kuo and golnaragh automatic control systems 8ed - solutions manual

377 Pages·1970·10 MB·english

by  Benjamin Kuo

Chapter 2 MATHEMATICAL FOUNDATION

2-1 (a) Poles: s = 0, 0, - 1, - 10; (b) Poles: s = - 2, - 2;
- ¥ ¥ ¥
Zeros: s = 2, , , . Zeros: s = 0.
-
The pole and zero at s = 1 cancel each other.

(c) Poles: s = 0, - 1 + j, - 1 - j; (d) Poles: s = 0, - 1, - 2, ¥ .
-
Zeros: s = 2.

2-2 (a) (b) (c)
G(s)= (s+55)2 G(s) = (s24+s4) +s+12 G(s)= s2 +44s +8
(d) (e)
¥
= 1 =(cid:229) kT(s+5) = 1
G(s) G(s) e
2 + - - T(s+5)
s 4 k=0 1 e

2-3 (a)
gt(u)(tu)=tut2uss t(1--+)--- 2+(2)s 2(3) s L
Gs(e)12= 1ss(e -+2 e-- se 2-+ - 2s= 3s L) (11-+e--ss)

gtu(tu)(t)u=t 2 (1--+)-£(2£) t 0 2
T ss s
Gse(e)12= 1( -+= --ess - 12 ) 1- (s )2
T
s s
1
g(t)=(cid:229) ¥ g (t - 2k)u (t - 2k) G(s)=(cid:229) ¥ 1(1- e- s)2e- 2ks = 1- e- s
T s + - s
k=0 k=0 s s(1 e )
(b)
gt(t)utt=u2---+tt(-----u)t4ts(us0+.5t) (0.5)4(1) (1s)4(1.5) (1.5) s L
Gs()e1e2=2-+s-+2s22(0e.52=e -- 0.51ss.5 - s L) 2((11-+e--0.5ss))
g (t) =2tu (t)- 4(t - 0.5)u (t - 0.5) +2(t - 1)u (t - 1) 0£ t £ 1
T s s s
Gse(e)12=-+2 ( = e-- 0.50s.s5- 1 ) - 2 ( s )2
T 2 2
s s
gt(g)()(=)t--=k((cid:229))u-¥ tkGse 1 = e (cid:229) ¥ 2 ( - 0.5sk)s2 - 2((1- e- 0.5s))
k=0 T s k=0ss220.5 e 1+ - s

2-4
= + - - - - - - - - + - - + -
g(t) (t 1)u (t) (t 1)u (t 1) 2u (t 1) (t 2)u (t 2) (t 3)u (t 3) u (t 3)
s s s s s s

Gs()e11=- 21ee(e-++ -es - --- 23sss+- )s 1( 3 )
2
s s

2-5 (a) Taking the Laplace transform of the differential equation, we get
1 1 1 1 1
2 + + = = = + -
(s 5s 4)F(s) F(s)
+ + + + + + +
s 2 (s 1)( s 2)( s 4) 6(s 4) 3(s 1) 2( s 2)
= 1 - 4t +1 - t - 1 - 2t ‡
f (t) e e e t 0
6 3 2
(b) sX (s)- x (0)= X (s) x (0)=1 sX (s)- x (0) =- 2X (s)- 3X (s)+1 x (0) =0
1 1 2 1 2 2 1 2 2
s
Solving for X(s) and X(s), we have
1 2
2 + +
s 3s 1 1 1 1
= = + -
X (s)
1 + + + +
s(s 1)( s 2) 2s s 1 2( s 2)

- -
1 1 1
X (s) = = +
2 + + + +
(s 1)(s 2) s 1 s 2
Taking the inverse Laplace transform on both sides of the last equation, we get
= + - t - - 2t ‡ =- - t + - 2t ‡
x (t) 0.5 e 0.5e t 0 x (t) e e t 0
1 2

2
2-6 (a)
= 1 - 1 + 1 = 1 - 1 - 2t +1 - 3t ‡
G(s) g(t) e e t 0
+ +
3s 2(s 2) 3(s 3) 3 2 3

(b)
-
= 2.5 + 5 + 2.5 =- - t + - t + - 3t ‡
G(s) g(t) 2.5e 5te 2.5e t 0
+ + 2 +
s 1 (s 1) s 3
(c) ( )
Gs(e)(g)t5e=0t20530020--=c3-o0s220(1)5tsins--- 2+(1) --- s- [ (1t) ]u (t1)-
ss +1s 2 +4 s
(d)
-
G(s)= 1 - s 1 = 1+ 1 - s Taking the inverse Laplace transform,
2 + + 2 + + 2 + +
s s s 2 s s s 2 s s 2
gt(e)tte=1t+1+.0-=6+9sin1.-3t02.53ots0[i.n51.32369.311-.44(7sin1.3 )] - t ( 23cos1.323 ) t ‡ 0
(e) g(t)=0.5t2e- t t ‡ 0

2-7
Ø - 1200 0 øØ ø
Œ œŒ œ Ø u (t)ø
AB=-=Œ 02310( ) œŒ uœ = t Œ 1 œ
º u (tß)
Œº - 1 - 31- 0 œŒßº 1 œß 2

2-8 (a) (b)
Ys()3 1 s + Y()s 5
= =
Rs()25 s36++s 2 + s Rs(s)1s0 s4 ++5+2
(c) (d)
Ys(s)(2s) + Ys()1 2 +e - s
= =
Rs(s)1ss02 4s3+ 22++ + Rs(s)2 s 2 +5 +
3
4
5
6
7
8
9