View source for HW4.2 Jeff Kubascik ECE301Fall2008mboutin - Rhea

Given the following periodic DT signal

<math>\,x[n]=\sum_{k=-\infty}^{\infty}\delta[n-4k] + \pi\delta[n-1-4k] - 3\delta[n-2-4k] + \sqrt[e]{\frac{\pi^j}{\ln(j)}}\delta[n-3-4k]\,</math>

which is an infinite sum of shifted copies of a non-periodic signal, compute its Fourier series coefficients.

== Answer ==
The equation for determining the Fourier coefficients of a DT signal is

<math>\,a_k=\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi}{N}n}\,</math>


The function has a fundamental period of 4 (it can be easily shown that <math>\,x[n]=x[n+5], \forall n\in\mathbb{Z}\,</math>), so <math>\,N=4\,</math>.  Therefore, we get

<math>\,a_k=\frac{1}{4}\sum_{n=0}^{3}x[n]e^{-jk\frac{2\pi}{4}n}\,</math>

<math>\,a_k=\frac{1}{4}(x[0]e^{0} + x[1]e^{-jk\frac{2\pi}{4}1} + x[2]e^{-jk\frac{2\pi}{4}2} + x[3]e^{-jk\frac{2\pi}{4}3})\,</math>

<math>\,a_k=\frac{1}{4}(1 + \pi e^{-jk\frac{\pi}{2}} - 3e^{-jk\pi} + \sqrt[e]{\frac{\pi^j}{\ln(j)}}e^{-jk\frac{3\pi}{2}})\,</math>


Using this equation, we can find all Fourier coefficients <math>\,a_k, k=0,1,2,3\,</math> of the signal <math>\,x[n]\,</math>.  They are:

<math>\,
a_0=\frac{1}{4}(1 + \pi e^{0} - 3e^{0} + \sqrt[e]{\frac{\pi^j}{\ln(j)}}e^{0})
   = \frac{1}{4}(1 + \pi - 3 + \sqrt[e]{\frac{\pi^j}{\ln(j)}})
   = \frac{1}{4}(\pi - 2 + \sqrt[e]{\frac{\pi^j}{\ln(j)}})
\,</math>

<math>\,
a_1=\frac{1}{4}(1 + \pi e^{-j\frac{\pi}{2}} - 3e^{-j\pi} + \sqrt[e]{\frac{\pi^j}{\ln(j)}}e^{-j\frac{3\pi}{2}})
   = \frac{1}{4}(1 - j\pi + 3 + j\sqrt[e]{\frac{\pi^j}{\ln(j)}})
   = \frac{1}{4}(4 - j\pi + j\sqrt[e]{\frac{\pi^j}{\ln(j)}})
\,</math>