自动控制原理及其应用(第2版)黄坚第二章习题课.ppt

求: R(s) E(s) R(s) C(s) C(s) E(s) G1 G2 G3 _ _ + R(s) 第二章习题课 　(2-13) 结构图变 换成： G1 G2 + - E(s) G3 - R(s) G1 + - E(s) R(s) 1+G2 G3G2 - E(s) - R(s) 1+G2 G2G3 G1G2G3 1+G2 系统的传递函数: ) E(s) = R(s) 1+ 1 (1- G1G2G3 1+G2 1+G2 G2G3 1+G2+G1G2G3 = 1+G2-G2G3 (a) (b) 求: R(s) E(s) R(s) C(s) 第二章习题课 　(2-13) C(s) E(s) G1 G2 - - + R(s) G3 G4 G5 解: L1=-G3G4 L2=-G2G3G5 Δ1=1 P1=G1G5 P1=G1G5 Δ1=1 P2=1 Δ2=1+G3G4 P2=G2G3G5 Δ2=1 1+G2G3G5+G3G4 = R(s) C(s) G1G2G5+G1G5 1+G2G3G5+G3G4 = R(s) E(s) G1G5+(1+G1G5 ) R(s) G4(s) + C(s) E(s) G1(s) G2(s) - + G3(s) D(s) X(s) 2-14 C(s) R(s) = 1+G3(G1+G2) (G1+G2)(G3+G4) 解: L1=-G1G3 L2=-G2G3 Δ1=1 P1=G1G3 P2=G2G3 Δ2=1 P3=G1G4 Δ3=1 P4=G2G4 Δ4=1 E(s) R(s) = 1+G3(G1+G2) 1 Δ1=1 P1=1 D(s) C(s) =1 =G2(s) E(s) X(s) 第二章习题课 　(2-14) R(s) G4(s) + C(s) E(s) G1(s) G2(s) - + G3(s) 第二章习题课 　(2-14) D(s) X(s) 解: D(s)=0 结构图变换为 R(s) G4(s) + C(s) G1(s) G2(s) - + G3(s) G3(s) (G1+G2) (G3+G4) 1+(G1+G2)G3 G1+G2 C(s) R(s) = 1+G3(G1+G2) (G1+G2)(G3+G4) 求: R(s) C(s) 2-14 D(s) 第二章习题课 　(2-14) R(s) + - E(s) G3 G2 G1 E(s) R(s) = 1+G3(G1+G2) 1 R(s) G4(s) + C(s) E(s) G1(s) G2(s) - + G3(s) X(s) 求: R(s) E(s) 2-14 解: D(s)=0 结构图变换为 G3(G1+G2) D(s) C(s) D(s) R(s) C(s) G4(s) + E(s) G1(s) G2(s) - + G3(s) X(s) 求: 2-14 解: R(s)=0 D(s) C(s) =1 E(s) X(s) =G2(s) E(s) X(s) 第二章习题课 　(2-14) 第二章习题课 　(2-15) + G1 G2 G3 C1(s) R1(s) + - H2 H1 G4 G5 - G6 C2(s) R2(s) 解: 结构图变换为 + G1 G2 G3 C1(s) R1(s) - H2 H1 G4 G5 - 1+G4 G4G5H1H2 1+G4 G4G5H1H2 1+ G1 G1G2 1+G4+G1G4G5H1H2-G1G2-G1G2G4 G1G2G3(1+G4 ) = C1(s) R1(s) 1+G4+G1G4G5H1H2 G1G2(1+G4 ) = C1(s) R1(s) 求: 2-15 G3 C1(s) R1(s) + + G1 G2 - H2 H1 G4 G5 G6 - C2(s) R2(s) 解: 结构图变换为 + G4 G5 G6 C2(s) R2(s) - H1 H2 G2 G1 - 第二章习题课 　(2-15) 1+G4 G4G5 G1H1H2 1-G1G2 1+G4+G1G4G5H1H2-G1G2-G1G2G4 G4G5G6(1-G1G2) = = 1+ 1+G4 G6 G4G5 G1H1H2 1-G2G2 1+G4 G4G5 C2(s) R2(s) 求: 2-14 C2(s) R2(s) C2(s) G4 G5 G6 - + + G1 G2 - H2 H1 R1(s) G3 C1(s) R2(s) 解: 结构图变换为 G4 G5 G6 - + - G1 G2 H2 H1 R1(s) C2(s) 第二章习题课 　(2-15) 1+G4 H2G4G5 G1 1-G1G2 1+G4+G1G4G5H1H2-G1G2-G1G2G4 G1G4G5G6H2 = = 1+ G6 H1 C2(s) R