[signals and systems] (XXI) Laplace transform and complex frequency domain analysis -- Laplace transform and its properties

Laplace transform and its properties

The Fourier transform ：$jw$
Laplace transform ：$s=\sigma +jw$

1 Definition of bilateral Laplace transform

Some functions do not satisfy the absolute integrability condition , It is difficult to solve Fourier transform . So , An attenuation factor can be used $e^{-\sigma t}$（$\sigma$ Is a real constant ） Multiply signal $f(t)$, Choose... Appropriately $\sigma$ Value , Make the product signal $f(t)e^{-\sigma t}$ When $t\to \infty$ The signal amplitude approaches 0 , So that $f(t)e^{-\sigma t}$ The Fourier transform of exists .

The corresponding inverse Fourier transform is ：

2 Convergence domain

Only choose the right $\sigma$ Value to make the integral converge , The signal $f(t)$ The bilateral Laplace transform of exists .

Convergence domain ： send $f(t)$ The existence of Laplace transform $\sigma$ Value range .

2.1 Causal signals

The convergence domain of causal signal is to the right of a straight line .

2.2 Anti causal signals

The convergence domain of anti causal signal is to the left of a certain straight line .

2.3 Bilateral signals

The convergence region of bilateral signal is between two straight lines .

Conclusion ：

(1) For the bilateral Laplace transform ,$F_b(s)$ Together with the convergence domain , It can be uniquely determined that $f(t)$. namely ：

(2) Different signals can have the same $F_b(s)$, But the convergence domain is different .

3 （ Causal signals ） Definition of unilateral Laplace transform

The signals we usually encounter have initial moments , Let's set its initial time as the coordinate origin . such ,$t<0$ when ,$f(t)=0$. So the Laplace transformation is written as
$$F(s)={\int }_{0_{-}}^{\infty }f(t){\mathrm{e}}^{-st}\mathrm{d}t$$

It is called unilateral Laplace transform . Laplace transform for short . Its convergence region must be $Re[s]>\alpha$, It can be omitted .

$\epsilon (t)$： unilateral ,$t$ Less than zero $f(t)$ Value is zero .

4 The relationship between unilateral Laplace transform and Fourier transform

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remarks ：
$$\int \frac{1}{a^2+x^2}dx=\frac{1}{a}\arctan \frac{x}{a}+C$$

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When $w\ne 0$ when ,
$$\underset{\sigma \to 0}{\lim }\frac{\sigma }{{\sigma }^2+{\omega }^2}=0=\pi \delta (\omega )$$

When $w=0$ when , The limit value is infinity , Equivalent to impulse function $\delta (w)$, Area is $\pi$：
$${\int }_{-\infty }^{\infty }\frac{\sigma }{{\sigma }^2+{\omega }^2}d\omega =\arctan (\frac{x}{\sigma }){|}_{-\infty }^{+\infty }=\pi$$
namely $\pi \delta (w)$.

5 Laplace transform of common signals

6 Properties of Laplace transform

6.1 linear 、 Scale transformation

6.2 Time shift 、 Complex frequency shift characteristics

6.3 Calculus characteristics in time domain and complex frequency domain

We can find the Laplace transform of the original function by finding the Laplace transform of the reciprocal of the original function .

6.4 Convolution theorem

6.5 initial value 、 The final value theorem

7 Inverse Laplace transform

Directly use the definition to find the inverse transformation — Integral of complex variable function , More difficult . Common methods :
（1） Look up the table ;

（2） Utilization property ;

（3） Partial fraction expansion ----- combination

Image function $F(s)$ yes $s$ Rational fraction of , Can be written as

if $m\ge n$ （ False fraction ）, The image function can be divided by polynomial $F(s)$ Decompose into rational polynomials $P(s)+YesThe\; reason\; isreallybranchtype$

for example ：

$P(s)$ The inverse Laplace transform consists of the impulse function ($\delta$) And its derivatives (${\delta }^{\prime }$,${\delta }^{\prime \prime }$…) constitute .

The following mainly discusses the rational proper fraction .

Partial fraction expansion

$$\frac{1}{s-(-\alpha +j\beta )}\to e^{(-\alpha +j\beta )t}$$

$$\frac{1}{s-(-\alpha -j\beta )}\to e^{(-\alpha -j\beta )t}$$

By Euler formula , obtain ：

$$e^{j\theta }\cdot e^{(-\alpha +j\beta )t}+e^{-j\theta }\cdot e^{(-\alpha -j\beta )t}$$
$$=e^{-\alpha }(e^{j(\theta +\beta t)}+e^{-j(\theta +\beta t)})$$
$$=2e^{-\alpha }\cos (\beta t+\theta )$$

True fraction ：


False fraction ：

University of China MOOC： Signals and systems , Xi'an University of Electronic Science and Technology , Bao Long Guo , Zhu JUANJUAN