Nyquist Sampling Rate

  • #1
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I have tried calculating sampling rates for signals like sinc(200pi t). It was simple and I thought I understood until I cam across signals like sinc(200πt)*sinc2(400πt).

I need help with finding sampling rate of these kind of signals and get a clear concept on the topic.

Thanks in advance.
 

Answers and Replies

  • #2
You need to sample twice as fast as the highest frequency you expect to see. You also have multiplication of signals which makes their bandwidths add. So, in this case, I think you have to sample at ##2\left( 200\pi + 400\pi + 400\pi\right) = 2000\pi ##rad/s, or 1000 Hz.
 
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  • #3
Here is a problem set with solutions that may help.

http://www.ee.columbia.edu/~mvp/e3801/hwk8_sols.pdf [Broken]
 
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  • #4
You need to sample twice as fast as the highest frequency you expect to see. You also have multiplication of signals which makes their bandwidths add. So, in this case, I think you have to sample at ##2\left( 200\pi + 400\pi + 400\pi\right) = 2000\pi ##rad/s, or 1000 Hz.
But the adding of frequencies didnt worked for 10cos3 (2π102t). For this signal the sampling frequency is 400 hz
 
  • #5
But the adding of frequencies didnt worked for 10cos3 (2π102t). For this signal the sampling frequency is 400 hz
Who is more likely to be wrong - Nyquist or the guy who wrote the answer to the set problem? I guess that is what you are quoting, here. The answer in the book must be to another problem (quite possible). What answer did you get, btw?
There are examples when sub-nyquist sampling can work without impairment but this is not one of them.
 
  • #6
https://www.physicsforums.com/threads/do-not-post-homework-or-coursework-questions-here-in-the-ee-forum.224442/
 

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