Difference between revisions of "HW4.2 Zachary Curosh ECE301Fall2008mboutin" - Rhea

Revision as of 10:58, 25 September 2008

Define a Periodic DT Signal and Compute the Fourier Series Coefficients

I am going to choose a sine signal, since there have been many cosines done already.

DT signal: $ x[n] = 2\sin(\pi n + \frac{\pi}{2}) + 4\sin(\frac{\pi}{2} n + \pi)\, $

Now, each sine has its own period, and the fundamental period of the function is the greater of the separate periods.

$ N_2sin = \frac{2\pi}{\pi} k = \frac{2}{1} k $

$ N_4sin = \frac{2\pi}{\frac{\pi}{2}} k = \frac{2}{\frac{1}{2}} k $

Take $ k = 1\, $,

$ N_2sin = 2\, $

$ N_4sin = 4\, $, so the overall fundamental period is

$ N = 4\, $

In order to find the coefficients, we must first calculate the values of $ x[n]\, $ for four consecutive integer values of $ n\, $. By plugging values of $ n\, $ into the given signal, we find that

$ x[0] = 2\, $

$ x[1] = -6\, $

$ x[2] = 2\, $

$ x[3] = 2\, $

$ x[4] = 2\, $

$ x[5] = -6\, $

$ x[6] = 2\, $

$ x[7] = 2\, $, which continue to repeat in this way every 4 integers.

$ a_k = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-jk\frac{\pi}{2} n}\, $

$ a_0 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^0\, $

$ = \frac{1}{4}\sum^{3}_{n = 0} x[n]\, $

$ = \frac{1}{4}(2 - 6 + 2 + 2)\, $

$ a_0 = 0\, $

$ a_1 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{\pi}{2} n}\, $

$ = \frac{1}{4}(x[0] + x[1]e^{-j\frac{\pi}{2}} + x[2]e^{-j\pi} + x[3]e^{-j\frac{3\pi}{2}})\, $

$ = \frac{1}{4}(2 - 6(-j) + 2(-1) + 2(j))\, $

$ = \frac{1}{4}(8j)\, $

$ a_1 = 2j\, $

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

$ = \frac{1}{4}(x[0] + x[1]e^{-j\pi} + x[2]e^{-j2\pi} + x[3]e^{-j3\pi})\, $

$ = \frac{1}{4}(2 - 6(-1) + 2(1) + 2(-1))\, $

$ = \frac{1}{4}(8)\, $

$ a_2 = 2\, $

$ a_3 = \frac{1}{4}\sum^{3}_{n = 0} x[n] e^{-j\frac{3\pi}{2} n}\, $

$ = \frac{1}{4}(x[0] + x[1]e^{-j\frac{3\pi}{2}} + x[2]e^{-j3\pi} + x[3]e^{-j\frac{9\pi}{2}})\, $

$ = \frac{1}{4}(2 - 6(j) + 2(-1) + 2(-j))\, $

$ = \frac{1}{4}(-8j)\, $

$ a_3 = -2j\, $