关于拉普拉斯变换的作用,可参考知乎
总的来说,拉普拉斯变换就是迫使函数满足绝对可积条件的傅里叶变换。
常用信号的Laplace变换
参考信号与系统/陈后金, 胡健, 薛健. ——2版.——北京:清华大学出版社;北京交通大学出版社, 2005.7(2017.3重印)第218-219页。
序号 | 单边信号(f(t)) | Laplace变换( F ( s ) F(s) F(s)) | 收敛域 |
---|---|---|---|
1 | e − λ t u ( t ) e^{-\lambda t}u(t) e−λtu(t) | 1 s + λ \frac{1}{s+\lambda} s+λ1 | R e ( s ) > − λ Re(s)>-\lambda Re(s)>−λ |
2 | e j ω 0 t u ( t ) e^{j\omega_0 t}u(t) ejω0tu(t) | 1 s − j ω 0 \frac{1}{s-j\omega_0} s−jω01 | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
3 | c o s ( ω 0 t ) u ( t ) cos(\omega_0 t)u(t) cos(ω0t)u(t) | s s 2 + ω 0 2 \frac{s}{s^2+\omega_0^2} s2+ω02s | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
4 | s i n ( ω 0 T ) u ( t ) sin(\omega_0 T)u(t) sin(ω0T)u(t) | ω 0 s 2 + ω 0 2 \frac{\omega_0}{s^2+\omega_0^2} s2+ω02ω0 | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
5 | e − σ 0 t c o s ( ω t ) u ( t ) e^{-\sigma_0 t}cos(\omega t)u(t) e−σ0tcos(ωt)u(t) | s + σ 0 ( s + σ 0 ) 2 + ω 0 \frac{s+\sigma_0}{(s+\sigma_0)^2+\omega_0} (s+σ0)2+ω0s+σ0 | R e ( s ) > − σ 0 Re(s)>-\sigma_0 Re(s)>−σ0 |
6 | e − σ 0 t s i n ( ω t ) u ( t ) e^{-\sigma_0 t}sin(\omega t)u(t) e−σ0tsin(ωt)u(t) | ω 0 ( s + σ 0 ) 2 + ω 0 \frac{\omega_0}{(s+\sigma_0)^2+\omega_0} (s+σ0)2+ω0ω0 | R e ( s ) > − σ 0 Re(s)>-\sigma_0 Re(s)>−σ0 |
7 | δ ( t ) \delta(t) δ(t) | 1 | R e ( s ) > − ∞ Re(s)>-\infty Re(s)>−∞ |
8 | δ ( n ) ( t ) \delta^{(n)}(t) δ(n)(t) | s n ( 1 , 2 , … ) s^n(1,2,\dots) sn(1,2,…) | R e ( s ) > − ∞ Re(s)>-\infty Re(s)>−∞ |
9 | u ( t ) u(t) u(t) | 1 s \frac{1}{s} s1 | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
10 | t u ( t ) tu(t) tu(t) | 1 s 2 \frac{1}{s^2} s21 | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
11 | t n u ( t ) t^n u(t) tnu(t) | n ! s n + 1 \frac{n!}{s^{n+1}} sn+1n! | R e ( s ) > 0 Re(s)>0 Re(s)>0 |
12 | t e − λ t u ( t ) te^{-\lambda t}u(t) te−λtu(t) | 1 ( s + λ ) 2 \frac{1}{(s+\lambda)^2} (s+λ)21 | R e ( s ) > − λ Re(s)>-\lambda Re(s)>−λ |
1. e − λ t u ( t ) e^{-\lambda t}u(t) e−λtu(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
lamb = symbols('\lambda')
F1 = laplace_transform(exp(-lamb*t)*Heaviside(t),t,s)
F1[0]
1 λ + s \displaystyle \frac{1}{\lambda + s} λ+s1
F1[1]
− λ \displaystyle - \lambda −λ
2. e j ω 0 t u ( t ) e^{j\omega_0 t}u(t) ejω0tu(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
w0 = symbols('\omega_0')
F1 = laplace_transform(exp(np.complex(0,1)*w0*t)*Heaviside(t),t,s)
F1[0]
1 − 1.0 i ω 0 + s \displaystyle \frac{1}{- 1.0 i \omega_{0} + s} −1.0iω0+s1
3. c o s ( ω 0 t ) u ( t ) cos(\omega_0 t)u(t) cos(ω0t)u(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
w0 = symbols('\omega_0')
F1 = laplace_transform(cos(w0*t)*Heaviside(t),t,s)
F1[0]
s ω 0 2 + s 2 \displaystyle \frac{s}{\omega_{0}^{2} + s^{2}} ω02+s2s
4. s i n ( ω 0 T ) u ( t ) sin(\omega_0 T)u(t) sin(ω0T)u(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
w0 = symbols('\omega_0')
F1 = laplace_transform(sin(w0*t)*Heaviside(t),t,s)
F1[0]
ω 0 ω 0 2 + s 2 \displaystyle \frac{\omega_{0}}{\omega_{0}^{2} + s^{2}} ω02+s2ω0
5. e − σ 0 t c o s ( ω t ) u ( t ) e^{-\sigma_0 t}cos(\omega t)u(t) e−σ0tcos(ωt)u(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
w0,sigma0 = symbols('\omega_0 \sigma_0')
F1 = laplace_transform(exp(-sigma0*t)*cos(w0*t)*Heaviside(t),t,s)
F1[0]
σ 0 + s ( − i ω 0 + σ 0 + s ) ( i ω 0 + σ 0 + s ) \displaystyle \frac{\sigma_{0} + s}{\left(- i \omega_{0} + \sigma_{0} + s\right) \left(i \omega_{0} + \sigma_{0} + s\right)} (−iω0+σ0+s)(iω0+σ0+s)σ0+s
6. e − σ 0 t s i n ( ω t ) u ( t ) e^{-\sigma_0 t}sin(\omega t)u(t) e−σ0tsin(ωt)u(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
w0,sigma0 = symbols('\omega_0 \sigma_0')
F1 = laplace_transform(exp(-sigma0*t)*sin(w0*t)*Heaviside(t),t,s)
F1[0]
ω 0 ( − i ω 0 + σ 0 + s ) ( i ω 0 + σ 0 + s ) \displaystyle \frac{\omega_{0}}{\left(- i \omega_{0} + \sigma_{0} + s\right) \left(i \omega_{0} + \sigma_{0} + s\right)} (−iω0+σ0+s)(iω0+σ0+s)ω0
7. δ ( t ) \delta(t) δ(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
F1 = laplace_transform(DiracDelta(t),t,s)
F1[0]
1 \displaystyle 1 1
8. δ ( n ) ( t ) \delta^{(n)}(t) δ(n)(t)
n次求导无法计算,用多个数字代替,总结规律(不严谨) 结果看不懂
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
delta,n = symbols('\delta n')
F1 = laplace_transform(DiracDelta(t,2),t,s)
F1
L t [ δ ( 2 ) ( t ) ] ( s ) \displaystyle \mathcal{L}_{t}\left[\delta^{\left( 2 \right)}\left( t \right)\right]\left(s\right) Lt[δ(2)(t)](s)
9. u ( t ) u(t) u(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
delta,n = symbols('\delta n')
F1 = laplace_transform(Heaviside(t),t,s)
F1[0]
1 s \displaystyle \frac{1}{s} s1
10. t u ( t ) tu(t) tu(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
delta,n = symbols('\delta n')
F1 = laplace_transform(t*Heaviside(t),t,s)
F1[0]
1 s 2 \displaystyle \frac{1}{s^{2}} s21
11. t n u ( t ) t^n u(t) tnu(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
delta,n = symbols('\delta n')
F1 = laplace_transform(t**n*Heaviside(t),t,s)
F1[0]
s − n Γ ( n + 1 ) s \displaystyle \frac{s^{- n} \Gamma\left(n + 1\right)}{s} ss−nΓ(n+1)
12. t e − λ t u ( t ) te^{-\lambda t}u(t) te−λtu(t)
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
lamb,n = symbols('\lambda n')
F1 = laplace_transform(t*exp(-lamb*t)*Heaviside(t),t,s)
F1[0]
1 ( λ + s ) 2 \displaystyle \frac{1}{\left(\lambda + s\right)^{2}} (λ+s)21
至于
Γ
\Gamma
Γ函数
Γ
(
x
)
=
∫
0
+
∞
t
x
−
1
e
−
t
d
t
\Gamma(x)=\int_0^{+\infty}t^{x-1}e^{-t}dt
Γ(x)=∫0+∞tx−1e−tdt
当
x
x
x为正整数时,
Γ
(
n
)
=
(
n
−
1
)
!
\Gamma(n)=(n-1)!
Γ(n)=(n−1)!
逆Laplace变换
逆Laplace变换与Laplace变换函数用法大致一致,举一个简单例子,对于
e
−
λ
t
u
(
t
)
e^{-\lambda t}u(t)
e−λtu(t)的拉普拉斯变换,输出结果为
1
λ
+
s
\frac{1}{\lambda+s}
λ+s1,那么对于
1
λ
+
s
\frac{1}{\lambda+s}
λ+s1的逆拉普拉斯变换,代码为文章来源:https://www.toymoban.com/news/detail-471419.html
from sympy import *
from sympy.integrals import laplace_transform,inverse_laplace_transform
lamb,s,t = symbols('\lambda s t')
F1 = inverse_laplace_transform(1/(lamb+s),s,t)
F1
e
−
t
re
(
λ
)
−
i
t
im
(
λ
)
θ
(
t
)
\displaystyle e^{- t \operatorname{re}{\left(\lambda\right)} - i t \operatorname{im}{\left(\lambda\right)}} \theta\left(t\right)
e−tre(λ)−itim(λ)θ(t)
结果的
θ
(
t
)
\theta(t)
θ(t)就是
u
(
t
)
u(t)
u(t)文章来源地址https://www.toymoban.com/news/detail-471419.html
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