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

Revision as of 13:34, 25 September 2008

Given the following periodic DT signal

$ \,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]\, $

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

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


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

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

$ \,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})\, $

$ \,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}})\, $


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

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva