DSP - DFT例題解析

例題1

驗證序列$x(n) = \frac{1^n}{4}u(n)$的Parseval定理

解 − $\displaystyle\sum\limits_{-\infty}^\infty|x_1(n)|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|X_1(e^{j\omega})|^2d\omega$

左邊 $\displaystyle\sum\limits_{-\infty}^\infty|x_1(n)|^2$

$= \displaystyle\sum\limits_{-\infty}^{\infty}x(n)x^*(n)$

$= \displaystyle\sum\limits_{-\infty}^\infty(\frac{1}{4})^{2n}u(n) = \frac{1}{1-\frac{1}{16}} = \frac{16}{15}$

右邊 $X(e^{j\omega}) = \frac{1}{1-\frac{1}{4}e^{-j\omega}} = \frac{1}{1-0.25\cos \omega+j0.25\sin \omega}$

$\Longleftrightarrow X^*(e^{j\omega}) = \frac{1}{1-0.25\cos \omega-j0.25\sin \omega}$

計算 $X(e^{j\omega}).X^*(e^{j\omega})$

$= \frac{1}{(1-0.25\cos \omega)^2+(0.25\sin \omega)^2} = \frac{1}{1.0625-0.5\cos \omega}$

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{1.0625-0.5\cos \omega}d\omega$

$\frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{1.0625-0.5\cos \omega}d\omega = 16/15$

我們可以看到，左邊 = 右邊。(證畢)

例題2

計算$x(n) = 3\delta (n)$的N點DFT

解 − 我們知道，

$X(K) = \displaystyle\sum\limits_{n = 0}^{N-1}x(n)e^{\frac{j2\Pi kn}{N}}$

$= \displaystyle\sum\limits_{n = 0}^{N-1}3\delta(n)e^{\frac{j2\Pi kn}{N}}$

$ = 3\delta (0)\times e^0 = 3$

所以$x(k) = 3,0\leq k\leq N-1$… 答案

例題3

計算$x(n) = 7\delta(n-n_0)$的N點DFT

解 − 我們知道，

$X(K) = \displaystyle\sum\limits_{n = 0}^{N-1}x(n)e^{\frac{j2\Pi kn}{N}}$

代入x(n)的值，

$\displaystyle\sum\limits_{n = 0}^{N-1}7\delta (n-n_0)e^{-\frac{j2\Pi kn}{N}}$

$= 7e^{-j2\pi kn_0/N}$… 答案