(4 intermediate revisions by the same user not shown)
Line 5: Line 5:
 
<math>y = cos(n)\,</math>
 
<math>y = cos(n)\,</math>
 
<br>
 
<br>
When I took a sample at 1 hz, the signal was non-periodic and ugly.
+
When I took a sample at 1 hz, the signal was ugly and random.
 
<br>
 
<br>
 
Therefore it is not periodic.
 
Therefore it is not periodic.
Line 15: Line 15:
 
[[Image:Periodiccnphilli_ECE301Fall2008mboutin.JPG]]
 
[[Image:Periodiccnphilli_ECE301Fall2008mboutin.JPG]]
 
<br>
 
<br>
 +
 +
== Part A-2 ==
 +
<br>
 +
I chose the signal y=log(x)
 +
<br>
 +
This code is not periodic, but taking a piece of it and<br>
 +
Repeating it will make it periodic<br>
 +
[[Image:LogA-2_ECE301Fall2008mboutin.JPG]]
 +
<pre>
 +
Here is the code I used:
 +
 +
dt = .00001;
 +
periodlength = 5;
 +
periods = 10;
 +
 +
t = [dt:dt:periodlength];
 +
y = log(t);
 +
x = [];
 +
for i=1:periods
 +
  x = [x,y];
 +
end
 +
t = [dt:dt:periodlength*periods];
 +
plot(t,x)
 +
</pre>

Latest revision as of 14:54, 11 September 2008

Homework 2 Part A-1


I chose the signal:
$ y = cos(n)\, $
When I took a sample at 1 hz, the signal was ugly and random.
Therefore it is not periodic.

File:Non-periodic.jpg ECE301Fall2008mboutin
Changing the function to:$ y = cos(n*pi/2)\, $
made the signal look periodic.
Periodiccnphilli ECE301Fall2008mboutin.JPG

Part A-2


I chose the signal y=log(x)
This code is not periodic, but taking a piece of it and
Repeating it will make it periodic
LogA-2 ECE301Fall2008mboutin.JPG

Here is the code I used:

dt = .00001;
periodlength = 5;
periods = 10;

t = [dt:dt:periodlength];
y = log(t);
x = [];
for i=1:periods
   x = [x,y];
end
t = [dt:dt:periodlength*periods];
plot(t,x)

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang