(New page: == Computing the Fourier series coefficients for a Discrete Time signal x[n] == === The System === <math>y[n] = x[n] + x[n-1] + x[n-2] + x[n-3]\,</math> === Unit Impulse Response === <m...)
 
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== Computing the Fourier series coefficients for a Discrete Time signal x[n] ==
 
== Computing the Fourier series coefficients for a Discrete Time signal x[n] ==
  
=== The System ===
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== 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>
  
 
+
== Unit Impulse Response ==
=== Unit Impulse Response ===
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<math>x[n] = \delta[n]\,</math>
 
<math>x[n] = \delta[n]\,</math>
  
 
<math>h[n] = \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3]\,</math>
 
<math>h[n] = \delta[n] + \delta[n-1] + \delta[n-2] + \delta[n-3]\,</math>
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== Frequency Response ==
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<math>y[n] = \sum^{\infty}_{\infty} h[n] * x[n] dn\,</math> where <math>x[n] = e^{jwn} \,</math>
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<math>y[n] = \sum^{\infty}_{-\infty} (\delta[n] + \delta[n-1]+ \delta[n-2] + \delta[n-3]) e^{jwn} \,</math>
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<math>y[n] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{jw(n-m)} \,</math>
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<math>y[n] = e^{jwn} \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \,</math>
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<math>H[z] = \sum^{\infty}_{-\infty} (\delta[m] + \delta[m-1]+ \delta[m-2] + \delta[m-3]) e^{-jwm} \,</math>
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<math>H[z] = e^{-jw0} + e^{-jw1}+ e^{-jw2}+ e^{-jw3}\,</math>
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<math>H[z] = 1 + e^{-jw1}+ e^{-jw2} + e^{-jw3}\,</math>

Revision as of 13:01, 25 September 2008

Computing the Fourier series coefficients for a Discrete Time signal x[n]

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}\, $

Alumni Liaison

Recent Math PhD now doing a post-doctorate at UC Riverside.

Kuei-Nuan Lin