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
)
,
o
r
X
(
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
+
2
s
2
+
4
s
+
3
=
s
+
2
(
s
+
1
)
(
s
+
3
)
=
1
2
[
1
s
+
1
+
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{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
)
}
=
1
2
e
−
t
+
1
2
e
−
3
t
x(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)}=21e−t+21e−3t因式分解的过程可以用留数定理来做。设
X
(
s
)
=
s
+
2
s
2
+
4
s
+
3
=
s
+
2
(
s
+
1
)
(
s
+
3
)
=
A
s
+
1
+
B
s
+
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
+
2
A\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
,
B
A, B
A,B:
1) 当
s
=
−
1
s=-1
s=−1时,
A
(
−
1
+
3
)
+
B
(
−
1
+
1
)
=
−
1
+
2
A\left( -1+3 \right) + B \left( -1+1 \right) = -1+2
A(−1+3)+B(−1+1)=−1+2得到
A
=
1
2
A = \frac{1}{2}
A=21;
2) 当
s
=
−
3
s=-3
s=−3时,
A
(
−
3
+
3
)
+
B
(
−
3
+
1
)
=
−
3
+
2
A\left( -3+3 \right) + B \left( -3+1 \right) = -3+2
A(−3+3)+B(−3+1)=−3+2得到
B
=
1
2
B = \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 , B A, B A,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
−
s
1
)
(
s
−
s
2
)
⋯
(
s
−
s
n
)
B(s) = (s-s_1) (s-s_2) \cdots (s - s_n)
B(s)=(s−s1)(s−s2)⋯(s−sn)时,共
n
n
n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
=
1
n
A
(
s
)
B
′
(
s
k
)
e
s
k
t
(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∑nB′(sk)A(s)eskt(1)举例:对于
X
(
s
)
=
s
+
2
s
2
+
5
s
+
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
s
1
=
−
4
,
s
2
=
−
1
s_1 = -4, s_2 = -1
s1=−4,s2=−1。代入式(1):
x
(
t
)
=
∑
k
=
1
n
A
(
s
)
B
′
(
s
k
)
e
s
k
t
=
A
(
s
1
)
B
′
(
s
1
)
e
s
1
t
+
A
(
s
2
)
B
′
(
s
2
)
e
s
2
t
=
s
1
+
2
2
s
1
+
5
e
s
1
t
+
s
2
+
2
2
s
2
+
5
e
s
2
t
=
2
3
e
−
4
t
+
1
3
e
−
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∑nB′(sk)A(s)eskt=B′(s1)A(s1)es1t+B′(s2)A(s2)es2t=2s1+5s1+2es1t+2s2+5s2+2es2t=32e−4t+31e−t
2) 没有重根,但是具有零极点
具有如下形式
X
(
s
)
=
A
(
s
)
s
B
(
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
=
1
n
A
(
s
k
)
s
k
B
′
(
s
k
)
e
s
k
t
(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∑nskB′(sk)A(sk)eskt(2)举例:
X
(
s
)
=
100
s
(
s
2
+
10
s
+
100
)
=
A
(
s
)
s
B
(
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
s
1
,
2
=
−
5
±
5
3
j
s_{1,2} = -5 \pm 5 \sqrt{3} j
s1,2=−5±53j。
由于
B
′
(
s
)
=
2
s
+
10
B'(s) = 2s+10
B′(s)=2s+10,故
B
′
(
s
1
)
=
10
3
j
,
B
′
(
s
2
)
=
−
10
3
j
B' \left( s_1 \right) = 10 \sqrt{3}j, B' \left( s_2 \right) = -10 \sqrt{3}j
B′(s1)=103j,B′(s2)=−103j。代入(2)有
x
(
t
)
=
A
(
0
)
B
(
0
)
+
∑
k
=
1
n
A
(
s
k
)
s
k
B
′
(
s
k
)
e
s
k
t
=
100
100
+
A
(
s
1
)
s
1
B
′
(
s
1
)
e
s
1
t
+
A
(
s
2
)
s
2
B
′
(
s
2
)
e
s
2
t
=
1
+
100
(
−
5
+
5
3
j
)
⋅
10
3
j
e
(
−
5
+
5
3
j
)
t
+
100
(
−
5
−
5
3
j
)
⋅
(
−
10
3
j
)
e
(
−
5
−
5
3
j
)
t
=
1
−
2
3
j
+
3
e
(
−
5
+
5
3
j
)
t
+
2
3
j
−
3
e
(
−
5
−
5
3
j
)
t
=
1
−
2
3
j
+
3
e
−
5
t
(
cos
5
3
t
+
j
sin
5
3
t
)
+
2
3
j
−
3
e
−
5
t
(
cos
5
3
t
−
j
sin
5
3
t
)
=
1
+
e
−
5
t
cos
5
3
t
−
1
3
e
−
5
t
sin
5
3
t
\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∑nskB′(sk)A(sk)eskt=100100+s1B′(s1)A(s1)es1t+s2B′(s2)A(s2)es2t=1+(−5+53j)⋅103j100e(−5+53j)t+(−5−53j)⋅(−103j)100e(−5−53j)t=1−3j+32e(−5+53j)t+3j−32e(−5−53j)t=1−3j+32e−5t(cos53t+jsin53t)+3j−32e−5t(cos53t−jsin53t)=1+e−5tcos53t−31e−5tsin53t
3) 有重根的情况
此时
X
(
s
)
X(s)
X(s)分母为
B
(
s
)
=
(
s
−
s
1
)
μ
1
(
s
−
s
2
)
μ
2
⋯
(
s
−
s
r
)
μ
r
B(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
s
i
s_i
si的重数为
μ
i
\mu_i
μi。共
r
r
r个根
s
1
,
s
2
,
⋯
,
s
r
s_1, s_2, \cdots, s_r
s1,s2,⋯,sr,且阶数
μ
1
+
μ
2
+
⋯
+
μ
r
=
n
\mu_1 + \mu_2 + \cdots + \mu_r = n
μ1+μ2+⋯+μr=n。
则时域为
x
(
t
)
=
∑
k
=
1
r
∑
j
=
1
μ
k
A
j
k
t
μ
k
−
j
(
μ
k
−
j
)
!
e
s
k
t
(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∑rj=1∑μkAjk(μk−j)!tμk−jeskt(3)其中系数
A
j
k
A_{jk}
Ajk满足
A
j
k
=
1
(
j
−
1
)
!
d
j
−
1
d
s
j
−
1
[
(
s
−
s
k
)
μ
k
X
(
s
)
]
s
=
s
k
(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)!1dsj−1dj−1[(s−sk)μkX(s)]s=sk(4)例:
X
(
s
)
=
s
2
(
s
−
1
)
3
(
s
+
1
)
3
X(s) = \frac{s^2}{(s-1)^3 (s+1)^3}
X(s)=(s−1)3(s+1)3s2极点
s
1
=
1
,
s
2
=
−
1
s_1 = 1, s_2 = -1
s1=1,s2=−1,其重数分别为
μ
1
=
3
,
μ
2
=
3
\mu_1 = 3, \mu_2 = 3
μ1=3,μ2=3。代入(3):
x
(
t
)
=
∑
k
=
1
r
∑
j
=
1
μ
k
A
j
k
t
μ
k
−
j
(
μ
k
−
j
)
!
e
s
k
t
=
A
11
t
μ
1
−
1
(
μ
1
−
1
)
!
e
s
1
t
+
A
21
t
μ
1
−
2
(
μ
1
−
2
)
!
e
s
1
t
+
A
31
t
μ
1
−
3
(
μ
1
−
3
)
!
e
s
1
t
+
A
12
t
μ
2
−
1
(
μ
2
−
1
)
!
e
s
2
t
+
A
22
t
μ
2
−
2
(
μ
2
−
2
)
!
e
s
2
t
+
A
32
t
μ
2
−
3
(
μ
2
−
3
)
!
e
s
2
t
=
1
2
A
11
t
2
e
t
+
A
21
t
e
t
+
A
31
e
t
+
1
2
A
12
t
2
e
−
t
+
A
22
t
e
−
t
+
A
32
e
−
t
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} \\ = 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∑rj=1∑μkAjk(μk−j)!tμk−jeskt=A11(μ1−1)!tμ1−1es1t+A21(μ1−2)!tμ1−2es1t+A31(μ1−3)!tμ1−3es1t+A12(μ2−1)!tμ2−1es2t+A22(μ2−2)!tμ2−2es2t+A32(μ2−3)!tμ2−3es2t=21A11t2et+A21tet+A31et+21A12t2e−t+A22te−t+A32e−t下面来算
A
j
k
A_{jk}
Ajk。
A
11
=
1
(
1
−
1
)
!
d
1
−
1
d
s
1
−
1
[
(
s
−
s
1
)
μ
1
X
(
s
)
]
s
=
s
1
=
s
2
(
s
+
1
)
3
∣
s
=
1
=
1
8
A
21
=
1
(
2
−
1
)
!
d
2
−
1
d
s
2
−
1
[
(
s
−
s
1
)
μ
1
X
(
s
)
]
s
=
s
1
=
d
d
s
[
s
2
(
s
+
1
)
3
]
s
=
1
=
2
s
−
s
2
(
s
+
1
)
4
∣
s
=
1
=
1
16
A
31
=
1
(
3
−
1
)
!
d
3
−
1
d
s
3
−
1
[
(
s
−
s
1
)
μ
1
X
(
s
)
]
s
=
s
1
=
d
2
d
s
2
[
s
2
(
s
+
1
)
3
]
s
=
1
=
s
2
−
4
s
+
1
(
s
+
1
)
5
∣
s
=
1
=
−
1
16
A
12
=
1
(
1
−
1
)
!
d
1
−
1
d
s
1
−
1
[
(
s
−
s
2
)
μ
2
X
(
s
)
]
s
=
s
2
=
s
2
(
s
−
1
)
3
∣
s
=
−
1
=
−
1
8
A
22
=
1
(
2
−
1
)
!
d
2
−
1
d
s
2
−
1
[
(
s
−
s
2
)
μ
2
X
(
s
)
]
s
=
s
2
=
d
d
s
[
s
2
(
s
−
1
)
3
]
s
=
−
1
=
−
s
2
−
2
s
(
s
−
1
)
4
∣
s
=
−
1
=
1
16
A
32
=
1
(
3
−
1
)
!
d
3
−
1
d
s
3
−
1
[
(
s
−
s
2
)
μ
2
X
(
s
)
]
s
=
s
2
=
d
2
d
s
2
[
s
2
(
s
−
1
)
3
]
s
=
−
1
=
s
2
+
4
s
+
1
(
s
−
1
)
5
∣
s
=
−
1
=
1
16
A_{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)!1ds1−1d1−1[(s−s1)μ1X(s)]s=s1=(s+1)3s2
s=1=81A21=(2−1)!1ds2−1d2−1[(s−s1)μ1X(s)]s=s1=dsd[(s+1)3s2]s=1=(s+1)42s−s2
s=1=161A31=(3−1)!1ds3−1d3−1[(s−s1)μ1X(s)]s=s1=ds2d2[(s+1)3s2]s=1=(s+1)5s2−4s+1
s=1=−161A12=(1−1)!1ds1−1d1−1[(s−s2)μ2X(s)]s=s2=(s−1)3s2
s=−1=−81A22=(2−1)!1ds2−1d2−1[(s−s2)μ2X(s)]s=s2=dsd[(s−1)3s2]s=−1=(s−1)4−s2−2s
s=−1=161A32=(3−1)!1ds3−1d3−1[(s−s2)μ2X(s)]s=s2=ds2d2[(s−1)3s2]s=−1=(s−1)5s2+4s+1
s=−1=161故时域为
x
(
t
)
=
1
2
A
11
t
2
e
t
+
A
21
t
e
t
+
A
31
e
t
+
1
2
A
12
t
2
e
−
t
+
A
22
t
e
−
t
+
A
32
e
−
t
=
1
16
t
2
e
t
+
1
16
t
e
t
−
1
16
e
t
−
1
16
t
2
e
−
t
+
1
16
t
e
−
t
+
1
16
e
−
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)=21A11t2et+A21tet+A31et+21A12t2e−t+A22te−t+A32e−t=161t2et+161tet−161et−161t2e−t+161te−t+161e−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。