m
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
[[Category:2010 Fall ECE 438 Boutin]]
 
 
----
 
----
−
== Solution to Q4 of Week 13 Quiz Pool ==
+
 
 +
== Solution to Q4 of Week 13 Quiz Pool ==
 +
 
 
----
 
----
  
−
a. y[m,n] = h[m,n] ** x[m,n]
+
a. y[m,n] = h[m,n] ** x[m,n]  
  
−
Using definition of convolution, <br/>
+
Using definition of convolution, <br> <math>\begin{align}
−
<math>
+
−
\begin{align}
+
 
y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\
 
y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\
−
\end{align}
+
\end{align}</math>  
−
</math>
+
  
−
Expanding, <br/>
+
Expanding, <br> y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]  
−
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]  
+
  
−
Sub values of h[m,n] from table, zero terms go away,<br/>
+
Sub values of h[m,n] from table, zero terms go away,<br>  
  
−
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,1] x[m+1,n-1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,1] x[m-1,n-1]
+
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,1] x[m+1,n-1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,1] x[m-1,n-1] y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
 +
 
 +
b. We can rewrite h[m,n] as
 +
 
 +
{| width="20%" cellspacing="2" cellpadding="2" border="1" class="wikitable" style="text-align: center;"
 +
|+ m
 +
|-
 +
! n
 +
! -1
 +
! 0
 +
! 1
 +
|-
 +
! -1
 +
| 0.5
 +
| 0
 +
| -0.5
 +
|-
 +
! 0
 +
| 0
 +
| 1
 +
| 0
 +
|-
 +
! 1
 +
| -0.5
 +
| 0
 +
| 0.5
 +
|}
 +
 
 +
We compute the output: <br> y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1] <br> by considering 3X3 portions of x[m,n], where the element at m,n corresponds to 0,0 in h[m,n], so we would look at neighboring elements (if they exist) and multiply with corresponding neighbors in h[m,n] and then sum them to form y[m,n].
 +
 
 +
Example - (Indexed starting from 0) y[3,3] = 0.5 x[4,4] - 0.5 x[4,2] + x[3,3] - 0.5 x[2,4] + 0.5 x[2,2] x[2,2] = 0 x[2,4] = 1 x[3,3] = 1 x[4,2] = 1 x[4,4] = 1 so y[3,3] = 0.5 - 0.5 + 1 - 0.5 + 0 = 0.5
 +
 
 +
Similarly calculating values sequentially, results in y[m,n] - <br>
 +
 
 +
{| width="50%" cellspacing="3" cellpadding="2" border="1"
 +
|-
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0
 +
| -0.5
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| -0.5
 +
| -0.5
 +
| 0
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
| 0
 +
| 0
 +
|-
 +
| 0
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1.5
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
| 0
 +
|-
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
 +
| 1.5
 +
| 1.5
 +
| -0.5
 +
| -0.5
 +
|-
 +
| 0.5
 +
| 1
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1.5
 +
| 1
 +
| -0.5
 +
|-
 +
| 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
 +
|-
 +
| -0.5
 +
| 0.5
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1
 +
| 1.5
 +
| 0.5
 +
|-
 +
| -0.5
 +
| -0.5
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0
 +
| 0.5
 +
| 0.5
 +
|}
 +
 
 +
c. From the difference equation -
 
y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
 
y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
  
−
b. under construction...
+
Taking Fourier Transform on both sides, <br/>
 +
<math>
 +
\begin{align}
 +
Y(\mu,\nu) &= \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{-j\nu} - \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{j\nu} + X(\mu,\nu) - \frac{1}{2}X(\mu,\nu) e^{j\mu}e^{-j\nu} + \frac{1}{2}X(\mu,\nu)e^{j\mu}e^{j\nu} \\
  
 +
\frac{Y(\mu,\nu)}{X(\mu,\nu)} &= \frac{1}{2}e^{-j\mu}e^{-j\nu} - \frac{1}{2}e^{-j\mu}e^{j\nu} + 1 - \frac{1}{2} e^{j\mu}e^{-j\nu} + \frac{1}{2}e^{j\mu}e^{j\nu} \\
  
 +
H(\mu,\nu) &= 1 + \frac{1}{2} ( e^{-j(\mu + \nu)} + e^{j(\mu + \nu)} ) - \frac{1}{2} ( e^{-j(\mu - \nu)} + e^{j(\mu - \nu)} ) \\
 +
 +
H(\mu,\nu) &= 1 + cos(\mu + \nu) - cos(\mu - \nu)
 +
 +
\end{align}
 +
</math>
 
----
 
----
−
Back to [[ECE438_Week13_Quiz|Lab Week 13 Quiz Pool]]
 
  
−
Back to [[ECE438_Lab_Fall_2010|ECE 438 Fall 2010 Lab Wiki Page]]
+
Back to [[ECE438 Week13 Quiz|Lab Week 13 Quiz Pool]]
 +
 
 +
Back to [[ECE438 Lab Fall 2010|ECE 438 Fall 2010 Lab Wiki Page]]
 +
 
 +
Back to [[2010 Fall ECE 438 Boutin|ECE 438 Fall 2010]]  
  
−
Back to [[2010_Fall_ECE_438_Boutin|ECE 438 Fall 2010]]
+
[[Category:2010_Fall_ECE_438_Boutin]]

Latest revision as of 18:54, 17 November 2010

Solution to Q4 of Week 13 Quiz Pool

a. y[m,n] = h[m,n] ** x[m,n]

Using definition of convolution,
$ \begin{align} y[m,n] &= \sum_{k=-1}^{1} \sum_{l=-1}^{1} h[k,l] x[m-k,n-l] \\ \end{align} $

Expanding,
y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,0] x[m+1,n] + h[-1,1] x[m+1,n-1] + h[0,-1] x[m,n+1] + h[0,0] x[m,n] + h[0,1] x[m,n-1] + h[1,-1] x[m-1,n+1] + h[1,0] x[m-1,n] + h[1,1] x[m-1,n-1]

Sub values of h[m,n] from table, zero terms go away,

y[m,n] = h[-1,-1] x[m+1,n+1] + h[-1,1] x[m+1,n-1] + h[0,0] x[m,n] + h[1,-1] x[m-1,n+1] + h[1,1] x[m-1,n-1] y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]

b. We can rewrite h[m,n] as

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

We compute the output:
y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]
by considering 3X3 portions of x[m,n], where the element at m,n corresponds to 0,0 in h[m,n], so we would look at neighboring elements (if they exist) and multiply with corresponding neighbors in h[m,n] and then sum them to form y[m,n].

Example - (Indexed starting from 0) y[3,3] = 0.5 x[4,4] - 0.5 x[4,2] + x[3,3] - 0.5 x[2,4] + 0.5 x[2,2] x[2,2] = 0 x[2,4] = 1 x[3,3] = 1 x[4,2] = 1 x[4,4] = 1 so y[3,3] = 0.5 - 0.5 + 1 - 0.5 + 0 = 0.5

Similarly calculating values sequentially, results in y[m,n] -

0 0 0 0 0.5 0 -0.5 0 0 0 0
0 0 0 0.5 0.5 1 -0.5 -0.5 0 0 0
0 0 0.5 0.5 0.5 1 1.5 -0.5 -0.5 0 0
0 0.5 0.5 0.5 0.5 1 1.5 1.5 -0.5 -0.5 0
0.5 0.5 0.5 0.5 1 1 1 1.5 1.5 -0.5 -0.5
0.5 1 0.5 1 1 1 1 1 1.5 1 -0.5
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
-0.5 0.5 1 1 1 1 1 1 1 1.5 0.5
-0.5 -0.5 0 0 0 0 0 0 0 0.5 0.5

c. From the difference equation - y[m,n] = 0.5 x[m+1,n+1] - 0.5 x[m+1,n-1] + x[m,n] - 0.5 x[m-1,n+1] + 0.5 x[m-1,n-1]

Taking Fourier Transform on both sides,
$ \begin{align} Y(\mu,\nu) &= \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{-j\nu} - \frac{1}{2}X(\mu,\nu)e^{-j\mu}e^{j\nu} + X(\mu,\nu) - \frac{1}{2}X(\mu,\nu) e^{j\mu}e^{-j\nu} + \frac{1}{2}X(\mu,\nu)e^{j\mu}e^{j\nu} \\ \frac{Y(\mu,\nu)}{X(\mu,\nu)} &= \frac{1}{2}e^{-j\mu}e^{-j\nu} - \frac{1}{2}e^{-j\mu}e^{j\nu} + 1 - \frac{1}{2} e^{j\mu}e^{-j\nu} + \frac{1}{2}e^{j\mu}e^{j\nu} \\ H(\mu,\nu) &= 1 + \frac{1}{2} ( e^{-j(\mu + \nu)} + e^{j(\mu + \nu)} ) - \frac{1}{2} ( e^{-j(\mu - \nu)} + e^{j(\mu - \nu)} ) \\ H(\mu,\nu) &= 1 + cos(\mu + \nu) - cos(\mu - \nu) \end{align} $

Back to Lab Week 13 Quiz Pool

Back to ECE 438 Fall 2010 Lab Wiki Page

Back to ECE 438 Fall 2010

Retrieved from "https://www.projectrhea.org/rhea/index.php?title=ECE438_Week13_Quiz_Q4sol&oldid=42093"
Shortcuts
Help Main Wiki Page Random Page Special Pages Log in
Search

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010