Difference between revisions of "HW4.2 Wei Jian Chan ECE301Fall2008mboutin" - Rhea

Revision as of 15:23, 24 September 2008

CT Periodic Signal : $ x[n] = 5\cos(6\pi n + \pi) + 7\cos(3\pi n)\, $

$ N_1 = \frac{2\pi}{6\pi} k = \frac{1}{3} k $

$ N_2 = \frac{2\pi}{3\pi} k = \frac{2}{3} k $

$ k\, $ that makes both $ N_1 , N_2 \, $ integer is $ k = 3\, $

As $ N_2 > N_1\, $ thus $ N = 2\, $

Through Matlab, the following values can be calculated:

$ x[0] = 2\, $

$ x[1] = -12\, $

$ x[2] = 2\, $

$ x[3] = -12\, $

$ a_k = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^{-jk\pi n}\, $

$ a_0 = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^0\, $

$ = \frac{1}{2}\sum^{1}_{n = 0} x[n]\, $

$ = \frac{1}{2}(2 - 12)\, $

$ = -5\, $

$ a_1 = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^{-j\pi n}\, $

$ = \frac{1}{2}(x[0] + x[1]e^{-j\pi})\, $

$ = \frac{1}{2}(2 - 12(-1))\, $

$ = \frac{1}{2}(14)\, $

$ = 7\, $

We can write the function as follows:

$ x(t) = \sum^{1}_{k = 0} a_k e^{jk\pi n}\, $ where

$ a_0 = -5\, $

$ a_1 = 7\, $

$ a_k = 0 , k \neq 0,1\, $

Correction:

Overlookd the properties of a DT system. $ a_k\, $ shouldn't be 0, but will repeat itself.

$ a_k = -5 , k = 2,4,6,8,10....\, $

$ a_k = 7 , k = 1,3,5,7,9,11....\, $