MATLAB Answers

# How is pspectrum() different from FFT() magnitude plotted in dB?

21 views (last 30 days)
Jay Vaidya on 20 Dec 2020
Commented: Mathieu NOE on 21 Dec 2020
I am having difficulty in finding a high-resolution frequency spectrum even though I have a pretty large number of time domain data samples (1999999 data points). The fundamental frequency peak in FFT is with a resolution of 50 Hz when I use fft(). I had also put this post here with the screenshots, code, and data.
I am thinking of using pspectrum() or zoomFFT(). The latter seems to be more relevant to me. But after creating the object of dsp.zoomFFT(), I am not sure how to proceed because z = zfft(data(:,2)) gives me error of
Fs = 40e3;
CF = 4e3;
BW = 1e3;
D = Fs/BW;
fftlen = 64;
L = D * fftlen;
zfft = dsp.ZoomFFT(D,CF,Fs,'FFTLength',fftlen);
z = zfft(data(:,2));
The number of rows in the input signal must be a multiple of 48 (the decimation factor).
Error in plot_fft_v8 (line 124)
Also, I am thnking that probably pspectrum() can help because
pspectrum(xTable,'FrequencyResolution',10)
Error using pspectrum>validateFrequencyResolution (line 1005)
'FrequencyResolution' value is not achievable for the specified parameters. 'FrequencyResolution' value must
be within [8.1702e-07, 1.634] (*pi radians/sample).
Error in pspectrum>parseAndValidateInputs (line 811)
validateFrequencyResolution(opts.FrequencyResolution, opts.Leakage,...
Error in pspectrum (line 246)
opts = parseAndValidateInputs(x,varargin);
can set my frequency resolution. ##### 0 CommentsShowHide -1 older comments

Sign in to comment.

### Answers (1)

Mathieu NOE on 20 Dec 2020
hello
if your data lenght is very long, there is no reason why you could not getter better resolution with a standard fft
remember that freq resolution df = Fs/NFFT
so if like in this demo I choose NFFT = Fs = 40e3, my resolution is 1 Hz, still you could even go beyong as you have more than 1 second of data (nb of samples is > Fs)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% dummy data
Fs = 40e3;
samples = 1999999;
t = (0:samples-1)'*1/Fs;
signal = cos(2*pi*50*t)+cos(2*pi*100*t)+1e-3*rand(samples,1);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 40e3; %
Overlap = 0.75;
w = hanning(NFFT); % Hanning window / Use the HANN function to get a Hanning window which has the first and last zero-weighted samples.
%% notch filter section %%%%%%
% H(s) = (s^2 + 1) / (s^2 + s/Q + 1)
fc = 50; % notch freq
wc = 2*pi*fc;
Q = 5; % adjust Q factor for wider (low Q) / narrower (high Q) notch
% at f = fc the filter has gain = 0
w0 = 2*pi*fc/Fs;
alpha = sin(w0)/(2*Q);
b0 = 1;
b1 = -2*cos(w0);
b2 = 1;
a0 = 1 + alpha;
a1 = -2*cos(w0);
a2 = 1 - alpha;
% analog notch (for info)
num1=[1/wc^2 0 1];
den1=[1/wc^2 1/(wc*Q) 1];
% digital notch (for info)
num1z=[b0 b1 b2];
den1z=[a0 a1 a2];
freq = linspace(fc-1,fc+1,200);
[g1,p1] = bode(num1,den1,2*pi*freq);
g1db = 20*log10(g1);
[gd1,pd1] = dbode(num1z,den1z,1/Fs,2*pi*freq);
gd1db = 20*log10(gd1);
figure(1);
plot(freq,g1db,'b',freq,gd1db,'+r');
title(' Notch: H(s) = (s^2 + 1) / (s^2 + s/Q + 1)');
legend('analog','digital');
xlabel('Fréquence (Hz)');
ylabel(' dB')
% now let's filter the signal
signal_filtered = filtfilt(num1z,den1z,signal);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display : averaged FFT spectrum before / after notch filter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[freq,fft_spectrum] = myfft_peak(signal, Fs, NFFT, Overlap);
sensor_spectrum_dB = 20*log10(fft_spectrum);% convert to dB scale (ref = 1)
[freq,fft_spectrum_filtered] = myfft_peak(signal_filtered, Fs, NFFT, Overlap);
sensor_spectrum_filtered_dB = 20*log10(fft_spectrum_filtered);% convert to dB scale (ref = 1)
figure(2),semilogx(freq,sensor_spectrum_dB,'b',freq,sensor_spectrum_filtered_dB,'r');grid
title(['Averaged FFT Spectrum / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
legend('before notch filter','after notch filter');
xlabel('Frequency (Hz)');ylabel(' dB')
function [freq_vector,fft_spectrum] = myfft_peak(signal, Fs, nfft, Overlap)
%FFT peak spectrum of signal (example sinus amplitude 1 = 0 dB after fft).
% signal - input signal,
% Fs - Sampling frequency (Hz).
% nfft - FFT window size
% Overlap - buffer overlap % (between 0 and 0.95)
samples = length(signal);
% fill signal with zeros if its length is lower than nfft
if samples<nfft
s_tmp = zeros(nfft,1);
s_tmp((1:samples)) = signal;
signal = s_tmp;
end
% window : hanning
window = hanning(nfft);
window = window(:);
% compute fft with overlap
offset = fix((1-Overlap)*nfft);
spectnum = 1+ fix((samples-nfft)/offset); % Number of windows
% % for info is equivalent to :
% noverlap = Overlap*nfft;
% spectnum = fix((samples-noverlap)/(nfft-noverlap)); % Number of windows
% main loop
fft_spectrum = 0;
for i=1:spectnum
start = (i-1)*offset;
sw = signal((1+start):(start+nfft)).*window;
fft_spectrum = fft_spectrum + (abs(fft(sw))*4/nfft); % X=fft(x.*hanning(N))*4/N; % hanning only
end
fft_spectrum = fft_spectrum/spectnum; % to do linear averaging scaling
% one sidded fft spectrum % Select first half
if rem(nfft,2) % nfft odd
select = (1:(nfft+1)/2)';
else
select = (1:nfft/2+1)';
end
fft_spectrum = fft_spectrum(select);
freq_vector = (select - 1)*Fs/nfft;
freq_vector = freq_vector(:);
fft_spectrum = fft_spectrum(:);
end
##### 2 CommentsShowHide 1 older comment
Mathieu NOE on 21 Dec 2020
hello
from the attached picture, it seems that you are looking for a peak in the 4 kHz
your sampling at 500MS/s is overkill , regarding sampling (Shannon), 20 kS/s would suffice
that should allow you to increase the acquisition time by the same amount , so you could quite significantly improve your frequency resolution without having a huge file size;

Sign in to comment.

R2020b

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!