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[[Category:2010 Fall ECE 438 Boutin]] | [[Category:2010 Fall ECE 438 Boutin]] | ||
| + | [[Category:Problem_solving]] | ||
| + | [[Category:ECE438]] | ||
| + | [[Category:digital signal processing]] | ||
== Quiz Questions Pool for Week 12 == | == Quiz Questions Pool for Week 12 == | ||
---- | ---- | ||
Q1. Consider a causal FIR filter of length M = 2 with impulse response | Q1. Consider a causal FIR filter of length M = 2 with impulse response | ||
| − | :<math>h[n]=\delta[n]+\delta[n- | + | :<math>h[n]=\delta[n-1]+\delta[n-2]\,\!</math> |
a) Provide a closed-form expression for the 9-pt DFT of <math>h[n]</math>, denoted <math>H_9[k]</math>, as a function of <math>k</math>. Simplify as much as possible. | a) Provide a closed-form expression for the 9-pt DFT of <math>h[n]</math>, denoted <math>H_9[k]</math>, as a function of <math>k</math>. Simplify as much as possible. | ||
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}\left(\frac{\pi}{3}n\right)(u[n]-u[n-9])\,\!</math> | + | :<math>x[n]=\text{cos}\left(\frac{2\pi}{3}n\right)(u[n]-u[n-9])\,\!</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>. | ||
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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]\text{ ,where} \;\; W_N=e^{-j\frac{2\pi}{N}}</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> | ||
| + | |||
| + | * Same as HW8, Q2 available [[ECE438_HW8_Solution|here]]. | ||
---- | ---- | ||
| − | Q3. | + | Q3. Consider the signal |
| − | * [[ | + | <math>x[n] = \begin{cases} |
| + | cos(\pi n / 8), & n < 0 \\ | ||
| + | cos(\pi n / 3), & \mbox{else} | ||
| + | \end{cases}</math> | ||
| + | |||
| + | and assume a rectangular window | ||
| + | |||
| + | <math>w[n] = \begin{cases} | ||
| + | 1, & |n| < 25 \\ | ||
| + | 0, & \mbox{else} | ||
| + | \end{cases}</math> | ||
| + | |||
| + | The STDFT is defined as | ||
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | X(\omega,n) &= \sum_{k} x[k]w[n-k]e^{-j\omega k} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | Compute the STDTFT for the following cases: <br/> | ||
| + | i. n < -25 <br/> | ||
| + | ii. n > 25 <br/> | ||
| + | iii. n = 0 <br/> | ||
| + | |||
| + | * [[Media:Qpw12ece438fa10.pdf|Solution]]. | ||
---- | ---- | ||
| − | Q4. | + | Q4. Consider the STDTFT defined as |
| + | |||
| + | <math>X(\omega ,n)=\sum_k x[k]w[n-k]e^{-j\omega k}</math> | ||
| + | |||
| + | where x[n] is the speech signal and w[n] is the window sequence. Prove the following properties: | ||
| + | |||
| + | a. Linearity – if <math>v[n]=ax[n]+by[n]</math> ,then <math>V(\omega ,n)=aX(\omega ,n)+bY(\omega, n)</math>. | ||
| + | |||
| + | b. Modulation – if <math>v[n]=x[n]e^{j\omega_0n}</math> ,then <math>V(\omega ,n)=X(\omega -\omega_0,n)</math>. | ||
* [[ECE438_Week12_Quiz_Q4sol|Solution]]. | * [[ECE438_Week12_Quiz_Q4sol|Solution]]. | ||
---- | ---- | ||
| − | Q5. | + | Q5. Suppose we have two 4-pt sequences x[n] and h[n] described as follows: |
| + | |||
| + | <math> | ||
| + | \begin{align} | ||
| + | x[n] &= cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3} \\ | ||
| + | h[n] &= 2^n\text{ ,n=0,1,2,3} | ||
| + | \end{align} | ||
| + | </math> | ||
| + | |||
| + | a. Compute 4-pt DFT X[k]; | ||
| + | |||
| + | b. Compute 4-pt DFT H[k]; | ||
| + | |||
| + | c. Compute 4-pt circular convolution directly of <math>y[n]=x[n]\circledast_4 h[n]</math>;(You may use plot to explain your answer) | ||
| + | |||
| + | d. Multiply DFT result of x[n] and h[n]. Then using IDFT to compute y[n] in question c. | ||
* [[ECE438_Week12_Quiz_Q5sol|Solution]]. | * [[ECE438_Week12_Quiz_Q5sol|Solution]]. | ||
Latest revision as of 09:43, 11 November 2011
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]\text{ ,where} \;\; W_N=e^{-j\frac{2\pi}{N}} $.
c) Compute six-point circular convolution between x[n] and the signal
- $ h[n]=\delta[n]+\delta[n-1]+\delta[n-2]. $
- Same as HW8, Q2 available here.
Q3. Consider the signal
$ x[n] = \begin{cases} cos(\pi n / 8), & n < 0 \\ cos(\pi n / 3), & \mbox{else} \end{cases} $
and assume a rectangular window
$ w[n] = \begin{cases} 1, & |n| < 25 \\ 0, & \mbox{else} \end{cases} $
The STDFT is defined as
$ \begin{align} X(\omega,n) &= \sum_{k} x[k]w[n-k]e^{-j\omega k} \end{align} $
Compute the STDTFT for the following cases:
i. n < -25
ii. n > 25
iii. n = 0
Q4. Consider the STDTFT defined as
$ X(\omega ,n)=\sum_k x[k]w[n-k]e^{-j\omega k} $
where x[n] is the speech signal and w[n] is the window sequence. Prove the following properties:
a. Linearity – if $ v[n]=ax[n]+by[n] $ ,then $ V(\omega ,n)=aX(\omega ,n)+bY(\omega, n) $.
b. Modulation – if $ v[n]=x[n]e^{j\omega_0n} $ ,then $ V(\omega ,n)=X(\omega -\omega_0,n) $.
Q5. Suppose we have two 4-pt sequences x[n] and h[n] described as follows:
$ \begin{align} x[n] &= cos(\frac{\pi n}{2})\text{ ,n=0,1,2,3} \\ h[n] &= 2^n\text{ ,n=0,1,2,3} \end{align} $
a. Compute 4-pt DFT X[k];
b. Compute 4-pt DFT H[k];
c. Compute 4-pt circular convolution directly of $ y[n]=x[n]\circledast_4 h[n] $;(You may use plot to explain your answer)
d. Multiply DFT result of x[n] and h[n]. Then using IDFT to compute y[n] in question c.
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