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