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Solutions to Skill-Assessment Exercises
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Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright © 2004 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitte...

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  • July 26, 2024
  • 72
  • 2023/2024
  • Exam (elaborations)
  • Questions & answers

Subjects

  • mesh analysis
  • inverting
  • nodal analysis

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  • Systems Engineering
  • Systems Engineering

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Solutions to
Skill-Assessment
Exercises
To Accompany
Control Systems Engineering
4th Edition
By
Norman S. Nise
John Wiley & Sons Copyright © 2004 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any from or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the
Copyright Clearance Center, 222 Rosewood Drive, Danvers, MA 01923, (978) 750-
8400, fax (978) 750-4470. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street,
Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, e-mail: PERMREQ@WILEY.COM. To order books please call 1 (800) 225-5945. ISBN 0-471-44577-0 John Wiley & Sons, Inc. 111 River Street Hoboken, NJ 07030 USA Solutions to Skill-Assessment
Exercises
Chapter 2
2.1.
The Laplace transform of t is 1
s2 using Table 2.1, Item 3. Using Table 2.2, Item 4,
F(s)=1
(s+5)2.
2.2. Expanding F(s) by partial fractions yields:
F(s)=A
s+B
s+2+C
(s+3)2+D
(s+3)
where,
AssS=++=
→10
235
92
0() () B=10
s(s+3)2
S→−2=−5C=10
s(s+2)S→−3=10
3, and
D=(s+3)2dF(s)
ds s→−3=40
9
Taking the inverse Laplace transform yields,
f(t)=5
9−5e−2t+10
3te−3t+40
9e−3t
2.3. Taking the Laplace transform of the differential equation assuming zero initial
conditions yields:
s3C(s) + 3s2C(s) + 7sC(s) + 5C(s) = s2R(s) + 4sR(s) + 3R(s)
Collecting terms,
(s3+3s2+7s+5)C(s)=(s2+4s+3)R(s)
Thus, 2 Solutions to Skill-Assessment Exercises
C(s)
R(s)=s2+4s+3
s3+3s2+7s+5
2.4.
G(s)=C(s)
R(s)=2s+1
s2+6s+2
Cross multiplying yields,
d2c
dt2+6dc
dt+2c=2dr
dt+r
2.5.
C(s)=R(s)G(s)=1
s2*s
(s+4)(s+8)=1
s(s+4)(s+8)=A
s+B
(s+4)+C
(s+8)
where
A=1
(s+4)(s+8)S→0=1
32B=1
s(s+8)S→−4=−1
16, and C=1
s(s+4)S→−8=1
32
Thus,
c(t)=1
32−1
16e−4t+1
32e−8t
2.6.
Mesh Analysis
Transforming the network yields,
Now, writing the mesh equations,

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