solutions to skill assessment problems

8/2/2019 Solutions to Skill Assessment Problems

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SOLUTIONS TOSKILL-ASSESSMENT EXERCISES

CHAPTER 2

2.1

The Laplace transform of t is1

s2using Table 2.1, Item 3. Using Table 2.2, Item 4,

Fs 1s 52.

2.2

Expanding Fs by partial fractions yields:F

s

A

s B

s 2 C

s 32

D

s 3where,

A 10s 2s 32S!0

59

B 10ss 32

S!2 5

C 10ss 2

S!3 10

3; andD s 32 dFs

dss!3

409

Taking the inverse Laplace transform yields,

ft 59

5e2t 103

te3t 409

e3t

2.3

Taking the Laplace transform of the differential equation assuming zero initial con-ditions yields:

s3Cs 3s2Cs 7sCs 5Cs s2Rs 4sRs 3Rs

1


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Collecting terms,

s3 3s2 7s 5Cs s2 4s 3RsThus,

C

s

Rs s2

4s

3

s3 3s2 7s 5

2.4

Gs CsRs

2s 1s2 6s 2

Cross multiplying yields,

d2c

dt2 6 dc

dt 2c 2 dr

dt r

2.5

Cs RsGs 1s2

ss 4s 8 1

ss 4s 8 A

s Bs 4

C

s 8where

A 1s 4s 8 S!0 1

32B 1

ss 8

S!4

116

; and

C 1ss 4 S!8 1

32

Thus,

ct 1

32 1

16 e4t

1

32 e8t

2.6

Mesh Analysis

Transforming the network yields,

+V(s)

I1(s) I 2(s)

V1(s)

V2(s)

I9(s)

+

1 1

s s

s

_

_

2 Solutions to Skill-Assessment Exercises


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Now, writing the mesh equations,

s 1I1s sI2s I3s VssI1s 2s 1I2s I3s 0

I1

s

I2

s

s

2

I3

s

0

Solving the mesh equations for I2(s),

I2s

s 1 Vs 1s 0 11 0 s 2

s 1 s 1s 2s 1 11 1 s 2

s

2 2s 1Vsss2 5s 2

But, VLs sI2sHence,

VL

s

s2 2s 1Vs

s2

5s 2or

VLsVs

s2 2s 1s2 5s 2

Nodal Analysis

Writing the nodal equations,

1

s 2

V1s VLs Vs

V1s 2s

1

VLs 1s

VsSolving for VL(s),

VLs

1

s 2

Vs

1 1s

Vs

1

s 2

1

1 2s

1

s2 2s 1Vss2 5s 2

or

VLsVs

s2 2s 1s2 5s 2

2.7

Inverting

Gs Z2sZ1s

100000105=s s

Chapter 2 Solutions to Skill-Assessment Exercises 3


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Noninverting

Gs Z1s Z2sZ1

s

105

s 105

105

s s

1

2.8

Writing the equations of motion,

s2 3s 1X1s 3s 1X2s Fs3s 1X1s s2 4s 1X2s 0

Solving for X2(s),

X2s

s2 3s 1 Fs3s 1 0

s

2

3s

1

3s

13s 1 s2 4s 1

3s 1Fs

s

s3

7s2

5s

1

Hence,

X2sFs

3s 1ss3 7s2 5s 1

2.9


s2 s 1u1s s 1u2s Tss 1u1s 2s 2u2s 0

where u1s is the angular displacement of the inertia.Solving for u2

s

,

u2s

s2 s 1 Tss 1 0

s2 s 1 s 1s 1 2s 2

s 1Fs

2s3 3s2 2s 1

From which, after simplification,

u2s 12s2 s 1

2.10

Transformingthe networkto onewithout gears by reflecting the 4 N-m/rad spring to the

left and multiplying by (25/50)

2

, we obtain,

1 kg

1 N-m-s/rad

1 N-m/rad

T(t)

qa(t)q1(t)



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s2 su1s suas Tssu1s s 1uas 0

where u1s is the angular displacement of the 1-kg inertia.Solving for uas,

uas s2 s Ts

s 0

s2 s s

s s 1

sTs

s3 s2 s

From which,

uasTs

1

s2 s 1But, u2s 1

2uas:

Thus,

u2sT

s

1=2

s2

s

1

2.11

First find the mechanical constants.

Jm Ja JL 15 1

4

2 1 400 1

400

2

Dm Da DL 15 1

4

2 5 800 1

400

7

Now find the electrical constants. From the torque-speed equation, setvm 0 to findstall torque and set Tm 0 to find no-load speed. Hence,

Tstall 200vnoload 25

which,

Kt

Ra Tstall

Ea 200

100 2

Kb Eavnoload

10025

4Substituting all values into the motor transfer function,

umsEas

KT

RaJm

s s 1Jm

Dm KTKb

Ra

1s s 15

2

where u

ms

is the angular displacement of the armature.

Now uLs 120

ums. Thus,uLsEas

1=20

s s 152



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2.12

Letting

u1s v1s=su2

s

v2

s

=s

in Eqs. 2.127, we obtain

J1s D1 Ks

v1s K

sv2s Ts

Ksv1s J2s D2 K

s

v2s

Fromthese equations we can draw both series and parallel analogs by considering these

to be mesh or nodal equations, respectively.

+

J1

J1

J2

J2

D1

D1

D2

D2

T(t)

T(t)

w1(t)

w1(t)

w2(t)

w2(t)

1 1

K

1

1

Series analog

Parallel analog

K

2.13

Writing the nodal equation,

Cdv

dt ir 2 it

But,

C 1v vo dvir evr ev evodv

Substituting these relationships into the differential equation,

dvo dvdt

evodv 2 it (1)

We now linearize ev

.The general form is

fv fvo% d fdv

vo

dv



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Substituting the function, fv ev, with v vo dv yields,

evodv evo % dev

dv

vo

dv

Solving for evodv,

evodv evo dev

dv

vo

dv evo evodv

Substituting into Eq. (1)

ddv

dt evo evodv 2 it (2)

Setting it 0 and letting the circuit reach steady state, the capacitor acts like an opencircuit. Thus, vo vr with ir 2. But, ir evr or vr ln ir.

Hence, vo ln 2 0:693. Substituting this value ofvo into Eq. (2) yieldsddv

dt 2dv it

Taking the Laplace transform,

s 2dvs IsSolving for the transfer function, we obtain

dvsIs

1

s 2or

VsIs

1

s 2 about equilibrium.

CHAPTER 3

3.1

Identifying appropriate variables on the circuit yields

+

+

C1

iL iC2

iC1iR

C2

R

L vo(t)v1(t)

Writing the derivative relations

C1dvC1

dt iC1

LdiL

dt vL

C2dvC2

dt iC2

(1)



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Using Kirchhoffs current and voltage laws,

iC1 iL iR iL 1

RvL vC2

vL vC1 viiC2 iR

1

RvL vC2

Substituting these relationships intoEqs. (1)and simplifying yieldsthe state equationsas

dvC1dt

1RC1

vC1 1

C1iL 1

RC1vC2

1

RC1vi

diL

dt 1

LvC1

1

Lvi

dvC2dt

1RC2

vC1 1

RC2vC2

1

RC2vi

where the output equation is

vo vC2Putting the equations in vector-matrix form,

_x

1

RC1

1

C1 1

RC1

1L

0 0

1RC2

0 1RC2

266666664

377777775x

1

RC1

1

L

1

RC2

266666664

377777775vit

y 0 0 1 x

3.2

Writing the equations of motion

s2 s 1X1s sX2s Fs

sX1

s

s2

s

1

X2

s

X3

s

0

X2s s2 s 1X3s 0Taking the inverse Laplace transform and simplifying,

x1 _x1 x1 _x2 fx2 _x1 _x2 x2 x3x3 _x3 x3 x2

Defining state variables, zi,

z1 x1; z2 _x1; z3 x2; z4 _x2; z5 x3; z6 _x3Writing the state equations using the definition of the state variables and the inverse

transform of the differential equation,

_z1 z2_z

2 x

1 _x

1 x

1 _x

2 f

z

2 z

1 z

4 f

_z3 _x2 z4_z4 x2 _x1 _x2 x2 x3 z2 z4 z3 z5_z5 _x3 z6_z6 x3 _x3 x3 x2 z6 z5 z3



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The output is z5. Hence, y z5. In vector-matrix form,

_z

0 1 0 0 0 0

1 1 0 1 0 00 0 0 1 0 0

0 1

1

1 1 0

0 0 0 0 0 1

0 0 1 0 1 1

2

6666664

3

7777775z

0

1

0

0

0

0

2

6666664

3

7777775f

t

; y

0 0 0 0 1 0

z

3.3

First derive the state equations for the transfer function without zeros.

XsRs

1

s2 7s 9Cross multiplying yields

s2 7s 9Xs RsTaking the inverse Laplace transform assuming zero initial conditions, we get

x 7_x 9x rDefining the state variables as,

x1 xx2 _x

Hence,

_x1 x2_x2 x 7_x 9x r 9x1 7x2 r

Using the zeros of the transfer function, we find the output equation to be,

c 2 _x x x1 2x2Putting all equation in vector-matrix form yields,

_x 0 19 7 !

x 01

!r

c 1 2 x

3.4

The state equation is converted to a transfer function using

Gs CsI A1B 1where

A 4 1:54 0

!; B 2

0

!; and C 1:5 0:625 :

Evaluating sI A yields

sI A s 4 1:54 s !



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Taking the inverse we obtain

sI A1 1s2 4s 6

s 1:54 s 4

!

Substituting all expressions into Eq. (1) yields

Gs 3s 5s2 4s 6

3.5

Writing the differential equation we obtain

d2x

dt2 2x2 10 dft (1)

Letting x xo dx and substituting into Eq. (1) yieldsd2xo dx

dt2 2xo dx2 10 dft (2)

Now, linearize x2.

xo dx2 x2o dx2

dx

xo

dx 2xodx

from which

xo dx2 x2o 2xodx (3)SubstitutingEq. (3) intoEq. (1) andperforming theindicateddifferentiationgivesus thelinearized intermediate differential equation,

d2dx

dt2 4xodx 2x2o 10 dft (4)

The force of the spring at equilibrium is 10 N. Thus, since F

2x2; 10

2x2

o

from

which

xo ffiffiffi

5p

Substituting this value of xo into Eq. (4) gives us the final linearized differential

equation.

d2dx

dt2 4

ffiffiffi5

pdx dft

Selecting the state variables,

x1 dxx2 _dx

Writing the state and output equations

_x1 x2_x2 dx 4

ffiffiffi5

px1 dft

y x1



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Converting to vector-matrix form yields the final result as

_x 0 14ffiffiffi

5p

0

!x 0

1

!dft

y

1 0

x

CHAPTER 4

4.1

For a step input

Cs 10s 4s 6ss 1s 7s 8s 10

A

s B

s 1 C

s 7 D

s 8 E

s 10Taking the inverse Laplace transform,

ct A Bet Ce7t De8t Ee10t

4.2

Since a 50; Tc 1a

150

0:02 s; Ts 4a

450

0:08s; and Tr 2:2a

2:250

0:044s.

4.3

a. Since poles are at 6 j19:08; ct A Be6tcos19:08t f.b. Since poles are at 78:54 and 11:46; ct A Be78:54t Ce11:4t.c. Since poles are double on the real axis at 15 ct A Be15t Cte15t:d. Since poles are at j25; ct A B cos25t f.

4.4

a. vn ffiffiffiffiffiffiffiffi

400p

20 and 2zvn 12; ; z 0:3 and system is underdamped.b. vn

ffiffiffiffiffiffiffiffi900

p 30 and 2zvn 90; ; z 1:5 and systemis overdamped.

c. vn ffiffiffiffiffiffiffiffi

225p

15 and 2zvn 30; ; z 1 and systemis critically damped.d. vn

ffiffiffiffiffiffiffiffi625

p 25 and 2zvn 0; ; z 0 andsystemis undamped.

4.5

vn ffiffiffiffiffiffiffiffi

361p

19and2zvn 16; ; z 0:421:

Now, Ts 4

zvn 0:5 s and Tp p

vnffiffiffiffiffiffiffiffiffiffiffiffiffi

1 z2p 0:182s.

From Figure 4.16, vnTr 1:4998. Therefore, Tr 0:079 s.

Finally, %os ezpffiffi1

pz2 100 23:3%



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4.6

a. Thesecond-order approximationis 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 of1 and the higher-order pole is at 4, i.e. not more than five-times further.4.7

a. ExpandingG(s) by partial fractions yields Gs 1s 0:8942

s 20 1:5918

s 10 0:3023

s 6:5.But 0:3023 is not an order of magnitude less than residues of second-orderterms (term 2 and 3). Therefore, a second-order approximation is not valid.

b. Expanding G(s) by partial fractions yields Gs 1s 0:9782

s 20 1:9078

s 10 0:0704

s 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

a. Since sI A s 23 s 5

!; sI A1 1

s2 5s 6s 5 23 s

!: Also,

BUs 01=s 1

!.

The state vector is Xs sI A1x0 BUs 1s 1s 2s 3 2s2 7s 7s2 4s 6

!. The output is Ys 1 3 Xs 5s2 2s 4s 1s 2s 3

0:5s

1

12s

2

17:5s

3

. Taking the inverse Laplace transform yields yt

0:5et 12e2t17:5e3t.b. The eigenvalues are given by the roots of jsI Aj s2 5s 6, or 2 and 3.4.10

a. Since sI A s 22 s 5

!; sI A1 1

s2 5s 4s 5 22 s

!. Taking the

Laplace transform of each term, the state transition matrix is given by

Ft 4

3et 1

3e4t

2

3et 2

3e4t

23

et 23

e4t 13

et 43

e4t

2664

3775:

b. Since Ft t 4

3ett 1

3e4tt 2

3ett 2

3e4tt

23

ett 23

e4tt 13

ett 43

e4tt

2664

3775 and



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But 0e2t

!; Ft tBut

2

3etet 2

3e2te4t

13

etet 43

e2te4t

2664

3775:

Thus, xt Ftx0 Zt

0

Ft t

BUtdt10

3et e2t 4

3e4t

53

et e2t 83

e4t

2664

3775:

c. yt 2 1 x 5et e2t

CHAPTER 5

5.1

Combinethe parallel blocks in the forward path. Then, push1

sto theleftpast the pickoff

point.

1

s

s

s

ss2 +

1+

R(s) C(s)

Combine the parallel feedback paths and get 2s. Then, apply the feedback formula,

simplify, and get, Ts s3 1

2s4 s2 2s.

5.2

Find the closed-loop transfer function, Ts Gs1 GsHs

16

s2 as 16, whereand Gs 16

s

s

a

and Hs 1. Thus, vn 4 and 2zvn a, from which z a8

.

But, for 5% overshoot, zln %

100

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffip2 ln2 %

100

s 0:69. Since, z a8; a 5:52.



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5.3

Label nodes.

+ +

+

R(s)

s

sN1 (s) N2 (s) N3 (s) N4 (s)

N6 (s)N5 (s)

N7 (s)

sC(s)

1s

1s

Draw nodes.

R(s) N1(s) N2(s)

N5(s)

N7(s)

N6(s)

N3(s) N4(s) C(s)

Connect nodes and label subsystems.

R(s)C(s)1

1s

s

1

ss

1 1

1

1

1sN1 (s) N2 (s)

N5 (s) N6 (s)

N7 (s)

N3 (s) N4 (s)

Eliminate unnecessary nodes.

R(s) C(s)1 ss1s

1s

s

1



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5.4

Forward-path gains are G1G2G3 and G1G3.Loop gains are G1G2H1; G2H2; and G3H3.Nontouching loops are G1G2H1G3H3 G1G2G3H1H3 and

G2H2

G3H3

G2G3H2H3.

Also, D 1 G1G2H1 G2H2 G3H3 G1G2G3H1H3 G2G3H2H3:Finally, D1 1 and D2 1.

Substituting these values into Ts CsRs

k

TkDk

Dyields

Ts G1sG3s1 G2s1 G2sH2s G1sG2sH1s1 G3sH3s

5.5

The state equations are,_x1 2x1 x2_x2 3x2 x3_x3 3x1 4x2 5x3 ry x2

Drawing the signal-flow diagram from the state equations yields

1s

1s

1s11 1

1

5

4

23

3

r x1x2x3 y

5.6

From Gs 100s 5s2 5s 6 we draw the signal-flow graph in controller canonical form

and add the feedback.

1

5

6

100

500

1

y

r1s

1sx1 x2



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

x 105 5061 0

!x 1

0

!r

y 100 500 x

5.7

From the transformation equations,

P1 3 21 4

!Taking the inverse,

P 0:4 0:20:1 0:3

!Now,

P1AP 3 21 4

!1 3

4 6 !

0:4 0:20:1 0:3

! 6:5 8:5

9:5 11:5 !

P1B 3 21 4

!1

3

! 311 !

CP 1 4 0:4 0:20:1 0:3

! 0:8 1:4

Therefore,

_z 6:5 8:59:5 11:5

!z 311

!u

y 0:8 1:4 z

5.8

First find the eigenvalues.

jlI Aj l 00 l

! 1 34 6 !

l 1 34 l 6

l2 5l 6

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

1 3

4 6 !

x1x2

! l x1

x2

!

For l 2,3x1 3x2 0

4x

1 4x

2 0

Thus x1 x2For l 3

4x1 3x2 04x1 3x2 0



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Thus x1 x2 and x1 0:75x2; from which we let

P 0:707 0:60:707 0:8 !

Taking the inverse yields

P1 5:6577 4:24335 5

!

Hence,

D P1AP 5:6577 4:24335 5

!1 3

4 6 !

0:707 0:60:707 0:8

! 2 0

0 3 !

P1B 5:6577 4:24335 5

!1

3

! 18:39

20

!

CP 1 4 0:707 0:60:707 0:8 !

2:121 2:6

Finally,

_z 2 00 3

!z 18:39

20

!u

y 2:121 2:6 z

CHAPTER 6

6.1

Make a Routh table.

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 thes3

row. The even polynomial iscontained in the previous row as 6s4 0s2 6. Taking the derivative yields

s7 3 6 7 2

s6 9 4 8 6

s5 4.666666667 4.333333333 0 0

s4 4.35714286 8 6 0s3 12.90163934 6.426229508 0 0

s2 10.17026684 6 0 0

s1 1.18515742 0 0 0s0 6 0 0 0



18/53

24s3 0s. Replacing therow of zeros withthe coefficients of the derivativeyieldsthes3row. We also encounter a zero in the first column at the s2 row. We replace the zerowith e and continue the table. The final result is shown now as

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 imaginaryaxis poles. From the top of the table down to the even polynomial yields one signchange. Thus, the rest of the polynomial has one right half-plane root, and one lefthalf-plane root. The total for the system is two right half-plane poles, two left half-plane poles, and 2 imaginary poles.

6.3

Since Gs Ks 20ss 2s 3; Ts

Gs1 Gs

Ks 20s3 5s2 6 Ks 20K

Form the Routh table.

From the s1

row, K< 2. From the s0 row, K> 0. Thus, for stability, 0 < K< 2.

6.4

First find

jsI Aj s 0 0

0 s 0

0 0 s

24

35 2 1 11 7 1

3 4 5

24

35

s 2 1 11 s 7 1

3 4 s 5

s3 4s2 33s 51

Now form the Routh table.

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

s6 1 6 1 6s5

1 0 1 0s4 6 0 6 0

s3 24 0 0 0 ROZs2

e 6 0 0

s1 144=e 0 0 0

s0 6 0 0 0

s3 1 6 K

s2 5 20K

s1 30 15K

5s0 20K

s3 1 33s2

4 51S1 20:25

S0 51



19/53

CHAPTER 7

7.1

a. First check stability.

Ts Gs1 Gs

10s2 500s 6000s3 70s2 1375s 6000

10s 30s 20s 26:03s 37:89s 6:085

Poles are in the lhp. Therefore, the system is stable. Stability also could be checked viaRouth-Hurwitz using the denominator ofT(s). Thus,

15ut :estep1 151 lim

s!0Gs

15

1 1 0

15tut : eramp1 15lims!0

sGs 15

1020302535

2:1875

15t2ut : eparabola1 15lims!0

s2G

s

30

0 1; sinceL15t2 30

s3

b. First check stability.

Ts Gs1 Gs

10s2 500s 6000s5 110s4 3875s3 4:37e04s2 500s 6000

10s 30s 20s 50:01s 35s 25s2 7:189e 04s 0:1372From 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 denomi-

nator of T(s). Since the system is unstable, calculations about steady-state errorcannot be made.

7.2

a. The system is stable, since

Ts Gs1 Gs

1000s 8s 9s 7 1000s 8

1000s 8s2 1016s 8063

and is of Type 0. Therefore,

Kp lims!0

Gs 10008

79 127; Kv lim

s!0sGs 0;

and Ka

lims!0

s2G

s

0

b.

estep1 11 lim

s!0Gs

1

1 127 7:8e 03



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eramp1 1lims!0

sGs 1

0 1

eparabola1 1lims!0

s2G

s

1

0 1

7.3

System is stable for positive K. System is Type 0. Therefore, for a step input

estep1 11 Kp 0:1. Solving for Kp yields Kp 9 lims!0 Gs

12K

1418; from

which we obtain K 189.

7.4

System is stable. Since G1s 1000, and G2s s 2s 4,

eD11

lims!0

1

G2s lim G1s!0s

1

2 1000 9:98e 04

7.5

System is stable. Create a unity-feedback system, where Hes 1s 1 1

ss 1 :The system is as follows:

+

R(s) Ea(s) C(s)100

s + 4

s

s + 1

Thus,

Ges Gs1 GSHes

100

s 41 100ss 1s 4

100s 1S2 95s 4

Hence, the system is Type 0. Evaluating Kp yields

Kp

100

4

25

The steady-state error is given by

estep1 11 Kp

1

1 25 3:846e 02



21/53

7.6

Since Gs Ks 7s2 2s 10; e1

1

1 Kp 1

1 7K10

1010 7K:

Calculating the sensitivity, we get

Se:K Ke

@e

@K K

10

10 7K 107

10 7K2 7K

10 7K

7.7

Given

A 0 13 6 !

; B 01

!; C 1 1 ; Rs 1

s:

Using the final value theorem,

estep1 lims!0

sRs1 CsI A1B lims!0

1 1 1 s 13 s 6

!10

1

!" #

lims!0

1 1 1

s 6 13s s

" #

s2 6s 30

1

" #266664

377775 lims!0

s2 5s 2s2 6s 3

2

3

Using input substitution,

step1 1 CA1B 1 1 1 0 1

3

6 !

10

1 ! 1 1 1

6 13 0

!3

0

1

! 1 1 1

1

3

0

24

35 2

3

CHAPTER 8

8.1

a.

F7 j9 7

j9

2

7

j9

4

0:0339

7 j97 j9 37 j9 6 5

j9

3

j9

7 j94 j91 j9

66 j72944 j378 0:0339 j0:0899 0:096 < 110:7



22/53

b. The arrangement of vectors is shown as follows:

jw

s

s-plane

X X

24 136 57

X

M1 M2 M3 M4 M5

(7+j9)

0

From the diagram,

F7 j9 M2M4M1M3M5

3 j95 j91 j94 j97 j9 66 j72944 j378

0:0339 j0:0899 0:096


23/53

From the diagram,

angles 180 tan1 31

tan1 31

180108:43 108:43 180:

b. Since the angle is 1808, the point is on the root locus.

c. K P pole lengthsP zero lengths

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

12 32p

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

12 32p

1

10

8.3

First, find the asymptotes.

sa poles zeros#poles #zeros

2 4 6 03 0 4

ua 2k 1p3

p3; p;

5p

3

Next draw root locus following the rules for sketching.

8 7 6 5 4 3 2 1 0 1 2 35

4

3

2

1

0

1

2

3

4

5

Real Axis

ImagAxis

8.4

a.

j3

s

jw

s-plane

X

X

O

2 2

j3

0



24/53

b. Using the Routh-Hurwitz criteria, we first find the closed-loop transfer function.

Ts Gs1 Gs

Ks 2s2 K 4s 2K 13

Using the denominator ofTs, make a Routh table.

We get a row of zeros for K 4. From the s2 row with K 4; s2 21 0. Fromwhich we evaluate the imaginary axis crossing at

ffiffiffiffiffi21

p.

c. From part (b), K 4.d. Searchingfor theminimum gainto theleft of2 on the realaxisyields7atagainof

18. Thus the break-in point is at 7.e. First, draw vectors to a point e close to the complex pole.

jw

s

s-plane

X

X

2 20

j3

j3

At thepoint e close to the complex pole, the angles must add up to zero.Hence, anglefromzero angle frompole in4th quadrant angle from pole in 1st quadrant 180,or tan1

3

4

90 u 180. Solving for the angle of departure, u 233:1.

s2 1 2K 13

s1 K 40 0s0 2K 13 0



25/53

8.5

a.

jw

43

X

X

s-plane

o0

z= 0.5

j4

j4

s2

o

b. Search along the imaginary axis and find the 1808 point at s j4:06.c. For the result in part (b), K 1.d. Searching between 2 and 4 on thereal axis for the minimum gain yields the break-in

at s 2:89.e. Searching along z 0:5 for the 1808 point we find s 2:42 j4:18.f. For the result in part (e), K 0:108.g. Using the result from part (c) and the root locus, K< 1.

8.6

a.

s

jwz= 0.591

246XXX

0

s-plane



26/53

b. Searching along the z 0:591 (10% overshoot) line for the 1808 point yields2:028 j2:768 with K 45:55.

c. Ts 4

jRe

j

42:028

1:97 s; Tp p

jIm

j

p2:768

1:13 s; vnTr 1:8346 from therise-time chart and graph in Chapter 4. Since vn is the radial distance to the pole,

vn ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

2:0282 2:7682p 3:431. Thus, Tr 0:53 s; since the system is Type 0,Kp K

246 45:55

48 0:949. Thus,

estep1 11 Kp 0:51:

d. Searching the real axis to the left of6 for the point whose gain is 45.55, we find7:94. Comparing this value to the real part of the dominant pole,2:028, we findthat 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 yieldspi as the root locusvariable. Thus,

Ts Gs1 Gs

100

s2 pis 100 100

s2 100 pis 100

s2 1001 pis

s2 100Hence, KGsHs pis

s2 100. The following shows the root locus.

s

jw

j10X

O

s-plane

0

Xj10

8.8

Following the rules for plotting the root locus of positive-feedback systems, we obtain

the following root locus:



27/53

s

jw

24X

s-plane

1XXo

03

8.9

The closed-loop transfer function is Ts Ks 1s2 K 2s K. Differentiating the

denominator with respect to K yields

2s@s

@K K 2 @s

@K s 1 2s K 2 @s

@K s 1 0

Solving for@s

@K, we get@s

@K s

12s K 2. Thus, Ss:K

K

s

@s

@K K

s

1s2s K 2 :

Substituting K 20 yields Ss:K 10s 1ss 11 .

Now find the closed-loop poles when K 20. From the denominator ofTs; s1;2 21:05; 0:95, when K 20.For the pole at 21:05,

Ds sSs:KDKK

21:05 1021:05 121:0521:05 11

0:05 0:9975:

For the pole at 0:95,

Ds sSs:KDK

K 0:95 10

0:95

10:950:95 11

0:05 0:0025:

CHAPTER 9

9.1

a. Searchingalong the15% overshootline, we find the pointon therootlocus at3:5 j5:8 at a gain of K 45:84. Thus, for the uncompensated system,Kv lim

s!0sGs K=7 45:84=7 6:55.

Hence, eramp uncompensated1 1=Kv 0:1527.b. Compensator zero should be 20x further to the left than the compensator pole.

Arbitrarily select Gcs s 0:2s 0:01.

c. Insert compensator and search along the 15% overshoot lineand find the root locus at

3:4 j5:63 with a gain, K 44:64. Thus, for the compensated system,



28/53

Kv 44:640:270:01 127:5and eramp compensated11

Kv 0:0078.

d.eramp uncompensated

eramp compensated 0:1527

0:0078 19:58

9.2

a. Searchingalong the 15% overshoot line, wefind thepoint on the root locus at3:5 j5:8 at a gain ofK 45:84. Thus, for the uncompensated system,

Ts 4jRej 4

3:5 1:143 s:

b. Thereal part of the design point must be three times largerthan the uncompensated

poles real part.Thus thedesignpoint is 33:5 j 35:8 10:5 j 17:4. Theangular contribution of the plants poles and compensator zero at the design pointis 130:8. Thus, the compensator pole must contribute 180 130:8 49:2.Using the following diagram,

pc

s

jw

s-plane

10.5

j17.4

49.2

we find17:4

pc 10:5 tan 49:2, from which, pc 25:52. Adding this pole, we find

the gain at the design point to be K 476:3. A higher- order closed-loop pole isfoundtobeat

11

:54. This pole may not beclose enough to the closed-loopzero at

10. Thus, we should simulate thesystem to be sure thedesign requirementshavebeen met.

9.3

a. Searchingalong the 20% overshoot line, wefind thepoint on the root locus at3:5 6:83 at a gain of K 58:9. Thus, for the uncompensated system,

Ts 4jRej 4

3:5 1:143s:

b. For the uncompensated system, Kv lims!0

sGs K=7 58:9=7 8:41. Hence,

eramp uncompensated1 1=Kv 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 j 13:66. Adding the angles from theplants poles andthe compensators zero at 3 to the design point, we obtain 100:8. Thus, thecompensator pole must contribute 180 100:8 79:2. Using the following



29/53

diagram,

pc

s

jw

s-plane

79.2

7

j13.66

we find13:66

pc 7 tan79:2, from which,pc 9:61. Adding this pole,we find thegain

at the design point to be K 204:9.

Evaluating Kv for the lead-compensated system:

Kv lims!0

sGsGlead K3=79:61 204:93=79:61 9:138:Kv for the uncompensated system was 8.41. For a 10x improvement in steady-state error, Kv must be 8:4110 84:1. Since lead compensation gave usKv 9:138, we need an improvement of 84:1=9:138 9:2. Thus, the lag com-pensator zero should be 9.2x further to the left than the compensator pole

Arbitrarily select Gcs s 0:092s 0:01 .Using all plant and compensator poles, we find the gain at the design point to

be K 205:4. Summarizing the forward path with plant, compensator, andgainyields

Ge

s

205:4s 3s 0:092ss 79:61s 0:01

:

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

9.4

The configuration for the system is shown in the

figure below.

1

s(s+ 7)(s +10)

R(s) C(s)+K

+

Kfs



30/53

Minor-Loop Design:

For the minor loop,GsHs Kfs 7s 10. Using thefollowing diagram, we findthat the minor-looproot locus intersects the 0.7 dampingratio lineat8:5 j 8:67.Theimaginary part was found as follows: u

cos1z

45:57. Hence,

Im

8:5 tan 45:57,

from which Im 8:67.

s

jw

s-plane

7

z= 0.7

X X

10 8.5

(-8.5 + j8.67)

q

Im

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

Kf ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1:52 8:672p ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

1:52 8:672p

77:42

Major-Loop Design:Using the closed-loop poles of the minor loop, we have an equivalent forward-path

transfer function of

Ges Kss 8:5 j8:67s 8:5 j8:67

K

ss2 17s 147:4 :

Usingthe threepoles ofGe(s) asopen-loop poles toplota root locus,we searchalong

z 0:5 and find that the root locus intersects this damping ratio line at 4:34 j7:51 at a gain, K 626:3.

9.5

a. An active PID controller must be used. We use the circuit shown in the following

figure:

+

Z1(s)

Z2(s)

I1(s)

V1(s)

Vo(s)

Vi(s)

Ia(s)

I2(s)

where the impedances are shown below as follows:



31/53

C1

R1

Z1(s) Z2(s)

C2R2

Matching the given transfer function with the transfer function of the PID controller

yields

Gcs s 0:1s 5s

s2 5:1s 0:5

s s 5:1 0:5

8

"

R2

R1 C1

C2

R2C1s

1R1C2

s

#

Equating coefficients

1

R1C2 0:5 (1)

R2C1 1 (2)

R2

R1 C1

C2

5:1 (3)

In Eq. (2) we arbitrarily let C1 105. Thus, R2 105. Using these values along withEqs. (1) and (3) we find C2 100mF and R1 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

C1R2

C2

+

+

vo(t)vi(t)

Matching thegiven transfer functionwith thetransfer function of thepassive lag-lead

compensator yields

Gcs s 0:1s 2s 0:01s 20 s 0:1s 2

s2 20:01s 0:2

s 1

R1C1

s 1

R2C2

s2 1

R1C1 1

R2C2 1

R2C1

s 1

R1R2C1C2



32/53

Equating coefficients

1

R1C1 0:1 (1)

1

R2C2 0:1 (2)

1

R1C1 1

R2C2 1

R2C1

20:01 (3)

Substituting Eqs. (1) and (2) in Eq. (3) yields

1

R2C1 17:91 (4)

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

CHAPTER 10

10.1

a.

Gs 1s 2s 4; Gjv 1

8 v2 j6v

Mv ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi8 v22 6v2

qFor v

solutions to skill assessment problems

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