05_结构图及其等效变换&信号流图&控制系统的传递函数

复习

典型环节及其传递函数

环节名称微分方程传递函数
比例环节c=K⋅rc=K·rc=KrKKK电位器, 放大器, 自整角器
惯性环节Tc˙+c=rT\dot c +c =rTc˙+c=r1Ts+1\frac{1}{Ts+1}Ts+11CR电路, 交、直流电动机
振荡环节T2c¨+2ξTc˙+c=rT^2\ddot c+2\xi T\dot c +c = rT2c¨+2ξTc˙+c=r1T2s2+2ξTs+1\frac{1}{T^2s^2+2\xi Ts+1}T2s2+2ξTs+11R-L-C电路, 弹簧质块阻尼器系统
积分环节c˙=r\dot c=rc˙=r1s\frac{1}{s}s1水箱(流量Q—液位h)
微分环节c=r˙c=\dot rc=r˙sss
一阶复合微分环节c=τr˙+rc=\tau\dot r+rc=τr˙+rτs+1\tau s + 1τs+1
二阶复合微分环节c=τ2r¨+2τξr˙+rc=\tau^2\ddot r+2\tau\xi\dot r+rc=τ2r¨+2τξr˙+rτ2s2+2τξs+1\tau^2s^2+2\tau\xi s+1τ2s2+2τξs+1

系统模型及其建立过程

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结构图

基本概念

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  • 信号线
  • 引出点(或测量点)
  • 比较点(或综合点)
  • 方框(或环节)

结构图的组成及绘制

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结构图等效变换规则

结构图等效变换规则

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信号流图

基本概念

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  • 源节点(输入节点): 只有信号输出

  • 阱节点(输出节点): 只有信号输入

  • 混合节点: 有信号输入和输出

  • 支路(环节)

  • 支路增益(传递函数)

  • 前向通路: 信号从输入节点到输出节点传递时, 每个节点只通过一次的通路

  • 回路: 起点和终点在同一节点(可以沿着箭头回到起点)

  • 互不接触回路:回路之间没有公共节点

  • 结构图和信号流图是等价的, 二者可以互相转化

梅森(Mason)增益公式

公式:
G(s)=1Δ∑k=1nPkΔk G(s)=\frac{1}{\Delta}\sum_{k=1}^nP_k\Delta_k G(s)=Δ1k=1nPkΔk

  • Δ\Delta\qquadΔ 特征式 Δ=1−∑La+∑LbLc−∑LdLeLf+...\Delta=1-\sum L_a+\sum L_bL_c-\sum L_dL_eL_f+...Δ=1La+LbLcLdLeLf+...
  • nn\qquadn 前向通路的条数
  • PkP_k\qquadPk 第k条前向通路的总收益
  • ∑La\sum L_a\quadLa 所有不同回路的增益之和
  • ∑LbLc\sum L_bL_cLbLc 两两互不接触回路的回路增益乘积之和
  • ∑LdLeLf\sum L_dL_eL_fLdLeLf 互不接触回路中, 每次取其中三个的回路增益乘积之和
  • Δk\Delta_kΔk 第k条前向通路的余子式(把与第k条前向通路接触的回路去除, 剩余回路构成的子特征式)

例1: 求传递函数C(s)/R(s)

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Δ=1−(−G2G3H2−G3G4H4−G4G5H3−G1G2G3G4G5G6H1)+(G2G3H2)(G4G5H3)=1+G2G3H2+G3G4H4+G4G5H3+G1G2G3G4G5G6H1+G2G3G4G5H2H3P1=G1G2G3G4G5G6Δ1=1Φ(s)=1Δ(P1Δ1)=G1G2G3G4G5G61+G2G3H2+G4G5H3+G1G2G3G4G5G6H1+G2G3G4G5H2H3 \begin{aligned} \Delta&=1-(-G_2G_3H_2-G_3G_4H_4-G_4G_5H_3-G_1G_2G_3G_4G_5G_6H_1)+(G_2G_3H_2)(G_4G_5H_3)\\ &=1+G_2G_3H_2+G_3G_4H_4+G_4G_5H_3+G_1G_2G_3G_4G_5G_6H_1+G_2G_3G_4G_5H_2H_3\\ P_1&=G_1G_2G_3G_4G_5G_6\qquad\Delta_1=1\\ \Phi(s)&=\frac{1}{\Delta}(P_1\Delta_1)=\frac{G_1G_2G_3G_4G_5G_6}{1+G_2G_3H_2+G_4G_5H_3+G_1G_2G_3G_4G_5G_6H_1+G_2G_3G_4G_5H_2H_3} \end{aligned} ΔP1Φ(s)=1(G2G3H2G3G4H4G4G5H3G1G2G3G4G5G6H1)+(G2G3H2)(G4G5H3)=1+G2G3H2+G3G4H4+G4G5H3+G1G2G3G4G5G6H1+G2G3G4G5H2H3=G1G2G3G4G5G6Δ1=1=Δ1(P1Δ1)=1+G2G3H2+G4G5H3+G1G2G3G4G5G6H1+G2G3G4G5H2H3G1G2G3G4G5G6
求解步骤:

  • 找回路,根据公式写出特征式
  • 找前向通道
  • 找前向通道对应的余子式
  • 第二条和第三条重复n次(有几条前向通道重复几次)
  • 根据公式计算出传递函数

例2 求C(s)/R(s)

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Δ=1−(−G1G2H1−G2G3H2−G1G2G3−G1G4−G4H2)=1+G1G2H1+G2G3H2+G1G2G3+G1G4+G4H2P1=G1G2G3Δ1=1P2=G1G4Δ2=1Φ(s)=G1G2G3+G1G41+G1G2H1+G2G3H2+G1G2G3+G1G4+G4H2 \begin{aligned} \Delta&=1-(-G_1G_2H_1-G_2G_3H_2-G_1G_2G_3-G_1G_4-G_4H_2)\\ &=1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3+G_1G_4+G_4H_2\\ P_1&=G_1G_2G_3\qquad\Delta_1=1\\ P_2&=G_1G_4\qquad\quad\Delta_2=1\\ \Phi(s)&=\frac{G_1G_2G_3+G_1G_4}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3+G_1G_4+G_4H_2} \end{aligned} ΔP1P2Φ(s)=1(G1G2H1G2G3H2G1G2G3G1G4G4H2)=1+G1G2H1+G2G3H2+G1G2G3+G1G4+G4H2=G1G2G3Δ1=1=G1G4Δ2=1=1+G1G2H1+G2G3H2+G1G2G3+G1G4+G4H2G1G2G3+G1G4

例3 求传递函数C(s)/R(s)

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Δ=1−(−1R⋅1Cs−1Cs⋅1R−1R⋅1Cs−1Cs⋅1R−1R⋅1Cs)+(6⋅1R2C2s2)+1R3C3s3=1+5RCs+6R2C2s2+1R3C3s3P1=1R3C3s3Δ1=1Φ(s)=P1Δ1Δ=1(RCs)3+5(RCs)2+6RCs+1 \begin{aligned} \Delta&=1-(-\frac{1}{R}·\frac{1}{Cs}-\frac{1}{Cs}·\frac{1}{R}-\frac{1}{R}·\frac{1}{Cs}-\frac{1}{Cs}·\frac{1}{R}-\frac{1}{R}·\frac{1}{Cs})+(6·\frac{1}{R^2C^2s^2})+\frac{1}{R^3C^3s^3}\\ &=1+\frac{5}{RCs}+\frac{6}{R^2C^2s^2}+\frac{1}{R^3C^3s^3}\\ P_1&=\frac{1}{R^3C^3s^3}\qquad \Delta_1=1\\ \Phi(s)&=\frac{P_1\Delta_1}{\Delta}=\frac{1}{(RCs)^3+5(RCs)^2+6RCs+1} \end{aligned} ΔP1Φ(s)=1(R1Cs1Cs1R1R1Cs1Cs1R1R1Cs1)+(6R2C2s21)+R3C3s31=1+RCs5+R2C2s26+R3C3s31=R3C3s31Δ1=1=ΔP1Δ1=(RCs)3+5(RCs)2+6RCs+11

例4 求传递函数C(s)/R(s) 注意G3和H1G_3和H_1G3H1两两不接触

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Δ=1−(−G1G2−G1−H1−G2+G3−G3)−G3H1=1+G1G2+G1+G2+H1−G3H1P1=G1G2Δ1=1P2=−G3Δ2=1+H1Φ(s)=G1G2−G3(1+H1)1+G1G2+G1+G2+H1−G3H1 \begin{aligned} \Delta&=1-(-G_1G_2-G_1-H_1-G_2+G_3-G_3)-G_3H_1\\ &=1+G_1G_2+G_1+G_2+H_1-G_3H_1\\ P_1&=G_1G_2\qquad \Delta_1=1\\ P_2&=-G_3\qquad\quad\Delta_2=1+H_1\\ \Phi(s)&=\frac{G_1G_2-G_3(1+H_1)}{1+G_1G_2+G_1+G_2+H_1-G_3H_1} \end{aligned} ΔP1P2Φ(s)=1(G1G2G1H1G2+G3G3)G3H1=1+G1G2+G1+G2+H1G3H1=G1G2Δ1=1=G3Δ2=1+H1=1+G1G2+G1+G2+H1G3H1G1G2G3(1+H1)

例5 求传递函数C(s)/R(s) 注意P6P_6P6

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Δ=1−(G2H2−G1G2G3G4H1−G1G2G4H1)=1−G2H2+G1G2G3G4H1+G1G2G4H1P1=G1G2G3G4Δ1=1P2=G1G2G4Δ2=1P3=G5G2G3G4Δ3=1P4=G5G2G4Δ4=1P5=−G6G3G4Δ5=1P6=−G6H2G2G4Φ(s)=G1G2G4+G2G4G5−G3G4G6+G1G2G3G4+G2G4G4G5−G2G4G6H21−G2H2+G1G2G4H1+G1G2G3G4 \begin{aligned} \Delta&=1-(G_2H_2-G_1G_2G_3G_4H_1-G_1G_2G_4H_1)\\ &=1-G_2H_2+G_1G_2G_3G_4H_1+G_1G_2G_4H_1\\ P_1&=G_1G_2G_3G_4\qquad\Delta_1=1\\ P_2&=G_1G_2G_4\qquad\quad\Delta_2=1\\ P_3&=G_5G_2G_3G_4\qquad\Delta_3=1\\ P_4&=G_5G_2G_4\qquad\quad\Delta_4=1\\ P_5&=-G_6G_3G_4\qquad\quad\Delta_5=1\\ P_6&=-G_6H_2G_2G_4\\ \Phi(s)&=\frac{G_1G_2G_4+G_2G_4G_5-G_3G_4G_6+G_1G_2G_3G_4+G_2G_4G_4G_5-G_2G_4G_6H_2}{1-G_2H_2+G_1G_2G_4H_1+G_1G_2G_3G_4} \end{aligned} ΔP1P2P3P4P5P6Φ(s)=1(G2H2G1G2G3G4H1G1G2G4H1)=1G2H2+G1G2G3G4H1+G1G2G4H1=G1G2G3G4Δ1=1=G1G2G4Δ2=1=G5G2G3G4Δ3=1=G5G2G4Δ4=1=G6G3G4Δ5=1=G6H2G2G4=1G2H2+G1G2G4H1+G1G2G3G4G1G2G4+G2G4G5G3G4G6+G1G2G3G4+G2G4G4G5G2G4G6H2

例6 求传递函数C(s)/R(s), C(s)/N(s)

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Δ=1−(−G1G2−G2H−G1G3)+(G1G2G3H)=1+G1G2+G2H+G1G3+G1G2G3HPR1=G1G2ΔR1=1PR2=G1G3ΔR2=1+G2HΦR(s)=G1G2+G1G3(1+G2H)1+G1G2+G2H+G1G3+G1G2G3HPN1=−1ΔN1=1+G2HPN2=G4G1G2ΔN2=1PN3=G4G1G3ΔN3=1+G2HΦN(s)=G1G2G4+(−1+G1G3G4)(1+G2H))1+G1G2+G2H+G1G3+G1G2G3H \begin{aligned} \Delta&=1-(-G_1G_2-G_2H-G_1G_3)+(G_1G_2G_3H)\\ &=1+G_1G_2+G_2H+G_1G_3+G_1G_2G_3H\\ P_{R1}&=G_1G_2\qquad\quad\Delta_{R1}=1\\ P_{R2}&=G_1G_3\qquad\quad\Delta_{R2}=1+G_2H\\ \Phi_R(s)&=\frac{G_1G_2+G_1G_3(1+G_2H)}{1+G_1G_2+G_2H+G_1G_3+G_1G_2G_3H}\\ P_{N1}&=-1\qquad\quad\Delta_{N1}=1+G_2H\\ P_{N2}&=G_4G_1G_2\qquad\Delta_{N2}=1\\ P_{N3}&=G_4G_1G_3\qquad\Delta_{N3}=1+G_2H\\ \Phi_N(s)&=\frac{G_1G_2G_4+(-1+G_1G_3G_4)(1+G_2H))}{1+G_1G_2+G_2H+G_1G_3+G_1G_2G_3H} \end{aligned} ΔPR1PR2ΦR(s)PN1PN2PN3ΦN(s)=1(G1G2G2HG1G3)+(G1G2G3H)=1+G1G2+G2H+G1G3+G1G2G3H=G1G2ΔR1=1=G1G3ΔR2=1+G2H=1+G1G2+G2H+G1G3+G1G2G3HG1G2+G1G3(1+G2H)=1ΔN1=1+G2H=G4G1G2ΔN2=1=G4G1G3ΔN3=1+G2H=1+G1G2+G2H+G1G3+G1G2G3HG1G2G4+(1+G1G3G4)(1+G2H))

  • 不会因N(s)的输入增加回路,故特征式相同

控制系统的传递函数

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开环传递函数

G(s)N(s)=B(s)E(s)=G1(s)G2(s)H(s) G(s)N(s)=\frac{B(s)}{E(s)}=G_1(s)G_2(s)H(s) G(s)N(s)=E(s)B(s)=G1(s)G2(s)H(s)

  • 将主反馈打断(负号不要),令前向通道传递函数和反馈通道传递函数相乘即为该系统对应的开环传递函数
  • 开环增益,将开环传递函数化为尾一标准形式,其对应参数即为开环增益

输入r(t)作用下的闭环传递函数

Φ(s)=C(s)R(s)=G1(s)G2(s)1+G1(s)G2(s)H(s)Φe(s)=E(s)R(s)=11+G1(s)G2(s)H(s)误差传递函数 \begin{aligned} \Phi(s)&=\frac{C(s)}{R(s)}=\frac{G_1(s)G_2(s)}{1+G_1(s)G_2(s)H(s)}\\ \Phi_e(s)&=\frac{E(s)}{R(s)}=\frac{1}{1+G_1(s)G_2(s)H(s)} \qquad误差传递函数 \end{aligned} Φ(s)Φe(s)=R(s)C(s)=1+G1(s)G2(s)H(s)G1(s)G2(s)=R(s)E(s)=1+G1(s)G2(s)H(s)1误差传递函数

干扰n(t)作用下的闭环传递函数

Φn(s)=C(s)N(s)=G2(s)1+G1(s)G2(s)H(s)Φen(s)=E(s)N(s)=−G2(s)H(s)1+G1(s)G2(s)H(s) \begin{aligned} \Phi_n(s)&=\frac{C(s)}{N(s)}=\frac{G_2(s)}{1+G_1(s)G_2(s)H(s)}\\ \Phi_{en}(s)&=\frac{E(s)}{N(s)}=\frac{-G_2(s)H(s)}{1+G_1(s)G_2(s)H(s)}\\ \end{aligned} Φn(s)Φen(s)=N(s)C(s)=1+G1(s)G2(s)H(s)G2(s)=N(s)E(s)=1+G1(s)G2(s)H(s)G2(s)H(s)

系统的总输出C(s)及总误差E(s)

C(s)=Φ(s)⋅R(s)+Φn(s)⋅N(s)=G1(s)G2(s)R(s)1+G1(s)G2(s)H(S)+G2(s)N(s)1+G1(s)G2(s)H(s)E(s)=Φe(s)⋅R(s)+Φen(s)⋅N(s)=R(s)1+G1(s)G2(s)H(s)−G2(s)H(s)N(s)1+G1(s)G2(s)H(s) \begin{aligned} C(s)&=\Phi(s)·R(s)+\Phi_n(s)·N(s)=\frac{G_1(s)G_2(s)R(s)}{1+G_1(s)G_2(s)H(S)}+\frac{G_2(s)N(s)}{1+G_1(s)G_2(s)H(s)}\\ E(s)&=\Phi_e(s)·R(s)+\Phi_{en}(s)·N(s)=\frac{R(s)}{1+G_1(s)G_2(s)H(s)}-\frac{G_2(s)H(s)N(s)}{1+G_1(s)G_2(s)H(s)}\\ \end{aligned} C(s)E(s)=Φ(s)R(s)+Φn(s)N(s)=1+G1(s)G2(s)H(S)G1(s)G2(s)R(s)+1+G1(s)G2(s)H(s)G2(s)N(s)=Φe(s)R(s)+Φen(s)N(s)=1+G1(s)G2(s)H(s)R(s)1+G1(s)G2(s)H(s)G2(s)H(s)N(s)

例7 系统结构图如图所示

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(1)求当输入r(t)=1(t)时系统的响应cr(t)c_r(t)cr(t)

(2)求当干扰n(t)=δ(t)\delta (t)δ(t)时系统的响应cn(t)c_n(t)cn(t)

(3)求当初条件{c(0)=−1c′(0)=0\begin{cases}c(0)=-1\\c'(0)=0\end{cases}{c(0)=1c(0)=0时系统的自由响应c0(t)c_0(t)c0(t)

(4)求在上述三个因素同时作用下系统的总输出c(t)c(t)c(t)

(5)求在上述三个因素同时作用下系统的总偏差e(t)e(t)e(t)

(1) 求当输入r(t)=1(t)时系统的响应cr(t)c_r(t)cr(t)

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Φr(s)=Cr(s)R(s)=2s(s+3)1+2s(s+3)=2s(s+3)+2=2s2+3s+2=2(s+1)(s+2)Cr(s)=Φr(s)R(s)=2s(s+1)(s+2)=C1s+C2s+1+C3s+2由留数法C1=lim⁡s→0s⋅2s(s+1)(s+2)=1C2=lim⁡s→−1(s+1)⋅2s(s+1)(s+2)=−2C3=lim⁡s→−2(s+2)⋅2s(s+1)(s+2)=1∴Cr(s)=1s−2s+1+1s+2∴cr(t)=1−2e−t+e−2t \begin{aligned} \Phi_r(s)&=\frac{C_r(s)}{R(s)}=\frac{\frac{2}{s(s+3)}}{1+\frac{2}{s(s+3)}}=\frac{2}{s(s+3)+2}=\frac{2}{s^2+3s+2}=\frac{2}{(s+1)(s+2)}\\ C_r(s)&=\Phi_r(s)R(s)=\frac{2}{s(s+1)(s+2)}=\frac{C_1}{s}+\frac{C_2}{s+1}+\frac{C_3}{s+2}\\ 由留数法&\\ \qquad C_1&=\lim_{s\rightarrow0}s·\frac{2}{s(s+1)(s+2)}=1\\ \qquad C_2&=\lim_{s\rightarrow-1}(s+1)·\frac{2}{s(s+1)(s+2)}=-2\\ \qquad C_3&=\lim_{s\rightarrow-2}(s+2)·\frac{2}{s(s+1)(s+2)}=1\\ \quad\therefore\quad C_r(s)&=\frac{1}{s}-\frac{2}{s+1}+\frac{1}{s+2}\\ \quad\therefore\quad c_r(t)&=1-2e^{-t}+e^{-2t\\} \end{aligned} Φr(s)Cr(s)由留数法C1C2C3Cr(s)cr(t)=R(s)Cr(s)=1+s(s+3)2s(s+3)2=s(s+3)+22=s2+3s+22=(s+1)(s+2)2=Φr(s)R(s)=s(s+1)(s+2)2=sC1+s+1C2+s+2C3=s0limss(s+1)(s+2)2=1=s1lim(s+1)s(s+1)(s+2)2=2=s2lim(s+2)s(s+1)(s+2)2=1=s1s+12+s+21=12et+e2t

(2) 求当干扰n(t)=δ(t)\delta (t)δ(t)时系统的响应cn(t)c_n(t)cn(t)

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Φn(s)=Cn(s)N(s)=1s+31+2s(s+3)=ss(s+3)+2=ss2+3s+2Cn(s)=Φn(s)⋅N(s)=s(s+1)(s+2)=C1s+1+C2s+2由留数法C1=lim⁡s→−1(s+1)⋅s(s+1)(s+2)=−1C2=lim⁡s→−2(s+2)⋅s(s+1)(s+2)=2∴Cn(s)=−1s+1+2s+2∴cn(t)=−e−t+2e−2t \begin{aligned} \Phi_n(s)&=\frac{C_n(s)}{N(s)}=\frac{\frac{1}{s+3}}{1+\frac{2}{s(s+3)}}=\frac{s}{s(s+3)+2}=\frac{s}{s^2+3s+2}\\ C_n(s)&=\Phi_n(s)·N(s)=\frac{s}{(s+1)(s+2)}=\frac{C_1}{s+1}+\frac{C_2}{s+2}\\ 由留数法&\\ \qquad C_1&=\lim_{s\rightarrow-1}(s+1)·\frac{s}{(s+1)(s+2)}=-1\\ \qquad C_2&=\lim_{s\rightarrow-2}(s+2)·\frac{s}{(s+1)(s+2)}=2\\ \quad\therefore\quad C_n(s)&=\frac{-1}{s+1}+\frac{2}{s+2}\\ \quad\therefore\quad c_n(t)&=-e^{-t}+2e^{-2t\\} \end{aligned} Φn(s)Cn(s)由留数法C1C2Cn(s)cn(t)=N(s)Cn(s)=1+s(s+3)2s+31=s(s+3)+2s=s2+3s+2s=Φn(s)N(s)=(s+1)(s+2)s=s+1C1+s+2C2=s1lim(s+1)(s+1)(s+2)s=1=s2lim(s+2)(s+1)(s+2)s=2=s+11+s+22=et+2e2t

(3) 求当初条件{c(0)=−1c′(0)=0\begin{cases}c(0)=-1\\c'(0)=0\end{cases}{c(0)=1c(0)=0时系统的自由响应c0(t)c_0(t)c0(t)

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  • 在自动控制原理中,自由响应(Free Response)是指在没有外部输入或干扰的情况下,控制系统对于初态或初始条件的响应。它是系统在没有受到任何控制输入或干扰的情况下,根据系统的初始状态自行演化的过程。

Φ(s)=C0(s)R(s)=2s2+3s+2C0(s)⋅(s2+3s+2)=2R(s)c′′+3c′+2=2r令r=0,系统的齐次微分方程为c′′+3c′+2=0[s2C0(s)−sc(0)−c′(0)]+3[sC(s)−c(0)]+2=0[s2+3s+2]C(s)+(s+3)=0C0(s)=−(s+3)s2+3s+2=C1s+1+C2s+2由留数法C1=lim⁡s→−1(s+1)⋅−(s+3)(s+1)(s+2)=−2C2=lim⁡s→−2(s+2)⋅−(s+3)(s+1)(s+2)=1∴C0(s)=−2s+1+1s+2∴c0(t)=−2e−t+e−2t \begin{aligned} \Phi(s)&=\frac{C_0(s)}{R(s)}=\frac{2}{s^2+3s+2}\\ C_0(s)&·(s^2+3s+2)=2R(s)\\ c''&+3c'+2=2r\\ 令r=0&,系统的齐次微分方程为\\ c''&+3c'+2=0\\ [s^2C_0(s)&-sc(0)-c'(0)]+3[sC(s)-c(0)]+2=0\\ [s^2&+3s+2]C(s)+(s+3)=0\\ C_0(s)&=\frac{-(s+3)}{s^2+3s+2}=\frac{C_1}{s+1}+\frac{C_2}{s+2}\\ 由留数法&\\ \qquad C_1&=\lim_{s\rightarrow-1}(s+1)·\frac{-(s+3)}{(s+1)(s+2)}=-2\\ \qquad C_2&=\lim_{s\rightarrow-2}(s+2)·\frac{-(s+3)}{(s+1)(s+2)}=1\\ \quad\therefore\quad C_0(s)&=\frac{-2}{s+1}+\frac{1}{s+2}\\ \quad\therefore\quad c_0(t)&=-2e^{-t}+e^{-2t} \end{aligned} Φ(s)C0(s)c′′r=0c′′[s2C0(s)[s2C0(s)由留数法C1C2C0(s)c0(t)=R(s)C0(s)=s2+3s+22(s2+3s+2)=2R(s)+3c+2=2r,系统的齐次微分方程为+3c+2=0sc(0)c(0)]+3[sC(s)c(0)]+2=0+3s+2]C(s)+(s+3)=0=s2+3s+2(s+3)=s+1C1+s+2C2=s1lim(s+1)(s+1)(s+2)(s+3)=2=s2lim(s+2)(s+1)(s+2)(s+3)=1=s+12+s+21=2et+e2t

(4) 求在上述三个因素同时作用下系统的总输出c(t)c(t)c(t)

c(t)=cr(t)+cn(t)+c0(t)=1−5e−t+4e−2t \begin{aligned} c(t)&=c_r(t)+c_n(t)+c_0(t)\\ &=1-5e^{-t}+4e^{-2t} \end{aligned} c(t)=cr(t)+cn(t)+c0(t)=15et+4e2t

(5) 求在上述三个因素同时作用下系统的总偏差e(t)e(t)e(t)

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e(t)=r(t)−c(t)=1−(1−5e−t+4e−2t)=5e−t−4e−2t \begin{aligned} e(t)&=r(t)-c(t)\\ &=1-(1-5e^{-t}+4e^{-2t})\\ &=5e^{-t}-4e^{-2t} \end{aligned} e(t)=r(t)c(t)=1(15et+4e2t)=5et4e2t

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