DFT detemine frequencies

Question

I am just begining learning about DFT and I am bit unsure on what is happening mathematically during DFT.

I've sampled a signal , a sine wave sin(1000*2*pi*t). And performed DFT to calculate the Ck-coefficients, by plotting the Ck coefficients with different k values.

The frequency resolution is 8000/205 where 205 is the amount of datapoint/samples and 8000 sample frequency.

I can by using the plot read, where |Ck| is highest and then using my frequency resolution, detemine the frequency.

I see that the amount of samples have alot to say about determining the frequency. If i use a lower amount of samples, or does not use all the samples, will the estimated frequency either be lower, or higher. Why does is do that?

ex.

k := Round[1/((8000/100)/1000)]
freqres := 8000/100 // N
estimated freq = 468.293

k := Round[1/((8000/205)/1000)]
freqres := 8000/205// N
estimated freq = 992

k := Round[1/((8000/500)/1000)]
freqres := 8000/500 
estimated freq = 992

Answer 1

The FFT get a discrete spectrum, as a result do you have a finite number of frequency or bins, this means that FFT resolves a time waveform into its sinusoidal components, the frequency resolution is explicitly linked to the FFT size and the sample rate, the resolution frequency is:

$$ \frac{Fs}{ N}$$

Where Fs is your sample rate and N is the size of your FFT, The larger the size of the FFT more spectral component you have and it makes its results more accurate, this tells me that to get lower frequency resolution you need a large size FFT (2048, 4096).

Follow your example 8000/205 this tells me that for each FFT bin returned you may be losing precision in order 39.02439024390244Hz

Answer 2

If you are estimating the location of a spectral frequency peak, then, depending on the S/N ratio, interpolation of the DFT result may provide a finer frequency resolution of a peak than just the DFT bin spacing. (Parabolic, spline, or Sinc kernel interpolation)

In general, the more samples fed to DFTs, the more "information" there is to use for spectral estimations. Thus, for stationary signals, greater resolution usually comes with greater lengths of data.