v ¼ 0 to v ¼ 1. Thus, the resultant rotates 180 while its magnitude goes to zero. The result is shown below. Im

w=∞ 0

w=0

1 8

Re

10.4 a. The frequency response is 1/48 at an angle of zero degrees at v ¼ 0. Each pole rotates 90 in going from v ¼ 0 to v ¼ 1. Thus, the resultant rotates 270 while its magnitude goes to zero. The result is shown below. Im

w = 6.63 w = ∞ – 1 480

0

w=0 1 48

Re

1 1 and ¼ 3 2 ðs þ 2Þðs þ2 4Þðs þ 6Þ s 3 þ 12s þ 44s þ 48 48  12v  j 44v  v simplifying, we obtain GðjvÞ ¼ 6 . The Nyquist diagram v þ 56v4 þ 784v2 þ 2304 crosses the real axis when the imaginary part of GðjvÞ is zero. Thus, the Nyquist pffiffiffiffiffi diagram crosses the real axis at v2 ¼ 44; or v ¼ 44 ¼ 6:63 rad=s. At this fre1 quency GðjvÞ ¼  . Thus, the system is stable for K < 480. 480

b. Substituting jv into GðsÞ ¼

33

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Solutions to Skill-Assessment Exercises

10.5 If K ¼ 100, the Nyquist diagram will intersect the real axis at 100=480. Thus, 480 GM ¼ 20 log ¼ 13:62 dB. From Skill-Assessment Exercise Solution 10.4, the 100  180 frequency is 6.63 rad/s. 10.6 a. –60 –80 –100 20 log M

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10

100

1000

100

1000

Frequency (rad/s)

0 –50 Phase (degrees)

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–100 –150 –200 –250 –300

1

10 Frequency (rad/s)

b. The phase angle is 180 at a frequency of 36.74 rad/s. At this frequency the gain is 99:67 dB. Therefore, 20 logK ¼ 99:67, or K ¼ 96; 270. We conclude that the system is stable for K < 96; 270. c. For K ¼ 10; 000, the magnitude plot is moved up 20log10; 000 ¼ 80 dB. Therefore, the gain margin is 99:67  80 ¼ 19:67 dB. The 180 frequency is 36.7 rad/s. The gain curve crosses 0 dB at v ¼ 7:74 rad=s, where the phase is 87:1. We calculate the phase margin to be 180  87:1 ¼ 92:9. 10.7 lnð%=100Þ Using z ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi, we find z ¼ 0:456, which corresponds to 20% overp2 þ ln2 ð%100Þ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 2 4 2 shoot. Using T s ¼ 2; vBW ¼ ð1  2z Þ þ 4z  4z þ 2 ¼ 5:79 rad=s. T sz

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Chapter 10 Solutions to Skill-Assessment Exercises

10.8

    160 6750000  101250v2 þ j1350 v2  1350 v .  For both parts find that GðjvÞ ¼ 27 v6 þ 2925v4 þ 1072500v2 þ 25000000 For a range of values for v, superimpose GðjvÞ on the a. M and N circles, and on the b. Nichols chart. a. Im 3 G-plane Φ = 20°

2

M = 1.3

25°

M = 1.0

30° 1.4

40° 1.5

1

M = 0.7

50°

1.6

70°

1.8

0.6

2.0 0.4

0.5

Re

–70° –1 –50° –40° –30° –25° –2

–20°

–3 –3

–4

–2

1

–1

2

b.

Nichols Charts 0 dB

0.25 dB 0.5 dB 1 dB 3 dB 6 dB

0

Open-Loop Gain (dB)

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–1 dB –3 dB –6 dB –12 dB –20 dB –40 dB

–50

–60 dB –80 dB –100 dB

–100

–120 dB –140 dB

–150

–160 dB –180 dB

–200

–200 dB –220 dB –350

–300

–250

–200 –150 Open-Loop Phase (deg)

–100

–50

0

–240 dB

35

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Solutions to Skill-Assessment Exercises

Plotting the closed-loop frequency response from a. or b. yields the following plot: 0 –20 –40

20 log M

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–60 –80 –100 –120 1

10

100

1000

Frequency (rad/s) 0 –50 Phase (degrees)

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1

10

100

1000

Frequency (rad/s)

10.9 The open-loop frequency response is shown in the following figure:

Bode Diagrams 40 20 0 Phase (deg); Magnitude (dB)

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–20 –40

–100 –120 –140 –160

10–1

100

101 Frequency (rad/sec)

102

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Chapter 10 Solutions to Skill-Assessment Exercises

The open-loop frequency response is 7 at v ¼ 14:5 rad=s. Thus, the estimated bandwidth is vWB ¼ 14:5 rad=s. The open-loop frequency response plot goes through zero dB at a frequency of 9.4 rad/s, where the phase is 151:98. Hence, the phase margin is 180  151:98 ¼ 28:02. This phase margin corresponds to z ¼ 0:25: Therefore; %OS ¼ e

Ts ¼

Tp ¼

4 vBW z



pffiffiffiffiffiffiffi2ffi

zp=

1z

 100 ¼ 44:4%

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ð1  2z Þ þ 4z4  4z2 þ 2 ¼ 1:64 s and

p pffiffiffiffiffiffiffiffiffiffiffiffiffi vBW 1  z2

qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  2z2 Þ þ 4z4  4z2 þ 2 ¼ 0:33 s

10.10 The initial slope is 40 dB/dec. Therefore, the system is Type 2. The initial slope intersects 0 dB at v ¼ 9:5 rad=s. Thus, K a ¼ 9:52 ¼ 90:25 and K p ¼ K v ¼ 1. 10.11 10 10 ¼ , from which the zero dB frejvðjv þ 1Þ vðv þ jÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi 10 quency is found as follows: M ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1. Solving for v; v v2 þ 1 ¼ 10, or v v2 þ 1 after squaring both sides and rearranging, v4 þ v2  100 ¼ 0. Solving for the roots, v2 ¼ 10:51; 9:51. Taking the square root of the positive root, we find the 0 dB frequency to be 3.08 rad/s. At this frequency, the phase angle, f ¼ ffðv þ jÞ ¼ ffð3:08 þ jÞ ¼ 162 . Therefore the phase margin is 180 162 ¼ 18 . b. With a delay of 0.1 s, a. Without delay, GðjvÞ ¼

f ¼ ffðv þ jÞ  vT ¼ ffð3:08 þ jÞ  ð3:08Þð0:1Þð180=piÞ ¼ 162  17:65 ¼ 179:65 Therefore the phase margin is 180  179:65 ¼ 0:35. Thus, the system is table. c. With a delay of 3 s, f ¼ ffðv þ jÞ  vT ¼ ffð3:08 þ jÞ  ð3:08Þð3Þð180=piÞ ¼ 162  529:41 ¼  691:41 ¼ 28:59 deg: Therefore the phase margin is 28:59  180 ¼ 151:41 deg. Thus, the system is unstable. 10.12 Drawing judicially selected slopes on the magnitude and phase plot as shown below yields a first estimate.

37

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Solutions to Skill-Assessment Exercises Experimental 20

Gain(dB)

10 0 –10 –20 –30 –40

Phase(deg)

38

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–45 –50 –55 –60 –65 –70 –75 –80 –85 –90 –95 1

2

3

4 5 6 7 8 10

20 30 40 50 70 100

200 300

500

1000

Frequency (rad/sec)

We see an initial slope on the magnitude plot of 20 dB/dec. We also see a final 20 dB/dec slope with a break frequency around 21 rad/s. Thus, an initial estimate is 1 . Subtracting G1 ðsÞ from the original frequency response yields the G 1 ðsÞ ¼ sðs þ 21Þ frequency response shown below. Experimental Minus 1/s(s+21) 90 80 Gain(dB)

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70 60 50 40 100 80

Phase(deg)

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60 40 20 0 1

2

3

4 5 6 7 8 10

20

30 40 50

Frequency (rad/sec)

70 100

200 300

500

1000

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Chapter 11 Solutions to Skill-Assessment Exercises

Drawing judicially selected slopes on the magnitude and phase plot as shown yields a final estimate. We see first-order zero behavior on the magnitude and phase plots with a break frequency of about 5.7 rad/s and a dc gain of about 44 dB ¼ 20logð5:7K Þ, or K ¼ 27:8. Thus, we estimate G2 ðsÞ ¼ 27:8ðs þ 7Þ. Thus, 27:8ðs þ 5:7Þ . It is interesting to note that the original problem GðsÞ ¼ G1 ðsÞG2 ðsÞ ¼ sðs þ 21Þ 30ðs þ 5Þ was developed from GðsÞ ¼ . sðs þ 20Þ CHAPTER 11 11.1 The Bode plot for K ¼ 1 is shown below. Bode Diagrams –60 –80 –100 –120

Phase (deg); Magnitude (dB)

E1SM

–140 –160 –180

–100

–150

–200

–250 10–1

100

101 Frequency (rad/sec)

102

103

  % log 100 A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi   ¼ 0:456. This damping ratio implies a % p2 þ log2 100 phase margin of 48.10, which is obtained when the phase angle ¼ 1800 þ 48:10 ¼ 131:9 . This phase angle occurs at v ¼ 27:6 rad=s. The magnitude at this frequency 1 is 5:15  106. Since the magnitude must be unity K ¼ ¼ 194; 200. 5:15  106 11.2 To meet the steady-state error requirement, K ¼ 1; 942; 000. The Bode plot for this gain is shown below.

39

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Solutions to Skill-Assessment Exercises Bode Diagrams 60 40 20 0 Phase (deg); Magnitude (dB)

E1SM

–20 –40

–100

–150

–200

–250 10–1

100

101

102

103

Frequency (rad/sec)



 % log 100 A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:456. This damping ratio % 2 p2 þ log 100   implies a phase margin of 48:1 . Adding 10 to compensate for the phase angle contribution of the lag, we use 58:1. Thus, we look for a phase angle of 180 þ 58:1 ¼ 129:9 . The frequency at which this phase occurs is 20.4 rad/s. At this frequency the magnitude plot must go through zero dB. Presently, the magnitude plot is 23.2 dB. Therefore draw the high frequency asymptote of the lag compensator at 23:2 dB. Insert a break at 0:1ð20:4Þ ¼ 2:04 rad=s. At this frequency, draw 23:2 dB/dec slope until it intersects 0 dB. The frequency of intersection will be the low frequency break or 0.141 rad/s. Hence the compensator is ðs þ 2:04Þ , where the gain is chosen to yield 0 dB at low frequencies, G c ðsÞ ¼ K c ðs þ 0:141Þ or K c ¼ 0:141=2:04 ¼ 0:0691. In summary, Gc ðsÞ ¼ 0:0691 11.3

ðs þ 2:04Þ 1;942;000 and GðsÞ ¼ ðs þ 0:141Þ sðs þ 50Þðs þ 120Þ 

 % log 100 A 20% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:456. The required bandwidth % 2 p2 þ log 100 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 ð1  2z2 Þ þ 4z4  4z2 þ 2 ¼ 57:9 rad/s. In order is then calculated as vBW ¼ T sz K to meet the steady-state error requirement of K v ¼ 50 ¼ , we calculate ð50Þð120Þ K ¼ 300; 000. The uncompensated Bode plot for this gain is shown below.

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Chapter 11 Solutions to Skill-Assessment Exercises Bode Plot for K = 300000 40 20 0 –20 Phase (deg); Magnitude (dB)

E1SM

–40 –60

–100

–150

–200

–250 10–1

100

101

102

103

Frequency (rad/sec)

The uncompensated system’s phase margin measurement is taken where the magnitude plot crosses 0 dB. We find that when the magnitude plot crosses 0 dB, the phase angle is 144:8. Therefore, the uncompensated system’s phase margin is 180 þ 144:8 ¼ 35:2 . The required phase margin based on the required damping 2z   ratio is FM ¼ tan1 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 48:1 . Adding a 10 correction factor, the 2 4 2z þ 1 þ 4z required phase margin is 58:1. Hence, the compensator must contribute 1b 1  sinfmax fmax ¼ 58:1  35:2 ¼ 22:9 . Using fmax ¼ sin1 ¼ 0:44. , b¼ 1 þ sinfmax 1þb 1 The compensator’s peak magnitude is calculated as Mmax ¼ pffiffiffi ¼ 1:51. Now find b the frequency at which the uncompensated system has a magnitude 1=Mmax, or 3:58 dB. From the Bode plot, this magnitude occurs at vmax ¼ 50 rad=s. The 1 1 compensator’s zero is at zc ¼ . vmax ¼ pffiffiffi Therefore, zc ¼ 33:2. T T b 1 zc The compensator’s pole is at Pc ¼ ¼ ¼ 75:4. The compensator gain is b bT chosen to yield unity gain at dc. ðs þ 33:2Þ Hence, K c ¼ 75:4=33:2 ¼ 2:27. Summarizing, Gc ðsÞ ¼ 2:27 , and ðs þ 75:4Þ 300;000 GðsÞ . sðs þ 50Þðs þ 120Þ

41

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11.4

 % log 100 A 10% overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:591. The required bandwidth % 2 p2 þ log 100 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi p is then calculated as vBW ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ð1  2z2 Þ þ 4z4  4z2 þ 2 ¼ 7:53 rad=s: T p 1  z2 K In order to meet the steady-state error requirement of K v ¼ 10 ¼ , we ð8Þð30Þ calculate K ¼ 2400. The uncompensated Bode plot for this gain is shown below.



Bode Diagrams 40 20 0 –20 –40 Phase (deg); Magnitude (dB)

E1SM

–60 –80 –100

–100

–150

–200

–250 10–1

100

101 Frequency (rad/sec)

102

103

Let us select a new phase-margin frequency at 0:8vBW ¼ 6:02 rad=s. The required phase margin based on the required damping ratio is FM ¼ tan1 2z   qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 58:6 . Adding a 5 correction factor, the required phase 2z2 þ 1 þ 4z4 margin is 63:6. At 6.02 rad/s, the new phase-margin frequency, the phase angle is–which represents a phase margin of 180  138:3 ¼ 41:7. Thus, the lead compensator must contribute fmax ¼ 63:6  41:7 ¼ 21:9. 1b 1  sinfmax ,b¼ Using fmax ¼ sin1 ¼ 0:456. 1 þ sinfmax 1þb We now design the lag compensator by first choosing its higher break frequency one decade below the new phase-margin frequency, that is, zlag ¼ 0:602 rad=s. The lag compensator’s pole is plag ¼ bzlag ¼ 0:275. Finally, the lag compensator’s gain is K lag ¼ b ¼ 0:456.

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Chapter 12 Solutions to Skill-Assessment Exercises

Now we design the lead compensator. The lead zero is the product of the new pffiffiffi pffiffiffi zlead phase margin frequency and b, or zlead ¼ 0:8vBW b ¼ 4:07. Also, plead ¼ b 1 ¼ 8:93. Finally, K lead ¼ ¼ 2:19. Summarizing, b Glag ¼ ðsÞ ¼ 0:456

ðs þ 0:602Þ ðs þ 4:07Þ ; Glead ðsÞ ¼ 2:19 ; ðs þ 0:275Þ ðs þ 8:93Þ

and k ¼ 2400:

CHAPTER 12 12.1 We first find  the  desired characteristic equation. A 5% overshoot requires % log p 100 ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi s z¼  ffi ¼ 0:69. Also, vn ¼ pffiffiffiffiffiffiffiffiffiffiffiffi2ffi ¼ 14:47 rad=s. Thus, the charTp 1  z % p2 þ log2 100 acteristic equation is s2 þ 2zvn s þ v2n ¼ s2 þ 19:97s þ 209:4. Adding a pole at 10 to the zero  at 10 yields the desired characteristic equation, cancel s2 þ 19:97s þ 209:4 ðs þ 10Þ ¼ s3 þ 29:97s2 þ 409:1s þ 2094. The compensated system matrix in phase-variable form is 2 6 A  BK ¼ 4

0

1

0

0

0

1

3 7 5. The characteristic equation for this

ðk1 Þ ð36 þ k2 Þ ð15 þ k3 Þ system is jsI  ðA  BKÞj ¼ s3 þ ð15 þ k3 Þs2 þ ð36 þ k2 Þs þ ðk1 Þ. Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as K ¼ ½ k1

k2

k3  ¼ ½ 2094 373:1

12.2 The controllability matrix is CM ¼ B

14:97 : 2

AB

2

6 2 A B ¼ 41

1 4

3 1 7 9 5. Since

1 1 16 jCM j ¼ 80, CM is full rank, that is, rank 3. We conclude that the system is controllable. 12.3



First check controllability. The controllability matrix is CMz ¼ B AB A2 B ¼ 3 2 0 0 1 7 6 1 17 5. Since jCMz j ¼ 1, CMz is full rank, that is, rank 3. We conclude that 40 1 9 81 the system is controllable. We now  find  the desired characteristic equation. A 20% % log 4 100 overshoot requires z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:456. Also, vn ¼ zT ¼ 4:386 rad=s. s % p2 þ log2 100

43

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Solutions to Skill-Assessment Exercises

Thus, the characteristic equation is s2 þ 2zvn s þ v2n ¼ s2 þ 4s þ 19:24. Adding a pole at 6 to cancel the zero at 6 yields the resulting desired characteristic equation, 2  s þ 4s þ 19:24 ðs þ 6Þ ¼ s3 þ 10s2 þ 43:24s þ 115:45: ð s þ 6Þ sþ6 , we can write the phase¼ 3 2 ðs þ 7Þðs þ 8Þðs þ 9Þ s þ 24s 2 þ 191s þ 504 3 2 3 0 1 0 0 6 7 6 7 0 1 5 ; Bp ¼ 4 0 5; Cp ¼ variable representation as Ap ¼ 4 0 504 191 24 1 ½ 6 1 0 . The compensated system matrix in phase-variable form is Ap  Bp Kp ¼ 2 3 0 1 0 6 7 0 0 1 4 5. The characteristic equation for this Since GðsÞ ¼

ð504 þ k1 Þ ð191 þ k2 Þ ð24 þ k3 Þ   system is jsI  Ap  Bp Kp j ¼ s3 þ ð24 þ k3 Þs2 þ ð191 þ k2 Þs þ ð504 þ k1 Þ. Equating coefficients of this equation with the coefficients of the desired characteristic equation yields the gains as Kp ¼ ½ k1 k2 k3  ¼ ½ 388:55 147:76 14 . We now develop the transformation matrix to transform back to the z-system. 2 3 0 0 1 h i 6 7 CMz ¼ Bz Az Bz A2z B z ¼ 4 0 1 17 5 and 9

1 h

CMp ¼ Bp

Ap Bp

A2p Bp

i

2

0 6 ¼ 40

81 3 1 7 24 5:

0 1

1 24

385

Therefore, 2

P ¼ CMz C1 Mx

0 6 ¼ 40

0 1

1 9

32 1 191 24 76 17 54 24 1 81

1

0

3 2 1 1 7 6 05 ¼ 4 7

0 1

3 0 7 05

0

15

1

56 2

Hence, Kz ¼ Kp P1 ¼ ½ 388:55 147:76

1

0

6 14 4 7

12.4

62:24

15 1

14 . 2

ð24 þ l1 Þ

3

7 05

1

49 ¼ ½ 40:23

0

1

0

3

7 6 For the given system ex_ ¼ ðA  LCÞex ¼ 4 ð191 þ l2 Þ 0 1 5ex . The characteristic ð504 þ l3 Þ 0 0 polynomial is given by j½sI  ðA  LCÞj ¼ s3 þ ð24 þ l1 Þs2 þ ð191 þ l2 Þs þ ð504 þ l3 Þ. Now we find the desired characteristic  equation. The  dominant poles from Skill-Assessment Exercise 12.3 come from s2 þ 4s þ 19:24 . Factoring yields ð2 þ j3:9Þ and ð2  j3:9Þ. Increasing these poles by a factor of 10 and adding a third pole 10 times the real part of the dominant second-order poles yields the

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Chapter 12 Solutions to Skill-Assessment Exercises

desired characteristic polynomial, ðs þ 20 þ j39Þðs þ 20  j39Þðs þ 200Þ ¼ s3 þ 240s2 þ 9921s þ 384200. Equating coefficients of the 2 desired3characteristic equation to the 216 6 7 system’s characteristic equation yields L ¼ 4 9730 5. 383696 12.5 The

2 observability 2

25

28

matrix 32

is

3

C

3

2

4

6 7 6 OM ¼ 4 CA 5 ¼ 4 64 CA2 674

6

8

3

7 80 78 5, 848 814

where

7 6 A2 ¼ 4 7 4 11 5. The matrix is of full rank, that is, rank 3, since 77 95 94 jOM j ¼ 1576. Therefore the system is observable. 12.6 The system is represented in cascade form by the following state and output equations: 2 3 2 3 7 1 0 0 6 7 6 7 z_ ¼ 4 0 8 1 5z þ 4 0 5u 0 0 9 1

The observability matrix is OMz 2

49

6 where A2z ¼ 4 0 0 ¼

s3

þ

24s2

15 64 0

1

y ¼ ½ 1 0 0 z 2 3 2 Cz 1 6 7 6 ¼ 4 Cz Az 5 ¼ 4 7 Cz A2z 49

0 1

3 0 7 0 5,

15 1

3

7 17 5. Since GðsÞ ¼ 81

1 ð s þ 7Þ ð s þ 8 Þ ð s þ 9Þ

1 , we can write the observable canonical form as þ 191s þ 504 2 3 2 3 24 1 0 0 6 7 6 7 x_ ¼ 4 191 0 1 5x þ 4 0 5u 504 0 0 1 y ¼ ½1

0

0 x 2

The observability matrix for this form is OMx

3

2

1 0 6 7 6 ¼ 4 Cx Ax 5 ¼ 4 24 1 2 Cx Ax 385 24 Cx

3 0 7 0 5, 1

45

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Solutions to Skill-Assessment Exercises

where 2

24

385

6 A2x ¼ 4 4080 191 12096 504

1

3

7 0 5: 0

We next find  the  desired characteristic equation. A 10% overshoot requires % log 4 100 z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:591. Also, vn ¼ zT ¼ 67:66 rad=s. Thus, the characteristic s % p2 þ log2 100 equation is s2 þ 2zvn s þ v2n ¼ s2 þ 80s þ 4578:42. Adding a pole at 400, or 10 times the real part of the dominant second-order poles, yields the resulting desired character  istic equation, s2 þ 80s þ 4578:42 ðs þ 400Þ ¼ s3 þ 480s2 þ 36580s þ 1:831x106 . For the in observable canonical form ex_ ¼ ðAx  Lx Cx Þ ex ¼ 2 system represented 3 ð24 þ l1 Þ 1 0 6 7 The characteristic polynomial is given by 4 ð191 þ l2 Þ 0 1 5ex . ð504 þ l3 Þ 0 0 j½sI  ðAx  Lx Cx Þj ¼ s3 þ ð24 þ l1 Þs2 þ ð191 þ l2 Þs þ ð504 þ l3 Þ. Equating coefficients of the2 desired characteristic equation to the system’s characteristic equation 3 456 7 6 yields Lx ¼ 4 36; 389 5. 1; 830; 496 Now, develop the transformation matrix between the observer canonical and cascade forms. 2

1

6 6 P ¼ O1 O ¼ 6 7 Mz Mx 4 49 2 1 6 6 ¼6 7 4 56 2 1 6 6 ¼ 6 17 4 81

0

0

31 2

1

0

0

3

7 6 7 7 6 7 0 7 6 24 1 07 5 4 5 15 1 385 24 1 32 3 0 0 1 0 0 76 7 76 7 1 0 76 24 1 07 54 5 15 1 385 24 1 3 0 0 7 7 1 07 5 9 1 1

Finally, 2

1 6 Lz ¼ PLx ¼ 4 17 81

0 1 9

32 3 2 3 2 3 0 456 456 456 76 7 6 7 6 7 0 54 36; 389 5 ¼ 4 28; 637 5  4 28; 640 5. 1 1; 830; 496 1; 539; 931 1; 540; 000

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Chapter 12 Solutions to Skill-Assessment Exercises

12.7 We first find the desired characteristic equation. A 10% overshoot requires   % log 100 z ¼ sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  ffi ¼ 0:591 % p2 þ log2 100 p pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1:948 rad=s. Thus, the characteristic equation is s2 þ T p 1  z2 2 2zvn s þ vn ¼ s2 þ 2:3s þ 3:79. Adding a pole at 4, which corresponds to the system’szero location, yields the resulting desired characteristic equation, original 2 s þ 2:3s þ 3:79 ðs þ 4Þ ¼ s3 þ 6:3s2 þ 13s þ 15:16.



ðA  BKÞ BK e 0 x_ x x ¼ þ , Now, r; and y ¼ ½ C 0  _xN C 0 1 xN xN Also, vn ¼

where A  BK ¼



0 1  ½ k1 1 9

0 7



0 ¼  ð 7 þ k1 Þ

k2  ¼

1 ð9 þ k2 Þ

0 7

0 1  k1 9

0 k2





C ¼ ½4 1



0 0 ke ¼ Bke ¼ 1 ke Thus, 3 2 2 0 x_1 7 6 6 4 x_2 5 ¼ 4 ð7 þ k1 Þ 4 x_N

32

0

 ð9 þ k 2 Þ 1

Finding the characteristic equation of
2





s


sI  ðA  BKÞ BK e
¼

6 40
C 0


0
2
s

6 ¼
4 ð7 þ k1 Þ

4

3



0 76 7 r; y ¼ ½ 4 ke 5 4 x 2 5 þ 1 0 xN

1

x1

this system yields 3 2 0 0 0 7 6 s 0 5  4 ð7 þ k1 Þ 0

s

4

2 1

x1

3

6 7 0  4 x2 5 . xN

3
1 0

7
ð9 þ k2 Þ ke 5

1 0


3


7
s þ ð9 þ k2 Þ ke 5
¼ s3 þ ð9 þ k2 Þs2 þ ð7 þ k1 þ ke Þs þ 4ke
1 s
1

0

Equating this polynomial to the desired characteristic equation, s3 þ 6:3s2 þ 13s þ 15:16 ¼ s3 þ ð9 þ k2 Þs2 þ ð7 þ k1 þ ke Þs þ 4ke Solving for the k’s, K ¼ ½ 2:21

2:7  and ke ¼ 3:79:

47

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Solutions to Skill-Assessment Exercises

CHAPTER 13 13.1 f ðtÞ ¼ sinðvkT Þ; f  ðtÞ ¼

1 P

sinðvkT Þdðt  kT Þ;  jvkT  1 1 X X e  ejvkT ekTs F  ðsÞ ¼ sinðvkT ÞekTs ¼ 2j k¼0 k¼0 1     k 1 X T ðsjvÞ k ¼ e  eT ðsþjvÞ 2j k¼0 k¼0

But,

1 P

xk ¼

k¼0

1 1  x1

Thus,



1 1 1 1 eTs ejvT  eTs ejvT ¼  2j 1  eT ðsjvÞ 1  eT ðsþjvÞ 2j 1  ðeTs ejvT  eTs ejvT Þ þ e2Ts

sinðvT Þ z1 sinðvT Þ ¼ ¼ eTs Ts 2Ts 1  e 2cosðvT Þ þ e 1  2z1 cosðvT Þ þ z2

F  ðsÞ ¼

13.2 F ðzÞ ¼

zðz þ 1Þðz þ 2Þ ðz  0:5Þðz  0:7Þðz  0:9Þ

F ðzÞ zðz þ 1Þðz þ 2Þ ¼ z ðz  0:5Þðz  0:7Þðz  0:9Þ ¼ 46:875 F ðzÞ ¼ 46:875

z 1 z  114:75 þ 68:875 z  0:5 z  0:7 z  0:9 z z z  114:75 þ 68:875 ; z  0:5 z  0:7 z  0:9

f ðkT Þ ¼ 46:875ð0:5Þk  114:75ð0:7Þk þ 68:875ð0:9Þk 13.3   Since GðsÞ ¼ 1  eTs

8 , s ð s þ 4Þ         8 z1 A B z1 2 2 1 ¼ z þ ¼ z þ : G ð zÞ ¼ 1  z z s ð s þ 4Þ z s sþ4 z s sþ4

2 2 þ . Therefore, g2 ðtÞ ¼ 2  2e4t , or g2 ðkT Þ ¼ 2  2e4kT . s sþ4   2z 1  e4T 2z 2z .  ¼ Hence, G2 ðzÞ ¼ z  1 z  e4T ðz  1Þðz  e4T Þ Let G2 ðsÞ ¼

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Chapter 13 Solutions to Skill-Assessment Exercises

  2 1  e4T z1 . G2 ðzÞ ¼ ðz  e4T Þ z 1 1:264 . For T ¼ s, GðzÞ ¼ 4 z  0:3679

Therefore, GðzÞ ¼

13.4 Add phantom samplers to the input, feedback after H(s), and to the output. Push G1(s)G2(s), along with its input sampler, to the right past the pickoff point and obtain the block diagram shown below. R(s)

+ –

G1(s)G2(s)

C(s)

H(s)G1(s)G2(s)

Hence, T ðzÞ ¼

G1 G2 ðzÞ . 1 þ HG1 G2 ðzÞ

13.5 20 GðsÞ 20 4 4 . Let G2 ðsÞ ¼ ¼ ¼  . Taking the inverse sþ5 s s ð s þ 5Þ s s þ 5 Laplace transform and letting t ¼ kT, g2 ðkT Þ ¼ 4  4e5kT . Taking the z-transform   4z 1  e5T 4z 4z yields G2 ðzÞ ¼ . ¼  z  1 z  e5T ðz  1Þðz  e5T Þ   4 1  e5T z1 Now, GðzÞ ¼ .  G2 ðzÞ ¼ ðz  e5T Þ z   4 1  e5T G ð zÞ . ¼ Finally, T ðzÞ ¼ 1 þ GðzÞ z  5e5T þ 4 The pole of the closed-loop system is at 5e5T  4. Substituting values of T, we find that the pole is greater than 1 if T > 0:1022 s. Hence, the system is stable for 0 < T < 0:1022 s.

Let GðsÞ ¼

13.6 sþ1 into DðzÞ ¼ z3  z2  0:5z þ 0:3, we obtain DðsÞ ¼ s3  8s2 s1 27s  6. The Routh table for this polynomial is shown below. Substituting z ¼

s3 s

2

s1 s0

1

27

8

6

27:75 6

0 0

Since there is one sign change, we conclude that the system has one pole outside the unit circle and two poles inside the unit circle. The table did not produce a row of zeros and thus, there are no jv poles. The system is unstable because of the pole outside the unit circle.

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Solutions to Skill-Assessment Exercises

13.7 Defining G(s) as G1(s) in cascade with a zero-order-hold,



    ð s þ 3Þ 1=4 2=5 Ts Ts 3=20 ¼ 20 1  e þ  : GðsÞ ¼ 20 1  e sðs þ 4Þðs þ 5Þ s ðs þ 4Þ ðs þ 5Þ Taking the z-transform yields

  ð3=20Þz ð1=4Þz ð2=5Þz 5 ð z  1Þ 8 ð z  1Þ GðzÞ ¼ 20 1  z1 ¼3þ þ   : 4T 5T z1 ze ze z  e4T z  e5T Hence for T ¼ 0:1 second, K p ¼ lim GðzÞ ¼ 3, and K v ¼ z!1

1 lim ðz  1ÞGðzÞ ¼ 0, and T z!1

1 lim ðz  1Þ2 GðzÞ ¼ 0. Checking for stability, we find that the system is T 2 z!1 GðzÞ 1:5z  1:109 has poles stable for T ¼ 0:1 second, since T ðzÞ ¼ ¼ 1 þ GðzÞ z2 þ 0:222z  0:703 inside the unit circle at 0:957 and þ 0:735. Again, checking for stability, we find that G ð zÞ ¼ the system is unstable for T ¼ 0:5 second, since T ðzÞ ¼ 1 þ G ð zÞ 3:02z  0:6383 has poles inside and outside the unit circle at þ0:208 and 2 z þ 2:802z  0:6272 3:01, respectively.

Ka ¼

13.8 Draw the root locus superimposed over the z ¼ 0:5 curve shown below. Searching along a 54:3 line, which intersects the root locus and the z ¼ 0:5 curve, we find the point 0:587ff54:3 ¼ ð0:348 þ j 0:468Þ and K ¼ 0:31. z-Plane Root Locus

1.5

1

(0.348 + j 0.468) K = 0.31

0.5 Imag Axis

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54.3°

0

–0.5

–1

–1.5

–3

–2.5

–2

–1.5

–1

–0.5 Real Axis

0

0.5

1

1.5

2

13.9 Let Ge ðsÞ ¼ GðsÞGc ðsÞ ¼

100K 2:38ðs þ 25:3Þ 342720ðs þ 25:3Þ ¼ : sðs þ 36Þðs þ 100Þ ðs þ 60:2Þ sðs þ 36Þðs þ 100Þðs þ 60:2Þ

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Chapter 13 Solutions to Skill-Assessment Exercises

The following shows the frequency response of Ge ðjvÞ.

Bode Diagrams 40 20 0 –20 Phase (deg); Magnitude (dB)

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–40 –60

–100

–150

–200

–250 10–1

100

101 Frequency (rad/sec)

102

103

We find that the zero dB frequency, vFM , for Ge ðjvÞ is 39 rad/s. Using Astrom’s guideline the value of T should be in the range, 0:15=vFM ¼ 0:0038 second to 0:5=vFM ¼ 0:0128 second. Let us use T ¼ 0:001 second. Now find the Tustin 2 ð z  1Þ transformation for the compensator. Substituting s ¼ into Gc ðsÞ ¼ T ðz  1Þ 2:38ðs þ 25:3Þ with T ¼ 0:001 second yields ðs þ 60:2Þ Gc ðzÞ ¼ 2:34

ðz  0:975Þ : ðz  0:9416Þ

13.10  X ðzÞ 1899z2  3761z þ 1861 . Cross-multiply and obtain z2  1:908z þ ¼ 2 EðzÞ  z  1:908z þ 0:9075 0:9075X ðzÞ ¼ 1899z2  3761z þ 1861 EðzÞ. Solve for the highest power of z   operating on the output, X(z), and obtain z2 X ðzÞ ¼ 1899z2  3761z þ 1861 EðzÞ  ð1:908z þ 0:9075ÞX ðzÞ. Solving for X(z) on the left-hand side yields Gc ðzÞ ¼

51

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Solutions to Skill-Assessment Exercises

    X ðzÞ ¼ 1899  3761z1 þ 1861z2 EðzÞ 1:908z1 þ 0:9075z2 X ðzÞ. Finally, we implement this last equation with the following flow chart: x*(t)

e*(t)

1899 Delay 0.1 second

Delay 0.1 second + e*(t-0.1)

–3761

–1.9.08

+

–

+ +

– Delay 0.1 second

Delay 0.1 second e*(t-0.2)

x*(t-0.1)

1861

x*(t-0.2) 0.9075

ONLINEFFIRS

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