How Long to Sample for Accurate 60Hz FFT Resolution?

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To accurately resolve a 60Hz wave using FFT, the sampling duration must be sufficiently long to ensure that the frequency resolution is adequate, which is determined by the relationship 1/T, where T is the sampling duration. A higher sampling rate above the Nyquist frequency is necessary, but the number of samples taken is crucial for resolving frequencies accurately. Sampling for a short duration can introduce artifacts and blurring in the frequency domain, complicating the representation of the signal. The frequency resolution is also affected by how the signal is distributed among frequency bins, which can lead to inaccuracies if not properly managed. Ultimately, a careful balance between sampling duration and rate is essential for effective frequency analysis.
dmorris619
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I am wondering what the minimum sampling duration must be in order to accurately resolve a 60Hz wave. I know the FFT works with periodic waves about the number N but I am not sure how this relates to say a sampling duration that is 1 half the sampling period of a 60 Hz wave. I'm worried that the duration will be so short that it cannot accurately resolve the wave. Also the nyquist rate will be 1MSPS so it far above the nyquist frequency.

Any guidance would be great.
 
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i'm pretty sure you need to sample at a frequency of over two times the original frequency to prevent "aliasing"
 
Im not talking aliasing or sampling rate I am talking duration of sampling or number of samples. Our sampling rate is high enough above the nyquist frequency I am just concerned we don't have enough samples to resolve a 60hz wave using an fft given the periodic nature of of the fft.
 
Presumably your sample window is equal to or shorter than 1/sample-rate, no? Then you can't squeeze any more information out of a signal at the Nyquist frequency. Also you can't resolve phase information below 180deg at the Nyquist frequency.

If your sample-rate is some-many-multiples of 60Hz you will be able to get a more accurate representation of a 60Hz input signal, including harmonics and phase.

But that may not be what you are asking about?
 
I think you'll have to come up with an objective measure of what you mean by 'accurately resolve'.

The number of frequency 'bins' that come out of the discrete Fourier transform is equal to the number of samples you take. So your frequency resolution will be Fs/N, where Fs is your sampling frequency and N is the number of samples you take. However, if you replace N with Fs*T, with T representing the length of time over which you sample the signal, you can see that your frequency resolution is actually 1/T (independent of the sampling frequency).

When I say frequency resolution, I mean the spacing between the frequency 'bins' in Fourier transformed signal. So, its not very simple to say how long you need to sample to acquire your signal, in order to 'accurately' resolve its frequency. In general, your signal won't fall neatly into one bin and so it will be distributed among multiple bins, which may or may not be a problem for you.

To complicate matters, the fact that you are taking a finite chunk of the signal means that you'll be introducing artifacts into the spectrum. An abrupt stop in sampling is equivalent to multiplying the 'true' signal by a rectangle function. That means you're convolving the spectrum with a sinc function (i.e. blurring it) in the frequency domain. The shorter your sampling time, the narrower the rectangle function will be, and correspondingly, the wider the sinc function will be in the frequency domain. So you can imagine if your signal is a perfect 60Hz sine wave, and you sample it over a very short period, you'll end up with a sinc function (as opposed to a delta) in the frequency domain, whose width gets larger as you lower the sampling time.
 
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