Does the number of samples matter for FFT, and how to get a specific frequency visible?

this is John BG [email protected]

1.- Let be the following example

N1=740; fs=1e3;dt=1/fs % time reference t=[0:dt:N1*dt]; A=1 % signal f0=26 s0=A*cos(2*pi*f0*t); n0=wgn(N1+1,1,-36,'complex')'; % -6dBmW noise n0=wgn(N1+1,1,-16,'complex')'; % 24dBm noise s=s0+real(n0); figure(1);plot(t,s);grid on 

Nfft=2^10; S=fft(s,Nfft) figure(2); Sabs=abs(S); Sang=angle(S); subplot(2,1,1);plot(Sabs);grid on;title('|S|') subplot(2,1,2);plot(Sang);grid on;title('angle(S)') 

Regarding your 2nd quetion

How do I get it appear properly in the frequency spectrum?

the way to translate bin position in the FFT to Hz is as follows:

f0_measure=28/(Nffs/2)*fs/2 = 27.3437 

the peak on the FFT bin 28 is equivalent to frequency 27.3437 Hz.

The added noise is complex, so it's going to affect both phase and amplitude.

The error on the frequency measurement is

abs(f0-f0_measure) = 1.343750000000000 

2.- Get MATLAB to find the peaks, instead of manually placing the marker on the peak and get the bin # 28:

nmax=find(Sabs==max(Sabs)) nmax=nmax(1) % remove the FFT mirror 

replace 28 with nmax in the above expression translating FFT bin to frequency.

3.- Now your 1st part of the 1st question

the frequency resolution does not depend on the number of samples I have, only FFT length and sampling frequency?

if one increases the amount of FFT bins for the same amount of input time samples, yes the FFT looks the same, but that doesn't mean that the frequency measurement is correct.

Let be

f0=.75*fs 

Now there's alias, the sampling frequency is not fast enough and one doesn't notice sampling below the Nyquist limit, then this happens, repeating the above

now

f0_measure=257/(1024/2)*fs/2 f0_measure = 2.509765625000000e+02 

the peak should be somewhere around 26Hz, yet you see 250.9Hz

And the 2nd part of the 1st question

If I had 1480 samples, the resolution would be still the same?

No, if you increase the amount of time samples, by this understanding that for the same signal, not a longer observation window, but more samples within the same observation window, thus higher sampling frequency fs, then if you also increase the amount of FFT bins, then you improve the frequency resolution, see the following example.

Let be a 2 tone example, one at 26Hz, the frequency of interest you mentioned, and another one at 28Hz, half amplitude of the 1st one:

N1=740; fs=1e3;dt=1/fs % time reference t=[0:dt:N1*dt]; A=1 f01=26 f02=28 s01=A*cos(2*pi*f01*t); s02=A*cos(2*pi*f02*t+pi/2); n0=wgn(N1+1,1,-16,'complex')'; % 24dBm s=s01+s02+real(n0); figure(1);plot(t,s);grid on 

repeating the above FFT lines,

Nfft=2^10; S=fft(s,Nfft); figure(2); Sabs=abs(S); Sang=angle(S); subplot(2,1,1);plot(Sabs);grid on;title('|S|') subplot(2,1,2);plot(Sang);grid on;title('angle(S)') 

one can tell there's not enough frequency resolution

But important, observe that merely increasing the amount of bins, from 2^10 to 2^20

Nfft=2^20; S=fft(s,Nfft); figure(2); Sabs=abs(S); Sang=angle(S); subplot(2,1,1);plot(Sabs);grid on;title('|S|') subplot(2,1,2);plot(Sang);grid on;title('angle(S)') 

It DOES improve somehow the frequency resolution, without additionally increasing the amount of input time samples:

It's still not clear whether there are 2 or more tones, but it's clear in comparison to the 2^10 bins FFT that there's something it has to be refined, that would have been ignored with a lower amount of FFT bins, with the same amount of input time samples.

The closer to MATLAB default upper limit of matrix size,

 Nfft=2^26; .. Nfft=2^28; .. 

Above certain amount of FFT bins, there's no significant improvement, following an overlap of FFTs with same amount of input time samples

 for k=1:1:5 N2=N1*2^(k-1) t=[0:dt:N2*dt]; s01=A*cos(2*pi*f01*t); s02=A*cos(2*pi*f02*t+pi/2); n0=wgn(N2+1,1,-16,'complex')'; % 24dBm s=s01+s02+real(n0); Nfft=2^10; S=fft(s,Nfft) Sabs=abs(S); Sang=angle(S); figure(5);plot(Sabs);grid on;title('|S|') hold all end axis( 1.0e+02 *[ 0.041474654377880 0.649769585253456 0 4.440962099125365]) 

and the larger the amount of FFT samples to calculate, the longer it takes to get each FFT.

The result is still slightly better than with a lower amount of FFT bins, yet the amplitude of the 2 tones now resolved, is still not accurate, because we know that they do not have equal amplitude, yet it shows as if they were.

So, YES, increasing the amount of FFT bins for same amount of time input signals, may to a certain extend improve the frequency resolution.

Conclusion:

But the best thing a do, is to increase both at the same time: amount of time samples and amount of FFT bins:

for k=1:1:5 N2=N1*2^(k-1) t=[0:dt:N2*dt]; s01=A*cos(2*pi*f01*t); s02=A*cos(2*pi*f02*t+pi/2); n0=wgn(N2+1,1,-16,'complex')'; % 24dBm s=s01+s02+real(n0); Nfft=2^(10+k-1); S=fft(s,Nfft) Sabs=abs(S); Sang=angle(S); figure(6);plot(Sabs);grid on;title('|S|') hold all end axis=( 1.0e+03 *[ 0.271758436944937 0.815275310834813 0 5.952380952380953]) 

To the question originator, if you find this answer useful would you please be so kind to consider marking my answer as Accepted Answer?

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thanks in advance for time and attention

John BG

[email protected]

The "information content" is related to the number of samples. The length of the FFT merely interpolates the spectral frequency curve represented by the number of samples. If an FFT result bin doesn't end up exactly centered on some desired frequency, you can interpolate it from the nearest bins, using a high-quality (Sinc kernel, et.al.) interpolator, since FFT bins have a greater-than-zero bandwidth (the shape of the transform of the default rectangular or other window). Using a longer zero-padded FFT gives you more interpolation points, and thus a denser looking point plot, but no better spectral separation resolution than provided by the number of actual samples.

Regarding your first question, your frequency granularity is simply $\frac{f_s}{N}$, where $f_s$ is your sampling frequency in Hz, and $N$ is the length of your FFT. (Remember that you always want to take an FFT length of at least equal to your data record length).

Next, suppose that the signal I have has an interesting frequency at 26 Hz (strictly that one frequency). How do I get it appear properly in the frequency spectrum (my frequency bins are spaced 1000/1024 = 0.9766 Hz apart, so am I at dead end?).

You can simply zero-pad your signal, and this will have the effect of smoothing out your fourier transform estimate. It wont give you better resolution in the engineering sense, but it will give you better "visual resolution" so to speak, so that you can better ascertain a peak.