K :6 1+KK 2/3 6.(1) (s) (s+1)G1(s)+K1 s+(K1K2+1)s+K1 D(s)=s2+(K1K2+1)s+K1=(s+5+j5s+5-j5)=s2+10s+50 得 K1=50 K2=9/50 (2)由 Φ,()=(s+l)s-(K2s+1)G(S s+(K1K2+1)s+K1 G1(s)= K2S+1 (3)由 (S+1)(K2S+1)-s+G2(s) +(K1K2+1)s+K 得 7 1+G(s)s-+2C0,s+ 当n(=1(时c(1)=2e-2-e s+6 E(S)=- s+2s+4s2+6s+8 E(s)=Φ2(s)R(s)=-Φ(S 所以 s-+220 250,s+o 2.828 所以 K,=lims(s)=% 所以r()=t时, es=l/K=0.75·104· (2) 3 2 K p , Kv 6 2/3 4 0 1 v v p p ss K A K A e 6.(1) 1 2 1 2 1 1 ( 1) ( 1) ( ) ( ) s K K s K s G s K s 令 ( ) ( 1) ( 5 5)( 5 5) 10 50 2 1 2 1 2 D s s K K s K s j s j s s 得 K1=50, K2=9/50 (2)由 0 ( 1) ( 1)[ ( 1) ( )] ( ) 1 2 1 2 2 1 s K K s K s s K s G s s er 得 1 ( ) 2 1 K s s G s (3)由 0 ( 1) ( 1)( 1)[ ( )] ( ) 1 2 1 2 2 2 s K K s K s K s s G s s en 得 G2(s)=s 7. 2 2 2 2 2 1 ( ) 1 ( ) n n n e s s s s G s s 当 r(t)=1(t)时 6 8 6 4 1 2 2 ( ) 2 , ( ) 2 2 4 s s s s s e t e e E s t t 而 ( ) 1 ( ) ( ) ( ) s s E s s R s e e 所以 sE(s) (s) e 即 2 2 2 2 2 2 6 8 6 n n n s s s s s s s s 所以 1.061 2.828 n 又 3 4 lim ( ) 0 K sG s s v 所以 r(t)=t 时, ess=1/Kv=0.75