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<math>h[n] = \delta[n-5] + \delta[n-3]\,</math>
 
<math>h[n] = \delta[n-5] + \delta[n-3]\,</math>
  
Then we find the frequency response:<br><br>
+
Then we find the frequency response:
 +
 
 
<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math>
 
<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\,</math>
  
<math>F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} \,</math>
+
find m value to make the value inside the bracket zero
 +
 
 +
m = -5 for the first set and 3 for the second set
 +
 
 +
<math>F(z) = e^{-5j\omega} + e^{3j\omega} \,</math>
 +
 
 +
 
  
 
=Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal=
 
=Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal=

Revision as of 08:45, 26 September 2008

Obtain the input impulse response h[n] and the system function H(z) of your system

Defining a DT LTI: $ y[n] = x[n+5] + x[n-3]\, $
So, we have the unit impulse response: $ h[n] = \delta[n-5] + \delta[n-3]\, $

Then we find the frequency response:

$ F(z) = \sum^{\infty}_{m=-\infty} h[m+5]e^{jm\omega} + h[m-3]e^{jm\omega}\, $

find m value to make the value inside the bracket zero

m = -5 for the first set and 3 for the second set

$ F(z) = e^{-5j\omega} + e^{3j\omega} \, $


Compute the response of your system to the signal you defined in Question 1 using H(z) and the Fourier series coefficients of your signal

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang