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Can someone tell me if the way I solved this problem is right?

Suppose I have a signal $X_a(t)= 0.5cos(700πt)+0.6cos(720πt)+0.1cos(780πt)$ where $F_s=8000Hz$

I want to determine the minimum samples which must be available in order the frequency resolution of DFT be ok.

I did this:

  • Firstly calculate each frequency of the signal: $F_1=350Hz$, $F_2=360Hz$, $F_3=390Hz$
  • Calculate the differences: $F_2$-$F_1$=$10Hz$ ,$F_3$-$F_2$= $30Hz$ , $F_3$-$F_1$= $40Hz$
  • $ΔF_{min}$=$10Hz$
  • $ΔF$ = $ \frac{f_s}{N} $, so $N\geq\frac{8000Hz}{10}$=800 samples

Can someone tell me if my solution is right and explain why we choose the minimum difference

Thanks

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    $\begingroup$ It could be argued that you need 5 Hz resolution rather than 10 Hz, otherwise F1, F2 will appear in adjacent bins and will be indistinguishable from a single peak that straddles both bins. $\endgroup$ – Paul R Jan 27 '15 at 22:07
  • $\begingroup$ @PaulR I don't understand my answer is right? $\endgroup$ – Aris Chrisak Jan 27 '15 at 22:21
  • $\begingroup$ It depends on what you mean by "right". ;-) I would say that your peaks in the frequency domain need to be at least two bins apart, in order for them to be distinct, in which case you would need 5 Hz resolution, not 10 Hz. $\endgroup$ – Paul R Jan 27 '15 at 22:43
  • $\begingroup$ @jojek I have seen you 've experience in this topic for similar threads which you have give answers, could you help me If my thought is wrong/right? $\endgroup$ – Aris Chrisak Jan 28 '15 at 14:41
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It's usually considered difficult to resolve a pair of adjacent frequencies unless there is dip between the two FFT peaks. Depending on the window (rectangular or otherwise) and how much of a dip you require between the two peaks to consider them resolved (above some floor), and whether or what kind of result interpolation you use, you will need from a bit less than double to triple or more of your suggested 800 FFT result bins.

Of course in a-priori zero noise, your signal of 3 pure sinusoids has 9 free parameters, so you might be able to determine your 3 input frequencies with perhaps as few as 9 or a bit more non-aliased samples. In any case, much less than 800.

The resolution of a single isolated frequency peak above a low enough noise floor to the nearest 10Hz can likely be done using far fewer than 800 samples by using interpolation methods on the FFT result, depending on the SNR.

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  • $\begingroup$ Thanks for your answer, so if I want the minimum samples which must be available in order the frequency resolution of DFT is sufficient is my answer right? I have found this formula in a signal processing book, in book is used the minimum diference but don't explain why.. that is I don't understand which frequency must put in denominator $\endgroup$ – Aris Chrisak Jan 28 '15 at 9:33

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