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| − | + | Given the following periodic DT signal | |
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| + | <math>\,x[n]=\sum_{k=-\infty}^{\infty}\delta[n-4k] + \pi\delta[n-1-4k] - 3\delta[n-2-4k] + \sqrt[e]{\frac{\pi^3}{\ln(5)}}\delta[n-3-4k]\,</math> | ||
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| + | which is an infinite sum of shifted copies of a non-periodic signal, compute its Fourier series coefficients. | ||
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| + | == Answer == | ||
| + | The equation for determining the Fourier coefficients of a DT signal is | ||
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| + | <math>\,a_k=\frac{1}{N}\sum_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi}{N}n}\,</math> | ||
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| + | The function has a fundamental period of 4 (it can be easily shown that <math>\,x[n]=x[n+4], \forall n\in\mathbb{Z}\,</math>), so <math>\,N=4\,</math>. Therefore, we get | ||
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| + | <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> | ||
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| + | <math>\,a_k=\frac{1}{4}(1 + \pi e^{-jk\frac{\pi}{2}} - 3e^{-jk\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}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: | ||
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| + | <math>\,a_0=\frac{1}{4}(1 + \pi e^{0} - 3e^{0} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{0})\,</math> | ||
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| + | <math>\,a_0=\frac{1}{4}(1 + \pi - 3 + \sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_0=\frac{1}{4}(-2 + \pi + \sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_1=\frac{1}{4}(1 + \pi e^{-j\frac{\pi}{2}} - 3e^{-j\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j\frac{3\pi}{2}})\,</math> | ||
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| + | <math>\,a_1=\frac{1}{4}(1 - j\pi + 3 + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_1=\frac{1}{4}(4 - j\pi + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_2=\frac{1}{4}(1 + \pi e^{-j\pi} - 3e^{-j2\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j3\pi})\,</math> | ||
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| + | <math>\,a_2=\frac{1}{4}(1 - \pi - 3 - \sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_2=\frac{1}{4}(-2 - \pi - \sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_3=\frac{1}{4}(1 + \pi e^{-j\frac{3\pi}{2}} - 3e^{-j3\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j\frac{9\pi}{2}})\,</math> | ||
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| + | <math>\,a_3=\frac{1}{4}(1 + j\pi + 3 - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
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| + | <math>\,a_3=\frac{1}{4}(4 + j\pi - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\,</math> | ||
Latest revision as of 14:10, 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^3}{\ln(5)}}\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+4], \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^3}{\ln(5)}}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:
$ \,a_0=\frac{1}{4}(1 + \pi e^{0} - 3e^{0} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{0})\, $
$ \,a_0=\frac{1}{4}(1 + \pi - 3 + \sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_0=\frac{1}{4}(-2 + \pi + \sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_1=\frac{1}{4}(1 + \pi e^{-j\frac{\pi}{2}} - 3e^{-j\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j\frac{3\pi}{2}})\, $
$ \,a_1=\frac{1}{4}(1 - j\pi + 3 + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_1=\frac{1}{4}(4 - j\pi + j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_2=\frac{1}{4}(1 + \pi e^{-j\pi} - 3e^{-j2\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j3\pi})\, $
$ \,a_2=\frac{1}{4}(1 - \pi - 3 - \sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_2=\frac{1}{4}(-2 - \pi - \sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_3=\frac{1}{4}(1 + \pi e^{-j\frac{3\pi}{2}} - 3e^{-j3\pi} + \sqrt[e]{\frac{\pi^3}{\ln(5)}}e^{-j\frac{9\pi}{2}})\, $
$ \,a_3=\frac{1}{4}(1 + j\pi + 3 - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
$ \,a_3=\frac{1}{4}(4 + j\pi - j\sqrt[e]{\frac{\pi^3}{\ln(5)}})\, $
