ASN10 - Quadrature sampling demonstration: Difference between revisions
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[[Jodi Hodge|Back to my home page]] |
[[Jodi Hodge|Back to my home page]] |
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Assignment was actually done |
Assignment was actually done in class by professor. |
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[http://www.example.com link title] |
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<br><b>Problem Statement</b><br><br> |
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Plot <math> \ \frac{2}{T} \sum_{n=1}^\infty sin\bigg(\frac{2 \pi nt}{T}\bigg) \!</math><br><br> |
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<b>Solution</b><br> |
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While we can't sum to infinity in the computer, we can get a close approximation summing over a large enough range of <math> n \!</math><br> |
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I found summing over <math> n = 1:1000 \!</math> was about the most the computer could handle reasonably.<br> |
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The following script was written in MATLAB to produce the desired plot. <br> |
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<pre> |
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clear all; |
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close all; |
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sum = 0; |
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T = 1; |
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t = -T:0.0001:T; |
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N = 1000; |
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for n = 1:N; |
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if n==0 |
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h = 0; |
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else |
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h = 2/T; |
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end |
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sum = sum+h*sin(2*pi*n*t/T); |
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end |
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plot(t,sum) |
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title('Quadrature Sampling Waveform') |
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xlabel('time(T)') |
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ylabel('Sampling Waveform') |
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</pre><br> |
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Running the MATLAB script above gives us the following plot.<br> |
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[[Image:Quadrature sampling.jpg]]<br> |
Revision as of 11:10, 3 December 2009
Assignment was actually done in class by professor.
link title
Problem Statement
Plot
Solution
While we can't sum to infinity in the computer, we can get a close approximation summing over a large enough range of
I found summing over was about the most the computer could handle reasonably.
The following script was written in MATLAB to produce the desired plot.
clear all; close all; sum = 0; T = 1; t = -T:0.0001:T; N = 1000; for n = 1:N; if n==0 h = 0; else h = 2/T; end sum = sum+h*sin(2*pi*n*t/T); end plot(t,sum) title('Quadrature Sampling Waveform') xlabel('time(T)') ylabel('Sampling Waveform')
Running the MATLAB script above gives us the following plot.