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* [[ECE438_Week12_Quiz_Q1sol|Solution]]. | * [[ECE438_Week12_Quiz_Q1sol|Solution]]. | ||
---- | ---- | ||
| − | Q2. | + | Q2. Consider the discrete-time signal |
| + | |||
| + | <math>x[n]=6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5].</math> | ||
| + | |||
| + | a) Obtain the 6-point DFT X[k] of x[n]. | ||
| + | |||
| + | b) Obtain the signal y[n] whose DFT is <math>W_6^{-2k} X[k]</math>. | ||
| + | |||
| + | c) Compute six-point circular convolution between x[n] and the signal | ||
| + | |||
| + | <math>h[n]=\delta[n]+\delta[n-1]+\delta[n-2].</math> | ||
* [[ECE438_Week12_Quiz_Q2sol|Solution]]. | * [[ECE438_Week12_Quiz_Q2sol|Solution]]. | ||
Revision as of 11:27, 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-1]+\delta[n-2]\,\! $
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{2\pi}{3}n\right)(u[n]-u[n-9])\,\! $
$ 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.
Q2. Consider the discrete-time signal
$ x[n]=6\delta[n]+5 \delta[n-1]+4 \delta[n-2]+3 \delta[n-3]+2 \delta[n-4]+\delta[n-5]. $
a) Obtain the 6-point DFT X[k] of x[n].
b) Obtain the signal y[n] whose DFT is $ W_6^{-2k} X[k] $.
c) Compute six-point circular convolution between x[n] and the signal
$ h[n]=\delta[n]+\delta[n-1]+\delta[n-2]. $
Q3.
Q4.
Q5.
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