一、序列傅里叶变换实例

x ( n ) = R N ( n )      ① x(n) = R_N(n) \ \ \ \ ① x(n)=RN(n)    ①

1、傅里叶变换

X ( e j ω ) = ∑ n = − ∞ + ∞ x ( n ) e − j ω n      ② X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x(n) e^{-j \omega n} \ \ \ \ ② X(ejω)=n=−∞∑+∞x(n)e−jωn    ②

X ( e j ω ) = ∑ n = − ∞ + ∞ x ( n ) e − j ω n = ∑ n = 0 N − 1 e − j ω n X(e^{j\omega}) = \sum_{n=-\infty}^{+\infty} x(n) e^{-j \omega n} = \sum_{n=0}^{N-1} e^{-j \omega n} X(ejω)=n=−∞∑+∞x(n)e−jωn=n=0∑N−1e−jωn

X ( e j ω ) = 1 − e − j ω n 1 − e − j ω X(e^{j\omega}) = \cfrac{1-e^{-j\omega n}}{1-e^{-j\omega}} X(ejω)=1−e−jω1−e−jωn

X ( e j ω ) = e − j ω N − 1 2 sin ⁡ ( ω N 2 ) sin ⁡ ( ω 2 ) X(e^{j\omega}) = e^{-j\omega \cfrac{N-1}{2}} \cfrac{ \sin( \cfrac{\omega N}{2} ) }{ \sin( \cfrac{\omega }{2} )} X(ejω)=e−jω2N−1sin(2ω)sin(2ωN)

S F T [ R N ( n ) ] = N      ω = 0 SFT[R_N(n)] = N \ \ \ \ \omega = 0 SFT[RN(n)]=N    ω=0

S F T [ R N ( n ) ] = 0      ω = 2 π k N , k = ± 1 , ± 2 , ⋯ SFT[R_N(n)] = 0 \ \ \ \ \omega = \cfrac{2\pi k}{N} , k = \pm1 , \pm2 , \cdots SFT[RN(n)]=0    ω=N2πk,k=±1,±2,⋯

• 当 ω = 0 \omega = 0 ω=0 时 , S F T [ R 4 ( n ) ] = 4 SFT[R_4(n)] = 4 SFT[R4​(n)]=4
• 当 ω = 2 π k N = 2 π k 4 = π k 2 \omega = \cfrac{2\pi k}{N} = \cfrac{2\pi k}{4} = \cfrac{\pi k}{2} ω=N2πk​=42πk​=2πk​ 时 , S F T [ R 4 ( n ) ] = 0 SFT[R_4(n)] = 0 SFT[R4​(n)]=0 , 第一个点是 π 2 \cfrac{\pi}{2} 2π​ , 第二个点是 π \pi π , 如下图所示 ;

3、傅里叶变换相频特性

N N N 如果确定了 , N − 1 2 \cfrac{N-1}{2} 2N−1 是常数 , 因此整个曲线是线性的 ,