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[[Category:2010 Fall ECE 438 Boutin]]
 
[[Category:2010 Fall ECE 438 Boutin]]
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[[Category:Problem_solving]]
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[[Category:ECE438]]
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[[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-1]\,\!</math>
+
:<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]].
 
----
 
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Q2.  
+
Q2. Consider the discrete-time signal
  
* [[ECE438_Week12_Quiz_Q2sol|Solution]].
+
:<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
  
* [[ECE438_Week12_Quiz_Q3sol|Solution]].
+
<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]].
 
----
 
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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 10:43, 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.


Back to ECE 438 Fall 2010 Lab Wiki Page

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