GGG+GGGGH,+GGG+GG.GGH+GGG+GGG -, G I+HGA+H3G5-GH,G H+GAGS H2 4.两条前向通道:P1=G1G2G3,P2=G1 五个回路:L=G1G2,L2=-G2G3,L3=-1,L4=-G1GG3,L5=-G1 Δ=1-(L1+L2+L3+L4+L5)=2+G2G3-G1G2+G1+G1G2G3 A,=1 Cs=1∑P,△ G, +GGG 2+G+G.GG+GG-gg 5.L1=-G3H1,L2=-G2G3H2,L3=-G1G2G3G4H3,L4=G1G2G4G5H3 △=1-(L1+L2+L3+L4)+L1L4 1+G3H1+G2G3 H2+G,G2G3 G4H3-G1G2G4G5H3-GIG2G3 G4GsH, H3 ①求C(s)R(s)时,两条前向通道: P1=G1G2G3G4,△1=1;P2=-GGGG,Δ2=1+G3H1 G,G,, G4-GG2G4G(1+G3H) R(s 1+G3H+G2G3H2+GG2GG4H3-G,G,GGSH3-G,,H3 ②求B(s(s)时,从M(s)到E(s)的前向通路有两条: P1=-G4H3,A1=1,P2=-H2G2GsG4H,△2=1 E(S) G4H3-H2GGG4H3 N()1+GH,+G2G3H2+G,G,G3G4H3-GG2G4GSH3-GG2G3G4GSHH3 R, Cs 6. E (S) 1/CS+R R,Cs+ ER(S) R, E0(s) R2+R3 因为EA(s)-ES)K=Es)且K》1,所以EA(s)=EB(s),得 R, Eo(s)R,+R, RCS R3 E(S) R R, CS+ s+ R, C·100· 1 4 3 5 1 2 8 1 4 5 1 2 2 4 6 2 4 6 5 2 3 5 7 3 5 7 4 1 3 8 6 2 1 7 3 8 1 1 7 2 1 2 8 6 1 HG H G GH G H G G H H G G G G G G G H GGG GGG G H GGG G GG GG HGG G GH GG            4. 两条前向通道：P1＝G1G2G3，P2＝G1 五个回路：L1=G1G2，L2=－G2G3，L3＝－1，L4＝－G1G2G3，L5＝－G1 1 2 3 4 5 2 3 1 2 1 1 2 3   1 (L  L  L  L  L )  2  G G  G G  G  G G G 1 1  1 2  所以 1 1 2 3 3 2 1 2 1 1 2 3 2 1 2 1 G ( ) ( ) G G G G G G G G G G G P R s C s k k k           5. L1＝－G3H1，L2＝－G2G3H2，L3＝－G1G2G3G4H3，L4＝G1G2G4G5H3 1 2 3 4 1 4   1 (L  L  L  L )  L L ＝1＋G3H1+G2G3H2+G1G2G3G4H3－G1G2G4G5H3－G1G2G3G4G5H1H3 ① 求 C(s)/R(s)时，两条前向通道： P1=G1G2G3G4， 1 1  ；P2＝－G1G2G4G5， 2 3 1   1 G H 3 1 2 3 2 1 2 3 4 3 1 2 4 5 3 1 2 3 4 5 1 3 1 2 3 4 1 2 4 5 3 1 1 (1 ) G H G G H G G G G H G G G G H G G G G G H H G G G G G G G G G H        ② 求 E(s)/N(s)时，从 N(s)到 E(s)的前向通路有两条： P1＝－G4H3， 1 1  ，P2＝－H2G2G5G4H3， 1 2   ( ) ( ) N s E s 3 1 2 3 2 1 2 3 4 3 1 2 4 5 3 1 2 3 4 5 1 3 4 3 2 2 5 4 3 1 G H G G H G G G G H G G G G H G G G G G H H G H H G G G H        6. ( ) 1/ 1 ( ) 1 1 1 1     R Cs R Cs Cs R R E s E s i A ， ( ) ( ) 0 2 3 3 E s R R R E s B   因为[EA(s)－EB(s)]K=E0(s)且 K》1，所以 E (s) E (s) A  B ，得 R C s s R R R Cs R Cs R R R E s E s i 1 3 2 1 1 3 0 2 3 1 1 ( ) 1 ( )              ( ) ( ) R s C s