Laplace变换

Laplace Transform

Fourier

Proof: see “Harmonious Analysis” CHAPTER 3 P37 ~ P38\
https://drive.google.com/file/d/0B7t_mQHDlsRsdnZkSE9LbndfOHM/view

https://yannisparissis.wordpress.com/2011/03/10/dmat0101-notes-3-the-fourier-transform-on-l1/

Fourier tranform

$$\mathbb{F}[g(t)] \equiv \int_{-\infty}^{+\infty} g(t) e^{-j\omega t} dt = G(\omega)$$

Fourier Inversion

$$\mathbb{F^{-1}}[G(\omega)] \equiv \frac{1}{2\pi} \int_{-\infty}^{+\infty} G(\omega) e^{j\omega t} d\omega =g^{*}(t)$$

Here is the Inversion of g(t)

$$g^{*}(t) = \left\{ \begin{aligned} g(t) && \text{continuous at t}\\ \frac{g(t_{-}) + g(t_{+})}{2} && \text{Not continuous at t} \end{aligned} \right.$$

Laplace and Fourier

set $f(t) e^{-\beta t}u(t) \equiv g(t), \quad f^{\ast}(t) \equiv g^{\ast}(t) e^{\beta t}$\

(I) $\forall [a, b] \subset (-\infty, +\infty)$\
g(t) satisfy Dirichlet conditions for [a, b]

(II) $\int_{-\infty}^{+\infty} g(t) dt < M$

So, Fourier Transform, $s = \beta + j\omega$, we would have

$$\begin{aligned} \mathbb{F}[g(t)] &= \int_{-\infty}^{+\infty} g(t) e^{-j\omega t} dt = \int_{0_{-}}^{\infty} f(t) e^{-s t} dt\\ &= G(\omega) = G(\frac{s-\beta}{j}) = F(s)\\ \mathbb{F^{-1}}[G(\omega)] &= \frac{1}{2\pi} \int_{-\infty}^{+\infty} G(\omega) e^{j\omega t} d\omega = \frac{1}{2\pi j} \int_{\beta-j\infty}^{\beta+j\infty} F(s) e^{(s-\beta) t} ds\\ &=g^{*}(t) \equiv f^{*}(t) e^{-\beta t} \end{aligned}$$

Then we have Inversion

$$f^{*}(t) = \left\{ \begin{aligned} f(t) && \text{continuous at t } (t \geq 0)\\ \frac{f(t_{-}) + f(t_{+})}{2} && \text{Not continuous at t } (t \geq 0) \end{aligned} \right.$$

Define Laplace Transform

Definition of LT

$$F(s) = \mathbb{L}[f(t)] \equiv \int_{0_{-}}^{\infty} f(t) e^{-st} dt = G(\frac{s-\beta}{j})\\ f^{*}(t) = \mathbb{L^{-1}}[F(s)] \equiv \frac{1}{2\pi j} \int_{\beta-j\infty}^{\beta+j\infty} F(s) e^{s t} ds = g^{*}(t) e^{\beta t} \quad [t \geq 0]$$

可以选择$\beta$, 使$Re\{\text{poles of }F(s)\} < \beta$, 此时

$$f^{*}(t) = \mathbb{L^{-1}}[F(s)] \equiv \frac{1}{2\pi j} \int_{\beta-j\infty}^{\beta+j\infty} F(s) e^{s t} ds \\ = \sum \mathop{Res}\limits_{s=s_k}[F(s) e^{st}]$$