Fourier变换与Laplace变换

Fourier

  • 2021.9.30:备考,整理了两者的常用变换公式和一些基本定理,未完成。
  • 2021.10.1:整理了Laplace逆变换和简单应用。
  • 2022.2.12:整理完所有的内容

1. Fourier变换和逆变换公式

\(F(\omega)=\int_{-\infty}^{\infty}f(t)e^{-i\omega t} dt\)

\(f(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} F(\omega)e^{i\omega t} d{\omega}\)

2. 单位冲激函数、单位阶跃函数

  1. 单位冲激函数\(\delta(t)\)

    1. 定义1:

      1. \(t\neq0\)时,\(\delta(t)=0\)
      2. \(\int_{-\infty}^{+\infty}\delta(t)dt=1\)
    2. 定义2:

      \[\delta(t)=\mathop{lim}\limits_{\tau\rightarrow 0}\delta_\tau(t)\\ \delta_\tau (t)=\begin{cases} 0,\ \ \ \ \ t<0\\ \frac{1}{\tau},\ \ \ \ \ 0\le t\le \tau\\ 0,\ \ \ \ \ t>\tau \end{cases} \]

    3. 性质:

      1. 筛选:\(f(t_0)=\int_{-\infty}^{\infty}\delta(t-t_0)f(t)dt\)
      2. 积分:\(\int_{-\infty}^{\infty}\delta(t)dt=\int_{-\infty}^{\infty}\mathop{lim}\limits_{\tau\rightarrow0}\delta_{\tau}(t)dt=\mathop{lim}\limits_{\tau\rightarrow0}\int_{-\infty}^{\infty}\delta_{\tau}(t)dt=1\)
      3. 傅里叶变换:\(F[\delta(t)]=\int_{-\infty}^{\infty}\delta(t)e^{-i\omega t}dt=e^0=1\)
      4. 缩放:\(\delta(at)=\frac{1}{|a|}\delta(t)\rightarrow \delta(-t)=\delta(t)\),偶函数
      5. 导数:
        1. \(\int_{-\infty}^{+\infty}\delta'(t)f(t)dt=-f'(0)\)
        2. \(\int_{-\infty}^{\infty}\delta^{(n)}(t)f(t)dt=(-1)^nf^{(n)}(0)\)
  2. 单位阶跃函数\(u(t)\)

    1. 定义1:

      \[u(t)=\begin{cases} 1,\ \ \ \ \ t>0\\ 0,\ \ \ \ \ t\le0 \end{cases} \]

    2. 定义2:

      \[u(t)=\frac{1}{2}[1+sgn(t)] \]

    3. 性质:

      1. 与冲激函数的关系:
        1. \(\int_{-\infty}^{t}\delta(\tau)d\tau=u(t)\)
        2. \(\frac{du(t)}{dt}=\delta(t)\)
      2. 傅里叶变换:\(F[u(t)]=F\{\frac{1}{2}[1+sgn(t)]\}=\frac{1}{2}(2\pi\delta(\omega)+\frac{2}{i\omega})=\pi\delta(\omega)+\frac{1}{i\omega}\)

3. 常用变换公式

  1. \(\delta(t)\longrightarrow 1\)
  2. \(\delta(t-t_0)=e^{-i\omega t_0}\)
  3. \(1\longrightarrow 2\pi\delta(\omega)\)
  4. \(t\longrightarrow 2\pi i\frac{d\delta(\omega)}{d\omega}\)
  5. \(t^k\longrightarrow 2\pi i^k\frac{d^k\delta(\omega)}{d\omega^k}\)
  6. \(e^{i\omega_0t}\longrightarrow2\pi\delta(\omega-\omega_0)\)
  7. \(u(t)\longrightarrow\frac{1}{i\omega}+\pi\delta(\omega)\)
  8. \(cos(at)\longrightarrow\pi[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]\)
  9. \(sin(at)\longrightarrow i\pi[\delta(\omega+\omega_0)-\delta(\omega-\omega_0)]\)

4. 性质

  1. 线性性质

    1. \(F[\alpha f_1(t)+\beta f_2(t)]\longrightarrow \alpha F_1(\omega)+\beta F_2(\omega)\)

    2. \(F^{-1}[\alpha F_1(\omega)+\beta F_2(\omega)]\longrightarrow \alpha f_1(t)+\beta f_2(t)\)

  2. 尺度性质

    1. \(F[f(at)]\longrightarrow \frac{1}{|a|}F(\frac{\omega}{a})\)
    2. \(F^{-1}[F(a\omega)]\longrightarrow \frac{1}{|a|}f(\frac{t}{a})\)
  3. 位移性质

    1. \(F[f(t\pm\tau)]\longrightarrow e^{\pm i\omega \tau}F(\omega)\)

    2. \(F^{-1}[F(\omega\pm \omega_0)]\longrightarrow e^{\mp i\omega t}f(t)\)

  4. 微分性质

    1. \(F[f'(t)]\longrightarrow i\omega F(\omega)\)

      \(F[f^{(n)}(t)]\longrightarrow (i\omega)^nF(\omega)\)

    2. \(F'(\omega)\longrightarrow -iF[tf(t)]\)

      \(F^{(n)}(\omega)\longrightarrow (-i)^nF[t^nf(t)]\)

  5. 积分性质

    \(F[\int_{-\infty}^tf(t)dt]\longrightarrow \frac{1}{i\omega}F(\omega)\)

  6. 能量积分

    \(\int_{-\infty}^{+\infty}[f(t)]^2dt\longrightarrow \frac{1}{2\pi}\int_{-\infty}^{+\infty}|F(\omega)|^2d\omega\)

    其中\(S(\omega)=|F(\omega)|^2\)称为能量密度函数或能量谱函数

  7. 卷积

    1. 定义:\(f_1(t)*f_2(t)=\int_{-\infty}^{+\infty}f_1(\tau)f_2(t-\tau)d\tau\)
    2. 卷积运算规律:
      1. \(f_1(t)*f_2(t)=f_2(t)*f_1(t)\)
      2. \(f_1(t)*[f_2(t)*f_3(t)]=[f_1(t)*f_2(t)]*f_3(t)\)
      3. \(f_1(t)*[f_2(t)+f*3(t)]=f_1(t)*f_2(t)+f_1(t)*f_3(t)\)
      4. \(|f_1(t)*f_2(t)|\leq|f_1(t)|*|f_2(t)|\)
    3. 卷积定理:
      1. \(F[f_1(t)*f_2(t)]=F_1(\omega) \cdot F_2(\omega)\)
      2. \(F^{-1}[F_1(\omega)\cdot F_2(\omega)]=f_1(t)*f_2(t)\)
    4. 频谱卷积定理:
      1. \(F[f_1(t)f_2(t)]=\frac{1}{2\pi}F_1(\omega)*F_2(\omega)\)
      2. \(F[f_1(t)f_2(t)\cdots f_n(t)]=\frac{1}{(2\pi)^{n-1}}F_1(\omega)*F_2(\omega)*\cdots*F_n(\omega)\)

Laplace

1. 常用变换公式

  1. \(\delta(t) \longrightarrow 1\)
  2. \(1(t)\longrightarrow \frac{1}{s}\)
  3. \(u(t)\longrightarrow \frac{1}{s}\)
  4. \(t\longrightarrow \frac{1}{s^2}\)
  5. \(\frac{t^n}{n!} \longrightarrow \frac{1}{s^{n+1}}\)
  6. \(e^{-at}\longrightarrow\frac{1}{s+a}\)
  7. \(te^{-at}\longrightarrow\frac{1}{(s+a)^2}\)
  8. \(sin(at)\longrightarrow\frac{a}{s^2+a^2}\)
  9. \(cos(at)\longrightarrow\frac{s}{s^2+a^2}\)
  10. \(a^{\frac{t}{T}}\longrightarrow\frac{1}{s-\frac{1}{T}lna}\)

2. 性质

  1. 线性性质

    1. \(L[af(t)+bg(t)]=aF(s)+bG(s)\)

    2. \(L^{-1}[aF(s)+bG(s)]=af(t)+bg(t)\)

  2. 尺度性质

    1. \(L[f(at)]=\frac{1}{|a|}F(\frac{s}{a})\)
    2. \(L[f(at-b)]=\frac{1}{|a|}F(\frac{s}{a})e^{-s\frac{b}{a}}\)
  3. 延迟性质

    1. \(L[f(t-\tau)]=e^{-s\tau}F(s)\)

    2. \(L[f(t-\tau)u(t-\tau)]=e^{-s\tau}F(s)\)

    3. \(L^{-1}[e^{-s\tau}F(s)]=f(t-\tau)u(t-\tau)\)

  4. 位移性质

    1. \(L[e^{at}f(t)]=F(s-a)\)
    2. \(L^{-1}[F(s-a)]=e^{at}f(t)\)
  5. 微分性质

    1. \(L[f'(t)]=sF(s)-f(0)\)

      \(L[f^{(n)}(t)]=s^nF(s)-s^{n-1}f(0)-s^{n-2}f'(0)-...-f^{(n-1)}(0)\)

      其中\(f^{(k)}(0)\)应理解为\(\mathop{lim}\limits_{t\rightarrow0^+}f^{(k)}(t)\)

    2. \(F'(s)=-L[tf(t)]\)

      \(F^{(n)}(s)=(-1)^nL[t^nf(t)]\)

  6. 积分性质

    1. 性质

      1. \(L[\int_0^tf(\tau)d\tau]=\frac{1}{s}F(s)\)

      2. \(L[\frac{f(t)}{t}]=\int_s^{\infty}F(s)ds\)

        \(L[\frac{f(t)}{t^n}]=\int_s^{\infty}ds\int_s^{\infty}ds...\int_s^{\infty}F(s)ds\)(共积分n次)

      3. \(f(t)=tL^{-1}[\int_s^\infty F(s)ds]\)

    2. 用Laplace变换计算广义积分:

      1. \(F(s)=\int_0^{+\infty}f(t)e^{-st}dt\)
      2. \(F'(s)=-\int_0^{+\infty}tf(t)e^{-st}dt\)
      3. \(F(0)=\int_0^{+\infty}f(t)dt\)
      4. \(F'(0)=-\int_0^{+\infty}tf(t)dt\)
      5. \(\int_s^{\infty}F(s)ds=\int_0^{+\infty}\frac{f(t)}{t}e^{-st}dt\)
      6. \(\int_0^{\infty}F(s)ds=\int_0^{+\infty}\frac{f(t)}{t}dt\)
  7. 初值定理

    \(f(t)\)\(t\geq0\)时可微,\(f'(t)\)满足Laplace变换存在定理的条件,又设\(L[f(t)]=F(s),\mathop{lim}\limits_{s\rightarrow \infty}sF(s)\)存在,则:

    \[f(0)=\mathop{lim}\limits_{s\rightarrow \infty}sF(s) \]

  8. 终值定理

    \(L[f(t)]=F(s)\)\(sF(s)\)在包含虚轴的右半平面内解析,并且\(\mathop{lim}\limits_{s\rightarrow0}sF(s)\)存在,则:

    \[f(+\infty)=\mathop{lim}\limits_{t\rightarrow +\infty}f(t)=\mathop{lim}\limits_{s\rightarrow 0}sF(s) \]

  9. 卷积

    1. 定义:\(f_1(t)*f_2(t)=\int_{-\infty}^{+\infty}f_1(\tau)f_2(t-\tau)d\tau\)

      \(t<0\)时,\(f_1(t)=f_2(t)=0\),则有:

      \(f_1(t)*f_2(t)=\int_0^tf_1(\tau)f_2(t-\tau)d\tau,(t\geq0)\)

    2. 卷积运算规律:

      1. \(f_1(t)*f_2(t)=f_2(t)*f_1(t)\)
      2. \(f_1(t)*[f_2(t)*f_3(t)]=[f_1(t)*f_2(t)]*f_3(t)\)
      3. \(f_1(t)*[f_2(t)+f*3(t)]=f_1(t)*f_2(t)+f_1(t)*f_3(t)\)
      4. \(|f_1(t)*f_2(t)|\leq|f_1(t)|*|f_2(t)|\)
    3. 卷积定理:

      \(L[f_1(t)*f_2(t)]=F_1(s)\cdot F_2(s)\)

3. Laplace逆变换

若函数\(f(t)\)满足Laplace变换存在定理的条件,\(L[f(t)]=F(s)\),c为增长指数,则\(L^{-1}[F(s)]\)由下式给出:

  • t为f(t)连续点时

\[f(t)=\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+\infty}F(s)e^{st}ds\\s=\beta+i\omega\\t>0 \]

  • t为f(t)间断点时:

    \[\frac{1}{2}[f(t+0)+f(t-0)]=\frac{1}{2\pi i}\int_{\beta-i\infty}^{\beta+\infty}F(s)e^{st}ds\\ Re(s)=\beta>c \]

我们称(1)中1式为复反演积分公式,其中的积分应理解为:

\[\int_{\beta-i\infty}^{\beta+i\infty}F(s)e^{st}ds=\mathop{lim}\limits_{\omega\rightarrow +\infty}\int_{\beta-i\omega}^{\beta+i\omega}F(s)e^{st}ds \]

4. Laplace变换应用

  1. 解线性常微分方程

    \(ex1.\) 求微分方程\(y''(t)+4y(t)=0\)满足初始条件\(y|_{t=0}=-2,y'|_{t=0}=4\)的特解:

    \(L[y(t)]=Y(s)\),对方程两边取Laplace变换并带入条件得:

    \[s^2Y(s)-sy(0)-y'(0)+4Y(s)=0 \]

    解得:

    \[Y(s)=\frac{-2s+4}{s^2+4}=\frac{-2s}{s^2+4}+\frac{4}{s^2+4} \]

    取逆变换得:

    \[y(t)=L^{-1}[Y(s)]=-2cos(2t)+2sin(2t) \]

    \(ex2.\) 求方程\(y''-2y'+y=0\)满足初始条件\(y|_{t=0}=0,y|_{t=1}=2\)的特解:

    \(L[y(t)]=Y(s)\),对方程两边取Laplace变换得:

    \[s^2Y(s)-sy(0)-y'(0)-2sy(0)+Y(s)=0 \]

    于是:

    \[Y(s)=\frac{y'(0)}{(s-1)^2} \]

    取逆变换得:

    \[y(t)=L^{-1}[\frac{y'(0)}{(s-1)^2}]=y'(0)te^t \]

    \(t=1\)带入上式得\(y'(0)=2e^{-1}\),从而原方程得解为:

    \[y(t)=2te^{t-1} \]

  2. 解常系数线性常微分方程组

    \(ex3.\) 求方程组\(\begin{cases}x''-2y'-x=0\\x'-y=0\end{cases}\)满足初始条件\(x|_{t=0}=0,x'|_{t=0}=1,y|_{t=0}=1\)的特解:

    \(L[x(t)]=X(s),L[y(t)]=Y(s)\),对方程两边取Laplace变换,带入初始条件得:

    \[\begin{cases}s^2X(s)-sx(0)-x'(0)-2[sY(s)-y(0)]-X(s)=0\\ sX(s)-x(0)-Y(s)=0 \end{cases} \]

    整理化简后,得到:

    \[\begin{cases} (s^2-1)X(s)-2sY(s)+1=0\\ sX(s)-Y(s)=0 \end{cases} \]

    解方程组,得到:

    \[\begin{cases} X(s)=\frac{1}{s^2+1}\\ Y(s)=\frac{s}{s^2+1} \end{cases} \]

    取逆变换得:

    \[\begin{cases} x(t)=sint\\ y(t)=cost \end{cases} \]

  3. 解积分微分方程

    \(ex4.\) 求方程\(y'-4y+4\int_0^tydt=\frac{1}{3}t^3\)满足初始条件\(y|_{t=0}\)的特解:

    \(L[y(t)]=Y(s)\),对方程两边取Laplace变换,带入初始条件得:

    \[sY(s)-4Y(s)+\frac{4Y(s)}{s}=\frac{2}{s^4} \]

    解得:

    \[Y(s)=\frac{2}{s^3(s-2)^2} \]

    \(Y(s)\)表示成部分分式之和:

    \[Y(s)=\frac{3}{8}\cdot\frac{1}{s}+\frac{1}{2}\cdot\frac{1}{s^2}+\frac{1}{2}\cdot\frac{1}{s^3}-\frac{3}{8}\cdot\frac{1}{s-2}+\frac{1}{4}\cdot\frac{1}{(s-2)^2} \]

    取Laplace逆变换,即得原方程的解为:

    \[y(t)=\frac{3}{8}+\frac{1}{2}t+\frac{1}{4}t^2-\frac{3}{8}e^{2t}+\frac{1}{4}te^{2t} \]

posted @ 2021-09-30 21:35  chengerccj  阅读(553)  评论(0)    收藏  举报