I'm analysing a large hyrophone data set in MATLAB using signal processing toolbox and I find when i do the PSD or Third octave band frequency levels, they levels in dB are offset by more than 30 or 40 dB. Can someone suggest what are the major or primary parameters that one needs to check for this. I did enter all ref pressure of water, system end to end senstivity and all correctly. If there is something I'm missing, I would like to know. please find below part of code
xgrid = buffer(xbit,N,ceil(N*r),'nodelay').'; %buffer function breaks data into segments or frames as set by user
%grid whose rows are each (overlapped)
% segment for analysis
if xgrid(length(xgrid(:,1)),N) == 0 %remove final segment if not full
xgrid = xgrid(1:length(xgrid(:,1))-1,:);
end
M = length(xgrid(:,1)); %total number of data segments
xgrid = xgrid.*repmat(w/alpha,M,1);%multiply segments by Hann window function
%%computing DFT
X = abs(fft(xgrid.')).'; %DFT of each segment
%%computing power spectrum which is square of the amplitude
P = (X./N).^2;
%%Compute single-sided power spectrum
Pss = 2*P(:,2:floor(N/2)+1); %
f = floor(fs/2)*linspace(1/(N/2),1,N/2);
%calculate frequencies of DFT bins
flow = find(single(f) >= lcut,1,'first'); %low-frequency cut-off
fhigh = find(single(f) <= hcut,1,'last'); %high-frequency cut-off
f = f(flow:fhigh); %frequency bins in user-defined range
nf = length(f); %number of frequency bins
%% Compute noise power bandwidth and delta(f)
B = (1/N).*(sum((w/alpha).^2)); %noise power bandwidth
delf = fs/N;
%% Converting to dB
%for PSD calculation
APSD = 10*log10((1/(delf*B))*Pss(:,flow:fhigh)./(pref^2))-S;
Thanks,
Related
I have the below code which is adapted from other code I found here. It runs fine and the plot looks as expected except the dB levels are about 80dB too low. Note, the intended unit is dB SPL. I would appreciate any input on fixing this, I am thinking it has something to do with needing to include the 1uPa reference but I can't track back the issue and I'm just a poor biologist trying to learn to code!
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% FFT parameters
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
NFFT = 96000; %
NOVERLAP = round(0.75*NFFT);
w = hanning(NFFT);
% spectrogram dB scale
spectrogram_dB_scale = 100; % dB range scale (means , the lowest displayed level is XX dB below the max level)
gain = 20
sensitivity = - 165
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% load signal
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[data,Fs]=audioread('sample.wav'); %(newer matlab)
samples = length(data);
dt = 1/Fs;
t = (0:dt:(samples-1)*dt);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% display 1 : averaged FFT spectrum
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
[sensor_spectrum, freq] = pwelch(data,w,NOVERLAP,NFFT,Fs);
% convert to dB scale (ref = 1)
sensor_spectrum_dB = 20*log10(sensor_spectrum) - sensitivity - gain ;
figure(1),semilogx(freq,sensor_spectrum_dB);grid
title(['Averaged FFT Spectrum / Fs = ' num2str(Fs) ' Hz / Delta f = ' num2str(freq(2)-freq(1)) ' Hz ']);
xlabel('Frequency (Hz)');ylabel('Amplitude (dB (L))');
Regardless, you want to plot the output of pwelch using 10*log10 since you're dealing with a power spectrum.
Without knowing a little more about your data it's going to be hard to help... is it sound data? physiological? what is the sensitivity unit? What is gain ??
Scaling the PSD
I created a 6-bit quantizer and passed a signal through it, but when I plot the DFT, it peaks at 200 MHz and then stops; I'm not seeing the whole spectrum. What's preventing me in my code from getting the rest of the points at the higher frequencies?
Here is my code:
bits = 6; %6-bit
fs = 400e6; %sampling frequency
amp = 1; %amplitude
f = 200e6; %actual frequency
vpp = 2; peak-to-peak voltage
LSB = vpp/(2^bits); %least-significant bit
cycles = 1000;
duration = cycles/f;
values = 0:1/fs:duration;
party = LSB:LSB:(vpp-LSB); %partition
blocker = 0:1:(2^bits - 1); %codebook
biblocker = fliplr(decimaltobinary(blocker)); %I created a function that converts decimal to binary
qtone = amp + amp*sin(2*pi*f*values); %tone
[index, q] = quantization(qtone,party,blocker); %I created a quantizing function
ftq = fft(q)/length(q); % Fourier Transform (Scaled)
qf = linspace(0, 1, fix(length(q))/2+1)*(fs/2); % Frequency Vector
qi = 1:length(qf); % Index Vector
qa = abs(ftq(qi))*2/.7562;
figure
plot(qf/1e6, qa) % One-Sided Amplitude Plot
xlim([100 500]);
xlabel('Frequency [MHz]')
ylabel('Amplitude')
Here is what I get:
Since you selected your sampling frequency, fs = 400e6 i.e. 400 MHz, you can only observe the spectrum up to 200e6, half of the sampling frequency. You can read the theory behind it using Nyquist Sampling Theorem.
As a solution you need to set your twice the frequency you want to observe on the spectrum. It is impossible to observe whole frequency, you need to set a finite frequency limit.
For base-band sampled data, everything in the spectrum above half the sample rate is redundant, just aliases for the spectrum below half the sample rate. So there's no need to display the same spectrum repeated.
When sampling for a finite length of time (less than the age of the Earth, etc.), you have to sample at a rate higher than twice the highest frequency in the signal. 2X (400Msps for a 200MHz signal) often won't work.
qf = linspace(0, 1, fix(length(q)))*(fs);
I'm trying to find the peak frequency for two signals 'CA1' and 'PFC', within a specified range (25-140Hz).
In Matlab, so far I have plotted an FFT for each of these signals (see pictures below). These FFTs suggest that the peak frequency between 25-140Hz is different for each signal, but I would like to quantify this (e.g. CA1 peaks at 80Hz, whereas PFC peaks at 55Hz). However, I think the FFT is not smooth enough, so when I try and extract the peak frequencies it doesn't make sense as my code pulls out loads of values. I was only expecting a few values - one each time the FFT peaks (around 2Hz, 5Hz and ~60Hz).
I want to know, between 25-140Hz, what is the peak frequency in 'CA1' compared with 'PFC'. 'CA1' and 'PFC' are both 152401 x 7 matrices of EEG data, recorded
from 7 separate individuals. I want the MEAN peak frequency for each data set (i.e. averaged across the 7 test subjects for CA1 and PFC).
My code so far (based on Matlab help files and code I've scrabbled together online):
Fs = 508;
%notch filter
[b50,a50] = iirnotch(50/(Fs/2), (50/(Fs/2))/70);
CA1 = filtfilt(b50,a50,CA1);
PFC = filtfilt(b50,a50,PFC);
%FFT
L = length(CA1);
NFFT = 2^nextpow2(L);
%FFT for each of the 7 subjects
for i = 1:size(CA1,2);
CA1_FFT(:,i) = fft(CA1(:,i),NFFT)/L;
PFC_FFT(:,i) = fft(PFC(:,i),NFFT)/L;
end
%Average FFT across all 7 subjects - CA1
Mean_CA1_FFT = mean(CA1_FFT,2);
% Mean_CA1_FFT_abs = 2*abs(Mean_CA1_FFT(1:NFFT/2+1));
%Average FFT across all 7 subjects - PFC
Mean_PFC_FFT = mean(PFC_FFT,2);
% Mean_PFC_FFT_abs = 2*abs(Mean_PFC_FFT(1:NFFT/2+1));
f = Fs/2*linspace(0,1,NFFT/2+1);
%LEFT HAND SIDE FIGURE
plot(f,2*abs(Mean_CA1_FFT(1:NFFT/2+1)),'r');
set(gca,'ylim', [0 2]);
set(gca,'xlim', [0 200]);
[C,cInd] = sort(2*abs(Mean_CA1_FFT(1:NFFT/2+1)));
CFloor = 0.1; %CFloor is the minimum amplitude value (ignore small values)
Amplitudes_CA1 = C(C>=CFloor); %find all amplitudes above the CFloor
Frequencies_CA1 = f(cInd(1+end-numel(Amplitudes_CA1):end)); %frequency of the peaks
%RIGHT HAND SIDE FIGURE
figure;plot(f,2*abs(Mean_PFC_FFT(1:NFFT/2+1)),'r');
set(gca,'ylim', [0 2]);
set(gca,'xlim', [0 200]);
[P,pInd] = sort(2*abs(Mean_PFC_FFT(1:NFFT/2+1)));
PFloor = 0.1; %PFloor is the minimum amplitude value (ignore small values)
Amplitudes_PFC = P(P>=PFloor); %find all amplitudes above the PFloor
Frequencies_PFC = f(pInd(1+end-numel(Amplitudes_PFC):end)); %frequency of the peaks
Please help!! How do I calculate the 'major' peak frequencies from an FFT, and ignore all the 'minor' peaks (because the FFT is not smoothed).
FFTs assume that the signal has no trend (this is called a stationary signal), if it does then this will give a dominant frequency component at 0Hz as you have here. Try using the MATLAB function detrend, you may find this solves your problem.
Something along the lines of:
x = x - mean(x)
y = detrend(x, 'constant')
I am trying to compare the FFT of exp(-t^2) to the function's analytical fourier transform, exp(-(w^2)/4)/sqrt(2), over the frequency range -3 to 3.
I have written the following matlab code and have iterated on it MANY times now with no success.
fs = 100; %sampling frequency
dt = 1/fs;
t = 0:dt:10-dt; %time vector
L = length(t); %number of sample points
%N = 2^nextpow2(L); %necessary?
y = exp(-(t.^2));
Y=dt*ifftshift(abs(fft(y)));
freq = (-L/2:L/2-1)*fs/L; %freq vector
F = (exp(-(freq.^2)/4))/sqrt(2); %analytical solution
%Y_valid_pts = Y(W>=-3 & W<=3); %compare for freq = -3 to 3
%npts = length(Y_valid_pts);
% w = linspace(-3,3,npts);
% Fe = (exp(-(w.^2)/4))/sqrt(2);
error = norm(Y - F) %L2 Norm for error
hold on;
plot(freq,Y,'r');
plot(freq,F,'b');
xlabel('Frequency, w');
legend('numerical','analytic');
hold off;
You can see that right now, I am simply trying to get the two plots to look similar. Eventually, I would like to find a way to do two things:
1) find the minimum sampling rate,
2) find the minimum number of samples,
to reach an error (defined as the L2 norm of the difference between the two solutions) of 10^-4.
I feel that this is pretty simple, but I can't seem to even get the two graphs visually agree.
If someone could let me know where I'm going wrong and how I can tackle the two points above (minimum sampling frequency and minimum number of samples) I would be very appreciative.
Thanks
A first thing to note is that the Fourier transform pair for the function exp(-t^2) over the +/- infinity range, as can be derived from tables of Fourier transforms is actually:
Finally, as you are generating the function exp(-t^2), you are limiting the range of t to positive values (instead of taking the whole +/- infinity range).
For the relationship to hold, you would thus have to generate exp(-t^2) with something such as:
t = 0:dt:10-dt; %time vector
t = t - 0.5*max(t); %center around t=0
y = exp(-(t.^2));
Then, the variable w represents angular frequency in radians which is related to the normalized frequency freq through:
w = 2*pi*freq;
Thus,
F = (exp(-((2*pi*freq).^2)/4))*sqrt(pi); %analytical solution
I am very interested to know how the top right figure in :http://en.wikipedia.org/wiki/Spectrogram
is generated (the script) and how to analyse it i.e what information does it convey?I would appreciate a simplified answer with minimum mathematical jargons. Thank you.
The plot shows time along the horizontal axis, and frequency along the vertical axis. With pixel color showing the intensity of each frequency at each time.
A spectrogram is generated by taking a signal and chopping it into small time segments, doing a Fourier series on each segment.
here is some matlab code to generate one.
Notice how plotting the signal directly, it looks like garbage, but plotting the spectrogram, we can clearly see the frequencies of the component signals.
%%%%%%%%
%% setup
%%%%%%%%
%signal length in seconds
signalLength = 60+10*randn();
%100Hz sampling rate
sampleRate = 100;
dt = 1/sampleRate;
%total number of samples, and all time tags
Nsamples = round(sampleRate*signalLength);
time = linspace(0,signalLength,Nsamples);
%%%%%%%%%%%%%%%%%%%%%
%create a test signal
%%%%%%%%%%%%%%%%%%%%%
%function for converting from time to frequency in this test signal
F1 = #(T)0+40*T/signalLength; #frequency increasing with time
M1 = #(T)1-T/signalLength; #amplitude decreasing with time
F2 = #(T)20+10*sin(2*pi()*T/signalLength); #oscilating frequenct over time
M2 = #(T)1/2; #constant low amplitude
%Signal frequency as a function of time
signal1Frequency = F1(time);
signal1Mag = M1(time);
signal2Frequency = F2(time);
signal2Mag = M2(time);
%integrate frequency to get angle
signal1Angle = 2*pi()*dt*cumsum(signal1Frequency);
signal2Angle = 2*pi()*dt*cumsum(signal2Frequency);
%sin of the angle to get the signal value
signal = signal1Mag.*sin(signal1Angle+randn()) + signal2Mag.*sin(signal2Angle+randn());
figure();
plot(time,signal)
%%%%%%%%%%%%%%%%%%%%%%%
%processing starts here
%%%%%%%%%%%%%%%%%%%%%%%
frequencyResolution = 1
%time resolution, binWidth, is inversly proportional to frequency resolution
binWidth = 1/frequencyResolution;
%number of resulting samples per bin
binSize = sampleRate*binWidth;
%number of bins
Nbins = ceil(Nsamples/binSize);
%pad the data with zeros so that it fills Nbins
signal(Nbins*binSize+1)=0;
signal(end) = [];
%reshape the data to binSize by Nbins
signal = reshape(signal,[binSize,Nbins]);
%calculate the fourier transform
fourierResult = fft(signal);
%convert the cos+j*sin, encoded in the complex numbers into magnitude.^2
mags= fourierResult.*conj(fourierResult);
binTimes = linspace(0,signalLength,Nbins);
frequencies = (0:frequencyResolution:binSize*frequencyResolution);
frequencies = frequencies(1:end-1);
%the upper frequencies are just aliasing, you can ignore them in this example.
slice = frequencies<max(frequencies)/2;
%plot the spectrogram
figure();
pcolor(binTimes,frequencies(slice),mags(slice,:));
The inverse Fourier transform of the fourierResult matrix, will return the original signal.
Just to add to Suki's answer, here is a great tutorial that walks you through, step by step, reading Matlab spectrograms, touching on only enough math and physics to explain the main concepts intuitively:
http://www.caam.rice.edu/~yad1/data/EEG_Rice/Literature/Spectrograms.pdf