(Part 2)
(Part 2)
 
(One intermediate revision by the same user not shown)
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This is <math>x(t)=e^{-t}cos(20t)</math> as the nonperiodic signal.
 
This is <math>x(t)=e^{-t}cos(20t)</math> as the nonperiodic signal.
 
[[image:ctsignal3.jpg|300px|frame|center]]
 
[[image:ctsignal3.jpg|300px|frame|center]]
 +
 +
 +
I haven't played around with matlab enough to know how to create the signal needed with my own code, so I used the same code that Ben Laskowski used:
 +
 +
----
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"y=zeros(size(x));
 +
 +
for count=1:period/delta:(nperiod+8)*period/delta
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 +
    y(1,count:size(y,2))=y(1,count:size(y,2))+f(1,1:size(y,2)-count+1);
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 +
end
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 +
plot(x,y)"
 +
 +
----
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 +
[[image:ctsignal4.jpg|300px|frame|center]]

Latest revision as of 14:14, 12 September 2008

Part 1

Choose the signal $ x(t)=cos(t) $ which is periodic.

Sampling ever interger yields

This is $ x[n]=cos(n) $

but sampling at a frequency of $ \pi/4 $ yields

This is $ x[n]=cos(\pi/4n) $

Part 2

This is $ x(t)=e^{-t}cos(20t) $ as the nonperiodic signal.


I haven't played around with matlab enough to know how to create the signal needed with my own code, so I used the same code that Ben Laskowski used:


"y=zeros(size(x));

for count=1:period/delta:(nperiod+8)*period/delta

   y(1,count:size(y,2))=y(1,count:size(y,2))+f(1,1:size(y,2)-count+1);

end

plot(x,y)"


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