主要知识:离散傅里叶变换定理及其证明
适用对象:初学数字信号处理的同学以及考研备考的同学(尤其是目标院校是:南京理工大学)
重要程度: ★ ★ ★ ★ \bigstar \bigstar \bigstar \bigstar ★★★★ (满5颗星)
设序列 x [ n ] x[n] x[n] 、 g [ n ] g[n] g[n] 、 h [ n ] h[n] h[n] 的DTFT为 X ( e j ω ) X(e^{j\omega}) X(ejω) 、 G ( e j ω ) G(e^{j\omega}) G(ejω) 、 H ( e j ω ) H(e^{j\omega}) H(ejω) ,m为常数。
内容:
x [ n ] = g [ n − m ] X ( e j ω ) = G ( e j ω ) e − j ω m \begin{aligned} x\left[ n \right]&=g\left[ n-m \right] \\ X\left( { {e}^{j\omega }} \right)&=G\left( { {e}^{j\omega }} \right){ {e}^{-j\omega m}} \\ \end{aligned} x[n]X(ejω)=g[n−m]=G(ejω)e−jωm
证明:
X ( e j ω ) = ∑ n = − ∞ ∞ x [ n ] e − j ω n = ∑ n = − ∞ ∞ g [ n − m ] e − j ω n → n − m = k = ∑ n = − ∞ ∞ g [ k ] e − j ω ( m + k ) = e − j ω m G ( e j ω ) \begin{aligned} X\left( { {e}^{j\omega }} \right)&=\sum\limits_{n=-\infty }^{\infty }{x\left[ n \right]{ {e}^{-j\omega n}}} \\ & =\sum\limits_{n=-\infty }^{\infty }{g\left[ n-m \right]}{ {e}^{-j\omega n}} \\ \xrightarrow{n-m=k}&=\sum\limits_{n=-\infty }^{\infty }{g\left[ k \right]{ {e}^{-j\omega \left( m+k \right)}}} \\ & ={ {e}^{-j\omega m}}G\left( { {e}^{j\omega }} \right) \end{aligned} X(ejω)n−m=k=n=−∞∑∞x[n]e−jωn=n=−∞∑∞g[n−m]e−jωn=n=−∞∑∞g[k]e−jω(m+k)=e−jωmG(ejω)
内容:
x [ n ] = e j ω 0 n g [ n ] X ( e j ω ) = G ( e j ( ω − ω 0 ) ) \begin{gathered} x\left[ n \right] = {e^{j{\omega _0}n}}g\left[ n \right] \\ X\left( { {e^{j\omega }}} \right) = G\left( { {e^{j\left( {\omega - {\omega _0}} \right)}}} \right) \\ \end{gathered} x[n]=ejω0ng[n]X(ejω)=G(ej(ω−ω0))
证明:
X ( e j ω ) = ∑ n = − ∞ ∞ x [ n ] e − j ω n = ∑ n = − ∞ ∞ e j ω 0 n g [ n ] e − j ω n = ∑ n = − ∞ ∞ g [ n ] e − j ( ω − ω 0 ) n = G ( e j ( ω − ω 0 ) ) \begin{aligned} X\left( { {e^{j\omega }}} \right) &= \sum\limits_{n = - \infty }^\infty {x\left[ n \right]} {e^{ - j\omega n}} \\ & = \sum\limits_{n = - \infty }^\infty { {e^{j{\omega _0}n}}g\left[ n \right]{e^{ - j\omega n}}} \\ & = \sum\limits_{n = - \infty }^\infty {g\left[ n \right]{e^{ - j\left( {\omega - {\omega _0}} \right)n}}} \\ & = G\left( { {e^{j\left( {\omega - {\omega _0}} \right)}}} \right) \\ \end{aligned} X(ejω)=n=−∞∑∞x[n]e−jωn=n=−∞∑∞ejω0ng[n]e−jωn=n=−∞∑∞g[n]e−j(ω−ω0)n=G(ej(ω−ω0))
内容:
x [ n ] = g [ n ] ∗ h [ n ] X ( e j ω ) = G ( e j ω ) H ( e j ω ) \begin{aligned} x\left[ n \right] &= g\left[ n \right] * h\left[ n \right]\\ X\left( { {e^{j\omega }}} \right) &= G\left( { {e^{j\omega }}} \right)H\left( { {e^{j\omega }}} \right) \end{aligned} x[n]X(ejω)=g[n]∗h[n]=G(ejω)H(ejω)
证明:
X ( e j ω ) = ∑ n = − ∞ ∞ x [ n ] e − j ω n = ∑ n = − ∞ ∞ { g [ n ] ∗ h [ n ] } e − j ω n = ∑ n = − ∞ ∞ { ∑ m = − ∞ ∞ g [ m ] h [ n − m ] } e − j ω n = ∑ m = − ∞ ∞ g [ m ] { ∑ n = − ∞ ∞ h [ n − m ] e − j ω n } = ∑ m = − ∞ ∞ g [ m ] H ( e j ω ) e − j ω m = H ( e j ω ) ∑ m = − ∞ ∞ g [ m ] e − j ω m = H ( e j ω ) G ( e j ω ) \begin{aligned} X\left( { {e^{j\omega }}} \right) &= \sum\limits_{n = - \infty }^\infty {x\left[ n \right]{e^{ - j\omega n}}} \\ &= \sum\limits_{n = - \infty }^\infty {\left\{ {g\left[ n \right]*h\left[ n \right]} \right\}{e^{ - j\omega n}}} \\ & = \sum\limits_{n = - \infty }^\infty {\left\{ {\sum\limits_{m = - \infty }^\infty {g\left[ m \right]h\left[ {n - m} \right]} } \right\}{e^{ - j\omega n}}} \\ &= \sum\limits_{m = - \infty }^\infty {g\left[ m \right]\left\{ {\sum\limits_{n = - \infty }^\infty {h\left[ {n - m} \right]{e^{ - j\omega n}}} } \right\}} \\ &= \sum\limits_{m = - \infty }^\infty {g\left[ m \right]H\left( { {e^{j\omega }}} \right){e^{ - j\omega m}}} \\ & = H\left( { {e^{j\omega }}} \right)\sum\limits_{m = - \infty }^\infty {g\left[ m \right]{e^{ - j\omega m}}} \\ &= H\left( { {e^{j\omega }}} \right)G\left( { {e^{j\omega }}} \right) \end{aligned} X(ejω)=n=−∞∑∞x[n]e−jωn=n=−∞∑∞{ g[n]∗h[n]}e−jωn=n=−∞∑∞{ m=−∞∑∞g[m]h[n−m]}e−jωn=m=−∞∑∞g[m]{ n=−∞∑∞h[n−m]e−jωn}=m=−∞∑∞g[m]H(ejω)e−jωm=H(ejω)m=−∞∑∞g[m]e−jωm=H(ejω)G(ejω)
内容:
x [ n ] = g [ n ] h [ n ] X ( e j ω ) = 1 2 π ∫ − π π G ( e j θ ) H ( e j ( ω − θ ) ) d θ \begin{aligned} x\left[ n \right]&=g\left[ n \right]h\left[ n \right] \\ X\left( { {e}^{j\omega }} \right)&=\frac{1}{2\pi }\int_{-\pi }^{\pi }{G\left( { {e}^{j\theta }} \right)H\left( { {e}^{j\left( \omega -\theta \right)}} \right)d\theta } \\ \end{aligned} x[n]X(ejω)=g[n]h[n]=2π1∫−ππG(ejθ)H(ej(ω−θ))dθ
证明:
X ( e j ω ) = ∑ n = − ∞ ∞ g [ n ] h [ n ] e − j ω n = ∑ n = − ∞ ∞ [ 1 2 π ∫ − π π G ( e j ω 1 ) e j ω 1 n d ω 1 ] h [ n ] e − j ω n = 1 2 π ∫ − π π G ( e j ω 1 ) [ ∑ n = − ∞ ∞ h [ n ] e − j ( ω − ω 1 ) n ] d ω 1 = 1 2 π ∫ − π π G ( e j ω 1 ) H ( e j ( ω − ω 1 ) ) d ω 1 \begin{aligned} X\left( { {e}^{j\omega }} \right)&=\sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]h\left[ n \right]{ {e}^{-j\omega n}}} \\ & =\sum\limits_{n=-\infty }^{\infty }{\left[ \frac{1}{2\pi }\int_{-\pi }^{\pi }{G\left( { {e}^{j{ {\omega }_{1}}}} \right){ {e}^{j{ {\omega }_{1}}n}}d{ {\omega }_{1}}} \right]h\left[ n \right]{ {e}^{-j\omega n}}} \\ & =\frac{1}{2\pi }\int_{-\pi }^{\pi }{G\left( { {e}^{j{ {\omega }_{1}}}} \right)\left[ \sum\limits_{n=-\infty }^{\infty }{h\left[ n \right]{ {e}^{-j\left( \omega -{ {\omega }_{1}} \right)n}}} \right]}d{ {\omega }_{1}} \\ & =\frac{1}{2\pi }\int_{-\pi }^{\pi }{G\left( { {e}^{j{ {\omega }_{1}}}} \right)H\left( { {e}^{j\left( \omega -{ {\omega }_{1}} \right)}} \right)d{ {\omega }_{1}}} \end{aligned} X(ejω)=n=−∞∑∞g[n]h[n]e−jωn=n=−∞∑∞[2π1∫−ππG(ejω1)ejω1ndω1]h[n]e−jωn=2π1∫−ππG(ejω1)[n=−∞∑∞h[n]e−j(ω−ω1)n]dω1=2π1∫−ππG(ejω1)H(ej(ω−ω1))dω1
内容:
x [ n ] = n g [ n ] X ( e j ω ) = j d G ( e j w ) d ω \begin{aligned} x\left[ n \right]&=ng\left[ n \right] \\ X\left( { {e}^{j\omega }} \right)&=j\frac{dG\left( { {e}^{jw}} \right)}{d\omega } \\ \end{aligned} x[n]X(ejω)=ng[n]=jdωdG(ejw)
证明:
G ( e j ω ) = ∑ n = − ∞ ∞ g [ n ] e − j ω n d G ( e j ω ) d ω = − j ∑ n = − ∞ ∞ n g [ n ] e − j ω n n g [ n ] ⟷ D T F T I D T F T j d G ( e j ω ) d ω \begin{aligned} G\left( { {e}^{j\omega }} \right)&=\sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]{ {e}^{-j\omega n}}} \\ \frac{dG\left( { {e}^{j\omega }} \right)}{d\omega }&=-j\sum\limits_{n=-\infty }^{\infty }{ng\left[ n \right]{ {e}^{-j\omega n}}} \\ ng\left[ n \right]&\underset{IDTFT}{\overset{DTFT}{\longleftrightarrow}}j\frac{dG\left( { {e}^{j\omega }} \right)}{d\omega } \\ \end{aligned} G(ejω)dωdG(ejω)ng[n]=n=−∞∑∞g[n]e−jωn=−jn=−∞∑∞ng[n]e−jωnIDTFT⟷DTFTjdωdG(ejω)
内容:
{ ∑ n = − ∞ ∞ g [ n ] h ∗ [ n ] = 1 2 π ∫ − π π G ( e j ω ) H ∗ ( e j ω ) d ω ∑ n = − ∞ ∞ ∣ g [ n ] ∣ 2 = 1 2 π ∫ − π π ∣ G ( e j ω ) ∣ 2 d ω \left\{ \begin{aligned} & \sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]{ {h}^{*}}\left[ n \right]}=\frac{1}{2\pi }\int_{-\pi }^{\pi }{G\left( { {e}^{j\omega }} \right)H^*\left( { {e}^{j\omega }} \right)d\omega } \\ & \sum\limits_{n=-\infty }^{\infty }{ { {\left| g\left[ n \right] \right|}^{2}}=\frac{1}{2\pi }\int_{-\pi }^{\pi }{ { {\left| G\left( { {e}^{j\omega }} \right) \right|}^{2}}d\omega }} \\ \end{aligned} \right. ⎩⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎧n=−∞∑∞g[n]h∗[n]=2π1∫−ππG(ejω)H∗(ejω)dωn=−∞∑∞∣g[n]∣2=2π1∫−ππ∣∣G(ejω)∣∣2dω
证明:
h [ n ] = 1 2 π ∫ − π π H ( e j ω ) e j ω n d ω h ∗ [ n ] = 1 2 π ∫ − π π H ∗ ( e j ω ) e − j ω n d ω \begin{aligned} h\left[ n \right]&=\frac{1}{2\pi }\int_{-\pi }^{\pi }{H\left( { {e}^{j\omega }} \right){ {e}^{j\omega n}}d\omega } \\ { {h}^{*}}\left[ n \right]&=\frac{1}{2\pi }\int_{-\pi }^{\pi }{ { {H}^{*}}\left( { {e}^{j\omega }} \right){ {e}^{-j\omega n}}d\omega } \\ \end{aligned} h[n]h∗[n]=2π1∫−ππH(ejω)ejωndω=2π1∫−ππH∗(ejω)e−jωndω
∑ n = − ∞ ∞ g [ n ] h ∗ [ n ] = ∑ n = − ∞ ∞ g [ n ] { 1 2 π ∫ − π π H ∗ ( e j ω ) e − j ω n d ω } = 1 2 π ∫ − π π H ∗ ( e j ω ) [ ∑ n = − ∞ ∞ g [ n ] e − j ω n ] d ω = 1 2 π ∫ − π π H ∗ ( e j ω ) G ( e j ω ) d ω \begin{aligned} \sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]{ {h}^{*}}\left[ n \right]}&=\sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]}\left\{ \frac{1}{2\pi }\int_{-\pi }^{\pi }{ { {H}^{*}}\left( { {e}^{j\omega }} \right){ {e}^{-j\omega n}}d\omega } \right\} \\ & =\frac{1}{2\pi }\int_{-\pi }^{\pi }{ { {H}^{*}}\left( { {e}^{j\omega }} \right)\left[ \sum\limits_{n=-\infty }^{\infty }{g\left[ n \right]{ {e}^{-j\omega n}}} \right]}d\omega \\ & =\frac{1}{2\pi }\int_{-\pi }^{\pi }{ { {H}^{*}}\left( { {e}^{j\omega }} \right)}G\left( { {e}^{j\omega }} \right)d\omega \end{aligned} n=−∞∑∞g[n]h∗[n]=n=−∞∑∞g[n]{ 2π1∫−ππH∗(ejω)e−jωndω}=2π1∫−ππH∗(ejω)[n=−∞∑∞g[n]e−jωn]dω=2π1∫−ππH∗(ejω)G(ejω)dω