三大变换 公式总结

    科技2022-07-11  173

    三大变换

    傅里叶变换、拉普拉斯变换、Z变换

    1. 傅里叶变换

    2. 拉普拉斯变换

    F ( s ) = L [ f ( t ) ] = ∫ 0 ∞ f ( t ) e − s t d t F(s) = L[f(t)] = \int_0^\infty f(t)e^{-st}dt F(s)=L[f(t)]=0f(t)estdt s = σ + j ω s = \sigma+j\omega s=σ+jω 复变量 将实数域上的实变函数变成复数域上的复变函数

    一.经典函数的拉式变换

    单位阶跃函数(1函数) 1 ( t ) = { 1 , t < 0 0 , t ≥ 0 1(t) = \left \{\begin{array}{cc} 1, & t<0\\ 0, & t\ge0 \end{array}\right. 1(t)={1,0,t<0t0 L [ 1 ( t ) ] = ∫ 0 ∞ 1 ( t ) e − s t d t = 1 s L[1(t)] =\int_0^\infty 1(t)e^{-st}dt =\frac{1}{s} L[1(t)]=01(t)estdt=s1

    单位脉冲函数 δ ( t ) = { ∞ , t = 0 0 , t ≠ 0 \delta(t) = \left \{\begin{array}{cc} \infty, & t=0\\ 0, & t\neq0 \end{array}\right. δ(t)={,0,t=0t=0 单位脉冲函数满足 1) ∫ 0 ∞ δ ( t ) d t = 1 \int^\infty_0 \delta(t)dt = 1 0δ(t)dt=1 2) L [ δ ( t ) ] = ∫ − ∞ ∞ δ ( t ) e − s t d t = 1 L[\delta(t)] = \int^\infty_{-\infty} \delta(t)e^{-st}dt = 1 L[δ(t)]=δ(t)estdt=1 根据2)可得 L [ 1 ( t ) ] = ∫ 0 ∞ δ ( t ) e − s t d t = 1 L[1(t)] =\int_0^\infty \delta(t)e^{-st}dt =1 L[1(t)]=0δ(t)estdt=1

    单位斜坡函数 f ( t ) = { 0 t < 0 t t ≥ 0 f(t)= \left\{\begin{array}{cc} 0&t<0\\ t&t\geq0 \end{array} \right. f(t)={0tt<0t0 L [ f ( t ) ] = 1 s 2 L[f(t)] =\frac{1}{s^2} L[f(t)]=s21

    指数函数 f ( t ) = A e − a t , t ≥ 0 f(t)=Ae^{-at},t\geq0 f(t)=Aeat,t0 L [ f ( t ) ] = A s + a L[f(t)] =\frac{A}{s+a} L[f(t)]=s+aA

    正余弦函数 (推到需要欧拉公式和指数函数变换) L [ sin ⁡ ω t ] = ω s 2 + ω 2 L[\sin \omega t] = \frac{\omega}{s^2+\omega^2} L[sinωt]=s2+ω2ω L [ cos ⁡ ω t ] = s s 2 + ω 2 L[\cos \omega t] = \frac{s}{s^2+\omega^2} L[cosωt]=s2+ω2s

    二.拉氏变换定理

    L [ f ( t ) ] = F ( s ) L[f(t)] = F(s) L[f(t)]=F(s) 则:

    线性定理 L [ α f 1 ( t ) + β f 2 ( t ) ] = α F 1 ( s ) + β F 2 ( s ) L[\alpha f_1(t)+\beta f_2(t)] = \alpha F_1(s) +\beta F_2(s) L[αf1(t)+βf2(t)]=αF1(s)+βF2(s)位移定理 L [ e − a t f ( t ) ] = F ( s + a ) L[e^{-at}f(t)] = F(s+a) L[eatf(t)]=F(s+a)微分定理 L [ d f ( t ) d t ] = s F ( s ) − f ( 0 ) L[\frac {df(t)}{dt}] = sF(s)-f(0) L[dtdf(t)]=sF(s)f(0)积分定理 L [ ∫ f ( t ) d t ] = F ( s ) s + ∫ f ( 0 ) d t s L[\int f(t){dt}] = \frac{F(s)}{s}+ \frac{\int f(0){dt}}{s} L[f(t)dt]=sF(s)+sf(0)dt终值定理 (要求极限存在) lim ⁡ t → ∞ f ( t ) = lim ⁡ s → 0 s F ( s ) \lim_{t\rightarrow \infty} f(t) = \lim_{s\rightarrow 0}sF(s) tlimf(t)=s0limsF(s)初值定理 lim ⁡ t → 0 f ( t ) = lim ⁡ s → ∞ s F ( s ) \lim_{t\rightarrow 0} f(t) = \lim_{s\rightarrow \infty}sF(s) t0limf(t)=slimsF(s)卷积定理 L [ ∫ 0 ∞ f 1 ( t − τ ) ⋅ f 2 ( τ ) d τ ] = F 1 ( s ) ⋅ F 2 ( s ) L[\int^\infty_0f_1(t-\tau)\cdot f_2(\tau) d\tau]=F_1(s)\cdot F_2(s) L[0f1(tτ)f2(τ)dτ]=F1(s)F2(s)

    二.拉氏反变换

    L − 1 [ F ( s ) ] = f ( t ) L^{-1}[F(s)]=f(t) L1[F(s)]=f(t) 部分分式展开法: F ( s ) = A ( s ) B ( s ) = A ( s ) ( s + p 1 ) ( s + p 2 ) ⋯ ( s + p n ) F(s)=\frac{A(s)}{B(s)}=\frac{A(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)} F(s)=B(s)A(s)=(s+p1)(s+p2)(s+pn)A(s) p 1 , p 2 , ⋯   , p n 为 B ( s ) 的 根 , 又 称 F ( s ) 的 极 点 p_1,p_2,\cdots,p_n为B(s)的根,又称F(s)的极点 p1,p2,,pnB(s)F(s)

    无重根 F ( s ) = A ( s ) ( s + p 1 ) ( s + p 2 ) ⋯ ( s + p n ) = α 1 s + p 1 + α 2 s + p 2 + ⋯ + α n s + p n F(s)=\frac{A(s)}{(s+p_1)(s+p_2)\cdots(s+p_n)}= \frac{\alpha_1}{s+p_1}+\frac{\alpha_2}{s+p_2}+\cdots+\frac{\alpha_n}{s+p_n} F(s)=(s+p1)(s+p2)(s+pn)A(s)=s+p1α1+s+p2α2++s+pnαn α k 为 s = − p k 处 的 留 数 \alpha_k为s=-p_k处的留数 αks=pk α k 求 解 办 法 : 等 式 两 边 × ( s + p k ) , 再 带 入 s = − p k 即 \alpha_k求解办法:等式两边×(s+p_k), 再带入s=-p_k即 αk×(s+pk),s=pk α k = [ ( s + p k ) A ( s ) B ( s ) ] s = − p k \alpha_k=\left[(s+p_k)\frac{A(s)}{B(s)}\right]_{s=-p_k} αk=[(s+pk)B(s)A(s)]s=pk有重根 F ( s ) = A ( s ) ( s + p 1 ) 3 ( s + p 2 ) ⋯ ( s + p n ) = α 1 ( s + p 1 ) 3 + α 1 ( s + p 1 ) 2 + α 1 ( s + p 1 ) 1 + α 1 s + p 2 + ⋯ + α n s + p n F(s)=\frac{A(s)}{(s+p_1)^3(s+p_2)\cdots(s+p_n)}= \frac{\alpha_1}{(s+p_1)^3}+\frac{\alpha_1}{(s+p_1)^2}+\frac{\alpha_1}{(s+p_1)^1}+\frac{\alpha_1}{s+p_2}+\cdots+\frac{\alpha_n}{s+p_n} F(s)=(s+p1)3(s+p2)(s+pn)A(s)=(s+p1)3α1+(s+p1)2α1+(s+p1)1α1+s+p2α1++s+pnαn

    3. Z变换

    f ( t ) = { f(t)= \left\{\begin{array}{cc} \end{array} \right. f(t)={

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