T W O

Modeling in the

Frequency Domain

SOLUTIONS TO CASE STUDIES CHALLENGES

Antenna Control: Transfer Functions

Finding each transfer function:

Pot: θ

i

i

V (s)

(s) =

π

10;

Pre-Amp: p

i

V (s)

V (s) = K;

Power Amp:

a

p

E (s)

V (s) =

150

s 150

Motor: Jm = 0.05 + 5( 50

250)

2 = 0.25

Dm =0.01 + 3( 50

250)

2 = 0.13

t

a

K

R =

1

5

t b

a

K K

R =

1

5

Therefore: θm

a

(s)

E (s) =

t

a m

t bm

m a

K

R J

K K1s(s (D ))

J R

=0.8

s(s 1.32)

And:

θO

a

(s)

E (s) =

θm

a

(s)1

5 E (s) =

0.16

s(s 1.32)

Transfer Function of a Nonlinear Electrical Network

Writing the differential equation,

200

d(i i)2(i i) 5 v(t)

dt. Linearizing i

2 about i0,

δ δ δ δ δ

0

2 2 2 2

0 0 0 0 0 0i i

(i i) i 2i | i 2i i. Thus, (i i) i 2i i.

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2-2 Chapter 2: Modeling in the Frequency Domain

Substituting into the differential equation yields, δd i

dt + 2i0

2 + 4i0i - 5 = v(t). But, the

resistor voltage equals the battery voltage at equilibrium when the supply voltage is zero since

the voltage across the inductor is zero at dc. Hence, 2i02 = 5, or i0 = 1.58. Substituting into the linearized

differential equation, δd i

dt + 6.32i = v(t). Converting to a transfer function,

δi(s)

V(s) =

1

s 6.32 . Using

the linearized i about i0, and the fact that vr(t) is 5 volts at equilibrium, the linearized vr(t) is vr(t) = 2i2 =

2(i0+i)2 = 2(i02+2i0i) = 5+6.32i. For excursions away from equilibrium, vr(t) - 5 = 6.32i = vr(t).

Therefore, multiplying the transfer function by 6.32, yields, rV (s)

V(s)

=

6.32

s 6.32 as the transfer function

about v(t) = 0.

ANSWERS TO REVIEW QUESTIONS

1. Transfer function

2. Linear time-invariant

3. Laplace

4. G(s) = C(s)/R(s), where c(t) is the output and r(t) is the input.

5. Initial conditions are zero

6. Equations of motion

7. Free body diagram

8. There are direct analogies between the electrical variables and components and the mechanical variables

and components.

9. Mechanical advantage for rotating systems

10. Armature inertia, armature damping, load inertia, load damping

11. Multiply the transfer function by the gear ratio relating armature position to load position.

12. (1) Recognize the nonlinear component, (2) Write the nonlinear differential equation, (3) Select the

equilibrium solution, (4) Linearize the nonlinear differential equation, (5) Take the Laplace transform of

the linearized differential equation, (6) Find the transfer function.

SOLUTIONS TO PROBLEMS

1.

a.

00

1 1( ) st stF s e dt e

s s

b. 2 20

00

( 1)( ) ( 1)

stst

st

e stF s te dt st

s s e

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Solutions to Problems 2-3

Using L'Hopital's Rule

3 2

1( ) 0. Therefore, ( ) .

sttt

sF s F s

s e s

c. 2 2 2 2

0 0

( ) sin ( sin cos )st

st eF s t e dt s t t

s s

d. 2 2 2 2

0 0

( ) cos ( cos sin )st

st e sF s t e dt s t t

s s

2.

a. Using the frequency shift theorem and the Laplace transform of sin t, F(s) = ω

ω2 2(s+a) +.

b. Using the frequency shift theorem and the Laplace transform of cos t, F(s) = ω2 2

(s+a)

(s+a) +.

c. Using the integration theorem, and successively integrating u(t) three times, dt = t; tdt =

2t

2;

2t2

dt =

3t

6 , the Laplace transform of t3u(t), F(s) =

4

6

s .

3.

a. Taking the sum of the voltages around the loop and assuming zero initial conditions yields:

0

( ) 1( ) ( ) ( )

tdi t

Ri t L i d v tdt C

b. Applying Laplace transform and solving for I(s)/V(s) gives:

( ) 1 1

1 1( )( )

I s

RV sLs R L s

Cs L LCs

Substituting the values of R, L, and LC, we have:

2

( ) 2 2

16( ) 2 16( 2 )

I s s

V s s ss

s

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2-4 Chapter 2: Modeling in the Frequency Domain

Solving for I(s) and noting that V(s) = 1/s, we get:

2

2( )

2 16I s

s s

Observing that the denominator has complex roots, we re-write the above equation as:

2 2

2( )

( 1) ( 15)I s

s

Applying the frequency shift theorem to the Laplace transform of sin t u(t), we find that the

transform for ( ) sin( )atf t e t is 2 2

( )( )

F ss a

.

Comparing F(s) to I(s), we conclude that in the latter: a = 1 and 15 . Thus, the current, i(t),

may be given by:

2( ) 15 sin( 15 )

15

ti t e t

c.

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Solutions to Problems 2-5

4.

a. The Laplace transform of the differential equation, assuming zero initial conditions,

is, (s+7)X(s) = 2 2

5s

s 2. Solving for X(s) and expanding by partial fractions,

2 2

5 35 1 5 7 4

53 7 53( 7)( 4) 4

s s

ss s s

Or,

2 2

5 35 1 5 7 2 4

53 7 53( 7)( 4) 4

s s

ss s s

Taking the inverse Laplace transform, x(t) = -35

53 e-7t + (

35

53 cos 2t +

10

53 sin 2t).

b. The Laplace transform of the differential equation, assuming zero initial conditions, is,

(s2+6s+8)X(s) = 2

15

s 9.

Solving for X(s)

2 2

15X(s)

(s 9)(s 6s 8)

and expanding by partial fractions,

2

16s 9

3 3 1 15 19X(s)

65 10 s 4 26 s 2s 9

Taking the inverse Laplace transform,

4t 2t18 1 3 15x(t) cos(3t) sin(3t) e e

65 65 10 26

c. The Laplace transform of the differential equation is, assuming zero initial conditions,

(s2+8s+25)x(s) = 10

s. Solving for X(s)

2

10X(s)

s(s 8 s 25)

and expanding by partial fractions,

2

41(s 4) 9

2 1 2 9X(s) -

5 5 s 4 9s

Taking the inverse Laplace transform,

42 8 2( ) sin(3 ) cos(3 )

5 15 5

tx t e t t

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2-6 Chapter 2: Modeling in the Frequency Domain

5.

a. Taking the Laplace transform with initial conditions, s2X(s)-4s+4+2sX(s)-8+2X(s) = 2 2

2

s 2.

Solving for X(s),

X(s) =

3 2

2 2

4 4 16 18

( 4)( 2 2)

s s s

s s s

.

Expanding by partial fractions

2 2 2

1s 2

1 1 21(s 1) 22X(s)5 s 2 5 (s 1) 1

Therefore, 1 2 1

( ) 21 cos sin sin2 cos25 21 2

t tx t e t e t t t

b. Taking the Laplace transform with initial conditions, s2X(s)-4s-1+2sX(s)-8+X(s) = 5

s 2 +

2

1

s .

Solving for X(s), 4 3 2

2 2

4 17 23 2( )

( 1) ( 2)

s s s sX s

s s s

2 2

1 2 11 1 5( )

( 1) ( 2)( 1)X s

s s ss s

Therefore2( ) 2 11 5t t tx t t te e e .

c. Taking the Laplace transform with initial conditions, s2X(s)-s-2+4X(s) = 3

2

s . Solving for X(s),

4 3

3 2

2 3 2( )

( 4)

s sX s

s s

2 3

17 3*2

1/ 2 1/88 2( )4

s

X sss s

Therefore 217 3 1 1( ) cos2 sin2

8 2 4 8x t t t t .

6.

Program: syms t 'a' theta=45*pi/180 f=8*t^2*cos(3*t+theta); pretty(f) F=laplace(f); F=simple(F); pretty(F) 'b' theta=60*pi/180 f=3*t*exp(-2*t)*sin(4*t+theta); pretty(f)

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Solutions to Problems 2-7

F=laplace(f); F=simple(F); pretty(F)

Computer response:

ans =

a

theta =

0.7854

2 / PI \

8 t cos| -- + 3 t |

\ 4 /

1/2 2

8 2 (s + 3) (s - 12 s + 9)

------------------------------

2 3

(s + 9)

ans =

b

theta =

1.0472

/ PI \

3 t sin| -- + 4 t | exp(-2 t)

\ 3 /

1/2 2

1/2 1/2 3 3 s

12 s + 6 3 s - 18 3 + --------- + 24

2

------------------------------------------

2 2

(s + 4 s + 20)

7.

Program:

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2-8 Chapter 2: Modeling in the Frequency Domain

syms s

'a'

G=(s^2+3*s+10)*(s+5)/[(s+3)*(s+4)*(s^2+2*s+100)];

pretty(G)

g=ilaplace(G);

pretty(g)

'b'

G=(s^3+4*s^2+2*s+6)/[(s+8)*(s^2+8*s+3)*(s^2+5*s+7)];

pretty(G)

g=ilaplace(G);

pretty(g)

Computer response: ans =

a

2

(s + 5) (s + 3 s + 10)

--------------------------------

2

(s + 3) (s + 4) (s + 2 s + 100)

/ 1/2 1/2 \

| 1/2 11 sin(3 11 t) |

5203 exp(-t) | cos(3 11 t) - -------------------- |

20 exp(-3 t) 7 exp(-4 t) \ 57233 /

------------ - ----------- + ------------------------------------------------------

103 54 5562

ans =

b

3 2

s + 4 s + 2 s + 6

-------------------------------------

2 2

(s + 8) (s + 8 s + 3) (s + 5 s + 7)

/ 1/2 1/2 \

| 1/2 4262 13 sinh(13 t) |

1199 exp(-4 t) | cosh(13 t) - ------------------------ |

\ 15587 /

----------------------------------------------------------- -

417

/ / 1/2 \ \

| 1/2 | 3 t | |

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Solutions to Problems 2-9

| / 1/2 \ 131 3 sin| ------ | |

/ 5 t \ | | 3 t | \ 2 / |

65 exp| - --- | | cos| ------ | + ---------------------- |

\ 2 / \ \ 2 / 15 / 266 exp(-8 t)

---------------------------------------------------------- - -------------

4309 93

8. The Laplace transform of the differential equation, assuming zero initial conditions, is,

(s3+3s2+5s+1)Y(s) = (s3+4s2+6s+8)X(s).

Solving for the transfer function, ( )

( )

Y s

X s=

3 2

3 2

4 6 8

3 5 1

s s s

s s s

.

9.

a. Cross multiplying, (s2+5s+10)X(s) = 7F(s).

Taking the inverse Laplace transform,

2

2

d x

dt+ 5

dx

dt+ 10x = 7f.

b. Cross multiplying after expanding the denominator, (s2+21s+110)X(s) = 15F(s).

Taking the inverse Laplace transform,

2

2

d x

dt+ 21

dx

dt+ 110x =15f.

c. Cross multiplying, (s3+11s2+12s+18)X(s) = (s+3)F(s).

Taking the inverse Laplace transform,

3

3

d x

dt+ 11

2

2

d x

dt+ 12

dx

dt+ 18x =

df

dt+3f.

10.

The transfer function is ( )

( )

C s

R s=

5 4 3 2

6 5 4 3 2

2 4 4

7 3 2 5

s s s s

s s s s s.

Cross multiplying, (s6+7s

5+3s

4+2s

3+s

2+5)C(s) = (s

5+2s

4+4s

3+s

2+4)R(s).

Taking the inverse Laplace transform assuming zero initial conditions,

6

6

d c

dt+ 7

5

5

d c

dt+ 3

4

4

d c

dt+ 2

3

3

d c

dt+

2

2

d c

dt+ 5c =

5

5

d r

dt+ 2

4

4

d r

dt+ 4

3

3

d r

dt+

2

2

d r

dt+ 4r.

11.

The transfer function is ( )

( )

C s

R s=

4 3 2

5 4 3 2

2 5 1

3 2 4 5 2

s s s s

s s s s s.

Cross multiplying, (s5+3s

4+2s

3+4s

2+5s+2)C(s) = (s

4+2s

3+5s

2+s+1)R(s).

Taking the inverse Laplace transform assuming zero initial conditions,

5

5

d c

dc+ 3

4

4

d c

dt+ 2

3

3

d c

dt+ 4

2

2

d c

dt+ 5

dc

dt+ 2c =

4

4

d r

dt+ 2

3

3

d r

dt+ 5

2

2

d r

dt+

dr

dt+ r.

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2-10 Chapter 2: Modeling in the Frequency Domain

Substituting r(t) = t3,

5

5

d c

dc+ 3

4

4

d c

dt+ 2

3

3

d c

dt+ 4

2

2

d c

dt+ 5

dc

dt+ 2c

= 18(t) + (36 + 90t + 9t2 + 3t

3) u(t).

12.

Taking Laplace transform of the differential equation:

2 ( ) 1 4 ( ) 4 5 ( ) ( )s X s s sX s X s R s

Collecting terms: 2( 4 5) ( ) ( ) 3s s X s R s s

Solving for X(s), 2 2

( ) 3( )

4 5 4 5

R s sX s

s s s s

The block diagram is shown below, where R(s) = 1/s.

13.

Program: 'Factored'

Gzpk=zpk([-15 -26 -72],[0 -55 roots([1 5 30])' roots([1 27 52])'],5)

'Polynomial'

Gp=tf(Gzpk)

Computer response: ans =

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Solutions to Problems 2-11

Factored

Zero/pole/gain:

5 (s+15) (s+26) (s+72)

--------------------------------------------

s (s+55) (s+24.91) (s+2.087) (s^2 + 5s + 30)

ans =

Polynomial

Transfer function:

5 s^3 + 565 s^2 + 16710 s + 140400

--------------------------------------------------------------------

s^6 + 87 s^5 + 1977 s^4 + 1.301e004 s^3 + 6.041e004 s^2 + 8.58e004 s

14.

Program: 'Polynomial'

Gtf=tf([1 25 20 15 42],[1 13 9 37 35 50])

'Factored'

Gzpk=zpk(Gtf)

Computer response: ans =

Polynomial

Transfer function:

s^4 + 25 s^3 + 20 s^2 + 15 s + 42

-----------------------------------------

s^5 + 13 s^4 + 9 s^3 + 37 s^2 + 35 s + 50

ans =

Factored

Zero/pole/gain:

(s+24.2) (s+1.35) (s^2 - 0.5462s + 1.286)

------------------------------------------------------

(s+12.5) (s^2 + 1.463s + 1.493) (s^2 - 0.964s + 2.679)

15.

Program: numg=[-5 -70];

deng=[0 -45 -55 (roots([1 7 110]))' (roots([1 6 95]))'];

[numg,deng]=zp2tf(numg',deng',1e4);

Gtf=tf(numg,deng)

G=zpk(Gtf)

[r,p,k]=residue(numg,deng)

Computer response:

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2-12 Chapter 2: Modeling in the Frequency Domain

Transfer function:

10000 s^2 + 750000 s + 3.5e006

-------------------------------------------------------------------------------

s^7 + 113 s^6 + 4022 s^5 + 58200 s^4 + 754275 s^3 + 4.324e006 s^2 + 2.586e007 s

Zero/pole/gain:

10000 (s+70) (s+5)

------------------------------------------------

s (s+55) (s+45) (s^2 + 6s + 95) (s^2 + 7s + 110)

r =

-0.0018

0.0066

0.9513 + 0.0896i

0.9513 - 0.0896i

-1.0213 - 0.1349i

-1.0213 + 0.1349i

0.1353

p =

-55.0000

-45.0000

-3.5000 + 9.8869i

-3.5000 - 9.8869i

-3.0000 + 9.2736i

-3.0000 - 9.2736i

0

k =

[]

16.

Program: syms s

'(a)'

Ga=45*[(s^2+37*s+74)*(s^3+28*s^2+32*s+16)]...

/[(s+39)*(s+47)*(s^2+2*s+100)*(s^3+27*s^2+18*s+15)];

'Ga symbolic'

pretty(Ga)

[numga,denga]=numden(Ga);

numga=sym2poly(numga);

denga=sym2poly(denga);

'Ga polynimial'

Ga=tf(numga,denga)

'Ga factored'

Ga=zpk(Ga)

'(b)'

Ga=56*[(s+14)*(s^3+49*s^2+62*s+53)]...

/[(s^2+88*s+33)*(s^2+56*s+77)*(s^3+81*s^2+76*s+65)];

'Ga symbolic'

pretty(Ga)

[numga,denga]=numden(Ga);

numga=sym2poly(numga);

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Solutions to Problems 2-13

denga=sym2poly(denga);

'Ga polynimial'

Ga=tf(numga,denga)

'Ga factored'

Ga=zpk(Ga)

Computer response: ans =

(a)

ans =

Ga symbolic

2 3 2

(s + 37 s + 74) (s + 28 s + 32 s + 16)

45 -----------------------------------------------------------

2 3 2

(s + 39) (s + 47) (s + 2 s + 100) (s + 27 s + 18 s + 15)

ans =

Ga polynimial

Transfer function:

45 s^5 + 2925 s^4 + 51390 s^3 + 147240 s^2 + 133200 s + 53280

--------------------------------------------------------------------------------

s^7 + 115 s^6 + 4499 s^5 + 70700 s^4 + 553692 s^3 + 5.201e006 s^2 + 3.483e006 s

+ 2.75e006

ans =

Ga factored

Zero/pole/gain:

45 (s+34.88) (s+26.83) (s+2.122) (s^2 + 1.17s + 0.5964)

-----------------------------------------------------------------

(s+47) (s+39) (s+26.34) (s^2 + 0.6618s + 0.5695) (s^2 + 2s + 100)

ans =

(b)

ans =

Ga symbolic

3 2

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2-14 Chapter 2: Modeling in the Frequency Domain

(s + 14) (s + 49 s + 62 s + 53)

56 ----------------------------------------------------------

2 2 3 2

(s + 88 s + 33) (s + 56 s + 77) (s + 81 s + 76 s + 65)

ans =

Ga polynimial

Transfer function:

56 s^4 + 3528 s^3 + 41888 s^2 + 51576 s + 41552

--------------------------------------------------------------------------------

s^7 + 225 s^6 + 16778 s^5 + 427711 s^4 + 1.093e006 s^3 + 1.189e006 s^2

+ 753676 s + 165165

ans =

Ga factored

Zero/pole/gain:

56 (s+47.72) (s+14) (s^2 + 1.276s + 1.111)

---------------------------------------------------------------------------

(s+87.62) (s+80.06) (s+54.59) (s+1.411) (s+0.3766) (s^2 + 0.9391s + 0.8119)

17.

a. Writing the node equations,

0o i oo

V V VV

s s. Solve for

1

2

o

i

V

V s.

b. Thevenizing,

Using voltage division,

1

( )( )

1 12

2

io

V s sV s

ss

. Thus, 2

( ) 1

( ) 2 2

o

i

V s

V s s s

18.

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Solutions to Problems 2-15

a.

Writing mesh equations

1 2 i(2s 2)I (s) 2 I (s) V (s)

1 2-2I (s) (2s 4)I (s) 0

But from the second equation, 1 2I (s) (s 2)I (s) . Substituting this in the first equation yields,

2 2 i(2s 2)(s 2)I (s) 2 I (s) V (s)

or

2

2 iI (s) / V (s) 1/(2s 4s 2)

2

L 2 L iBut, V (s) sI (s). Therefore, V (s) / V (s) s /(2s 4s 2).

b.

1 2

1 2

4 2(4 ) ( ) (2 ) ( ) ( )

2 2(2 ) ( ) (4 2 ) ( ) 0

I s I s V ss s

I s s I ss s

Solving for I2(s):

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2-16 Chapter 2: Modeling in the Frequency Domain

2 2

2

4 4( )

(2 2)0

( )( )

4 4 (2 2) 4 6 2

(2 2) (2 4 2)

sV s

s

s

sV ssI s

s s s s

s s

s s s

s s

Therefore,

2 2

2

2 2

( ) 2 ( ) 2

( ) ( ) 4 6 2 2 3 1

LV s sI s s s

V s V s s s s s

19.

a.

Writing mesh equations,

(2s + 1)I1(s) – I2(s) = Vi(s)

-I1(s) + (3s + 1 + 2/s)I2(s) = 0

Solving for I2(s),

2

2

2 1 ( )

1 0( )

2 1 1

3 21

is V s

I ss

s s

s

Solving for I2(s)/Vi(s),

2

3 2

( )

( ) 6 5 4 2i

I s s

V s s s s

But Vo(s) = I2(s)3s. Therefore, G(s) = 3s2/(6s

3 + 5s

2 +4s + 2).

b. Transforming the network yields,

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Solutions to Problems 2-17

Writing the loop equations,

1 2 32 2( ) ( ) ( ) ( ) ( )

1 1i

s ss I s I s sI s V s

s s

1 2 32 2

1( ) ( 1 ) ( ) ( ) 0

1 1

s sI s I s I s

s s s

1 2 3( ) ( ) (2 1) ( ) 0sI s I s s I s

Solving for I2(s),

2

2 4 3 2

( 2 2)( ) ( )

2 3 3 2i

s s sI s V s

s s s s

But, Vo(s) = 2I (s)

s =

2

4 3 2

( 2 2)( )

2 3 3 2i

s sV s

s s s s. Therefore,

2

4 3 2

( ) 2 2

( ) 2 3 3 2

o

i

V s s s

V s s s s s

20. a. Writing the nodal equations yields,

( ) ( )( ) ( ) ( )0

2 1 3

1 1 1( ) ( ) 0

3 2 3

R CR i R

R C

V s V sV s V s V s

s s

V s s V ss s

Rewriting and simplifying,

2

6 5 1 1( ) ( ) ( )

6 3 2

1 3 2( ) ( ) 0

3 6

R C i

R C

sV s V s V s

s s s

sV s V s

s s

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2-18 Chapter 2: Modeling in the Frequency Domain

Solving for VR(s) and VC(s),

2

2 2

1 1 6 5 1( ) ( )

2 3 6 2

3 2 10 0

6 3( ) ; ( )

6 5 1 6 5 1

6 3 6 3

1 3 2 1 3 2

3 6 3 6

i i

R C

sV s V s

s s s s

s

s sV s V s

s s

s s s s

s s

s s s s

Solving for Vo(s)/Vi(s)

2

3 2

( ) ( ) ( ) 3

( ) ( ) 6 5 4 2

o R C

i i

V s V s V s s

V s V s s s s

b. Writing the nodal equations yields,

2

11 1

1

( ( ) ( )) ( 1)( ) ( ( ) ( )) 0

( ( ) ( ))( ( ) ( )) ( ) 0

io

o io o

V s V s sV s V s V s

s s

V s V sV s V s sV s

s

Rewriting and simplifying,

1

1

2 1( 1) ( ) ( ) ( )

1 1( ) ( 1) ( ) ( )

o i

o i

s V s V s V ss s

V s s V s V ss s

Solving for Vo(s)

Vo(s) =

2

4 3 2

( 2 2)( )

2 3 3 2i

s sV s

s s s s

.

Hence,

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Solutions to Problems 2-19

2

4 3 2

( ) ( 2 2)

( ) 2 3 3 2

o

i

V s s s

V s s s s s

21.

a.

Mesh:

(4+4s)I1(s) - (2+4s)I2(s) - 2I3(s) = V(s)

- (2+4s)I1(s) + (14+10s)I2(s) - (4+6s)I3(s) = 0

-2I1(s) - (4+6s)I2(s) + (6+6s+9

s)I3(s) = 0

Nodal:

11 1 ( ( ) ( ))( ( ) ( )) ( )0

2 2 4 4 6

oV s V sV s V s V s

s s

1( ( ) ( )) ( ) ( ( ) ( ))0

4 6 8 9 /

o o oV s V s V s V s V s

s s

or

2

12

6s + 12s + 5 1 1( ) ( ) ( )

6 4 212s 14 4oV s V s V s

ss

2

1

1 24s + 43s + 54( ) ( ) ( )

6 4 216 144 9o

sV s V s V s

s s

b.

Program: syms s V %Construct symbolic object for frequency

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2-20 Chapter 2: Modeling in the Frequency Domain

%variable 's' and V. 'Mesh Equations' A2=[(4+4*s) V -2 -(2+4*s) 0 -(4+6*s) -2 0 (6+6*s+(9/s))] %Form Ak = A2. A=[(4+4*s) -(2+4*s) -2 -(2+4*s) (14+10*s) -(4+6*s) -2 -(4+6*s) (6+6*s+(9/s))] %Form A. I2=det(A2)/det(A); %Use Cramer's Rule to solve for I2. Gi=I2/V; %Form transfer function, Gi(s) = I2(s)/V(s). G=8*Gi; %Form transfer function, G(s) = 8*I2(s)/V(s). G=collect(G); %Simplify G(s). 'G(s) via Mesh Equations' %Display label. pretty(G) %Pretty print G(s)

'Nodal Equations' A2=[(6*s^2+12*s+5)/(12*s^2+14*s+4) V/2 -1/(6*s+4) s*(V/9)] %Form Ak = A2. A=[(6*s^2+12*s+5)/(12*s^2+14*s+4) -1/(6*s+4) -1/(6*s+4) (24*s^2+43*s+54)/(216*s+144)] %Form A. Vo=simple(det(A2))/simple(det(A)); %Use Cramer's Rule to solve for Vo. G1=Vo/V; %Form transfer function, G1(s) = Vo(s)/V(s). G1=collect(G1); %Simplify G1(s). 'G(s) via Nodal Equations' %Display label. pretty(G1) %Pretty print G1(s)

Computer response:

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Solutions to Problems 2-21

ans =

Mesh Equations

A2 =

[ 4*s + 4, V, -2]

[ - 4*s - 2, 0, - 6*s - 4]

[ -2, 0, 6*s + 9/s + 6]

A =

[ 4*s + 4, - 4*s - 2, -2]

[ - 4*s - 2, 10*s + 14, - 6*s - 4]

[ -2, - 6*s - 4, 6*s + 9/s + 6]

ans =

G(s) via Mesh Equations

3 2

48 s + 96 s + 112 s + 36

----------------------------

3 2

48 s + 150 s + 220 s + 117

ans =

Nodal Equations

A2 =

[ (6*s^2 + 12*s + 5)/(12*s^2 + 14*s + 4), V/2]

[ -1/(6*s + 4), (V*s)/9]

A =

[ (6*s^2 + 12*s + 5)/(12*s^2 + 14*s + 4), -1/(6*s + 4)]

[ -1/(6*s + 4), (24*s^2 + 43*s + 54)/(216*s + 144)]

ans =

G(s) via Nodal Equations

3 2

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2-22 Chapter 2: Modeling in the Frequency Domain

48 s + 96 s + 112 s + 36

----------------------------

3 2

48 s + 150 s + 220 s + 117

22.

a.

5

1 6

5

2 6

1( ) 5 10

2 10

1( ) 10

2 10

Z s xx s

Z sx s

Therefore,

2

1

5( ) 1

( ) 5 1

sZ s

Z s s

b.

5 5

1

5 ( 5)( ) 10 1 10

sZ s

s s

5 5

2

5 ( 10)( ) 10 1 10

5 5

sZ s

s s

Therefore,

2

2

1

10( )

( ) 5

s sZ s

Z s s

23.

a.

5

1 6

5

2 6

1( ) 4 10

4 10

1( ) 1.1 10

4 10

Z s xx s

Z s xx s

Therefore,

1 2

1

( ) ( ) ( 0.98)( ) 1.275

( ) ( 0.625)

Z s Z s sG s

Z s s

b.

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Solutions to Problems 2-23

11

5

1 65

9

5

2 63

10

( ) 4 100.25 10

4 10

1027.5

( ) 6 100.25 10

110 10

sZ s xx

xs

sZ s xx

xs

Therefore,

2

1 2

2

1

( ) ( ) 2640 8420 4275

( ) 1056 3500 2500

Z s Z s s s

Z s s s

24. Writing the equations of motion, where x2(t) is the displacement of the right member of spring,

(5s2+4s+5)X1(s) -5X2(s) = 0

-5X1(s) +5X2(s) = F(s)

Adding the equations,

(5s2+4s)X1(s) = F(s)

From which, 1X (s) 1 1/ 5

F(s) s(5s 4) s(s 4 / 5)

.

25. Writing the equations of motion,

2

1 2

2

1 2

( 1) ( ) ( 1) ( ) ( )

( 1) ( ) ( 1) ( ) 0

s s X s s X s F s

s X s s s X s

Solving for X2(s),

2

2 2 22

2

( 1) ( )

( 1) 0 ( 1) ( )( )

( 2 2)( 1) ( 1)

( 1) ( 1)

s s F s

s s F sX s

s s ss s s

s s s

From which,

2

2 2

( ) ( 1)

( ) ( 2 2)

X s s

F s s s s

.

26.

Let X1(s) be the displacement of the left member of the spring and X3(s) be the displacement of the

mass. Writing the equations of motion, gives:

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2-24 Chapter 2: Modeling in the Frequency Domain

1 2

1 2 3

2

2 3

2 ( ) 2 ( ) ( )

2 ( ) (4 2) ( ) 4 ( ) 0

4 ( ) (8 6 ) ( ) 0

X s X s F s

X s s X s sX s

sX s s s X s

The third equation may be rewritten as: 2 3

2 ( ) (4 3) ( ) 0X s s X s

From which we get: 3 2

2( ) ( )

(4 3)X s X s

s

Substituting for X3(s) into the second equation and simplifying, gives the following

set of two equations:

1 2

2

1 2

2 ( ) 2 ( ) ( )

(4 3) ( ) (8 6 3) ( ) 0

X s X s F s

s X s s s X s

Solving for X2(s),

2 2

2

2 ( )

(4 3) 0 (4 3) ( )( )

2 2 2(8 6 3) 2(4 3)

(4 3) (8 6 3)

F x

s s F sX s

s s s

s s s

Thus,

27. 2

1 2

2

1 2

( 6 9) ( ) (3 5) ( ) 0

(3 5) ( ) (2 5 5) ( ) ( )

s s X s s X s

s X s s s X s F s

Solving for X1(s);

2

1 4 3 22

2

0 (3 5)

( ) (2 5 5) (3 5) ( )( )

2 17 44 45 20( 6 9) (3 5)

(3 5) (2 5 5)

s

F s s s s F sX s

s s s ss s s

s s s

Thus G(s) = X1(s)/F(s) =4 3 2

(3 5)

2 17 44 45 20

s

s s s s

28. Writing the equations of motion,

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Solutions to Problems 2-25

2

1 2

2

1 2 3

2

2 3

(4 2 6) ( ) 2 ( ) 0

2 ( ) (4 4 6) ( ) 6 ( ) ( )

6 ( ) (4 2 6) ( ) 0

s s X s sX s

sX s s s X s X s F s

X s s s X s

Solving for X3(s),

2

2

3 3 22

2

2

(4 2 6) 2 0

2 (4 4 6) ( )

0 6 0 3 ( )( )

(8 12 26 18)(4 2 6) 2 0

2 (4 4 6) 6

0 6 (4 2 6)

s s s

s s s F s

F sX s

s s s ss s s

s s s

s s

From which, 3

3 2

( ) 3

( ) (8 12 26 18)

X s

F s s s s s

.

29.

a.

2

1 2 3

2

1 2 3

1 2 3

(4s 8s 5)X (s) 8sX (s) 5X (s) F(s)

8sX (s) (4s 16s)X (s) 4sX (s) 0

5X (s) 4sX (s) (4s 5)X (s) 0

Solving for X3(s),

2

2 2

3

(4s 8s 5) -8s F(s)

8s (4s 16s) 0 8s (4s 16s)F(s)

5 -4s 0 5 4X (s)

s

or,

3

3 2

X (s) 13s 20

F(s) 4s(4s 25s 43s 15)

b.

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2-26 Chapter 2: Modeling in the Frequency Domain

2

1 2 3

2

1 2 3

1 2 3

(8s 4s 16)X (s) (4s 1)X (s) 15X (s) 0

(4s 1)X (s) (3s 20s 1)X (s) 16sX (s) F(s)

15X (s) 16sX (s) (16s 15)X (s) 0

Solving for X3(s),

2

2 2

3

(8s 4s 16) -(4s+1) 0

(4s+1) (3s 20s+1) F(s) (8s 4s 16) -(4s+1)-F(s)

15 -16s 0 15 16X (s)

s

or

3X (s)

F(s) =

3 2

5 4 3 2

128 64 316 15

384 1064 3476 165

s s s

s s s s

30.

Writing the equations of motion,

2

1 2 3

2

1 2 3

2

1 2

(4 4 8) ( ) 4 ( ) 2 ( ) 0

4 ( ) (5 3 4) ( ) 3 ( ) ( )

2 ( ) 3 ( ) (5 5 5) 0

s s X s X s sX s

X s s s X s sX s F s

sX s sX s s s

31.

Using the impedance method the two equations are:

1x : 2

1 1

mms k x x k F

mx : 1

m iso

x k Bs k x F

Solving both equations simultaneously, one gets

1

1 1 1

1 2 2 2 3

iso isoiso

F k

F Bs k F k F Bs k F FF Bs kx

k ms k Bs k k s mBs kms kB

k Bs k

ms k

32.

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Solutions to Problems 2-27

a.

Writing the equations of motion,

2

1 2

2

1 2

(5 9 9) ( ) ( 9) ( ) 0

( 9) ( ) (3 12) ( ) ( )

s s s s s

s s s s s T s

b.

Defining

1 1

2 1 1

3 3

4 2

( ) = rotation of

( ) = rotation between and

( ) = rotation of

( ) = rotation of right-hand side of

s J

s K D

s J

s K

the equations of motion are

2

1 1 1 1 2

1 1 1 1 2 1 3

2

1 2 2 1 2 3 2 4

2 3 2 2 3 4

( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( ) 0

( ) ( ( )) ( ) 0

J s K s K s T s

K s D s K s D s s

D s s J s D s K s K s

K s D s K K s

33. Writing the equations of motion,

2

1 2

1 2

( 2 1) ( ) ( 1) ( ) ( )

( 1) ( ) (2 1) ( ) 0

s s s s s T s

s s s s

Solving for 2( )s

2

2 2

( 2 1) ( )

( 1) 0 ( )( )

2 ( 1)( 2 1) ( 1)

( 1) (2 1)

s s T s

s T ss

s ss s s

s s

Hence,

2( ) 1

( ) 2 ( 1)

s

T s s s

34.

The corresponding impedance equations are:

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2-28 Chapter 2: Modeling in the Frequency Domain

1 : 2

1 21 1s s s T

2 : 2

1 21 2 0s s s

Solving for 1 one gets:

22

1 22 2 2

2

1

20 2

1 1 ( 1) 2 1

1 2

T s

T s ss s

s s s s s s s s

s s s

Simplifying: 2

1

4 3 2

2

2 3 1

s s

T s s s

35.

Reflecting impedances to 3,

(Jeqs2+Deqs)3(s) = T(s) ( 4 2

3 1

N N

N N)

Thus,

3 ( )

( )

s

T s

=

4 2

3 1

2

eq eq

N N

N N

J s D s

where

Jeq = J4+J5+(J2+J3) 4

3

N

N

2 + J1

4 2

3 1

N N

N N

2, and

2 24 4 2eq 4 5 2 3 1

3 3 1

N N ND (D D ) (D D )( ) D ( )

N N N

36.

Reflecting all impedances to 2(s),

{[J2+J1( 2

1

N

N)

2+J3 ( 3

4

N

N)

2]s2 + [f2+f1( 2

1

N

N)

2+f3( 3

4

N

N)

2]s + [K( 3

4

N

N)

2]}2(s) = T(s)

2

1

N

N

Substituting values,

{[1+2(3)2+16(1

4)

2]s2 + [2+1(3)2+32(

1

4)

2]s + 64(

1

4)

2}2(s) = T(s)(3)

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Solutions to Problems 2-29

Thus,

θ2 (s)

T(s) =

2

3

20s 13s 4

37.

Reflecting impedances across gears from the right hand side to the left hand side one gets:

2 25 5

3 100 150 925 20

eqJ

2 25 5

500 300 2325 50

eqD

25

3 300 650

eqK

So 29 23 6s s s T s . Since

2

2 1

10N

N, 2

29 23 6 1  0 s s s T s

2

2 2

1 0.011 0.011

0.29 2.2690 230 60 2.55 0.67

s

T s s ss s s s

38. Reflecting impedances and applied torque to respective sides of the spring yields the following

equivalent circuit:

Writing the equations of motion,

2(s) -2 3(s) = 4.231T(s)

-22(s) + (0.955s+2)3(s) = 0

Solving for 3(s),

3

2 4.231 ( )

2 0 8.462 ( ) 4.43 ( )( )

2 2 1.91

2 0.955 2

T s

T s T ss

s s

s

Hence,

3 ( ) 4.43

( )

s

T s s. But, 4 3( ) 0.192 ( )s s . Thus,

4 ( ) 0.851

( )

s

T s s.

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2-30 Chapter 2: Modeling in the Frequency Domain

39.

Reflecting the 0.02 Nm/rad damper towards the left we get

The corresponding impedance equations are:

1 : 2

1 2 12 2s s s T

2 : 1 2

2 2.32 2 0s s

Solving:

2

1

1

2 22

1 1

3 2 2 2 2

2

2

2 0 2

( 2 )(2.32 2) 42 2

2 2.32 2

2 2

2.32 2 4.64 4 4 2.32 2.64 4

s s T

s sT

s s s ss s s

s s

sT T

s s s s s s s

So

2

2

1

2

2.32 2.64 4T s s

Using the gear ratios we get 1

5 1

20 4

T

T and

2 10 1

40 4L

. It follows that

1

14

4 16

L

L L

T T T

. Finally

2 2

32 13.8

2.32 2.64 4 1.14 1.72

L

T s s s s

40.

Reflect all impedances on the right to the viscous damper and reflect all impedances and torques on the

left to the spring and obtain the following equivalent circuit:

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Solutions to Problems 2-31

Writing the equations of motion,

(J1eqs2+K)2(s) -K3(s) = Teq(s)

-K2(s)+(Ds+K)3(s) -Ds4(s) = 0

-Ds3(s) +[J2eqs2 +(D+Deq)s]4(s) = 0

where: J1eq = J2+(Ja+J1)( 2

1

N

N)

2 ; J2eq = J3+(JL+J4)(

3

4

N

N)

2 ; Deq = DL( 3

4

N

N)

2 ; 2(s) = 1(s)

1

2

N

N.

41. Reflect impedances to the left of J5 to J5 and obtain the following equivalent circuit:

Writing the equations of motion,

[Jeqs2+(Deq+D)s+(K2+Keq)]5(s) -[Ds+K2]6(s) = 0

-[K2+Ds]5(s) + [J6s2+2Ds+K2]6(s) = T(s)

From the first equation, θ

θ

6

5

(s)

(s) =

2

eq eq 2 eq

2

J s (D D)s (K K )

Ds K

. But,

θ

θ

5

1

(s)

(s) = 1 3

2 4

N N

N N.

Therefore,

θ

θ

6

1

(s)

(s) =

2

eq eq 2 eq1 3

2 4 2

J s (D D)s (K K )N N

N N Ds K

,

where Jeq = [J1( 4 2

3 1

N N

N N)

2 + (J2+J3)( 4

3

N

N)

2 + (J4+J5)], Keq = K1( 4

3

N

N)

2 , and

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2-32 Chapter 2: Modeling in the Frequency Domain

Deq = D[( 4 2

3 1

N N

N N)

2 + ( 4

3

N

N)

2 + 1].

42. Draw the freebody diagrams,

Write the equations of motion from the translational and rotational freebody diagrams,

(Ms2+2fv s+K2)X(s) -fvrs(s) = F(s)

-fvrsX(s) +(Js2+fvr2s)(s) = 0

Solve for (s),

θ

2

v 2

v v

3 2 2 2 2 22v v 2 v 2 vv 2 v

2 2

v v

Ms 2f s K F(s)

-f rs 0 f rF(s)(s)

JMs (2Jf Mf r )s (JK f r )s K f rMs 2f s K -f rs

-f rs Js f r s

From which, θ(s)

F(s) =

v

23 2 2 2 2

v v 2 v 2 v

f r

JMs (2Jf Mf r )s (JK f r )s K f r .

43. Draw a freebody diagram of the translational system and the rotating member connected to the

translational system.

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Solutions to Problems 2-33

From the freebody diagram of the mass, F(s) = (2s2+2s+3)X(s). Summing torques on the rotating

member,

(Jeqs2 +Deqs)(s) + F(s)2 = Teq(s). Substituting F(s) above, (Jeqs2 +Deqs)(s) + (4s2+4s+6)X(s) =

Teq(s). However, (s) = X(s)

2. Substituting and simplifying,

Teq = [( eqJ

2 +4)s2 +( eq

D

2 +4)s+6]X(s)

But, Jeq = 3+3(4)2 = 51, Deq = 1(2)2 +1 = 5, and Teq(s) = 4T(s). Therefore,

[ 59

2 s2 +

13

2s+6]X(s) = 4T(s). Finally,

X(s)

T(s) =

2

8

59 13 12s s .

44. Writing the equations of motion,

(J1s2+K1)1(s) - K12(s) = T(s)

-K11(s) + (J2s2+D3s+K1)2(s) +F(s)r -D3s3(s) = 0

-D3s2(s) + (J2s2+D3s)3(s) = 0

where F(s) is the opposing force on J2 due to the translational member and r is the radius of J2. But,

for the translational member,

F(s) = (Ms2+fvs+K2)X(s) = (Ms2+fvs+K2)r(s)

Substituting F(s) back into the second equation of motion,

(J1s2+K1)1(s) - K12(s) = T(s)

-K11(s) + [(J2 + Mr2)s2+(D3 + fvr2)s+(K1 + K2r2)]2(s) -D3s3(s) = 0

-D3s2(s) + (J2s2+D3s)3(s) = 0

Notice that the translational components were reflected as equivalent rotational components by the

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2-34 Chapter 2: Modeling in the Frequency Domain

square of the radius. Solving for 2(s),

2

1 3 32

( ) ( )( )

K J s D s T ss

where is the

determinant formed from the coefficients of the three equations of motion. Hence,

2

1 3 32 ( )( )

( )

K J s D ss

T s

Since

2

1 3 32

( )( )( ) ( ),

( )

rK J s D sX sX s r s

T s

45.

Reflecting through gears the inertia and damping from the load side to motor shaft one gets,

250

4 36 8150

mJ

and

250

50 36 54150

mD

Note from the motor load curve that 150

350

t stall

a a

K T

R e and

50 1

100 2

ab

no load

eK

.

Substituting all of the above, one gets

0.375

7.18751

t

a mm

a t bm

m a

K

R J

E s sK Ks s D

J R

Noting that 2

1

3m

L

N

N

0.125

7.1875

L

aE s s

46.

The parameters are:

5

15

t s

a a

K T

R E;

5 1

600 1 42

60

ab

EK ;

2 21 1

18 4 1 3.1254 2

mJ ;

21

36 2.254

mD

Thus,

1( ) 0.323.125

1 1( ) ( 0.8)( (2.25 (1)( )))

3.125 4

m

a

s

E s s ss s

Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

Solutions to Problems 2-35

Since: 2

1( ) ( )

4m

s s ; then:

2( ) 0.08

( ) ( 0.8)a

s

E s s s

47. The following torque-speed curve can be drawn from the data given:

Therefore, t

a

K

R= stall

a

T

E=

100

12 ; Kb = a

no load

E

= 12

1333.33. Also, Jm = 7+105( 1

6)

2 = 9.92; Dm =

3. Thus,

( )

( )

m

a

s

E s

=

100 1

12 9.92

1( (3.075))

9.92s s

= 0.84

( 0.31)s s . Since L(s) =

1

6m(s),

( )

( )

L

a

s

E s

=

0.14

( 0.31)s s .

48.

From Eqs. (2.45) and (2.46),

RaIa(s) + Kbs(s) = Ea(s) (1)

Also,

Tm(s) = KtIa(s) = (Jms2+Dms)(s). Solving for (s) and substituting into Eq. (1), and simplifying

yields

( )

( )

a

a

I s

E s=

( )1

m

m

a m b ta

a m

Ds

J

R D K KRs

R J

(2)

Using Tm(s) = KtIa(s) in Eq. (2),

Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

2-36 Chapter 2: Modeling in the Frequency Domain

( )

( )

m

a

T s

E s=

( )m

t m

a m b ta

a m

Ds

K J

R D K KRs

R J

49. For the rotating load, assuming all inertia and damping has been reflected to the load,

(JeqLs2+DeqLs)L(s) + F(s)r = Teq(s), where F(s) is the force from the translational system, r=2 is

the radius of the rotational member, JeqL is the equivalent inertia at the load of the rotational load and

the armature, and DeqL is the equivalent damping at the load of the rotational load and the armature.

Since JeqL = 1(2)2 +1 = 5, and DeqL = 1(2)2 +1 = 5, the equation of motion becomes, (5s2+5s)L(s) +

F(s)r = Teq(s). For the translational system, (s2+s)X(s) = F(s). Since X(s) = 2L(s), F(s) =

(s2+s)2L(s). Substituting F(s) into the rotational equation, (9s2+9s)L(s) = Teq(s). Thus, the

equivalent inertia at the load is 9, and the equivalent damping at the load is 9. Reflecting these back to

the armature, yields an equivalent inertia of 9

4 and an equivalent damping of

9

4. Finally, t

a

K

R = 1;

Kb = 1. Hence, θ

m

a

(s)

E (s) =

4

94 9

s(s ( 1))9 4

=

4

913

s(s )9

. Since L(s) = 1

2 m(s),

θL

a

(s)

E (s) =

2

913

s(s )9

. But X(s) = rL(s) = 2L(s). therefore,

a

X(s)

E (s) =

4

913

s(s )9

.

50.

The equations of motion in terms of velocity are:

1 2 21 1 3 1 2 3 3

2 21 2 2 4 2 4 3

3 1 4 2 3 3 4 3

[ ( ) ] ( ) ( ) ( ) 0

( ) [ ( ) ] ( ) ( ) ( )

( ) ( ) [ ] ( ) 0

v v v

v v v

v v V v

K K KM s f f V s V s f V s

s s s

K KV s M s f f V s f V s F s

s s

f V s f V s M s f f V S

For the series analogy, treating the equations of motion as mesh equations yields

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Solutions to Problems 2-37

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

For the parallel analogy, treating the equations of motion as nodal equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

51.

Writing the equations of motion in terms of angular velocity, (s) yields

1 11 1 1 1 2

1 1 21 1 2 1 2

2 22 2 3 2 4

33 2 3 2 4

( ) ( ) ( ) ( ) ( )

( )( ) ( ) ( ) ( ) 0

( ) ( ) ( ) ( ) 0

( ) ( ) ( ) 0

K KJ s D s D s T s

s s

K K KD s J s D s

s s

K Ks D s D s

s s

KJ s D s D s

s

For the series analogy, treating the equations of motion as mesh equations yields

Full file at http://TestBankSolutionManual.eu/Solution-Manual-for-Control-Systems-Engineering-7th-Edition-by-Nise

2-38 Chapter 2: Modeling in the Frequency Domain

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

For the parallel analogy, treating the equations of motion as nodal equations yields

In the circuit, resistors are in ohms, capacitors are in farads, and inductors are in henries.

52. An input r1 yields c1 = 5r1+7. An input r2 yields c2 = 5r2 +7. An input r1 +r2 yields, 5(r1+r2)+7 =

5r1+7+5r2 = c1+c2-7. Therefore, not additive. What about homogeneity? An input of Kr1 yields c =

5Kr1+7 ≠ Kc1. Therefore, not homogeneous. The system is not linear.

53.

a. Let x = x+0. Therefore,

δ δ δ δ

δ δ δ δ

δ δ δ δ δ δ δ

x 0 x 0

x 3 x 2 x sin (0 x)

d sin xBut, sin (0 x) sin 0 | x 0 cosx | x x

dx

Therefore, x 3 x 2 x x. Collecting terms, x 3 x x 0

.

b. Let x = x+ Therefore,

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Solutions to Problems 2-39

π π

δ δ δ π δ

π δ π δ δ δ

δ δ δ δ δ δ δ

x x

x 3 x 2 x sin( x)

d sin xBut, sin ( x) sin | x 0 cosx | x x

dx

Therefore, x 3 x 2 x - x. Collecting terms, x 3 x 3 x 0.

.

54.

The truncated Taylor series expansion of 53 0 0 3 15xf x e f f x x

Letting x x and substituting for f x one gets

3 2

3 210 20 15 3 15

d x d x d xx x

dtdt dt

Simplifying 3 2

3 210 20 30 3

d x d x d xx

dt dt dt

55. The given curve can be described as follows:

f(x) = -6 ; -∞<x<-3;

f(x) = 2x; -3<x<3;

f(x) = 6; 3<x<+∞

Thus,

a. 17 50 6

b. 17 50 2 or 17 48 0

c. 17 50 6

x x x

x x x x x x x

x x x

56.

The relationship between the nonlinear spring’s displacement, xs(t) and its force, fs(t) is

( )( ) 1 sf t

sx t e

Solving for the force, ( ) ln(1 ( ))s sf t x t (1)

Writing the differential equation for the system by summing forces,

2

2

( ) ( )2 2 ln(1 ( )) ( )

d x t dx tx t f t

dtdt (2)

Letting x(t) = x0 + x and f(t) = 1 + f, linearize ln(1 – x(t)).

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2-40 Chapter 2: Modeling in the Frequency Domain

0

0

ln(1 )ln(1 ) ln(1 )

x x

d xx x x

dx

Solving for ln(1 – x),

0

0 0

0

1 1ln(1 ) ln(1 ) ln(1 )

1 1x x

x x x x xx x

(3)

When f = 1, x = 0. Thus from Eq. (1), 1 = -ln(1 – x0 ).

Solving for x0, 1 – x0 = e-1

, or x0 = 0.6321.

Substituting x0 = 0.6321 into Eq. (3),

1

ln (1- ) =  ln(1 – 0.6321) -    x=-1-2.718 x1-0.6321

x

Placing this value into Eq. (2) along with x(t) = x0 + x and f(t) = 1 + f, yields the linearized differential

equation,

2

22 2 1 2.718 1

d x d xx f

dt dt

or 2

22 2 2.718

d x d xx f

dt dt

Taking the Laplace transform and rearranging yields the transfer function,

2

( ) 1

( ) 2 2 2.718

x s

f s s s

57.

a. The three equations are transformed into the Laplace domain:

0 SSs S k K C k S

( )M

Cs k S K C

2Ps k C

The three equations are algebraically manipulated to give:

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Solutions to Problems 2-41

0 Sk KS

S Cs k s k

M

SkC

s k K

2k

P Cs

By direct substitutions it is obtained that:

02 2

( )

(1 ) ( )

M

M M S

s k KS S

s k K s k K K

02 2( (1 ) ( ))M M S

kC S

s k K s k K K

2

02 2( (1 ) ( ))M M S

k kP S

s s k K s k K K

b.

0

( ) ( ) 0s

S Lim sS s

0( ) ( ) 0

sC Lim sC s

2 0 2 0020

2

( ) ( )( )

( )s

M SS S

k k S k SP Lim sP s S

kk K Kk K K

k

58.

Eliminate bal

T by direct substitution. This results in

2

2

0

( ) ( ) ( ) ( )

t

d

dJ kJ t J t J t dt T t

dt

Obtaining Laplace transform on both sides of this equation and eliminating terms one gets that:

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2-42 Chapter 2: Modeling in the Frequency Domain

59.

a.

We have that

L La Lm x m g

La T Lx x x

Lx L

From the second equation

La T L Tx x x v L g

Obtaining Laplace transforms on both sides of the previous equation

TsV Ls g from which

2( )T

sV g Ls

so that

2 2 22 0

1 1( )

T

s s ss

gV g Ls L L ss

L

b. Under constant velocity 0( )

T

VV s

s so the angle is

0

2 2

0

1( )

Vs

L s

Obtaining inverse Laplace transform

00

0

( ) sin( )V

t tL

, the load will sway with a frequency 0

.

c. From T T T Lm x f m g and Laplace transformation we get

22

2 2 2 2

0 0

1 1( ) ( )T T T L T L T T L T

s sm s X s F m g s F m g V F m g X

L Ls s

From which

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Solutions to Problems 2-43

2 2 2 2

0 0

2 2 2 2 2 2 22 0 0 0

2 2

0

1 1

1 ( ( ) ) ( )( )

T

LT TT LT

s sX

m gF ms m s m s s as m

L s

Where 1 L

T

ma

m

d. From part c 2 2

0

2 2

0

1

( )

T T

T T T

sV sX

F F m s s a

Let 0T

FF

s then

2 2

0 0

2 2 2 2 2 2

0 0

( )( )

T

T

F s A B Cs DV s

m s s a s s s a

After partial fraction expansions, so

0( ) ' 'cos( )

Tv t A t B C a t

From which it is clear that t

Tv

60.

a. Obtaining Laplace transforms on both sides of the equation

0( ) ( )sN s N KN s or

0( )N

N ss K

By inverse Laplace transformation

0( ) KtN t N e

b. Want to find the time at which

0 02KtN e N

Obtaining ln on both sides of the equation

ln 2t

K

61. The Laplace transform of the systems output is

2 2 2 2 2 2

2 2

4 4

ref ref refT T Ta f a f

T t T ss s s f s s s f

Dividing by the input one gets

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2-44 Chapter 2: Modeling in the Frequency Domain

2 2 2

2

4ref

T a f ss

U s T s f

62.

a. By direct differentiation (1 )

0

( )( ) ( )

atet tdV t

V e e e V tdt

b. (1 )

0 0( ) ( )

te

t t

V Lim V t Lim V e V e

c.

Lambda = 2.5;

alpha = 0.1;

V0=50;

t=linspace(0,100);

V=V0.*exp(Lambda.*(1-exp(-alpha.*t))/alpha);

plot(t,V)

grid

xlabel('t (days)')

ylabel('mm^3 X 10^-3')

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Solutions to Problems 2-45

d. From the figure 12( ) 3.5 10V X mm

3 X 10

-3

From part c

2.5

120.10

( ) 50 3.6 10V V e e X

mm3 X 10

-3

63. Using the impedance method the two equations are:

1 :x 2

1 1

mms k x x k F

:mx 1

m isox k Bs k x F

Solving both equations simultaneously, one gets

1

1 11

1 2 2 3

isoiso iso

F k

F Bs k F FF Bs k F Bs k F kx

k ms k Bs k k s mBs kms kB

k Bs k

2ms k

64. Opening the current source, we find the contribution of the voltage source, Va(s), to the ac current,

IacF1(s).

21

( ) ( )( ) ( )

1( ) 1

a aacF a

V s V s CsI s V s

Z s LCs RCsLs R

Cs

Short-circuiting the voltage source, Va(s), we find the contribution of the current source, IacR(s), to

the ac current, IacF2(s).

22

1

1( ) ( ) ( )

1 1acF acR acR

RRCsCsI s I s I s

LCs RCsLs R

Cs

Thus, the total current, IacF (s), is given by:

1 2 22

1( ) ( ) ( ) ( ) ( )

1 1acF acF acF acR a

RCs CsI s I s I s I s V s

LCs RCs LCs RCs

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2-46 Chapter 2: Modeling in the Frequency Domain

65. Writing the loop equation around the armature circuit for the motor in Figure 2.35:

a ma a a a b

di de t R i L K

dt dt

Taking the Laplace transform:

( ) ( ) ( ) ( )a a a a a b m

E s R I s L sI s K s s (1)

The torque developed at the motor is:

2 ( ) ( )

( ) ( )m mm m m m m

d t d tT t J D K t

dt dt

Taking the Laplace transform:

2 2( ) ( ) ( ) ( ) ( ) ( )m m m m m m m m m m m

T s J s s D s s K s J s D s K s

But ( ) ( )m t a

T s K I s . Solving for ( )a

I s and substituting for ( )m

T s

2( ) 1

( ) ( ) ( )ma m m m m

t t

T sI s J s D s K s

K K

Substituting in (1) for ( )a

I s and simplifying

21( ) ( )( ) ( )

a a a m m m t b m

t

E s R L s J s D s K K K s sK

Thus

3 2

( )

( ) ( ) ( )

tm

a m a m a m a m a m a t b m a

Ks

E s J L s J R D L s D R K L K K s K R

66.

a. Expressing 2eA

ghA

as a Taylor series around h0i

0

0 0

0

2 2 2 22

e e e e e

h

A A A A A ggh gh gh h gh h

A A h A A A gh

(1)

Also,

0h h h (2)

and

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Solutions to Problems 2-47

0q q q (3)

Substituting (1), (2), and (3) into the given nonlinear equation and eliminating

the equilibrium values yields the linear equation

02

eAd h g q

hdt A Agh

Thus the transfer function is

0

( ) 1 /

( )

2

e

H s A

Q s A gs

A gh

b. Substituting 2e e av

q A gh into 1( )

e av

de q eq Aeh

dt

12

e av av

dee q e A gh Ah

dt

Rearranging

12

av e av

deAh e A gh e q

dt

Simplifying,

12e

av

av

Adegh e e q

dt Ah

Taking the Laplace transform

1( ) 2 ( ) ( )e

av

av

AsE s gh E s e Q s

Ah

From which,

1( )

( )( 2 )e

av

av

eE s

AQ ss gh

Ah

67.

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2-48 Chapter 2: Modeling in the Frequency Domain

a. The first two equations are nonlinear because of the Tv products on their right hand side.

Otherwise the equations are linear.

b. To find the equilibria let

*

0dT dT dv

dt dt dt

Leading to

0s dT T

* 0Tv T

* 0kT cv

The first equilibrium is found by direct substitution. For the second equilibrium, solve the last two

equations for T*

* TvT

and

* cvT

k . Equating we get that

cT

k

Substituting the latter into the first equation after some algebraic manipulations we get that

ks dv

c . It follows that

* cv s cdT

k k .

68.

a. From w

m

F Fa

k m

, we have:

Ow m R L St mF F k m a F F F k m a (1)

Substituting for the motive force, F, and the resistances FRo, FL, and Fst using the equations given in

the problem, yields the equation:

2cos sin 0.5tot

whw m

PF f m g m g C A v v k m a

v

(2)

b. Noting that constant acceleration is assumed, the average values for speed and acceleration are:

aav = 20 (km/h)/ 4 s = 5 km/h.s = 5x1000/3600 m/s2 = 1.389 m/s

2

vav = 50 km/h = 50,000/3,600 m/s = 13.89 m/s

The motive force, F (in N), and power, P (in kW) can be found from eq. 2:

Fav = 0.011 x 1590 x 9.8 + 0.5 x 1.2 x 0.3 x 2 x 13.892 + 1.2 x 1590 x 1.389 = 2891 N

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Solutions to Problems 2-49

Pav = Fav. v / tot

= 2891 x 13.89 / 0.9 = 44, 617 N.m/s = 44.62 kW

To maintain a speed of 60 km/h while climbing a hill with a gradient = 5o, the car engine or

motor needs to overcome the climbing resistance:

sin 1590 9.8 sin5 1358St

F m g N

Thus, the additional power, Padd, the car needs after reaching 60 km/h to maintain its speed while

climbing a hill with a gradient = 5o is:

/add F vSt

P = 1358 x 60 x 1000/(3,600 x 0.9) = 25, 149 W = 25.15 kW

c. Substituting for the car parameters into equation 2 yields:

20.011 x 1590 x 9.8 0.5 x 1.2 x 0.3 x 2 1.2 x 1590 /F v dv dt

or 2( ) 171.4 0.36 1908 /F t v dv dt (3)

To linearize this equation about vo = 50 km/h = 13.89 m/s, we use the truncated taylor series: 2

2 2 ( )( ) 2 ( )

o

o o o o

v v

d vv v v v v v v

dv (4), from which we obtain:

2 2 22 27.78 13.89o ov v v v v (5)

Substituting from equation (5) into (3) yields:

( ) 171.4 10 69.46 1908 /F t v dv dt or

( ) ( ) ( ) 171.4 69.46 10 1908 /e Ro o

F t F t F F F t v dv dt (6)

Equation (6) may be represented by the following block-diagram:

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2-50 Chapter 2: Modeling in the Frequency Domain

d. Taking the Laplace transform of the left and right-hand sides of equation (6) gives,

( ) 10 (s) 1908 (s)eF s V sV (7)

Thus the transfer function, Gv(s), relating car speed, V(s) to the excess motive force, Fe(s), when the

car travels on a level road at speeds around vo = 50 km/h = 13.89 m/s under windless conditions is:

e

(s) 1( )

(s) 10 1908 v

VG s

F s

(8)

69.

a.

Since the system’s transfer function exhibits a pure time delay of T seconds, the

unit step response of the system is the unit step response of a first order system

delayed T seconds, namely

1t T

h t K e u t T

b.

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Solutions to Problems 2-51

c.

The output will he delayed T seconds, thus writing

1

sTH s K

eQ s s

Then cross-multiplying

1 sTH s s e KQ s

And obtaining the inverse Laplace transform, one gets:

d

h t T h t T Kq tdt

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ONLINEFFIRS 11/25/2014 13:29:37 Page 1

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Library of Congress Cataloging-in-Publication Data

Nise, Norman S.Control systems engineering / Norman S. Nise, California State Polytechnic University, Pomona. — Seventh edition.

1 online resource.Includes bibliographical references and index.Description based on print version record and CIP data provided by publisher; resource not viewed.ISBN 978-1-118-80082-9 (pdf) — ISBN 978-1-118-17051-9 (cloth : alk. paper)1. Automatic control–Textbooks. 2. Systems engineering–Textbooks. I. Title.TJ213629.8–dc23

2014037468

Printed in the United States of America

10 9 8 7 6 5 4 3 2 1

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