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<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}\sum_{n=0}^{3}x[n]e^{-jk\frac{2\pi}{4}n}\,</math> | ||
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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> | ||
Revision as of 13:27, 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})\, $
