Calculating Minimum Sampling Rate for Fourier Signal

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The discussion focuses on calculating the minimum sampling rate for a Fourier complex signal defined by three sine components. According to the Nyquist Theorem, the sampling rate must be at least twice the highest frequency present in the signal. The highest frequency component identified is 750 Hz, leading to a calculated Nyquist sampling rate of 1500 samples per second. Participants confirm the correctness of this calculation. The conclusion affirms the importance of adhering to the Nyquist criteria for accurate signal representation.
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.
 
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Looks right! :smile:
 

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