Solution manual for system dynamics 3rd edition by palm

This work is only for non-profit use by instructors in courses for whichthe textbook has been adopted.. This work is only for non-profit use by instructors in courses for whichthe textbo

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Solutions Manual c

to accompany System Dynamics, Third Edition

by William J Palm III University of Rhode Island

Solutions to Problems in Chapter Two

creserved No part of this manual may be displayed, reproduced, or distributed

in any form or by any means without the written permission of the publisher

or used beyond the limited distribution to teachers or educators permitted byMcGraw-Hill for their individual course preparation Any other reproduction

or translation of this work is unlawful

From https://testbankgo.eu/p/Solution-Manual-for-System-Dynamics-3rd-Edition-by-Palm

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2.1 a) Nonlinear because of the y ¨y term b) Nonlinear because of the sin y term c)Nonlinear because of the√y term d) Variable coefficient, but Linear e) Nonlinear because

of the sin y term f) Variable coefficient, but linear

c 2014 McGraw-Hill This work is only for non-profit use by instructors in courses for whichthe textbook has been adopted Any other use without publisher’s consent is unlawful.

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2.2 a)

4

Z x 2

dx = 3

Z t 0

dx = 2

Z t 0

e−4tdtx(t) = 3.1 − 0.1e−4tc) Let v = ˙x

3

Z v 7

dv = 5

Z t 0

dx =

Z t 0

dv = 7

Z t 0

dx =

Z t 0

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2.3 a)

Z x 3

dx

25 − 5x2 =

Z t 0

dt = t

Z x 3

dx

25 − 5x2 =

√525

"

arctanh

√5x5

!

− arctanh 3

√55

!#

= tLet

C = arctanh 3

√55

Z x 10

dx

36 + 4x2 =

Z t 0

dt = t1

12tan

−1x3

x 10

= t

x(t) = 3 tan(12t + C) C = tan−1 10

3c)

Z x 4

x dx5x + 25 =

Z t 0

dtx

5 − ln(x + 5)

x 4

So a closed form solution does not exist

(continued on the next page)

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Problem 2.3 continued:

d)

Z x 5

dx

x = −2

Z t 0

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2.4 From the transform definition, we have

L[mt] = lim

T →∞

"

Z T 0

te−stdt

#

The method of integration by parts states that

Z T 0

u dv = uv|T0 −

Z T 0

v duChoosing u = t and dv = e−stdt, we have du = dt, v = −e−st/s, and

L[mt] = m lim

T →∞

"

Z T 0

T 0

−

Z T 0

T 0

− e

−st

(−s)2

T 0

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2.5 From the transform definition, we have

L[t2] = lim

T →∞

"

Z T 0

t2e−stdt

#

The method of integration by parts states that

Z T 0

u dv = uv|T0 −

Z T 0

v duChoosing u = t2 and dv = e−stdt, we have du = 2t dt, v = −e−st/s, and

L[t2] = lim

T →∞

"

Z T 0

T 0

−

Z T 0

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X(s) = −dY (s)

dswhere y(t) = e−3tsin 5t Thus

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f (t) = 5us(t) − 7us(t − 6) + 2us(t − 14)Thus

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2.8 a)

2 sin 3tb)

4 cos 2t + 5

2sin 2tc)

2e−2tsin 3td)

5

3 −

5 e−3 t3e)

5 e−3 t

5 e−7 t2f)

e−3 t

3 e−7 t2

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2.9 a)

5 cos(3 t)b)

e3 t− e−3 tc)

5 − 15 t e−3 t− 5 e−3 td)

2

13−

2 e−2 t cos 3t + 2 sin 3 t3

13e)

5 − 5 cos 2tf)

5 t sin 2t

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2.10 a)

x(0+) = lim

s→∞s 53s + 7 =

53x(∞) = lim

s→0s 53s + 7 = 0b)

x(0+) = lim

s→∞s 103s2+ 7s + 4 = 0x(∞) = lim

s→0s 103s2+ 7s + 4 = 0

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s + 3x(t) = 5

1

s + 5x(t) = −1

1

s2 − 532

1

s +

532

1

s + 4x(t) = 5

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s−

1325

1

s + 5x(t) = 2

16

1

s + 3−

7916

1

s + 7x(t) = −31

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2.12 a)

X(s) = 7s + 2

(s + 3)2+ 52 = C1

5(s + 3)2+ 52 + C2

s + 3(s + 3)2+ 52

or

X(s) = −19

5

5(s + 3)2+ 52 + 7 s + 3

(s + 3)2+ 52

x(t) = −19

5 e

−3tsin 5t + 7e−3tcos 5tb)

5(s + 3)2+ 52 − 3

34

s + 3(s + 3)2+ 52

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s + 3(s + 3)2+ 52

328

4(s + 2)2+ 42 +19

82

s + 2(s + 2)2+ 42

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2.13 a) ˙x = 7t/5

Z x 3

dx = 75

Z t 0

t dt

x(t) = 7

10t

2+ 3b) ˙x = 3e−5t/4

Z x 4

dx = 34

Z t 0

˙x(t) − ˙x(0) = 4

7

Z t 0

t dt

˙x(t) = 4

14t

2+ 5

Z x 3

dx =

Z t 0

4

14t

2+ 5

dt

x(t) = 4

42t

3+ 5t + 3d) ¨x = 8e−4t/3

˙x(t) − ˙x(0) = 8

3

Z t 0

e−4tdt

˙x(t) = 17

dx =

Z t 0

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2.14 a) The root is −7/5 and the form is x(t) = Ce−7t/5 With x(0) = 4, C = 4 andx(t) = 4e−7t/5

b) The root is −7/5 and the form is x(t) = C1e−7t/5+ C2 At steady state, x = 15/7 =

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2.15 a) The roots are −7 and −3 The form is

x(t) = C1e−7t+ C2e−3tEvaluating C1 and C2 for the initial conditions gives

x(t) = e−7t+ 10te−7tc) The roots are −7 ± 3j The form is

x(t) = C1e−7t sin 3t + C2e−7t cos 3tEvaluating C1 and C2 for the initial conditions gives

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2.16 a)

x = 6 e−2 t− 3 e−5 t+ 2b)

x = 3 sin 4t − 4 cos 4t + 9d)

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2.17 a) The roots are −3 and −7 The form is

x(t) = C1e−7t+ C2te−7t+ C3

At steady state, x = 98/49 = 2 so C3 = 2 Evaluating C1 and C2 for the initial conditionsgives

x(t) = −2e−7t− 14te−7t+ 2c) The roots are −7 ± 3j The form is

x(t) = C1e−7t sin 3t + C2e−7t cos 3t + C3

At steady state, x = 174/58 = 3 so C3= 3 Evaluating C1 and C2 for the initial conditionsgives

x(t) = −7e−7t sin 3t − 3e−7t cos 3t + 3

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2.18 a)

s2+ 8s + 12

x = 15 e−2t− 15e−6tb)

s2+ 12s + 144

x = 16√3e−6t sin 6√3tc)

X(s) = 147

s2+ 49

x = 21 sin 7td)

s2+ 14s + 85

x = 85 e

−7tsin 6t3

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2.19 a)

6s(s + 5) =

65s −

65

1

s + 5x(t) = 6

5

1 − e−5tb)

4

s + 3)(s + 8) =

45

1

s + 3 −

45

1

s + 8x(t) = 4

5

e−3t− e−8tc)

8s + 52s2+ 20s + 48 =

12

8s + 5(s + 4)(s + 6) = −

274

1

s + 4 +

434

1

s + 6x(t) = −27

(s + 4)2+ 102 + C2 s + 4

(s + 4)2+ 102

= − 310

10(s + 4)2+ 102 + 4 s + 4

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2.20 a)

3s + 2

s2(s + 10) =

15

1

s2 + 725

1

s−

725

1

s + 10x(t) = 1

5t +

725

1 − e−10tb)

5(s + 4)2(s + 1) = −

159

1(s + 4)2 −5

9

1

s + 4+

59

1

s + 1x(t) = −15

1

s3 + 14

1

s2 + 38

1

s−

38

1

s + 2x(t) = 5

1

s2 +14

1

s−

14

1

s + 2x(t) = 1

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2s3+ 2s2+ 2/5

s3(s + 3/5) =

23

1

s3 −109

1

s2 + 1409

1

s−

8627

1

s + 3/5x(t) = 1

4

−2413

1(s + 5)2 − 96

169

1

s + 5+

18056845

1

s + 7/4 −

45

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c) This simple-looking problem actually requires quite a lot of algebra to find the tion, and thus it serves as a good motivating example of the convenience of using MATLAB.The algebraic complexity is due to a pair of repeated complex roots

solu-First obtain the transform of the forcing function Let f (t) = te−3tsin 5t From erty 8,

Prop-F (s) = −dY (s)

dswhere y(t) = e−3tsin 5t Thus

Thus

F (s) = 10s + 30

(s2+ 6s + 34)2 (1)(continued on the next page)

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Using the same technique, we find that the transform of te−3tcos 5t is

2s2+ 12s + 18(s2+ 6s + 34)2 − 1

s2+ 6s + 34 (2)This fact will be useful in finding the forced response

From the differential equation,

xfree(t) = −√4

3 sin

√3

2 t + 10 cos

√3

2 t = −2.3094 sin 0.866t + 10 cos 0.866t (3)The forced response is given by the second fraction, which can be expressed as

2.5s + 7.5[(s + 3)2+ 25]2(s2+ 3/4) (4)(continued on the next page)

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The roots of this are s = ±j√3/2 and the repeated pair s = −3 ± 5j Thus, referring

to (1), (2), and (3), we see that the form of the forced response will be

xforced(t) = C1te−3tsin 5t + C2te−3tcos 5t

+ C3e−3tsin 5t + C4e−3tcos 5t+ C5 sin

√3

2 t + C6 cos

√3

2 t (5)The forced response can be obtained several ways 1) You can substitute the form (5)into the differential equation and use the initial conditions to obtain equations for the Ci

coefficients 2) You can use (1) and (2) to create a partial fraction expansion of (4) in terms

of the complex factors 3) You can perform an expansion in terms of the six roots, of theform

√3A5/2

s2+ 3/4 +

A6s

s2+ 3/44) You can use the MATLAB residue function

The solution for the forced response is

xforced(t) = −0.0034te−3tsin 5t + 0.0066te−3tcos 5t

− 0.0026e−3tsin 5t + 2.308 × 10−4e−3tcos 5t+ 0.00796 sin 0.866t − 2.308 × 10−4 cos 0.866tThe initial condition ˙x(0) = 0 is not exactly satisfied by this expression because of thelimited number of digits used to display it

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2.22 The denominator roots are s = −3 and s = −5, which are distinct Factor thedenominator so that the highest coefficients of s in each factor are unity:

X(s) = 7s + 4

2s2+ 16s + 30 =

12

7s + 4(s + 3)(s + 5)

C1 = lim

s→−3

(s + 3) 7s + 4

= −174

C2 = lim

s→−5

(s + 5) 7s + 42(s + 3)(s + 5)

= lim

s→−5

7s + 42(s + 3)

= 314(continued on the next page)

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Using the LCD method we have

12

7s + 4(s + 3)(s + 5) =

= (C1+ C2)s + 5C1+ 3C2

(s + 3)(s + 5)Comparing numerators, we see that C1+ C2 = 7/2 and 5C1+ 3C2 = 4/2 = 2, which give

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2.23 a) The roots are −3 and −5 The form of the free response is

x(t) = A1e−3t+ A2e−5tEvaluating this with the given initial conditions gives

x(t) = 27e−3t− 17e−5tThe steady-state solution is xss= 30/15 = 2 Thus the form of the forced response is

x(t) = 2 + B1e−3t+ B2e−5tEvaluating this with zero initial conditions gives

x(t) = 2 − 5e−3t+ 3e−5tThe total response is the sum of the free and the forced response It is

x(t) = 2 + 22e−3t− 14e−5tThe transient response consists of the two exponential terms

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b) The roots are −5 and −5 The form of the free response is

x(t) = A1e−5t+ A2te−5tEvaluating this with the given initial conditions gives

x(t) = e−5t+ 9te−5tThe steady-state solution is xss= 75/25 = 3 Thus the form of the forced response is

x(t) = 3 + B1e−5t+ B2te−5tEvaluating this with zero initial conditions gives

x(t) = 3 − 3e−5t− 15te−5tThe total response is the sum of the free and the forced response It is

x(t) = 3 − 2e−5t− 6te−5tThe transient response consists of the two exponential terms

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c) The roots are ±5j The form of the free response is

x(t) = A1 sin 5t + A2 cos 5tEvaluating this with the given initial conditions gives

x(t) = 4

5 sin 5t + 10 cos 5tThe form of the forced response is

x(t) = B1+ B2 sin 5t + B3 cos 5tThus the entire forced response is the steady-state forced response There is no transientforced response Evaluating this function with zero initial conditions shows that B2 = 0and B3 = −B1 Thus

x(t) = B1− B1 cos 5tSubstituting this into the differential equation shows that B1 = 4 and the forced responseis

x(t) = 4 − 4 cos 5tThe total response is the sum of the free and the forced response It is

x(t) = 4 + 6 cos 5t + 4

5 sin 5tThe entire response is the steady-state response There is no transient response

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d) The roots are −4 ± 7j The form of the free response is

x(t) = A1e−4t sin 7t + A2e−4t cos 7tEvaluating this with the given initial conditions gives

x(t) = 2 − 8

7e

−4t sin 7t − 2e−4t cos 7tThe total response is the sum of the free and the forced response It is

x(t) = 2 + 36

7 e

−4t sin 7t + 8e−4t cos 7tThe transient response consists of the two exponential terms

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2.24 a) The root is s = 5/3, which is positive So the model is unstable.

b) The roots are s = 5 and −2, one of which is positive So the model is unstable.c) The roots are s = 3 ± 5j, whose real part is positive So the model is unstable.d) The root is s = 0, so the model is neutrally stable

e) The roots are s = ±2j, whose real part is zero So the model is neutrally stable.f) The roots are s = 0 and −5, one of which is zero and the other is negative So themodel is neutrally stable

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2.25 a) The system is stable if both of its roots are real and negative or if the roots arecomplex with negative real parts Assuming that m 6= 0, we can divide the characteristicequation by m to obtain

s = −a ±

√

a2− 4b2(continued on the next page)

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(s − r1)(s − r2) = s2− (r1+ r2)s + r1r2= 0Comparing the two forms shows that

r1r2 = b (1) and r1+ r1 = −a (2)

If b > 0, condition (1) shows that both roots have the same sign If a < 0, condition(2) shows that the roots must be negative Therefore, if the roots are distinct andreal, the roots will be negative if a > 0 and b > 0

b) Neutral stability occurs if either 1) both roots are imaginary or 2) one root is zerowhile the other root is negative Imaginary roots occur when a = 0 (the roots are s = ±√b)

In this case the free response is a constant-amplitude oscillation Case 2 occurs when b = 0and a > 0 (the roots are s = 0 and s = −a) In this case the free response decays to anon-zero constant

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2.26 a) τ = 5b) τ = 4c) τ = 3d) The roots is s = 3/8, so the model is unstable, so no time constant is defined.

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2.27 a) The root is s = −4/13, so the model is stable, and xss= 16/4 = 4 Since τ = 13/4,

it takes about 4τ = 13 to reach steady state

b) The root is s = −4/13, so the model is stable, and xss= 16/4 = 4 Since τ = 13/4,

it takes about 4τ = 13 to reach steady state

c) The root is s = 7/15, so the model is unstable, and no steady state exists

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2.25 a) The system is... publisher’s consent is unlawful.

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