Solucionario de control

Solutions to
Skill-Assessment
Exercises
To Accompany

Control Systems Engineering
rd
3 Edition
By
Norman S. Nise

John Wiley & Sons

Copyright © 2000 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 orotherwise, 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) 7508400, fax (978) 750-4470. Requests to the Publisher for permission should be
addressed tothe Permissions Department, John Wiley & Sons, Inc., 605 Third
Avenue, New York, NY 10158-0012, (212) 850-6011, fax (212) 850-6008, e-mail:
PERMREQ@WILEY.COM. To order books please call 1 (800) 225-5945.
ISBN 0-471-36601-3

John Wiley & Sons, Inc.
605 Third Avenue
New York, NY 10158
USA

Solutions to Skill-Assessment
Exercises
Chapter 2
2.1.
The Laplace transform of t is
F(s) =1
using Table 2.1, Item 3. Using Table 2.2, Item 4,
s2

1
.
( s + 5)2

2.2.
Expanding F(s) by partial fractions yields:
F(s) =

A
B
C
D
+
+
+
2
s s + 2 ( s + 3)
( s + 3)

where,

A=

10
( s + 2)(s + 3)2

D = ( s + 3)2

=
S→0

5
10
B=
s( s + 3)2
9

= −5 C =
S → −2

10
10
= , and
s( s + 2) S → −3 3

40
dF ( s )
=
ds s → −3 9

Taking the inverseLaplace transform yields,

f (t ) =

5
10
40
− 5e −2 t + te −3t + e −3t
9
3
9

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,

( s 3 + 3s 2 + 7s + 5)C( s ) = ( s 2 + 4 s + 3) R( s )
Thus,

2

Solutions to Skill-Assessment Exercises

s2 + 4s+ 3
C( s )
=3
R( s ) s + 3s 2 + 7s + 5

2.4.
G( s ) =

2s + 1
C( s )
=2
R( s ) s + 6 s + 2

Cross multiplying yields,
dc
dr
d 2c
+ 6 + 2c = 2 + r
2
dt
dt
dt
2.5.
C( s ) = R( s )G( s ) =

1
s
1
A
B
C
*
=
=+
+
2
s ( s + 4)( s + 8) s( s + 4)( s + 8) s ( s + 4) ( s + 8)

where

A=

1
1
1
1
1
1
=
B=
= − , and C =
=
( s + 4)( s + 8) S → 0 32
s( s +8) S → −4
16
s( s + 4) S → −8 32

Thus,

c( t ) =

1
1
1
− e −4 t + e −8t
32 16
32

2.6.
Mesh Analysis
Transforming the network yields,

Now, writing the mesh equations,

Chapter 2

( s + 1) I1 ( s ) − sI2 ( s ) − I3 ( s ) = V ( s )
− sI1 ( s ) + (2 s + 1) I2 ( s ) − I3 ( s ) = 0
− I1 ( s ) − I2 ( s ) + ( s + 2) I3 ( s ) = 0
Solving the mesh equations for I2(s),
( s + 1)V ( s )
−s
−1
I2 ( s ) =
( s + 1)
−s
−1

0
0
−s

−1
−1
( s + 2)
−1

=

( s 2 + 2 s + 1)V ( s )
s( s 2 + 5s + 2)

(2 s + 1)
−1
−1
( s + 2)

But, V L ( s ) = sI2 ( s )
Hence,
V L (s) =

( s 2 + 2 s + 1)V ( s )
( s 2 + 5s + 2)

or
V L (s) s 2 + 2 s + 1
=
V ( s ) s 2 + 5s + 2

Nodal Analysis
Writing the nodal equations,
1
( + 2)V1 ( s ) − V L ( s ) = V ( s )s
2
1
−V1 ( s ) + ( + 1)V L ( s ) = V ( s )
s
s
Solving for V L ( s ) ,

1
( + 2) V ( s )
s
1
V (s)
−1
( s 2 + 2 s + 1)V ( s )
s
V L (s) =
=
1
( s 2 + 5s + 2)
( + 2)
−1
s
2
−1
( + 1)
s
or
V L (s) s 2 + 2 s + 1
=
V ( s ) s 2 + 5s + 2

3

4

Solutions to Skill-Assessment Exercises

2.7.
Inverting
Z2 ( s ) −100000
=
= −s
Z1 ( s ) (10 5 / s )
NoninvertingG( s ) = −

10 5
+ 10 5 )
(
[ Z1 ( s ) + Z2 ( s )]
G( s ) =
=s5
= s +1
10
Z1 ( s )
(
)
s
2.8.
Writing the equations of motion,
( s 2 + 3s + 1) X1 ( s ) − (3s + 1) X2 ( s ) = F ( s )
−(3s + 1) X1 ( s ) + ( s 2 + 4 s + 1) X2 ( s ) = 0
Solving for X2 ( s ) ,

( s 2 + 3s + 1) F ( s )
−(3s + 1)
0
(3s + 1) F ( s )
X2 ( s ) = 2
=
3
s( s + 7s 2 + 5s + 1)
( s + 3s + 1)
−(3s...