Calculating Minimum Sampling Rate for Fourier Signal

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SUMMARY

The minimum sampling rate for the Fourier complex signal v(t) = 2/πsin(500πt) + 1/2sin(1000πt) + 1/3sin(1500πt) is determined using the Nyquist Theorem. The highest frequency component is 750 Hz, derived from the third harmonic (1500πt). Therefore, the required Nyquist sampling rate is 1500 samples per second, which is twice the highest frequency component. This conclusion confirms the correct application of the Nyquist Theorem in calculating the sampling rate.

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fran1942
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Hello, I have the following Fourier complex signal:
v(t) = 2/πsin(500πt)+1/2sin(1000πt)+1/3sin(1500πt)

I need to find the minimum sampling rate for this signal.
So, according to the Nyquist Theorem, the sampling rate must be twice the highest frequency component contained in the original signal.

So I assume my sampling rate would be twice the third harmonic (highest frequency component).

Now, because the format of a sine wave is expressed as 'sin(2πft)', I assume each component within brackets represents '2πft'. So for instance:
(500πt) = 2πft, so f = 250
(1000πt) = 2πft, so f = 500
(1500πt) = 2πft, so f = 750

Therefore, my Nyquist sampling rate would need to be twice the highest frequency component which would mean a sampling rate of 2 * 750 = 1500 samples per second ?

Any confirmation of whether I am correct here or not would be great.

Thanks kindly.
 
Last edited:
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Looks right! :smile:
 

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