03_拉氏反变换&&传递函数
复习
拉氏变换的定义: F ( s ) = ∫ 0 ∞ f ( t ) ⋅ e − s t d t F(s)=\int_0^\infty f(t)·e^{-st}dt F(s)=∫0∞f(t)⋅e−stdt
常见函数的拉普拉斯变换
| f ( t ) f(t) f(t) | F ( s ) F(s) F(s) | |
|---|---|---|
| 单位脉冲 | δ ( t ) \delta(t) δ(t) | 1 |
| 单位阶跃 | 1 ( t ) 1(t) 1(t) | 1 s \frac{1}{s} s1 |
| 单位斜坡 | t t t | 1 s 2 \frac{1}{s^2} s21 |
| 单位加速度 | t 2 2 \frac{t^2}{2} 2t2 | 1 s 3 \frac{1}{s^3} s31 |
| 指数函数 | e − a t e^{-at} e−at | 1 s + a \frac{1}{s+a} s+a1 |
| 正弦函数 | sin w t \sin wt sinwt | w s 2 + w 2 \frac{w}{s^2+w^2} s2+w2w |
| 余弦函数 | cos w t \cos wt coswt | s s 2 + w 2 \frac{s}{s^2+w^2} s2+w2s |
L变换重要定理
| 线性性质 | L [ a f 1 ( t ) ] ± b f 2 ( t ) = a F 1 ( s ) ± b F 2 ( s ) L[af_1(t)]\pm bf_2(t)=aF_1(s)\pm bF_2(s) L[af1(t)]±bf2(t)=aF1(s)±bF2(s) |
|---|---|
| 微分定理 | L [ f ′ ( t ) ] = s ⋅ F ( s ) − f ( 0 ) L[f'(t)]=s·F(s)-f(0) L[f′(t)]=s⋅F(s)−f(0) |
| 积分定理 | L [ ∫ f ( t ) d t ] = 1 s ⋅ F ( s ) + 1 s f ( − 1 ) ( 0 ) L[\int f(t)dt]=\frac{1}{s}·F(s)+\frac{1}{s}f^{(-1)}(0) L[∫f(t)dt]=s1⋅F(s)+s1f(−1)(0) |
| 实位移定理 | L [ f ( t − τ ) ] = e − τ s ⋅ F ( s ) L[f(t-\tau)]=e^{-\tau s}·F(s) L[f(t−τ)]=e−τs⋅F(s) |
| 复位移定理 | L [ e A t f ( t ) ] = F ( s − A ) L[e^{At}f(t)]=F(s-A) L[eAtf(t)]=F(s−A) |
| 初值定理 | lim t → 0 f ( t ) = lim s → ∞ s ⋅ F ( s ) \lim_{t\rightarrow0}f(t)=\lim_{s\rightarrow\infty}s·F(s) limt→0f(t)=lims→∞s⋅F(s) |
| 终值定理 | lim t → ∞ f ( t ) = lim s → 0 s ⋅ F ( s ) \lim_{t\rightarrow\infty}f(t)=\lim_{s\rightarrow0}s·F(s) limt→∞f(t)=lims→0s⋅F(s) |
拉氏反变换
基本概念
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反演公式 f ( t ) = 1 2 π j ∫ σ − j ∞ σ + j ∞ F ( s ) ⋅ e t s d s f(t)=\frac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}F(s)·e^{ts}ds f(t)=2πj1∫σ−j∞σ+j∞F(s)⋅etsds
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查表法(分解部分分式法) { 试凑法 系数比较法 留数法 \begin{cases}试凑法\\系数比较法\\留数法\end{cases} ⎩ ⎨ ⎧试凑法系数比较法留数法
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模态: 如果n阶微分方程的特征根是 λ 1 , λ 2 , . . . λ n \lambda_1,\lambda_2,...\lambda_n λ1,λ2,...λn且无重根，则把函数 e λ 1 t , e λ 2 t , . . . , e λ n t e^{\lambda_1t}, e^{\lambda_2t},...,e^{\lambda_nt} eλ1t,eλ2t,...,eλnt称为该微分方程所描述运动的模态, 也叫做振型
留数法分解部分分式
一般有 F ( s ) = B ( s ) A ( s ) = b m s m + b m − 1 s m − 1 + . . . + b 0 a n s n + a n − 1 s ( n − 1 ) + . . . a 0 ( n > m ) 设 A ( s ) = a n s n + a n − 1 s n − 1 + . . . a 0 = ( s − p 1 ) ( s − p 2 ) . . . ( s − p n ) \begin{aligned} 一般有\quad F(s)&=\frac{B(s)}{A(s)}=\frac{b_ms^m+b_{m-1}s^{m-1}+...+b_0}{a_ns^n+a_{n-1}s^{(n-1)}+...a_0}(n>m)\\设\quad A(s)&=a_ns^n+a_{n-1}s^{n-1}+...a_0=(s-p_1)(s-p_2)...(s-p_n) \end{aligned} 一般有F(s)设A(s)=A(s)B(s)=ansn+an−1s(n−1)+...a0bmsm+bm−1sm−1+...+b0(n>m)=ansn+an−1sn−1+...a0=(s−p1)(s−p2)...(s−pn)
Ⅰ. 当 A ( s ) = 0 A(s)=0 A(s)=0无重根时 F ( s ) F(s) F(s)可分解为 F ( s ) = C 1 s − p 1 + C 2 s − p 2 + . . . + C n s − p n = ∑ i = 1 n C i s − p i F(s)=\frac{C_1}{s-p_1}+\frac{C_2}{s-p_2}+...+\frac{C_n}{s-p_n}=\sum_{i=1}^{n}\frac{C_i}{s-p_i} F(s)=s−p1C1+s−p2C2+...+s−pnCn=∑i=1ns−piCi
其中 : { C i = lim s → p i ( s − p i ) F ( s ) = lim s → p i ( s − p i ) b m S m + b m − 1 s m − 1 + . . . + b o ( s − p 1 ) . . . ( s − p i ) . . . ( s − p n ) C i = B ( s ) A ′ ( s ) ∣ s = p i = b m S m + b m − 1 s m − 1 + . . . + b o lim s → p i ( s − p 1 ) . . . ( s − p i ) . . . ( s − p n ) ( s − p i ) f ( t ) = C 1 e p 1 t + C 2 e p 2 t + . . . + C n e p n t = ∑ i = t n C i e p i t \begin{aligned} 其中&:\begin{cases}C_i=\lim_{s\rightarrow p_i}(s-p_i)F(s)=\lim_{s\rightarrow p_i}(s-p_i)\frac{b_mS^m+b_{m-1}s^{m-1}+...+bo}{(s-p_1)...(s-p_i)...(s-p_n)}\\C_i=\frac{B(s)}{A'(s)}|_{s=p_i}=\frac{b_mS^m+b_{m-1}s^{m-1}+...+bo}{\lim_{s\rightarrow p_i}\frac{(s-p_1)...(s-p_i)...(s-p_n)}{(s-p_i)}}\end{cases}\\ f(t)&=C_1e^{p_1t}+C_2e^{p_2t}+...+C_ne^{p_nt}=\sum_{i=t}^nC_ie^{p_it} \end{aligned} 其中f(t):⎩ ⎨ ⎧Ci=lims→pi(s−pi)F(s)=lims→pi(s−pi)(s−p1)...(s−pi)...(s−pn)bmSm+bm−1sm−1+...+boCi=A′(s)B(s)∣s=pi=lims→pi(s−pi)(s−p1)...(s−pi)...(s−pn)bmSm+bm−1sm−1+...+bo=C1ep1t+C2ep2t+...+Cnepnt=i=t∑nCiepit
例2 已知 F ( s ) = s + 2 s 2 + 4 s + 3 , 求 f ( t ) = ? \quad 已知F(s)=\frac{s+2}{s^2+4s+3} ,求f(t)=? 已知F(s)=s2+4s+3s+2,求f(t)=?
解 . F ( s ) = s + 2 s 2 + 4 s + 3 = C 1 ( s + 1 ) + C 2 s + 3 由留数法 C 1 = lim s → − 1 ( s + 1 ) s + 2 ( s + 1 ) ( s + 3 ) = 1 2 C 2 = lim s → − 3 ( s + 3 ) s + 2 ( s + 1 ) ( s + 3 ) = 1 2 ∴ F ( s ) = 1 2 s + 1 + 1 2 s + 3 f ( t ) = 1 2 e − t + 1 2 e − 3 t \begin{aligned} 解.\quad F(s)=\frac{s+2}{s^2+4s+3}=\frac{C_1}{(s+1)}+\frac{C2}{s+3} \\ \begin{aligned} 由留数法\quad C_1&=\lim_{s\rightarrow-1}(s+1)\frac{s+2}{(s+1)(s+3)}=\frac{1}{2}\\ C_2&=\lim_{s\rightarrow-3}(s+3)\frac{s+2}{(s+1)(s+3)}=\frac{1}{2} \end{aligned}\\ \therefore\quad F(s)=\frac{\frac{1}{2}}{s+1}+\frac{\frac{1}{2}}{s+3} \quad f(t)=\frac{1}{2}e^{-t}+\frac{1}{2}e^{-3t} \end{aligned} 解.F(s)=s2+4s+3s+2=(s+1)C1+s+3C2由留数法C1C2=s→−1lim(s+1)(s+1)(s+3)s+2=21=s→−3lim(s+3)(s+1)(s+3)s+2=21∴F(s)=s+121+s+321f(t)=21e−t+21e−3t
例3 已知 F ( s ) = s 2 + 5 s + 5 s 2 + 4 s + 3 , 求 f ( t ) = ? \quad 已知F(s)=\frac{s^2+5s+5}{s^2+4s+3}, 求f(t)=? 已知F(s)=s2+4s+3s2+5s+5,求f(t)=?
解 . F ( s ) = ( s 2 + 4 s + 3 ) + s + 2 s 2 + 4 s + 3 = 1 + s + 2 ( s + 1 ) ( s + 3 ) = 1 + 1 2 s + 1 + 1 2 s + 3 f ( t ) = δ ( t ) + 1 2 e − t + 1 2 e − 3 t \begin{aligned} 解. \quad F(s)&=\frac{(s^2+4s+3)+s+2}{s^2+4s+3}=1+\frac{s+2}{(s+1)(s+3)}=1+\frac{\frac{1}{2}}{s+1}+\frac{\frac{1}{2}}{s+3}\\ f(t)&=\delta(t)+\frac{1}{2}e^{-t}+\frac{1}{2}e^{-3t} \end{aligned} 解.F(s)f(t)=s2+4s+3(s2+4s+3)+s+2=1+(s+1)(s+3)s+2=1+s+121+s+321=δ(t)+21e−t+21e−3t
例4 已知 F ( s ) = s + 3 s 2 + 2 s + 2 , 求 f ( t ) = ? \quad 已知F(s)=\frac{s+3}{s^2+2s+2},求f(t)=? 已知F(s)=s2+2s+2s+3,求f(t)=?
解一 . F ( s ) = s + 3 ( s + 1 + j ) ( s + 1 − j ) = C 1 s + 1 + j + C 2 s + 1 − j 由留数法 C 1 = lim s → − ( 1 + j ) ( s + 1 + j ) s + 3 ( s + 1 + j ) ( s + 1 − j ) = 2 − j − 2 j C 2 = lim s → ( j − 1 ) ( s + 1 − j ) s + 3 ( s + 1 + j ) ( s + 1 − j ) = 2 + j 2 j ∴ F ( s ) = 2 + j 2 j s + 1 − j − 2 − j 2 j s + 1 + j f ( t ) = 2 + j 2 j e − ( 1 − j ) t − 2 − j 2 j e − ( 1 + j ) t = e − t 2 j [ ( 2 + j ) e j t − ( 2 − j ) e − j t ] = e − t 2 j [ 2 j e j t + e − j t 2 + 4 j e j t − e − j t 2 j ] = e − t ⋅ [ cos t + 2 sin t ] 解一. \quad F(s)=\frac{s+3}{(s+1+j)(s+1-j)}=\frac{C_1}{s+1+j}+\frac{C_2}{s+1-j}\\ \begin{aligned} 由留数法 \quad C_1&=\lim_{s\rightarrow-(1+j)}(s+1+j)\frac{s+3}{(s+1+j)(s+1-j)}=\frac{2-j}{-2j}\\ C2&=\lim_{s\rightarrow(j-1)}(s+1-j)\frac{s+3}{(s+1+j)(s+1-j)}=\frac{2+j}{2j} \end{aligned}\\ \begin{aligned} \therefore F(s) &= \frac{\frac{2+j}{2j}}{s+1-j}-\frac{\frac{2-j}{2j}}{s+1+j}\quad f(t)=\frac{2+j}{2j}e^{-(1-j)t}-\frac{2-j}{2j}e^{-(1+j)t}\\ &=\frac{e^{-t}}{2j}[(2+j)e^{jt}-(2-j)e^{-jt}]=\frac{e^{-t}}{2j}[2j\frac{e^{jt}+e^{-jt}}{2}+4j\frac{e^{jt}-e^{-jt}}{2j}]\\&=e^{-t}·[\cos t+2\sin t] \end{aligned} 解一.F(s)=(s+1+j)(s+1−j)s+3=s+1+jC1+s+1−jC2由留数法C1C2=s→−(1+j)lim(s+1+j)(s+1+j)(s+1−j)s+3=−2j2−j=s→(j−1)lim(s+1−j)(s+1+j)(s+1−j)s+3=2j2+j∴F(s)=s+1−j2j2+j−s+1+j2j2−jf(t)=2j2+je−(1−j)t−2j2−je−(1+j)t=2je−t[(2+j)ejt−(2−j)e−jt]=2je−t[2j2ejt+e−jt+4j2jejt−e−jt]=e−t⋅[cost+2sint]
解二 . F ( s ) = s + 3 ( s + 1 ) 2 + 1 2 = s + 1 ( s + 1 ) 2 + 1 2 + 2 1 ( s + 1 ) 2 + 1 2 f ( t ) = e − t ⋅ [ cos t + 2 sin t ] \begin{aligned} 解二. \quad F(s)&=\frac{s+3}{(s+1)^2+1^2}=\frac{s+1}{(s+1)^2+1^2}+2\frac{1}{(s+1)^2+1^2}\\ f(t)&= e^{-t}·[\cos t+2\sin t] \end{aligned} 解二.F(s)f(t)=(s+1)2+12s+3=(s+1)2+12s+1+2(s+1)2+121=e−t⋅[cost+2sint]
Ⅱ. 当 A ( s ) = ( s − p 1 ) . . . ( s − p n ) = 0 A(s)=(s-p_1)...(s-p_n)=0 A(s)=(s−p1)...(s−pn)=0有重根时(设 p 1 p_1 p1为m重根, 其余为单根)
F ( s ) = C m ( s − p 1 ) m + C m − 1 ( s − p 1 ) m − 1 + . . . + C 1 s − p 1 + C m + 1 s − p m + 1 + . . . + C n s − p n F(s)=\frac{C_m}{(s-p_1)^m}+\frac{C_{m-1}}{(s-p_1)^{m-1}}+...+\frac{C_1}{s-p_1}+\frac{C_{m+1}}{s-p_{m+1}}+...+\frac{C_n}{s-p_n} F(s)=(s−p1)mCm+(s−p1)m−1Cm−1+...+s−p1C1+s−pm+1Cm+1+...+s−pnCn
例5 F ( s ) = s + 2 s ( s + 1 ) 2 ( s + 3 ) , 求 f ( t ) = ? \quad F(s)=\frac{s+2}{s(s+1)^2(s+3)}, 求f(t)=? F(s)=s(s+1)2(s+3)s+2,求f(t)=?
解 . F ( s ) = C 2 ( s + 1 ) 2 + C 1 s + 1 + C 3 s + C 4 s + 3 C 2 = lim s → − 1 ( s + 1 ) 2 s + 2 s ( s + 1 ) 2 ( s + 3 ) = − 1 2 C 1 = lim s → − 1 d d s [ ( s + 1 ) 2 s + 2 s ( s + 1 ) 2 ( s + 3 ) ] = lim s → − 1 s ( s + 3 ) − ( s + 2 ) ( 2 s + 3 ) s 2 ( s + 3 ) 2 = − 3 4 解.\qquad F(s)=\frac{C_2}{(s+1)^2}+\frac{C_1}{s+1}+\frac{C_3}{s}+\frac{C_4}{s+3}\\ \begin{aligned} C_2&=\lim_{s\rightarrow-1}(s+1)^2\frac{s+2}{s(s+1)^2(s+3)}=-\frac{1}{2}\\ C_1&=\lim_{s\rightarrow-1}\frac{d}{ds}[(s+1)^2\frac{s+2}{s(s+1)^2(s+3)}]=\lim_{s\rightarrow-1}\frac{s(s+3)-(s+2)(2s+3)}{s^2(s+3)^2}=-\frac{3}{4} \end{aligned} 解.F(s)=(s+1)2C2+s+1C1+sC3+s+3C4C2C1=s→−1lim(s+1)2s(s+1)2(s+3)s+2=−21=s→−1limdsd[(s+1)2s(s+1)2(s+3)s+2]=s→−1lims2(s+3)2s(s+3)−(s+2)(2s+3)=−43
例6 R − C R-C R−C 电路计算
由 K V L u r = R i + u c i = C d u c d t u r = R C d u c d t + u c 由拉氏变换 ( R C s + 1 ) U c ( s ) = U r ( s ) + R C u c ( 0 ) 设 u r ( t ) = E 0 ⋅ 1 ( t ) U r ( s ) = E 0 s U c ( s ) = U r ( s ) R C s + 1 + R C u c ( 0 ) R C s + 1 = E 0 s ( R C s + 1 ) + R C u c ( 0 ) R C s + 1 = E 0 R C s ( s + 1 R C ) + u c ( 0 ) s + 1 R C = C 0 s + C 1 s + 1 R C + u c ( 0 ) s + 1 R C 又 { C 0 = lim s → 0 s E 0 / R C s ( s + 1 R C ) = E 0 C 1 = lim s → − 1 R C ( s + 1 R C ) E 0 / R C s ( s + 1 R C ) = − E 0 ∴ U c ( s ) = E 0 s − E 0 s + 1 R C + u c ( 0 ) s + 1 R C ∴ u c ( t ) = E 0 − [ E 0 − u c ( 0 ) ] ⋅ e − 1 R C \begin{aligned} 由KVL \qquad u_r &= Ri + u_c\\ i &=C\frac{du_c}{dt}\\u_r&=RC\frac{du_c}{dt}+u_c\\由拉氏变换\qquad (RCs+1)U_c(s)&=U_r(s)+RCu_c(0)\\设 \qquad u_r(t)&=E_0·1(t)\qquad U_r(s)=\frac{E_0}{s}\\ U_c(s)&=\frac{U_r(s)}{RCs+1}+\frac{RCu_c(0)}{RCs+1}=\frac{E_0}{s(RCs+1)}+\frac{RCu_c(0)}{RCs+1}\\&=\frac{\frac{E_0}{RC}}{s(s+\frac{1}{RC})}+\frac{u_c(0)}{s+\frac{1}{RC}}\\&=\frac{C_0}{s}+\frac{C_1}{s+\frac{1}{RC}}+\frac{u_c(0)}{s+\frac{1}{RC}}\\又\qquad& \begin{cases}C_0=\lim_{s\rightarrow0}s\frac{E_0/RC}{s(s+\frac{1}{RC})}=E_0\\C_1=\lim_{s\rightarrow-\frac{1}{RC}}(s+\frac{1}{RC})\frac{E_0/RC}{s(s+\frac{1}{RC})}=-E_0\end{cases}\\\therefore\qquad U_c(s)&=\frac{E_0}{s}-\frac{E_0}{s+\frac{1}{RC}}+\frac{u_c(0)}{s+\frac{1}{RC}}\\ \therefore \qquad u_c(t)&=E_0-[E_0-u_c(0)]·e^{-\frac{1}{RC}} \end{aligned} 由KVLuriur由拉氏变换(RCs+1)Uc(s)设ur(t)Uc(s)又∴Uc(s)∴uc(t)=Ri+uc=Cdtduc=RCdtduc+uc=Ur(s)+RCuc(0)=E0⋅1(t)Ur(s)=sE0=RCs+1Ur(s)+RCs+1RCuc(0)=s(RCs+1)E0+RCs+1RCuc(0)=s(s+RC1)RCE0+s+RC1uc(0)=sC0+s+RC1C1+s+RC1uc(0){C0=lims→0ss(s+RC1)E0/RC=E0C1=lims→−RC1(s+RC1)s(s+RC1)E0/RC=−E0=sE0−s+RC1E0+s+RC1uc(0)=E0−[E0−uc(0)]⋅e−RC1
影响系统响应的因素
- 输入 u r ( t ) u_r(t) ur(t) —— 规定 r ( t ) = 1 ( t ) r(t)=1(t) r(t)=1(t)
- 初始条件 —— 规定0初始条件
- 系统的结构参数 —— 自身特性决定系统性能
传递函数
基本概念
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传递函数的定义: 在零初始条件下, 线性定常系统输出量拉氏变换与输入量拉氏变换之比
G ( s ) = C ( s ) R ( s ) \qquad G(s)=\frac{C(s)}{R(s)}\qquad \qquad\qquad\qquad\quad G(s)=R(s)C(s)
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传递函数的标准形式
- 首1标准型:
G ( s ) = K ∗ ∏ j = 1 m ( s − z j ) ∏ i = 1 n ( s − p i ) G(s)=\frac{K^*\prod_{j=1}^m(s-z_j)}{\prod_{i=1}^n(s-p_i)} G(s)=∏i=1n(s−pi)K∗∏j=1m(s−zj)
- 首1标准型:
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尾1标准型:
G ( s ) = K ∏ k = 1 m 1 ( τ k s + 1 ) ∏ l = 1 m 2 ( τ 1 2 s 2 + 2 ξ τ l s + 1 ) s v ∏ i = 1 n 1 ( T i s + 1 ) ∏ j = 1 n 2 ( T j 2 s 2 + 2 ξ T j s + 1 ) G(s)=K\frac{\prod_{k=1}^{m_1}(\tau_ks+1)\prod_{l=1}^{m_2}(\tau_1^2s^2+2\xi\tau_ls+1)}{s^v\prod_{i=1}^{n_1}(T_is+1)\prod_{j=1}^{n_2}(T_j^2s^2+2\xi T_js+1)} G(s)=Ksv∏i=1n1(Tis+1)∏j=1n2(Tj2s2+2ξTjs+1)∏k=1m1(τks+1)∏l=1m2(τ12s2+2ξτls+1)
例7 已知 G ( s ) = 4 s − 4 s 3 + 3 s 2 + 2 s 将其化为首 1 、尾 1 标准型，并确定其增益 \quad已知G(s)=\frac{4s-4}{s^3+3s^2+2s}将其化为首1、尾1标准型，并确定其增益 已知G(s)=s3+3s2+2s4s−4将其化为首1、尾1标准型，并确定其增益
解 . G ( s ) = 4 ( s − 1 ) s ( s + 1 ) ( s + 2 ) 首 1 标准型 G ( s ) = 2 ⋅ s − 1 s ( 1 2 s 2 + 3 2 s + 1 ) = 2 ⋅ s − 1 s ( 1 2 s + 1 ) ( s + 1 ) 尾 1 标准型 K = 2 \begin{aligned} 解.\qquad G(s)&=\frac{4(s-1)}{s(s+1)(s+2)}\qquad 首1标准型\\ G(s)&=2·\frac{s-1}{s(\frac{1}{2}s^2+\frac{3}{2}s+1)}=2·\frac{s-1}{s(\frac{1}{2}s+1)(s+1)}\quad 尾1标准型\\ K&=2 \end{aligned} 解.G(s)G(s)K=s(s+1)(s+2)4(s−1)首1标准型=2⋅s(21s2+23s+1)s−1=2⋅s(21s+1)(s+1)s−1尾1标准型=2
传递函数的性质
- G ( s ) G(s) G(s)是复数
- G ( s ) G(s) G(s)只与系统自身的结构参数有关
- G ( s ) G(s) G(s)与系统微分方程直接关联
- G ( s ) = L [ k ( t ) ] G(s)=L[k(t)] G(s)=L[k(t)]
- G ( s ) G(s) G(s)与 s s s平面上的零极点图相对应
传递函数的局限性
- 原则上不反映非零初始条件时系统的全部信息
- 适合于描述单输入/单输出系统
- 只能用于表示线性定常系统
参考资料
【（新版！最清晰！去噪不炸耳！）自动控制原理 西北工业大学 卢京潮】