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* [[ECE438_Week11_Quiz_Q1sol|Solution]]. | * [[ECE438_Week11_Quiz_Q1sol|Solution]]. | ||
---- | ---- | ||
| − | Q2. | + | Q2. Consider a causal FIR filter of length M = 2 with impulse response |
| + | :<math>h[n]=\delta[n]-\delta[n-1]\,\!</math> | ||
| + | a) Provide a closed-form expression for the 8-pt DFT of <math>h[n]</math>, denoted <math>H_8[k]</math>, as a function of <math>k</math>. Simplify as much as possible. | ||
| + | |||
| + | b) Consider the sequence <math>x[n]</math> of length 8 below, equal to a sum of several finite-length sinewaves. | ||
| + | :<math>x[n]=\left[\text{cos}\left(\frac{\pi}{2}n\right)+2\text{cos}(\pi n)\right](u[n]-u[n-8])</math> | ||
| + | <math>y_8[n]</math> is formed by computing <math>X_8[k]</math> as an 8-pt DFT of <math>x[n]</math>, <math>H_8[k]</math> as an 8-pt DFT of <math>h[n]</math>, and then <math>y_8[n]</math> as the 8-pt inverse DFT of <math>Y_8[k] = X_8[k]H_8[k]</math>. | ||
| + | |||
| + | Express the result <math>y_8[n]</math> as a weighted sum of finite-length sinewaves similar to how <math>x[n]</math> is written | ||
| + | above. | ||
* [[ECE438_Week11_Quiz_Q2sol|Solution]]. | * [[ECE438_Week11_Quiz_Q2sol|Solution]]. | ||
Revision as of 09:49, 2 November 2010
- Under construction --Zhao
Quiz Questions Pool for Week 11
Q1. Consider the two LTI systems, $ y[n]=T_1[x[n]] $ and $ y[n]=T_2[x[n]] $, with the following difference equations,
- $ y[n]=T_1[x[n]]=x[n]-x[n-1]\,\! $
- $ y[n]=T_2[x[n]]=\frac{1}{2}y[n-1]+x[n]\,\! $
Then, calculate the impulse response and difference equation of the combined system $ (T_1+T_2)[x[n]] $.
Q2. 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 8-pt DFT of $ h[n] $, denoted $ H_8[k] $, as a function of $ k $. Simplify as much as possible.
b) Consider the sequence $ x[n] $ of length 8 below, equal to a sum of several finite-length sinewaves.
- $ x[n]=\left[\text{cos}\left(\frac{\pi}{2}n\right)+2\text{cos}(\pi n)\right](u[n]-u[n-8]) $
$ y_8[n] $ is formed by computing $ X_8[k] $ as an 8-pt DFT of $ x[n] $, $ H_8[k] $ as an 8-pt DFT of $ h[n] $, and then $ y_8[n] $ as the 8-pt inverse DFT of $ Y_8[k] = X_8[k]H_8[k] $.
Express the result $ y_8[n] $ as a weighted sum of finite-length sinewaves similar to how $ x[n] $ is written above.
Q3.
Q4.
Q5.
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