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Solutions to
Skill-Assessment
Exercises
To Accompany

Control Systems Engineering
rd
3 Edition
By
Norman S. Nise

John Wiley & Sons

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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...