Line 66: Line 66:
 
b. Find the output image that results when this filter is applied to the input image shown below:<br/>
 
b. Find the output image that results when this filter is applied to the input image shown below:<br/>
  
0 0 0 0 0 0 0 0 0 0 0 <br/>
+
{| cellspacing="1" cellpadding="1" border="0" width="20%"
0 0 0 0 0 1 0 0 0 0 0 <br/>
+
|-
0 0 0 0 1 1 1 0 0 0 0 <br/>
+
| 0  
0 0 0 1 1 1 1 1 0 0 0 <br/>
+
| 0  
0 0 1 1 1 1 1 1 1 0 0 <br/>
+
| 0  
0 1 1 1 1 1 1 1 1 1 0 <br/>
+
| 0  
0 1 1 1 1 1 1 1 1 1 0 <br/>
+
| 0  
0 1 1 1 1 1 1 1 1 1 0 <br/>
+
| 0  
0 1 1 1 1 1 1 1 1 1 0 <br/>
+
| 0  
0 1 1 1 1 1 1 1 1 1 0 <br/>
+
| 0  
0 0 0 0 0 0 0 0 0 0 0 <br/>
+
| 0  
 +
| 0  
 +
| 0
 +
|-
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0
 +
| 1  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0
 +
|-
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0
 +
|-
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0  
 +
| 0  
 +
| 0
 +
|-
 +
| 0  
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0  
 +
| 0
 +
|-
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0
 +
|-
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0
 +
|-
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0
 +
|-
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0
 +
|-
 +
| 0  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 1  
 +
| 0
 +
|-
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0  
 +
| 0
 +
|}
  
 
c. Find a simple expression for the frequency response H(<math>\mu ,\nu</math>) of this filter.<br/>
 
c. Find a simple expression for the frequency response H(<math>\mu ,\nu</math>) of this filter.<br/>

Revision as of 11:05, 17 November 2010


Quiz Questions Pool for Week 13

  • Under construction --Zhao

Q1. Show that the DTFT of time-reversal, $ x[-n]\,\! $, is $ X(-\omega)\,\! $


Q2. Consider the discrete-time signal

$ x[n]=\delta[n]+5 \delta[n-1]+\delta[n-1]- \delta[n-2]. $

a) Determine the DTFT $ X(\omega) $ of x[n] and the DTFT of $ Y(\omega) $ of y[n]=x[-n].

b) Using your result from part a), compute

$ x[n]* y[n] $.

c) Consider the discrete-time signal

$ z[n]=\left\{ \begin{array}{ll}x[(-n)\mod 4],& 0\leq n < 3,\\ 0 & \text{else }\end{array} \right. $.

Obtain the 4-point circular convolution of x[n] and z[n].

d) When computing the N-point circular convolution of x[n] and the signal

$ z[n]=\left\{ \begin{array}{ll}x[(-n)\mod N],& 0\leq n < N-1,\\ 0 & \text{else }\end{array} \right. $.

how should N be chosen to make sure that the result is the same as the usual convolution between x[n] and z[n]?

  • Same as HW8 Q3 available here.

Q3. Consider the discrete-time signal

$ x[n]=\delta[n] $

a) Obtain the N-point DFT X[k] of x[n].

b) Obtain the signal y[n] whose DFT is $ (W_N^{k}+W_N^{2k}+W_N^{3k}) X[k] $.

c) Now fix $ N=5 $. Compute 5-point circular convolution between $ y[n] $ and the signal

$ h[n]=\delta[n]+2\delta[n-1]+3\delta[n-2]. $

Q4. Consider a 3X3 FIR filter with coefficients h[m,n]

m
n -1 0 1
1 -0.5 0 0.5
0 0 1 0
-1 0.5 0 -0.5

a. Find a difference equation that can be used to implement this filter.
b. Find the output image that results when this filter is applied to the input image shown below:

0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 1 1 1 0 0 0 0
0 0 0 1 1 1 1 1 0 0 0
0 0 1 1 1 1 1 1 1 0 0
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
0 1 1 1 1 1 1 1 1 1 0
0 0 0 0 0 0 0 0 0 0 0

c. Find a simple expression for the frequency response H($ \mu ,\nu $) of this filter.


Q5.


Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva