【学习笔记】傅里叶变换：方形函数，三角函数

方形函数：

F(ω)=∫−t0t0A⋅e−jωtdtF\left( \omega \right) =\int ^{t_{0}}_{-t_{0}}A\cdot e^{-j\omega t_{dt}}F(ω)=∫−t0t0A⋅e−jωtdt
=Ajω(ejωt0−e−jωt0)=\dfrac {A}{j\omega }\left( e^{j\omega t_{0}}-e^{-j\omega t_{0}}\right) =jωA(ejωt0−e−jωt0)
=Ajω⋅2jsin⁡ωt0=\dfrac {A}{j\omega }\cdot 2j\sin \omega t_{0}=jωA⋅2jsinωt0
=A⋅2t0sin⁡ωt0ωt0=A\cdot \dfrac {2t_{0}\sin \omega t_{0}}{\omega t_{0}}=A⋅ωt02t0sinωt0
=2At0⋅Sa(ωt0)=2At_{0}\cdot S_{a}\left( \omega t_{0}\right) =2At0⋅Sa(ωt0)
即：A[u(t+t0)−u(t−t0)]⇔2At0Sa(ωt0)\begin{aligned}A\left[ u\left( t+t_{0}\right) -u\left( t-t_{0}\right) \right] \Leftrightarrow 2At_{0}Sa\left( \omega t_{0}\right) \end{aligned}A[u(t+t0)−u(t−t0)]⇔2At0Sa(ωt0)

三角函数：

**

先特殊化：令t0t_{0}t0=1,A=1;
则：F(ω)=∫−10(1+t)e−jωtdt+∫01(1+t)e−jωtF\left( \omega \right) =\int ^{0}_{-1}\left( 1+t\right) e^{-j\omega t}dt+\int ^{1}_{0}\left( 1+t\right) e^{-j\omega t}F(ω)=∫−10(1+t)e−jωtdt+∫01(1+t)e−jωt
分部积分：
=[(1+t)⋅e−jωt−jω‾]−10+∫−10e−jωtjωdt+[(1−t)⋅e−jωt−jω‾]−10+∫01e−jωtjωd(1−t)=\begin{bmatrix} \left( 1+t\right) \cdot e^{-j\omega t} \\ \overline {-j\omega } \end{bmatrix}^{0}_{-1}+\int ^{0}_{-1}\dfrac {e^{-j\omega t}}{j\omega }dt+\begin{bmatrix} \left( 1-t\right) \cdot e^{-j\omega t} \\ \overline {-j\omega } \end{bmatrix}^{0}_{-1}+\int ^{1}_{0}\dfrac {e^{-j\omega t}}{j\omega }d\left( 1-t\right) =[(1+t)⋅e−jωt−jω]−10+∫−10jωe−jωtdt+[(1−t)⋅e−jωt−jω]−10+∫01jωe−jωtd(1−t)
=1−jω+1−ejωω2+1jω+1−e−jωω2=\dfrac {1}{-j\omega }+\dfrac {1-e^{j\omega }}{\omega ^{2}}+\dfrac {1}{j\omega }+\dfrac {1-e^{-j\omega }}{\omega ^{2}}=−jω1+ω21−ejω+jω1+ω21−e−jω
=2−ejω−e−jωω2=\dfrac {2-e^{j\omega }-e^{-j\omega }}{\omega ^{2}}=ω22−ejω−e−jω
又因为：cos⁡w=1−2sin2ω2\cos w=1-2sin^{2}\dfrac {\omega }{2}cosw=1−2sin22ω

所以化简得：F(ω)=Sa2ω2F\left( \omega \right) =Sa^{2}\dfrac {\omega }{2}F(ω)=Sa22ω

同理，一般化可得：F(ω)=At0Sa2ωt02F\left( \omega \right) =At_{0}Sa^{2}\dfrac {\omega t_{0}}{2}F(ω)=At0Sa22ωt0