由拉氏变换直接写出时域函数的简便方法

• 1. 因式分解
• 2. 简便方法
• • 1) 没有重根、没有零极点情况
  • 2) 没有重根，但是具有零极点
  • 3) 有重根的情况
• 3. 备注

众所周知，拉普拉斯变换可以将时域函数x(t)x(t)x(t)转换到频域内X(s)X(s)X(s)：
x(t)⟹X(s),orX(s)=L{x(t)}x(t) \Longrightarrow X(s), \qquad or \qquad X(s) = \mathcal{L} \left\lbrace x(t) \right\rbrace x(t)⟹X(s),orX(s)=L{x(t)}但当我们知道了输出信号的拉氏变换X(s)X(s)X(s)后，如何能快速得到其对应的时域形式呢？

1. 因式分解

一种常用的方法是因式分解，将X(s)X(s)X(s)分解为若干简单分式加和的形式，这些简单分式简单到可以直接写出其对应的时域形式。如：
X(s)=s+2s2+4s+3=s+2(s+1)(s+3)=12[1s+1+1s+3]\begin{aligned} X(s) &= \frac{s+2}{s^2 + 4s + 3} \\ &= \frac{s+2}{\left( s+1 \right) \left( s+3 \right)} \\ &= \frac{1}{2} \left[ \frac{1}{s+1} + \frac{1}{s+3} \right] \end{aligned} X(s)​=s2+4s+3s+2​=(s+1)(s+3)s+2​=21​[s+11​+s+31​]​则可以立即写出其时域形式
x(t)=L−1{X(s)}=12e−t+12e−3tx(t) = \mathcal{L}^{-1} \left\lbrace X(s) \right\rbrace = \frac{1}{2} e^{-t} + \frac{1}{2} e^{-3t} x(t)=L−1{X(s)}=21​e−t+21​e−3t因式分解的过程可以用留数定理来做。设
X(s)=s+2s2+4s+3=s+2(s+1)(s+3)=As+1+Bs+3=A(s+3)+B(s+1)(s+1)(s+3)\begin{aligned} X(s) &= \frac{s+2}{s^2 + 4s + 3} \\ &= \frac{s+2}{\left( s+1 \right) \left( s+3 \right)} \\ &= \frac{A}{s+1} + \frac{B}{s+3} \\ &= \frac{A\left( s+3 \right) + B \left( s+1 \right)}{\left( s+1 \right) \left( s+3 \right)} \end{aligned} X(s)​=s2+4s+3s+2​=(s+1)(s+3)s+2​=s+1A​+s+3B​=(s+1)(s+3)A(s+3)+B(s+1)​​则A(s+3)+B(s+1)=s+2A\left( s+3 \right) + B \left( s+1 \right) = s+2 A(s+3)+B(s+1)=s+2留数定理的原理是，依次把X(s)X(s)X(s)的根代入，以求得A,BA, BA,B：
1） 当s=−1s=-1s=−1时，A(−1+3)+B(−1+1)=−1+2A\left( -1+3 \right) + B \left( -1+1 \right) = -1+2A(−1+3)+B(−1+1)=−1+2得到A=12A = \frac{1}{2}A=21​；
2） 当s=−3s=-3s=−3时，A(−3+3)+B(−3+1)=−3+2A\left( -3+3 \right) + B \left( -3+1 \right) = -3+2A(−3+3)+B(−3+1)=−3+2得到B=12B = \frac{1}{2}B=21​。

这种方法的缺点很明显：因式分解的过程冗长复杂，系数A,BA, BA,B的求解繁琐。

2. 简便方法

幸运的是，笔者在这里列出3个公式，可以解决所有的由复数域向时域转变的问题，甚至只需要记其中一个即可。
设X(s)=A(s)B(s)X(s) = \frac{A(s)}{B(s)}X(s)=B(s)A(s)​。

1) 没有重根、没有零极点情况

当B(s)=(s−s1)(s−s2)⋯(s−sn)B(s) = (s-s_1) (s-s_2) \cdots (s - s_n)B(s)=(s−s1​)(s−s2​)⋯(s−sn​)时，共nnn个极点，且每个极点都不同，且没有重根。则时域函数为
x(t)=∑k=1nA(s)B′(sk)eskt(1)x(t) = \sum _{k=1}^n \frac{A(s)}{B' \left( s_k \right) } e^{s_k t} \tag{1} x(t)=k=1∑n​B′(sk​)A(s)​esk​t(1)举例：对于
X(s)=s+2s2+5s+4=s+2(s+4)(s+1)X(s) = \frac{s+2}{s^2 + 5s + 4} = \frac{s+2}{(s+4)(s+1)} X(s)=s2+5s+4s+2​=(s+4)(s+1)s+2​可见A(s)=s+2,B(s)=(s+4)(s+1)A(s) = s+2, \quad B(s) = (s+4)(s+1)A(s)=s+2,B(s)=(s+4)(s+1)，两个极点s1=−4,s2=−1s_1 = -4, s_2 = -1s1​=−4,s2​=−1。代入式(1)：
x(t)=∑k=1nA(s)B′(sk)eskt=A(s1)B′(s1)es1t+A(s2)B′(s2)es2t=s1+22s1+5es1t+s2+22s2+5es2t=23e−4t+13e−t\begin{aligned} x(t) &= \sum _{k=1}^n \frac{A(s)}{B' \left( s_k \right) } e^{s_k t} \\ &= \frac{A(s_1)}{B' \left( s_1 \right) } e^{s_1 t} + \frac{A(s_2)}{B' \left( s_2 \right) } e^{s_2 t} \\ &= \frac{s_1 + 2}{2s_1 + 5} e^{s_1 t} + \frac{s_2 + 2}{2s_2 + 5} e^{s_2 t} \\ &= \frac{2}{3} e^{-4t} + \frac{1}{3} e^{-t} \end{aligned} x(t)​=k=1∑n​B′(sk​)A(s)​esk​t=B′(s1​)A(s1​)​es1​t+B′(s2​)A(s2​)​es2​t=2s1​+5s1​+2​es1​t+2s2​+5s2​+2​es2​t=32​e−4t+31​e−t​

2) 没有重根，但是具有零极点

具有如下形式
X(s)=A(s)sB(s)X(s) = \frac{A(s)}{s B(s)} X(s)=sB(s)A(s)​则时域为
x(t)=A(0)B(0)+∑k=1nA(sk)skB′(sk)eskt(2)x(t) = \frac{A(0)}{B(0)} + \sum _{k=1}^n \frac{A \left( s_k \right)}{s_k B' \left( s_k \right)} e^{s_k t} \tag{2} x(t)=B(0)A(0)​+k=1∑n​sk​B′(sk​)A(sk​)​esk​t(2)举例：X(s)=100s(s2+10s+100)=A(s)sB(s)X(s) = \frac{100}{s \left( s^2 + 10s + 100 \right)} = \frac{A(s)}{s B(s)} X(s)=s(s2+10s+100)100​=sB(s)A(s)​极点为s1,2=−5±53js_{1,2} = -5 \pm 5 \sqrt{3} js1,2​=−5±53​j。
由于B′(s)=2s+10B'(s) = 2s+10B′(s)=2s+10，故B′(s1)=103j,B′(s2)=−103jB' \left( s_1 \right) = 10 \sqrt{3}j, B' \left( s_2 \right) = -10 \sqrt{3}jB′(s1​)=103​j,B′(s2​)=−103​j。代入(2)有
x(t)=A(0)B(0)+∑k=1nA(sk)skB′(sk)eskt=100100+A(s1)s1B′(s1)es1t+A(s2)s2B′(s2)es2t=1+100(−5+53j)⋅103je(−5+53j)t+100(−5−53j)⋅(−103j)e(−5−53j)t=1−23j+3e(−5+53j)t+23j−3e(−5−53j)t=1−23j+3e−5t(cos⁡53t+jsin⁡53t)+23j−3e−5t(cos⁡53t−jsin⁡53t)=1+e−5tcos⁡53t−13e−5tsin⁡53t\begin{aligned} x(t) &= \frac{A(0)}{B(0)} + \sum _{k=1}^n \frac{A \left( s_k \right)}{s_k B' \left( s_k \right)} e^{s_k t} \\ &= \frac{100}{100} + \frac{A \left( s_1 \right)}{s_1 B' \left( s_1 \right)} e^{s_1 t} + \frac{A \left( s_2 \right)}{s_2 B' \left( s_2 \right)} e^{s_2 t} \\ &= 1 + \frac{100}{\left( -5 + 5 \sqrt{3} j \right) \cdot 10 \sqrt{3} j} e^{ \left( -5 + 5 \sqrt{3} j \right) t} + \frac{100}{\left( -5 - 5 \sqrt{3} j \right) \cdot \left( -10 \sqrt{3} j \right)} e^{ \left( -5 - 5 \sqrt{3} j \right) t} \\ &= 1 - \frac{2}{\sqrt{3} j + 3} e^{\left( -5 + 5 \sqrt{3} j \right) t} + \frac{2}{\sqrt{3} j - 3} e^{\left( -5 - 5 \sqrt{3} j \right) t} \\ &= 1 - \frac{2}{\sqrt{3} j + 3} e^{-5t} \left( \cos 5\sqrt{3}t + j \sin 5 \sqrt{3}t \right) + \frac{2}{\sqrt{3} j - 3} e^{-5t} \left( \cos 5\sqrt{3}t - j \sin 5 \sqrt{3}t \right) \\ &= 1 + e^{-5t} \cos 5\sqrt{3}t - \frac{1}{3} e^{-5t} \sin 5 \sqrt{3}t \end{aligned} x(t)​=B(0)A(0)​+k=1∑n​sk​B′(sk​)A(sk​)​esk​t=100100​+s1​B′(s1​)A(s1​)​es1​t+s2​B′(s2​)A(s2​)​es2​t=1+(−5+53​j)⋅103​j100​e(−5+53​j)t+(−5−53​j)⋅(−103​j)100​e(−5−53​j)t=1−3​j+32​e(−5+53​j)t+3​j−32​e(−5−53​j)t=1−3​j+32​e−5t(cos53​t+jsin53​t)+3​j−32​e−5t(cos53​t−jsin53​t)=1+e−5tcos53​t−31​e−5tsin53​t​

3) 有重根的情况

此时X(s)X(s)X(s)分母为
B(s)=(s−s1)μ1(s−s2)μ2⋯(s−sr)μrB(s) = \left(s - s_1 \right) ^{\mu_1} \left(s - s_2 \right) ^{\mu_2} \cdots \left(s - s_r \right) ^{\mu_r} B(s)=(s−s1​)μ1​(s−s2​)μ2​⋯(s−sr​)μr​即：极点sis_isi​的重数为μi\mu_iμi​。共rrr个根s1,s2,⋯,srs_1, s_2, \cdots, s_rs1​,s2​,⋯,sr​，且阶数μ1+μ2+⋯+μr=n\mu_1 + \mu_2 + \cdots + \mu_r = nμ1​+μ2​+⋯+μr​=n。
则时域为
x(t)=∑k=1r∑j=1μkAjktμk−j(μk−j)!eskt(3)x(t) = \sum _{k=1}^r \sum _{j=1}^{\mu_k} A_{jk} \frac{t^{\mu_k - j}}{\left( \mu_k - j \right)!} e^{s_k t} \tag{3} x(t)=k=1∑r​j=1∑μk​​Ajk​(μk​−j)!tμk​−j​esk​t(3)其中系数AjkA_{jk}Ajk​满足
Ajk=1(j−1)!dj−1dsj−1[(s−sk)μkX(s)]s=sk(4)A_{jk} = \frac{1}{\left( j - 1 \right) !} \frac{d^{j-1}}{ds^{j-1}} \Big[ \left( s -s_k \right) ^{\mu_k} X(s) \Big] _{s = s_k} \tag{4} Ajk​=(j−1)!1​dsj−1dj−1​[(s−sk​)μk​X(s)]s=sk​​(4)例：
X(s)=s2(s−1)3(s+1)3X(s) = \frac{s^2}{(s-1)^3 (s+1)^3} X(s)=(s−1)3(s+1)3s2​极点s1=1,s2=−1s_1 = 1, s_2 = -1s1​=1,s2​=−1，其重数分别为μ1=3,μ2=3\mu_1 = 3, \mu_2 = 3μ1​=3,μ2​=3。代入(3)：
x(t)=∑k=1r∑j=1μkAjktμk−j(μk−j)!eskt=A11tμ1−1(μ1−1)!es1t+A21tμ1−2(μ1−2)!es1t+A31tμ1−3(μ1−3)!es1t+A12tμ2−1(μ2−1)!es2t+A22tμ2−2(μ2−2)!es2t+A32tμ2−3(μ2−3)!es2t=12A11t2et+A21tet+A31et+12A12t2e−t+A22te−t+A32e−tx(t) = \sum _{k=1}^r \sum _{j=1}^{\mu_k} A_{jk} \frac{t^{\mu_k - j}}{\left( \mu_k - j \right)!} e^{s_k t} \\ = A_{11} \frac{t^{\mu_1 - 1}}{\left(\mu_1 - 1 \right) !} e^{s_1 t} + A_{21} \frac{t^{\mu_1 - 2}}{\left(\mu_1 - 2 \right) !} e^{s_1 t} + A_{31} \frac{t^{\mu_1 - 3}}{\left(\mu_1 - 3 \right) !} e^{s_1 t} + A_{12} \frac{t^{\mu_2 - 1}}{\left(\mu_2 - 1 \right) !} e^{s_2 t} + A_{22} \frac{t^{\mu_2 - 2}}{\left(\mu_2 - 2 \right) !} e^{s_2 t} + A_{32} \frac{t^{\mu_2 - 3}}{\left(\mu_2 - 3 \right) !} e^{s_2 t} \\ = \frac{1}{2} A_{11} t^2 e^t + A_{21} t e^t + A_{31} e^t + \frac{1}{2} A_{12} t^2 e^{-t} + A_{22} t e^{-t} + A_{32} e^{-t} x(t)=k=1∑r​j=1∑μk​​Ajk​(μk​−j)!tμk​−j​esk​t=A11​(μ1​−1)!tμ1​−1​es1​t+A21​(μ1​−2)!tμ1​−2​es1​t+A31​(μ1​−3)!tμ1​−3​es1​t+A12​(μ2​−1)!tμ2​−1​es2​t+A22​(μ2​−2)!tμ2​−2​es2​t+A32​(μ2​−3)!tμ2​−3​es2​t=21​A11​t2et+A21​tet+A31​et+21​A12​t2e−t+A22​te−t+A32​e−t下面来算AjkA_{jk}Ajk​。
A11=1(1−1)!d1−1ds1−1[(s−s1)μ1X(s)]s=s1=s2(s+1)3∣s=1=18A21=1(2−1)!d2−1ds2−1[(s−s1)μ1X(s)]s=s1=dds[s2(s+1)3]s=1=2s−s2(s+1)4∣s=1=116A31=1(3−1)!d3−1ds3−1[(s−s1)μ1X(s)]s=s1=d2ds2[s2(s+1)3]s=1=s2−4s+1(s+1)5∣s=1=−116A12=1(1−1)!d1−1ds1−1[(s−s2)μ2X(s)]s=s2=s2(s−1)3∣s=−1=−18A22=1(2−1)!d2−1ds2−1[(s−s2)μ2X(s)]s=s2=dds[s2(s−1)3]s=−1=−s2−2s(s−1)4∣s=−1=116A32=1(3−1)!d3−1ds3−1[(s−s2)μ2X(s)]s=s2=d2ds2[s2(s−1)3]s=−1=s2+4s+1(s−1)5∣s=−1=116A_{11} = \frac{1}{\left( 1 - 1 \right) !} \frac{d^{1-1}}{ds^{1-1}} \Big[ \left( s -s_1 \right) ^{\mu_1} X(s) \Big] _{s = s_1} = \frac{s^2}{(s+1)^3} \Big\rvert _{s = 1} = \frac{1}{8} \\ A_{21} = \frac{1}{\left( 2 - 1 \right) !} \frac{d^{2-1}}{ds^{2-1}} \Big[ \left( s -s_1 \right) ^{\mu_1} X(s) \Big] _{s = s_1} = \frac{d}{ds} \left[ \frac{s^2}{(s+1)^3} \right] _{s=1}= \frac{2s-s^2}{(s+1)^4} \Bigg\rvert _{s = 1} = \frac{1}{16} \\ A_{31} = \frac{1}{\left( 3 - 1 \right) !} \frac{d^{3-1}}{ds^{3-1}} \Big[ \left( s -s_1 \right) ^{\mu_1} X(s) \Big] _{s = s_1} = \frac{d^2}{ds^2} \left[ \frac{s^2}{(s+1)^3} \right] _{s=1}= \frac{s^2-4s+1}{(s+1)^5} \Bigg\rvert _{s = 1} = - \frac{1}{16} \\ A_{12} = \frac{1}{\left( 1 - 1 \right) !} \frac{d^{1-1}}{ds^{1-1}} \Big[ \left( s -s_2 \right) ^{\mu_2} X(s) \Big] _{s = s_2} = \frac{s^2}{(s-1)^3} \Bigg\rvert _{s = -1} = - \frac{1}{8} \\ A_{22} = \frac{1}{\left( 2 - 1 \right) !} \frac{d^{2-1}}{ds^{2-1}} \Big[ \left( s -s_2 \right) ^{\mu_2} X(s) \Big] _{s = s_2} = \frac{d}{ds} \left[ \frac{s^2}{(s-1)^3} \right] _{s=-1} = \frac{-s^2-2s}{(s-1)^4} \Bigg\rvert _{s = -1} = \frac{1}{16} \\ A_{32} = \frac{1}{\left( 3 - 1 \right) !} \frac{d^{3-1}}{ds^{3-1}} \Big[ \left( s -s_2 \right) ^{\mu_2} X(s) \Big] _{s = s_2} = \frac{d^2}{ds^2} \left[ \frac{s^2}{(s-1)^3} \right] _{s=-1} = \frac{s^2+4s+1}{(s-1)^5} \Bigg\rvert _{s = -1} = \frac{1}{16} A11​=(1−1)!1​ds1−1d1−1​[(s−s1​)μ1​X(s)]s=s1​​=(s+1)3s2​​s=1​=81​A21​=(2−1)!1​ds2−1d2−1​[(s−s1​)μ1​X(s)]s=s1​​=dsd​[(s+1)3s2​]s=1​=(s+1)42s−s2​​s=1​=161​A31​=(3−1)!1​ds3−1d3−1​[(s−s1​)μ1​X(s)]s=s1​​=ds2d2​[(s+1)3s2​]s=1​=(s+1)5s2−4s+1​​s=1​=−161​A12​=(1−1)!1​ds1−1d1−1​[(s−s2​)μ2​X(s)]s=s2​​=(s−1)3s2​​s=−1​=−81​A22​=(2−1)!1​ds2−1d2−1​[(s−s2​)μ2​X(s)]s=s2​​=dsd​[(s−1)3s2​]s=−1​=(s−1)4−s2−2s​​s=−1​=161​A32​=(3−1)!1​ds3−1d3−1​[(s−s2​)μ2​X(s)]s=s2​​=ds2d2​[(s−1)3s2​]s=−1​=(s−1)5s2+4s+1​​s=−1​=161​故时域为
x(t)=12A11t2et+A21tet+A31et+12A12t2e−t+A22te−t+A32e−t=116t2et+116tet−116et−116t2e−t+116te−t+116e−t\begin{aligned} x(t) &= \frac{1}{2} A_{11} t^2 e^t + A_{21} t e^t + A_{31} e^t + \frac{1}{2} A_{12} t^2 e^{-t} + A_{22} t e^{-t} + A_{32} e^{-t} \\ &= \frac{1}{16} t^2 e^t + \frac{1}{16} t e^t - \frac{1}{16} e^t - \frac{1}{16} t^2 e^{-t} + \frac{1}{16} t e^{-t} + \frac{1}{16} e^{-t} \end{aligned} x(t)​=21​A11​t2et+A21​tet+A31​et+21​A12​t2e−t+A22​te−t+A32​e−t=161​t2et+161​tet−161​et−161​t2e−t+161​te−t+161​e−t​

3. 备注

实际上，简便方法中的方法一和方法二，都可以用方法三，即式(3)来解决，前两者只不过是方法三的特殊形式。而方法三本质也是基于留数定理得到的。所以，为了简化记忆，可以只记忆方法三。