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b) Consider the sequence <math>x[n]</math> of length 9 below,
 
b) Consider the sequence <math>x[n]</math> of length 9 below,
:<math>x[n]=\text{cos}(\frac{1}{3}\pi n)(u[n]-u[n-8])\,\!</math>
+
:<math>x[n]=\text{cos}\left(\frac{1}{3}\pi n\right)(u[n]-u[n-8])\,\!</math>
 
<math>y_9[n]</math> is formed by computing <math>X_9[k]</math> as an 9-pt DFT of <math>x[n]</math>, <math>H_9[k]</math> as an 9-pt DFT of <math>h[n]</math>, and then <math>y_9[n]</math> as the 9-pt inverse DFT of <math>Y_9[k] = X_9[k]H_9[k]</math>.  
 
<math>y_9[n]</math> is formed by computing <math>X_9[k]</math> as an 9-pt DFT of <math>x[n]</math>, <math>H_9[k]</math> as an 9-pt DFT of <math>h[n]</math>, and then <math>y_9[n]</math> as the 9-pt inverse DFT of <math>Y_9[k] = X_9[k]H_9[k]</math>.  
  

Revision as of 01:45, 9 November 2010


Quiz Questions Pool for Week 12


Q1. Consider a causal FIR filter of length M = 2 with impulse response

$ h[n]=\delta[n]+\delta[n-1]\,\! $

a) Provide a closed-form expression for the 9-pt DFT of $ h[n] $, denoted $ H_9[k] $, as a function of $ k $. Simplify as much as possible.

b) Consider the sequence $ x[n] $ of length 9 below,

$ x[n]=\text{cos}\left(\frac{1}{3}\pi n\right)(u[n]-u[n-8])\,\! $

$ y_9[n] $ is formed by computing $ X_9[k] $ as an 9-pt DFT of $ x[n] $, $ H_9[k] $ as an 9-pt DFT of $ h[n] $, and then $ y_9[n] $ as the 9-pt inverse DFT of $ Y_9[k] = X_9[k]H_9[k] $.

Express the result $ y_9[n] $ as a weighted sum of finite-length sinewaves similar to how $ x[n] $ is written above.


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Ryne Rayburn