Revision as of 17:41, 26 September 2008 by Li186 (Talk)

Guessing the periodic signal:

Hint

1. Period of x[n] is N=4.

2. $ a_0=3 $.

3.$ \sum_{n=4}^{7}(-1)^{n}x[n]=4 $.

4.x[n] has minimum power among all the signals that satisfy 1,2,3.

Solution

From (1), we can get $ omega_0=2\pi/N=\frac{1}{2}\pi $,so x[n]=$ \sum_{k=0}^{N-1}a_ke^{jk\frac{1}{2}\pi n} $.

From (2),x[n]=3+$ \sum_{k=1}^{3}a_ke^{jk\frac{1}{2}\pi n} $.

Since N=4, we have x[1]=x[5],x[2]=x[6],x[3]=x[7], and $ (-1)^n $ is periodic with 4,too

From (3),we can get $ a_2=\frac{1}{4}sum_{n=4}^{7}e^{-j\pi n}x[n]=\frac{1}{4}sum_{n=4}^{7}(-1)^nx[n]=1 $

So,

$ x[n]= 3+a_1e^{j\frac{\pi}{2}n}+e^{j\frac{2\pi}{2}n}+a_3e^{j\frac{3\pi}{2}n} $

Power of x[n] is

$ P=\frac{1}{4}\sum{n=0}^{3}|x[n]|^2 $

$ =\sum{k=0}^{3}|a_k|^2 $

$ =3^2+|a_1|^2+1^2+|a_3|^2 $

the minimum happens when $ |a_1|^2=0 $ and $ |a_3|^2=0 $

Answer:

$ <font=5>x[n]=3+e^{j\pi n} $

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