- #1

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^{2}(400πt).

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

Thanks in advance.

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- #1

- 2

- 0

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

Thanks in advance.

- #2

- 35

- 1

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

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Here is a problem set with solutions that may help.

http://www.ee.columbia.edu/~mvp/e3801/hwk8_sols.pdf [Broken]

http://www.ee.columbia.edu/~mvp/e3801/hwk8_sols.pdf [Broken]

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- #4

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But the adding of frequencies didnt worked for 10cos

- #5

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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?But the adding of frequencies didnt worked for 10cos^{3}(2π10^{2}t). For this signal the sampling frequency is 400 hz

There are examples when sub-nyquist sampling can work without impairment but this is not one of them.

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