PPT《第二章习题课 （2-1a）》

1 - R5 R4+ R5 UO R2 + R3 R3 SC +

1 - c(t) t

0 T K δ(t) 2-8 设有一个初始条件为零的系统,系统的输入、输出曲线如图,求G(s). c(t) t

0 T K δ(t) c(t)= K T t- (t-T) K T C(s)= K (1-e ) Ts2 -TS C(s)=G(S)

第二章习题课 (2-8) 解: 2-9 若系统在单位阶跃输入作用时,已知初始条件为零的条件下系统的输出响应,求系统的传递函数和脉冲响应. r(t)=I(t) c(t)=1-e +e -2t -t 解: R(s)=

1 s G(S)=C(s)/R(s)

1 s+2

1 s - C(s)=

1 s+1 + = s(s+1)(s+2) (s2+4s+2) = (s+1)(s+2) (s2+4s+2) C(s)= (s+1)(s+2) (s2+4s+2) 脉冲响应:

2 s+2 =1+

1 s+1 - c(t)=δ(t)+2e +e -2t -t

第二章习题课 (2-9) 2-10 已知系统的微分方程组的拉氏变换式,试画出系统的动态结构图并求传递函数. 解: X1(s)=R(s)G1(s)-G1(s)[G7(s)-G8(s)]C(s) X2(s)=G2(s)[X1(s)-G6(s)X3(s)] X3(s)=G3(s)[X2(s)-C(s)G5(s)] C(s)=G4(s)X3(s) G1 G2 G3 G5 - - - C(s) - R(s) G4 G6 G8 G7 X1(s)={R(s)-C(s)[G7(s)-G8(s)]}G1(s) C(s)[G7(s)-G8(s)] G6(s)X3(s) X1(s) X2(s) C(s)G5(s) X3(s) G1 G2 G3 G5 - - - C(s) - R(s) G4 G2G6 G8 G7 G1G2 G5 - C(s) - R(s) G7-G8 1+G3G2G6 G3G4 - C(s) R(s) G7-G8 1+G3G2G6 +G3G4G5 G1G2G3G4 1+G3G2G6 +G3G4G5+G1G2G3G4(G7 -G8) G1G2G3G4 R(s) C(s) =

第二章习题课 (2-10) 解: 2-11(a) G1(s) G2(s) G3(s) H1(s) _ _ + R(s) C(s) H2(s) G1(s) G2(s) H1(s) _ _ + R(s) C(s) H2(s) G3(s) 求系统的传递函数 1+G2H1 G2 G1+G3 1+G1H2 1+G2H1 G2 1+G2H1 G2 = 1+G2H1+G1G2H2 G2 R(s) C(s) = 1+G2H1+G1G2H2 G2G1+G2G3 G1(s) G2(s) G3(s) H1(s) _ _ + R(s) C(s) G 1(s)H2(s)

第二章习题课 (2-11a) 2-11(a) G1(s) G2(s) G3(s) H1(s) _ _ + R(s) C(s) H2(s) 求系统的传递函数 解: L1 L1=-G2H1 L2 L2=-G1G2H1 P1=G1G2 P2=G3G2 Δ1 =1 Δ2 =1 R(s) C(s) = Σ n k=1 PkΔk Δ Δ=1+G2H1+G1G2H2 1+G2H1+G1G2H2 G2G1+G2G3 =

第二章习题课 (2-11a) 解: 2-11(b) G1(s) G2(s) G3(s) G4(s) _ + + R(s) C(s) H(s) 求系统的传递函数 G1(s) G2(s) G3(s) G4H _ + R(s) C(s) H(s) 1+G4G1H G1 G2(s) G3(s) _ + R(s) C(s) H(s) 1+G4HG1 G1 G2 G3 _ + R(s) C(s) 1+G4HG1 G1 HG1 1+G4HG1 G1+G3 (1+HG1G4) 1+G4HG1 G2 (1+HG1G4) 1+G4G1H+G1G2H R(s) C(s) = 1+G1G2H+G1G4H G1G2+G2G3+G1G2G3G4 H

第二章习题课 (2-11b) 解: 2-11(b) G1(s) G2(s) G3(s) G4(s) _ + + R(s) C(s) H(s) 求系统的传递函数 R(s) C(s) = 1+G1G2H+G1G4H G1G2+G2G3+G1G2G3G4 H L1 L1=-G1G2H L1=-G1G4H L2 P1=G1G2 Δ1 =1 P2=G3G2 Δ=1+G4G2H+G1G2H Δ2=1+G1G4H

第二章习题课 (2-11b) H1 _ + + + G1 + C(s) R(s) G3 G2 2-11c 求系统的闭环传递函数 . 解: H1 _ + G1 + C(s) R(s) G3 G2 H1 R(s) C(s) 1+G1G2+G1H1CG3H1 G1G2 (1C G3H1) = _ G1 C(s) R(s) G2 H1+G2 1-G3H1

1

第二章习题课 (2-11c) H _ G1 + C(s) R(s) G2 2-11d 求系统的闭环传递函数 . 解: (1) _ G1 + C(s) R(s) G2 HG2 1+G2H

1 (G1+G2 ) R(s) C(s) = (2) L1 L1=-G2H P1=G1 Δ1 =1 P2=G2 Δ2 =1

第二章习题课 (2-11d) - _ G1 + C(s) R(s) G2 G3 G4 2-11e 求系统的闭环传递函数 . 解: (1) _ C(s) R(s) G1+G2 G3-G4 C(s) = R(s) 1+(G1+G2)(G3-G4) (G1+G2) 1+G1G3+G2G3CG1G4-G2G4 = (G1+G2)

第二章习题课 (2-11e) L1 L2 L3 L4 L2=G1G4 L3=-G2G3 L4=G2G4 (2) L1=-G1G3 P1=G1 Δ1 =1 P2=G2 Δ2 =1 1+G1G3+G2G3CG1G4-G2G4 = (G1+G2) C(s) R(s) _ G1 + C(s) R(s) G2 2-11f 求系统的闭环传递函数 . _ C(s) R(s) G1 1-G2 G2 C(s) = R(s) 1+ 1-G2 G1 G1G2 1+G1G2CG2 G1 (1C G2) =

第二章习题课 (2-11f) 解: (1) (2) L1 L1=-G1G2 L2 L2=G2 P1=G1 Δ1 =1-G2 Δ=1+G1G2-G2 C(s) R(s) 1+G1G2CG2 G1 (1C G2) = 2-12(a) R(s) G1(s) G2(s) H2(s) _ + C(s) H3(s) H1(s) _ + D(s) 解: 求: D(s) C(s) R(s) C(s) D(s)=0 1-G2H2 G2 G(s)= 1-G2H2 G1G2 C(s) = R(s) 1+ 1-G2H2 G1G2H3 1-G2H2 G1G2 1-G2H2+G1G2H3 G2G1 = R(s)=0 结构图变 换成: G2(s) H2(s) _ + C(s) G1H3 G1H1 _ D(s) 1-G2H2 G2 1-G1H1 C(s) = D(s) 1+ 1-G2H2 G2 1-G2H2 G2 G1H3 (1-G1H1 ) 1-G2H2+G1G2H3 G2(1-G1H1 ) =