自动控制原理黄坚第二版第二章习题答案.pptVIP

2-13(a) 求: R(s) E(s) R(s) C(s) C(s) E(s) G1 G2 G3 _ _ + R(s) 解: L1 L1=-G2 L2 L2=-G1G2G3 P1=G2G3 P2=G1G2G3 R(s) C(s) = 1+G2+G1G2G3 G2G3+G1G2G3 Δ1 =1 Δ2 =1 E(s) 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 第二章习题课 　(2-13a) 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) 2-14 求: R(s) C(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) 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) C1(s) R1(s) 第二章习题课 　(2-15) 求: 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 ) = C2(s) R2(s) G3 C1(s) R1(s) + + G1 G2 - H2 H1 G4 G5 G6 - C2(s) R2(s) 求: 2-14 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) C2(s) G4 G5 G6 - + + G1 G2 - H2 H1 R1(s) G3 C1(s) R2(s) 求: 2-14 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) C2(s) R1(s) 1+G4 H2G4G5 G1 1-G1G2 1+G4+G1G4G5H1H2-G1G2-G1G2G4 G1G4G5G6H2 = = 1+ G6 H1 C2(s) R1(s) G1 1-G1G2 1+G4 H2G4G5 G1 1-G1G2 1+G4 H2G4G5 第二章习题课 　(2-1a) uo i2= R2 C ＋ － ＋ － ui uo R1 R2 2-1(a) 试建立图所示电路的动态微分方程。 解： 输入量为ui，输出量为uo。 ui=u1+uo u1=i1R1 ic=C duc dt = dt d(ui-uo) i1=i2-ic u1=[ ]R1+uo uo -C d(ui-uo) dt R2 R2ui=uoR1-C R1R2+C R1R2+uoR2 dui dt dt duo uoR1+C R1R2+uoR2=R2ui+C R1R2 duo dui dt dt ＋ － ＋ － C uc