Difference between revisions of "HW4.5 Zachary Curosh ECE301Fall2008mboutin" - Rhea

Revision as of 12:56, 25 September 2008

Guess the Periodic Signal

A certain periodic signal has the following properties:

1. N = 4

2. $ \sum_{n=0}^{3}x[n] = 4 $

3. $ \sum_{n=1}^{4}(-1)^nx[n] = 2 $

4. For even $ k\, $'s, $ a_k = a_{k+1}\, $

Answer

From 1. we know that $ x[n] = \sum_{n=0}^{3}a_k e^{jk\frac{\pi}{2}n}\, $

Using 2., it is apparent that this is the formula for $ a_k\, $. Specifically, for $ a_0\, $, since the only thing under the sum is $ x[n]\, $. So,

Now that we know $ a_0\, $, we know that $ x[n] = 1 + \sum_{n=1}^{4}a_k e^{jk\frac{\pi}{2}n}\, $

Since $ \omega_0 = \frac{\pi}{2}\, $, let's try and find $ a_2\, $,

$ a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-2j\frac{\pi}{2}n} = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-j\pi n}\, $

Using the property that $ e^{-j\pi n} = (e^{-j\pi})^{n} = (-1)^{n} \, $, we can change the above equation to

$ a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n](-1)^{n}\, $

Since the function is periodic and the $ a_k\, $'s repeat every 4 integers, we are able to shift the bounds of summation by one.

According to 3. $ \sum_{n=1}^{4}(-1)^nx[n] = 2 $, and

$ a_2 = \frac{1}{4}\sum_{n=1}^{4}x[n](-1)^{n}\, $

$ a_2 = \frac{1}{4} * 2\, $, and

$ a_2 = \frac{1}{2}\, $

Since we know the 2 even $ a_k\, $'s in the fundamental period, by property 4, we can find the 2 odd $ a_k\, $'s.

$ a_0 = a_1 = 1\, $

$ a_2 = a_3 = \frac{1}{2}\, $