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== The System == | == The System == | ||
<math>y[n] = x[n] + x[n-1] + x[n-2] + x[n-3]\,</math> | <math>y[n] = x[n] + x[n-1] + x[n-2] + x[n-3]\,</math> | ||
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<math>H[z] = 1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}\,</math> | <math>H[z] = 1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}\,</math> | ||
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| + | == Response of y[n] to the signal I defined in Question 2 using H[z] and the Fourier series coefficients == | ||
| + | From Question 2: <math>x[n] = 7sin(7\pi n + \frac{\pi}{8})\,</math> where <math>a_k = 7 , k = 1,3,5,7,9,11....\,</math> | ||
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| + | <math>x[n] = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi n}\,</math> | ||
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| + | <math>y[t] = \sum^{\infty}_{k = -\infty} a_k H[z] e^{jk\pi n}\,</math> | ||
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| + | <math>y[t] = \sum^{\infty}_{k = -\infty} a_k (1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}) e^{jk\pi n}\,</math> | ||
Latest revision as of 13:11, 25 September 2008
Contents
The System
$ y[n] = x[n] + x[n-1] + x[n-2] + x[n-3]\, $
Unit Impulse Response
$ x[n] = \delta[n]\, $
$ h[n] = \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3]\, $
Frequency Response
$ y[n] = \sum^{\infty}_{\infty} h[n] * x[n] dn\, $ where $ x[n] = e^{jwn} \, $
$ y[n] = \sum^{\infty}_{-\infty} (\delta[n] + \delta[n-1]+ \delta[n-2] + \delta[n-3]) e^{jwn} \, $
$ y[n] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{jw(n-m)} \, $
$ y[n] = e^{jwn} \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \, $
$ H[z] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \, $
$ H[z] = e^{-jw0} + e^{-jw1}+ e^{-jw2}+ e^{-jw3}\, $
$ H[z] = 1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}\, $
Response of y[n] to the signal I defined in Question 2 using H[z] and the Fourier series coefficients
From Question 2: $ x[n] = 7sin(7\pi n + \frac{\pi}{8})\, $ where $ a_k = 7 , k = 1,3,5,7,9,11....\, $
$ x[n] = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi n}\, $
$ y[t] = \sum^{\infty}_{k = -\infty} a_k H[z] e^{jk\pi n}\, $
$ y[t] = \sum^{\infty}_{k = -\infty} a_k (1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}) e^{jk\pi n}\, $
