Solution automatic control systems 8ed - kuo and golnaraghi - solutions manual

Chapter 2 MATHEMATICAL FOUNDATION
2-1 (a)
Poles: s = 0, 0, Zeros: s =

−1, −10;

(b)

Poles: s =

−2, −2; −1 cancel each other.

−2, ∞, ∞, ∞.

Zeros: s = 0. The pole and zero at s =

(c)

Poles: s = 0,

−1 + j, −1 − j; −2.

(d)

Poles: s = 0,

−1, −2, ∞.

Zeros: s =

2-2 (a) G (s) = (d)
G ( s)

(b) 5

(c) G ( s) = (e)

( s + 5)
1 s
2

2

(s
=

4s
2+4
∞

)

+

1 s+ 2

G (s)

=

4 s
2

+ 4s +8

=

+4

G (s)

∑e
k =0

kT ( s + 5 )

=
1

1

−e

−T ( s+5 )

2-3 (a)
g ( t ) = u s ( t ) − 2u s (t − 1) + 2 u s( t − 2) − 2 u s ( t − 3) + L G (s ) = 1 s

(1 − 2e − s + 2e−2 s − 2e −3s + L ) =
1 s

s 1+ e

(

1−e

−s −s

)

gT (t ) = u s (t ) − 2us ( − 1) + us (t − 2) t GT (s ) =

0≤ t ≤ 2
2(1 − 2e − s + e −2s ) = ( 1 − e − s )
1 s

1

g(t )

=

∑
k =0

∞

g

T

(t

− 2k )us (t − 2k )

G (s)

=

∑s
k =0

∞

1 (1

−e

−s

) e

2

−2 ks

=

1− e

−s −s

s (1 + e

)

(b) g ( t) = 2tu s ( t ) − 4(t − 0.5) u s (t − 0.5) + 4(t − 1) us (t − 1) − 4(t − 1.5)us (t − 1.5) + L G ( s) =
g
T

2 s
2

(1 − 2e

−0.5 s

+ 2e

−s

−2e

−1.5 s

( − 0.5 s ) + L) = 2 −0.5 s s (1 + e )
2 1−e
2

(t )

= 2 tu s ( t ) − 4 ( t − 0 . 5) u s ( t − 0 . 5) + 2( t − 1 ) u s ( t − 1 ) 2 s
2

0

≤ t ≤1

GT ( s ) =
∞

(1 − 2e−0.5 s + e− s ) = s 2 (1 − e−0.5 s )
2
∞

g (t ) =

∑

k=0

g T ( t − k )us ( t − k )

G(s ) =

k=0

∑ s2 (
2

1 −e

−0.5 s

)

2

e

− ks

=

( −0.5 s ) 2 −0.5s s (1+ e )
2 1−e

2-4
g(t )

= ( t + 1 ) u s ( t ) − ( t − 1 ) u s ( t − 1 ) − 2 u s ( t − 1 ) − ( t − 2 ) u s ( t − 2 ) + ( t − 3) u s ( t − 3) + u s ( t − 3)

G ( s) =

1 s
2

(1 − e − s − e −2 s + e −3 s ) + s (1 −2e − s + e −3 s )
1
1 6( s 1 3( s 1 2( s

2-5 (a)

Taking the Laplace transform of the differential equation, we get 1 1 2 ( s + 5s + 4) F ( s) = F (s) = s +2 ( s + 1)(s + 2 )( s f (t )

+4)

=

+ 4)

+

+ 1)

−

+ 2)

=

1 6

e

−4 t

+

1 3

e

−t

−

1 2

e

−2 t

t

≥0
sX (s)

(b)

sX ( s )
1

− x1( 0 ) =

X (s)
2

x (0)
1

=1

2

− x 2 ( 0 ) = −2 X 1 ( s ) − 3 X 2 ( s ) +

1 s

x (0)
2

=0

Solving for X1 (s) and X2 (s), we have X 1 ( s)

= =

s s( s

2

+ 3 s +1 −1

+ 1 )( s+ 2 )

= =

1 2s

+

1 s

+1
1 s

−

1 2( s

+ 2)

X (s)
2

−1
s

(s

+ 1 )( s + 2 )
−t

+1

+

+2
x (t )
2

Taking the inverse Laplace transform on both sides of the last equation, we get x (t )
1

= 0 .5 + e

− 0 .5 e

−2 t

t

≥0

= −e

−t

+e

−2 t

t

≥0

2

2-6 (a)
G (s)

=

1 3s

−

1 2( s

+ 2)

+

1 3( s

+3)

g (t )

=

1 3

−

1 2

e

−2 t

+

1 3

e

−3 t

t

≥0

(b)
G (s)

=

−2 . 5
s

+1
50 s

+

5 (s

+ 1)

2

+

2 .5 s

+3

g(t )

= −2 . 5 e

−t

+ 5 te

−t

+ 2 .5 e

−3 t

t

≥0

(c)
G (s ) =

(
1 s

−

20 s +1
s

−

30s + 20 s +4
1 s
2

)
2

e

−s

g (t ) = 50 − 20e

[

− (t −1)

− 30cos2(t− 1) − 5sin2(t − 1)

] us (t − 1)

(d)
G (s)

=

−

−1
−0.5 t

s

2

+s+2

=

+

1 s

g (t ) = 1 + 1.069e

[ sin1.323t + sin (1.323t − 69.3o ) ] = 1 + e−0.5 t (1.447sin1.323t − cos1.323t )
t

+s +2

−

s s
2

+s+2

Taking the inverse Laplace transform,

t≥0

(e)

g(t )

= 0 .5 t

2

e

−t

≥0

2-7

 −1 2 0  A =  0 −2 3     −1 −3−1   

0 0  B = 1 0    0 1   

u (t ) =

 u1( t)   u ( t)   2 

2-8

(a) Y (s ) R (s ) (c) Y (s ) R (s ) = s ( s + 2) s + 10 s + 2 s + s + 2
4 3 2

(b) = 3s + 1 s + 2 s +5s + 6
(d)
3 2

Y (s) R (s )

=

5 s + 10 s + s + 5 1+ 2e
2 −s 4 2

Y (s ) R (s )

=

2s + s + 5

3

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Chapter 4...