?(s)?C(s)R(s)G4?G1G2G3(1?G1H1)(1?G3H3)?G2H2??G1G2G31?[G4?]H4(1?G1H1)(1?G3H3)?G2H2???G4[(1?G1H1)(1?G3H3)?G2H2]?G1G2G3(1?G1H1)(1?G3H3)?G2H2?G4[(1?G1H1)(1?G3H3)?G2H2]H4?G1G2G3H4G4[1?G1H1?G3H3?G1H1G3H3?G2H2]?G1G2G31?G1H1?G3H3?G1H1G3H3?G2H2?G4H4[1?G1H1?G3H3?G1H1G3H3?G2H2]?G1G2G3H4
4．化简结构图，求系统传递函数 5 ．
C(s)?? R(s)
共有4个单回路：
?L??Lii?1i?1n4i?L1?L2?L3?L4
??G1G2G3G4G5G6H1?G2G3H2?G4G5H3?G3G4H4只有II、III两个回路不接触：
?LL??L?LLL?0iijjk2L3?(?G2G3H2)(?G4G5H3)?G2G3G4G5H2H3
只有一条
? ??1-?Li??LiLj?1?G1G2G3G4G5G6H1?G2G3H2?G4G5H3?G3G4H4 ?G2G3G4G5H2H3前向通路 p1?G1G2G3G4G5G6 所有回路均与之接触
? ?(s)??1?1
G1G2G3G4G5G6p1?1 ??1?G1G2G3G4G5G6H1?G2G3H2?G4G5H3?G3G4H4?G2G3G4G5H2H36.
有五个回路：
?L??GGG?GGH?GGH?GH?GG?LL?0??1??L?1?GGG?GGH?GGH?GHii1231212324214ji1231212324
2?G1G4两条前向通路：
p1?G1G2G3 ;?1?1
p2?G1G4 ;?2?1? ?(s)?G1G2G3?G1G4p1?1?p2?2??1?G1G2G3?G1G2H1?G2G3H2?G4H2?G1G4
7.
L1?L2???L5??1CRs有五个单回路：并且
5? ? Li??CRs
可找出六组两两互不接触的回路：Ⅰ-Ⅱ; Ⅰ-Ⅲ;Ⅰ-Ⅴ; Ⅱ-Ⅲ; Ⅲ-Ⅳ; Ⅳ-Ⅴ
? ? Li Lj?6.(?11)(?)?6
(CRs)2CRsCRs有一组三个互不接触回路Ⅰ-Ⅱ-Ⅲ
? ? Li LjLk?(?13?1)?333CRsCRs561? ??1?? Li?? Li Lj?? Li LjLk?1??222?333CRsCRsCRs
前向通路一条：p1?1 ; ?1?1 333CRs1UC(s)p1?1(CRs)3? ?(s)???561UR(s)?1???CRs(CRs)2(CRs)3 1 ?(CRs)3?5(CRs)2?6CRs?18.
回路4个：?Li??G1H1?G2H2?G1G2H3?G3H3
两两不接触回路两个：Ⅰ-Ⅱ， Ⅱ-Ⅳ
?LiLj?(?G1H1)(G2H2)?(G2H2)(?G3H3)?L
iLjLk?0? ??1?G1H1?G2H2?G1G2H3?G3H3?G1G2H1H2?G2G3H2H3
前向通道两条：
p1?G1G2 ;?1?1p2?G3 ;?2?1-G2H2
? ?(s)?p1?1?p2?2G1G2?G3(1-G2H2)??1?G1H1?G2H2?G1G2H3?G3H3?G1G2H1H2?G2G3H2H39. 已知系统结构图，求
C(s)R(s)??
解：本结构图有2条前向通道，6个回路(其中I，V两回路不相交)
??1?{?H?G2?G1?G1G2?(?G3)?[?(?G3)]}?[?(?G3)].(?H) ?1?H?G2?G1?G1G2?G3Hp1?G1G2 ;?1?1 p2??G3 ;?2?1?H? ?(s)?G1G2?G3 (1?H)1?H?G2?G1?G1G2?G3H求
C(s)R(s)??
10.
解：共有3个单回路(全部有公共接触部分)
?L??L??GGGGH?GGGH?GHi3
i?1i12341124122? Δ?1??Li ?1? G1G2G3G4H1?G1G2G4H1?G2H2 i?13
前向通道共有6条：
p1?G1G2G3G4 ?1?1 p2?G1G2G4 ?2?1
p3?G5G2G3G4 ?3?1 p4?G5G2G4 ?4?1 p5??G6G3G4 ?5?1 p6??G6H2G2G4 ?6?1 G1G2G3G4H1?G1G2G4H1?G
由梅逊公式：
p1Δ1?p2Δ2?p3Δ3?p4Δ4?p5Δ5?p6Δ6 ??GGGG?G1G2G4?G5G2G3G4?G5G2G4?G6G3G4?G6H2G2G4 ?12341?G1G2G3G4H1?G1G2G4H1?G2H2i?1?(s)??pΔi6i?11. 已知系统结构图
1).画出系统信号流图 2).求
C(s)C(s) ,R1(s)R2(s)