Nise - Control Systems Engineering 6e solution skill PDF

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Solutions to Skill-Assessment Exercises

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

1 ðs 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 D C B F ðsÞ ¼ þ þ þ s s þ 2 ðs þ 3Þ 2 ðs þ 3Þ where,

   10  5  B¼ A¼ ¼ 5 ¼   sðs þ 3Þ2 S!2 ðs þ 2Þðs þ 3Þ 2 S!0 9    40 10  10 2 dF ðsÞ  ¼ C¼ ¼ ; and D ¼ ð s þ 3Þ ds s!3 9 sð s þ 2ÞS!3 3 10

Taking the inverse Laplace transform yields, f ðtÞ ¼

40 10 5  5e2t þ te3t þ e3t 3 9 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 3Rð sÞ Collecting terms,

Thus,

3    s þ 3s2 þ 7s þ 5 CðsÞ ¼ s2 þ 4s þ 3 RðsÞ s2 þ 4s þ 3 CðsÞ ¼ 3 RðsÞ s þ 3s2 þ 7s þ 5 1

2

Solutions to Skill-Assessment Exercises

2.4 GðsÞ ¼

2s þ 1 CðsÞ ¼ RðsÞ s2 þ 6s þ 2

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

1 1 B s C A ¼  þ ¼ þ s ð s þ 4Þ ðs þ 8Þ s2 ðs þ 4Þð s þ 8Þ sð s þ 4Þð s þ 8Þ

where A¼

  1 1  ¼  32 ðs þ 4Þðs þ 8Þ S!0

B¼

Thus,

cðtÞ ¼

  1  1  1 1 ; and C ¼ ¼  ¼   16 sð s þ 8Þ S!4 sð s þ 4Þ S!8 32 1 1 1  e4t þ e8t 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, ð s 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),    ðs þ 1Þ V ðsÞ 1    s 0 1    2   1 s þ 2s þ 1 V ðsÞ 0 ð s þ 2Þ  ¼ I 2 ðsÞ ¼  sð s2 þ 5s þ 2Þ s 1   ð s þ 1Þ  s  ð 2s þ 1 Þ 1    1 1 ðs þ 2Þ 

Chapter 2 Solutions to Skill-Assessment Exercises

But, VL ðsÞ ¼ sI 2 ðsÞ Hence, VL ðsÞ ¼

2  s þ 2s þ 1 V ðsÞ ð s2 þ 5s þ 2Þ

or VL ðsÞ s2 þ 2s þ 1 ¼ 2 V ðsÞ s þ 5s þ 2

Nodal Analysis

Writing the nodal equations,   1 þ 2 V 1 ðsÞ  VL ðsÞ ¼ V ðsÞ s   1 2 þ 1 VL ðsÞ ¼ V ðsÞ V 1 ðsÞ þ s s Solving for VL(s),

or

     1  V ðsÞ  þ 2   s   1      1 V ðsÞ  s2 þ 2s þ 1 V ðsÞ s  ¼ VL ðsÞ ¼    1  ð s2 þ 5s þ 2Þ   1 þ 2  s       2  1 þ 1   s VL ðsÞ s2 þ 2s þ 1 ¼ 2 s þ 5s þ 2 V ðsÞ

2.7 Inverting GðsÞ ¼  Noninverting

Z 2 ðsÞ 100000  ¼ s ¼ Z 1 ðsÞ 105 =s

105 þ 105 s ½Z 1 ðsÞ þ Z ðsÞ ! GðsÞ ¼ ¼ Z 1 ðsÞ 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

4

Solutions to Skill-Assessment Exercises

Solving for X2 (s),

Hence,

  2   s þ 3s þ 1 F ðsÞ    ð3s þ 1Þ ð3s þ 1ÞF ðsÞ 0   ¼ 3 X 2 ðsÞ ¼   2   s s ð þ 7s2 þ 5s þ 1Þ ð3s þ 1Þ   s þ 3s þ 1 2    ð3s þ 1Þ s þ 4s þ 1  ð3s þ 1Þ X 2 ðsÞ ¼ 3 F ðsÞ sð s þ 7s2 þ 5s þ 1Þ

2.9 Writing the equations of motion,  2  s þ s þ 1 u 1 ðsÞ  ð s þ 1Þu 2 ðsÞ ¼ T ðsÞ ð s þ 1Þu 1 ðsÞ þ ð2s þ 2Þu 2 ðsÞ ¼ 0

where u 1 ðsÞ is the angular displacement of the inertia. Solving for u 2 ðsÞ,  2    s þsþ1 T ðs Þ    ðs þ 1Þ ðs þ 1ÞF ðsÞ 0   ¼ 3 u 2 ðsÞ ¼   2  s þ s þ 1 ðs þ 1Þ  2s þ 3s2 þ 2s þ 1    ðs þ 1Þ ð2s þ 2Þ 

From which, after simplification,

u 2 ðsÞ ¼

1 2s2 þ s þ 1

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 u 1 ðsÞ  su a ðsÞ ¼ T ðsÞ su 1 ðsÞ þ ð s þ 1Þu a ðsÞ ¼ 0 where u 1 ðsÞ is the angular displacement of the 1-kg inertia. Solving for u a ðsÞ,   2   s þ s T ðsÞ     s sT ðsÞ 0   ¼ 3 u a ðsÞ ¼   2  s þs  s þ s2 þ s s    s ðs þ 1Þ 

Chapter 2 Solutions to Skill-Assessment Exercises

From which, 1 u a ðsÞ ¼ T ðsÞ s2 þ s þ 1 1 But, u 2 ðsÞ ¼ u a ðsÞ: 2 Thus, 1=2 u 2 ðsÞ ¼ T ðsÞ s2 þ s þ 1 2.11 First find the mechanical constants.     1 1 1 2 ¼ 1 þ 400 ¼2  Jm ¼ Ja þ JL 400 5 4     1 1 1 2 ¼ 5 þ 800 ¼7 Dm ¼ Da þ DL  400 5 4 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

vnoload ¼ 25 which, Kt T stall 200 ¼ ¼ ¼2 Ea Ra 100 Ea 100 Kb ¼ ¼ ¼4 vnoload 25 Substituting all values into the motor transfer function, KT u m ðsÞ 1 Ra J m ¼    ¼  15 1 KT Kb Ea ðsÞ s sþ s sþ Dm þ 2 Ra Jm where u m ðsÞ is the angular displacement of the armature. 1 Now u L ðsÞ ¼ u m ðsÞ. Thus, 20 1=20 u L ðsÞ  ¼  Ea ðsÞ 15 s sþ 2 2.12 Letting u 1 ðsÞ ¼ v1 ðsÞ=s

u 2 ðsÞ ¼ v2 ðsÞ=s

5

6

Solutions to Skill-Assessment Exercises

in Eqs. 2.127, we obtain   K K v1 ðsÞ  v2 ðsÞ ¼ T ðsÞ J 1 s þ D1 þ s s   K K v2 ðsÞ  v1 ðsÞ þ J 2 s þ D2 þ s s From these equations we can draw both series and parallel analogs by considering these to be mesh or nodal equations, respectively. D1

J1

T(t)

J2 1 K

– +

w1(t)

D2 w2(t)

Series analog 1 K

w 1( t ) 1 D1

J1

T(t)

w2(t)

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, dð vo þ dvÞ þ evo þdv  2 ¼ iðtÞ dt We now linearize ev. The general form is f ðvÞ  f ðvo 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

Chapter 3 Solutions to Skill-Assessment Exercises

Solving for evo þdv, e

vo þdv

Substituting into Eq. (1)

 dev  dv ¼ evo þ evo dv ¼e þ dv vo vo

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, ð s þ 2ÞdvðsÞ ¼ I ðsÞ Solving for the transfer function, we obtain dvðsÞ 1 ¼ sþ2 I ðsÞ or 1 V ðsÞ ¼ about equilibrium: I ðsÞ sþ2 CHAPTER 3 3.1 Identifying appropriate variables on the circuit yields C1

R iR

iC 1 v1(t) +

L

–

+

C2

vo (t) –

iL

iC 2

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

ð1Þ

7

8

Solutions to Skill-Assessment Exercises

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

1 ðvL  vC2 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 1 1 1 dvC1 1 vC1 þ iL  vC 2 þ ¼ vi dt RC1 C1 RC1 RC 1 diL 1 1 ¼  vC 1 þ vi L dt L 1 dvC2 1 1 vC 1  vC 2 vi ¼ dt RC2 RC 2 RC2 where the output equation is vo ¼ vC 2 Putting the equations in vector-matrix form, 2



1 RC1

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

0

1 C1 0 0

3 2 3 1 1 6 RC 1 7 RC1 7 7 6 7 7 6 7 7 6 1 7 7 vi ðtÞ 0 7x þ 6 7 6 L 7 7 6 7 4 1 5 1 5  RC2 RC 2 

1 x

3.2 Writing the equations of motion 

 sX 2 ðsÞ ¼ F ðsÞ s2 þ s þ 1 X 1 ðsÞ   X 3 ðsÞ ¼ 0 sX 1 ðsÞ þ s2 þ s þ 1 X 2 ðsÞ   X 2 ðsÞ þ s2 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

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, _z1 ¼ z2 _z2 ¼ x€1 ¼ x_1  x1 þ x_ 2 þ f ¼ z2  z1 þ z4 þ f _z3 ¼ x_ 2 ¼ z4 _z4 ¼ x€2 ¼ x_ 1  x_2  x2 þ x3 ¼ z2  z4  z3 þ z5 _ 5 ¼ x_3 ¼ z6 z _z6 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, 2 3 3 0 1 0 0 0 0 7 6 7 1 0 1 0 0 7 6 17 6 7 7 6 07 0 0 1 0 0 7 7 z þ 6 7 f ðtÞ; y ¼ ½ 0 6 07 7 1 1 1 1 0 7 6 7 6 7 7 4 05 0 0 0 0 1 5 0

1

0

1

0

0

0

1

0 z

0

1

3.3 First derive the state equations for the transfer function without zeros. 1 X ðsÞ ¼ RðsÞ s2 þ 7s þ 9 Cross multiplying yields 2  s þ 7s þ 9 X ðsÞ ¼ Rðs Þ

Taking the inverse Laplace transform assuming zero initial conditions, we get x€ þ 7 x_ 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 r x_ ¼ xþ 1 9 7 c ¼ ½ 1 2 x

9

10

Solutions to Skill-Assessment Exercises

3.4 The state equation is converted to a transfer function using GðsÞ ¼ CðsI  AÞ

1

ð1Þ

B

where A¼



4 4

1:5 0

   2 ;B ¼ ; and C ¼ ½ 1:5 0

0:625 :

Evaluating ðsI  AÞ yields 

ðsI  AÞ ¼

sþ4

1:5

4

s



Taking the inverse we obtain 1

ðsI  AÞ

¼

1 s2 þ 4s þ 6



s

1:5

4 sþ4



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  xo2 ¼ from which

  d x 2  dx ¼ 2xo dx dx xo

2 ðxo þ dxÞ ¼ xo2 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 2xo2 þ 10 þ df ðtÞ dt2

ð4Þ

The force of the spring at equilibrium is 10 N. Thus, since F ¼ 2x2 ; 10 ¼ 2xo2 from which pffiffiffi xo ¼ 5

Chapter 4 Solutions to Skill-Assessment Exercises

Substituting this value of xo into Eq. (4) gives us the final linearized differential equation. pffiffiffi d2 d x þ 4 5 dx ¼ df ðtÞ 2 dt

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 1 4 5 0 y ¼ ½1

0 x

CHAPTER 4 4.1 For a step input CðsÞ ¼

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

Taking the inverse Laplace transform, cðtÞ ¼ A þ Bet þ Ce7t þ De8t þ Ee10t 4.2 Since a ¼ 50; T c ¼ Tr ¼

1 4 4 1 ¼ 0:02s; T s ¼ 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. Since poles are at 6  j 19:08; cðtÞ ¼ A þ Be6t cosð19:08t þ fÞ. b. Since poles are at 78:54 and 11:46; cðtÞ ¼ A þ Be78:54t þ Ce11:4t . c. Since poles are double on the real axis at 15 cðtÞ ¼ A þ Be 15t þ Cte15t : d. Since poles are at j 25; cðtÞ ¼ A þ B cosð25t þ fÞ. 4.4 a. b. c. d.

pffiffiffiffiffiffiffiffi vn ¼ p400 ¼ 20 and 2zvn ffiffiffiffiffiffiffiffi vn ¼ 900 ¼ 30 and 2zvn pffiffiffiffiffiffiffiffi 225ffi ¼ 15 and 2zvn vn ¼ pffiffiffiffiffiffiffi vn 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.

11

12

Solutions to Skill-Assessment Exercises

4.5 vn ¼

pffiffiffiffiffiffiffiffi 361 ¼ 19 and 2zvn ¼ 16;

;z ¼ 0:421:

p 4 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. ffi p ffiffizp 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   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. a. Expanding G(s) by partial fractions yields GðsÞ ¼

4.8 See Figure 4.31 in the textbook for the Simulink block diagram and the output responses. 4.9 a. Since BUðsÞ ¼

 s sI  A ¼ " #3 0 1=ðs þ 1Þ

  sþ5 2 1 ; ðsI  AÞ 1 ¼ 2 s þ 5s þ 6 3 sþ5

2 s

 :

Also,

.

1 1  The state vector is XðsÞ ¼ ðsI  AÞ ½ xð0Þ þ BUðsÞ ¼ ð s þ 1 Þ ð s þ 2Þðs þ 3Þ "  #  2 s2 þ 7s þ 7 5s2 þ 2s  4 . The output is Y ðsÞ ¼ ½ 1 3 XðsÞ ¼ ¼ 2 s  4s  6 ðs þ 1Þðs þ 2Þðs þ 3Þ 17:5 12 0:5 þ  . Taking the inverse Laplace transform yields yðtÞ ¼  sþ1 sþ2 sþ3 t 2t 0:5e  12e þ 17:5e3t . b. The eigenvalues are given by the roots of jsI  Aj ¼ s2 þ 5s þ 6, or 2 and 3. 4.10   1 sþ5 2 . Taking the s2 þ 5s þ 4 2 s 2 sþ5 Laplace transform of each term, the state transition matrix is given by 3 2 2 t 2 4t 4 t 1 4t e  e 6 3e  3e 7 3 3 7: FðtÞ ¼ 6 4 2 5 4 1 2 4t t 4t t  e þ e  e þ e 3 3 3 3

a. Since ðsI  AÞ ¼



s

2

 1 ; ðsI  AÞ ¼

Chapter 5 Solutions to Skill-Assessment Exercises

2

4 ðttÞ 1 2 eðttÞ  2 e4ðttÞ 3  e4ðttÞ 6 3e 7 3 3 7 and 3 b. Since Fðt  tÞ ¼ 6 5 2 2 4 1 4  eðttÞ þ e4ðttÞ  eðttÞ þ e4ðttÞ 3 3 3 3 3 2 2 t t 2 2t 4t   7 6 3e e  3e e 0 7: BuðtÞ ¼ 2t ; Fðt  tÞBuðtÞ ¼ 6 5 4 1 4 e 2t 4t t t  e e þ e e 3 3 Rt Thus, xðtÞ ¼ FðtÞxð0Þ þ 0 Fðt  tÞ 3 2 10 t 4 4t 2t e e  e  7 6 3 3 7 BUðtÞdt ¼ 6 5 4 5 8 t 2t 4t  e þe þ e 3 3 c. yðtÞ ¼ ½ 2 1 x ¼ 5et  e2t

CHAPTER 5 5.1 Combine the parallel blocks in the forward path. Then, push pickoff point.

1 to the left past the s

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

16 GðsÞ ¼ , where 1 þ GðsÞHð sÞ s2 þ as þ 16 a 16 and H ðsÞ ¼ 1. Thus, vn ¼ 4 and 2zvn ¼ a, from which z ¼ . and GðsÞ ¼ 8 sðs þ aÞ   % 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

13

14

Solutions to Skill-Assessment Exercises

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. N1(s)

R(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

N6 (s)

s

N7 (s)

Eliminate unnecessary nodes. –1

R(s)

1

s

s

1 s –s

1 s

C(s)

C(s)

Chapter 5 Solutions to Skill-Assessment Exercises

5.4 Forward-path gains are G1 G2G 3 and G 1G 3. Loop gains are G1 G2 H 1 ; G2 H 2; and G3 H 3. Nontouching loops are ½G1 G2 H 1 ½ G3 H 3 ¼ G1 G2 G3 H 1 H 3 and ½ G2 H 2 ½G3 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 Rð sÞ

k

T k Dk yields D

G1 ðsÞG3 ðsÞ½1 þ G2 ðsÞ ½ 1 þ G2 ðsÞH 2 ðsÞ þ G1 ðsÞG2 ðsÞH 1 ðsÞ½ 1 þ G3 ðsÞH 3 ðsÞ

5.5 The state equations are, _x1 ¼ 2x1 þ x2 _x2 ¼ 3x2 þ x3 _x3 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

–5

1 s

1

x2

–3

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 Gð sÞ ¼

100

r

1

1 s

1 s

x1

x2

500 y

–5 –6

–1

15

16

Solutions to Skill-Assessment Exercises

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

"

105

506

1

0

y ¼ ½ 100

#

" # 1 r xþ 0

500 x

5.7 From the transformation equations, P1 ¼



3 2 1 4



Taking the inverse, 0:4 0:2 P¼ 0:1 0:3 



Now, 1

P AP ¼

"

3

2

1

4

#"

1

"

6:5

3

#"

0:4

0:2

#

¼ 4 6 0:1 0:3 " #" # " # 3 3 2 1 1 ¼ P B¼ 3 1 4 11 " # 0:4 0:2 CP ¼ ½ 1 4  ¼ ½ 0:8 1:4  0:1 0:3

"

6:5

8:5

9:5

11:5

#

Therefore, _z ¼

8:5

9:5 11:5 y ¼ ½ 0:8

5.8 First find the eigenva...