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
Related
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 want to find the natural frequencies and its damping ratio's of a test sample that is excited by use of an impact hammer and the response is measured by an accelerometer. So far I have got the Frequency Response Function (the Accelerance) by fourier transforming the input and output data. The Accelerance transfer function has a logarithmic scale. Now I got stuck what to do next. How do I obtain the natural frequencies and its damping ratio? The code so far:
N = 125000; %number of datapoints
fs = 12500; %sampling frequency
ts = 1/fs;
%set the x-axis of graph
x_max = (N-1)*ts;
x_axis = 0:ts:x_max;
%data
Time = data(1,:);
Force = data(2,:);
Acceleration = data(3,:);
Input = abs(fft(Force)); %FFT input force
Output = abs(fft(Acceleration)); %FFT output acceleration
%Accelerance transfer function
Accelerance = Output./Input;
Accelerance_dB = 20*log10(Accelerance); %Accelerance with logarithmic scale
I hope someone can help me!
I have a series of 2D measurements (time on x-axis) that plot to a non-smooth (but pretty good) sawtooth wave. In an ideal world the data points would form a perfect sawtooth wave (with partial amplitude data points at either end). Is there a way of calculating the (average) period of the wave, using OCTAVE/MATLAB? I tried using the formula for a sawtooth from Wikipedia (Sawtooth_wave):
P = mean(time.*pi./acot(tan(y./4))), -pi < y < +pi
also tried:
P = mean(abs(time.*pi./acot(tan(y./4))))
but it didn't work, or at least it gave me an answer I know is out.
An example of the plotted data:
I've also tried the following method - should work - but it's NOT giving me what I know is close to the right answer. Probably something simple and wrong with my code. What?
slopes = diff(y)./diff(x); % form vector of slopes for each two adjacent points
for n = 1:length(diff(y)) % delete slope of any two points that form the 'cliff'
if abs(diff(y(n,1))) > pi
slopes(n,:) = [];
end
end
P = median((2*pi)./slopes); % Amplitude is 2*pi
Old post, but thought I'd offer my two-cent's worth. I think there are two reasonable ways to do this:
Perform a Fourier transform and calculate the fundamental
Do a curve-fitting of the phase, period, amplitude, and offset to an ideal square-wave.
Given curve-fitting will likely be difficult because of discontinuities in saw-wave, so I'd recommend Fourier transform. Self-contained example below:
f_s = 10; # Sampling freq. in Hz
record_length = 1000; # length of recording in sec.
% Create noisy saw-tooth wave, with known period and phase
saw_period = 50;
saw_phase = 10;
t = (1/f_s):(1/f_s):record_length;
saw_function = #(t) mod((t-saw_phase)*(2*pi/saw_period), 2*pi) - pi;
noise_lvl = 2.0;
saw_wave = saw_function(t) + noise_lvl*randn(size(t));
num_tsteps = length(t);
% Plot time-series data
figure();
plot(t, saw_wave, '*r', t, saw_function(t));
xlabel('Time [s]');
ylabel('Measurement');
legend('measurements', 'ideal');
% Perform fast-Fourier transform (and plot it)
dft = fft(saw_wave);
freq = 0:(f_s/length(saw_wave)):(f_s/2);
dft = dft(1:(length(saw_wave)/2+1));
figure();
plot(freq, abs(dft));
xlabel('Freqency [Hz]');
ylabel('FFT of Measurement');
% Estimate fundamental frequency:
[~, idx] = max(abs(dft));
peak_f = abs(freq(idx));
peak_period = 1/peak_f;
disp(strcat('Estimated period [s]: ', num2str(peak_period)))
Which outputs a couple of graphs, and also the estimated period of the saw-tooth wave. You can play around with the amount of noise and see that it correctly gets a period of 50 seconds till very high levels of noise.
Estimated period [s]: 50
I know I can change frequency by whole numbers by changing the variable shift but how can I change the frequency using numbers with decimal places like .754 or 1.2345 or 67.456. If I change the variable 'shift' to a non-whole like number like 5.1 I get an error subscript indices must be either positive integers less than 2^31 or logicals from line mag2s = [mag2(shift+1:end), zeros(1,shift)];
Example Code below from question increase / decrease the frequency of a signal using fft and ifft in matlab / octave works with changing the variable shift (but it only works with whole numbers, I need it to work with decimals numbers also).
PS: I'm using octave 3.8.1 which is like matlab and I know I could change the frequency by adjusting the formula in the variable ya but ya will be a signal taken from an audio source (human speech) so it won't be an equation. The equation is just used to keep the example simple. And yes Fs is large due to the fact that signal files used are around 45 seconds long which is why I can't use resample because I get a out of memory error when used.
Here's a animated youtube video example of what I'm trying to get when I use the test equation ya= .5*sin(2*pi*1*t)+.2*cos(2*pi*3*t) and what I'm trying to get happen if I varied the variable shift from (0:0.1:5) youtu.be/pf25Gw6iS1U please keep in mind that ya will be an imported audio signal so I won't have an equation to easily adjust
clear all,clf
Fs = 2000000;% Sampling frequency
t=linspace(0,1,Fs);
%1a create signal
ya = .5*sin(2*pi*2*t);
%2a create frequency domain
ya_fft = fft(ya);
mag = abs(ya_fft);
phase = unwrap(angle(ya_fft));
ya_newifft=ifft(mag.*exp(i*phase));
% ----- changes start here ----- %
shift = 5; % shift amount
N = length(ya_fft); % number of points in the fft
mag1 = mag(2:N/2+1); % get positive freq. magnitude
phase1 = phase(2:N/2+1); % get positive freq. phases
mag2 = mag(N/2+2:end); % get negative freq. magnitude
phase2 = phase(N/2+2:end); % get negative freq. phases
% pad the positive frequency signals with 'shift' zeros on the left
% remove 'shift' components on the right
mag1s = [zeros(1,shift) , mag1(1:end-shift)];
phase1s = [zeros(1,shift) , phase1(1:end-shift)];
% pad the negative frequency signals with 'shift' zeros on the right
% remove 'shift' components on the left
mag2s = [mag2(shift+1:end), zeros(1,shift)];
phase2s = [phase2(shift+1:end), zeros(1,shift) ];
% recreate the frequency spectrum after the shift
% DC +ve freq. -ve freq.
magS = [mag(1) , mag1s , mag2s];
phaseS = [phase(1) , phase1s , phase2s];
x = magS.*cos(phaseS); % change from polar to rectangular
y = magS.*sin(phaseS);
yafft2 = x + i*y; % store signal as complex numbers
yaifft2 = real(ifft(yafft2)); % take inverse fft
plot(t,ya,'-r',t,yaifft2,'-b'); % time signal with increased frequency
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
You can do this using a fractional delay filter.
First, lets make the code ore workable by letting Matlab handle the conjugate symmetry of the FFT. Just make mag1 and phase1 go to the end . . .
mag1 = mag(2:end);
phase1 = phase(2:end);
Get rid of mag2s and phase2s completely. This simplifies lines 37 and 38 to . .
magS = [mag(1) , mag1s ];
phaseS = [phase(1) , phase1s ];
Use the symmetric option of ifft to get Matlb to handle the symmetry for you. You can then drop the forced real, too.
yaifft2 = ifft(yafft2, 'symmetric'); % take inverse fft
With that cleaned up, we can now think of the delay as a filter, e.g.
% ----- changes start here ----- %
shift = 5;
shift_b = [zeros(1, shift) 1]; % shift amount
shift_a = 1;
which can be applied as so . . .
mag1s = filter(shift_b, shift_a, mag1);
phase1s = filter(shift_b, shift_a, phase1);
In this mindset, we can use an allpass filter to make a very simple fractional delay filter
The code above gives the 'M Samples Delay' part of the circuit. You can then add on the fraction using a second cascaded allpass filter . .
shift = 5.5;
Nw = floor(shift);
shift_b = [zeros(1, Nw) 1];
shift_a = 1;
Nf = mod(shift,1);
alpha = -(Nf-1)/(Nf+1);
fract_b = [alpha 1];
fract_a = [1 alpha];
%// now filter as a cascade . . .
mag1s = filter(shift_b, shift_a, mag1);
mag1s = filter(fract_b, fract_a, mag1s);
Ok so the question as I understand it is "how do I shift my signal by a specific frequency?"
First let's define Fs which is our sample rate (ie samples per second). We collect a signal which is N samples long. Then the frequency change between samples in the Fourier domain is Fs/N. So taking your example code Fs is 2,000,000 and N is 2,000,000 so the space between each sample is 1Hz and shifting your signal 5 samples shifts it 5Hz.
Now say we want to shift our signal by 5.25Hz instead. Well if our signal was 8,000,000 samples then the spacing would be Fs/N = 0.25Hz and we would shift our signal 11 samples. So how do we get an 8,000,000 sample signal from a 2,000,000 sample signal? Just zero pad it! literally append zeros until it is 8,000,000 samples long. Why does this work? Because you are in essence multiplying your signal by a rectangular window which is equivalent to sinc function convolution in the frequency domain. This is an important point. By appending zeros you are interpolating in the frequency domain (you don't have any more frequency information about the signal you are just interpolating between the previous DTFT points).
We can do this down to any resolution you want, but eventually you'll have to deal with the fact that numbers in digital systems aren't continuous so I recommend just choosing an acceptable tolerance. Lets say we want to be within 0.01 of our desired frequency.
So lets get to actual code. Most of it doesn't change luckily.
clear all,clf
Fs = 44100; % lets pick actual audio sampling rate
tolerance = 0.01; % our frequency bin tolerance
minSignalLen = Fs / tolerance; %minimum number of samples for our tolerance
%your code does not like odd length signals so lets make sure we have an
%even signal length
if(mod(minSignalLen,2) ~=0 )
minSignalLen = minSignalLen + 1;
end
t=linspace(0,1,Fs); %our input signal is 1s long
%1a create 2Hz signal
ya = .5*sin(2*pi*2*t);
if (length(ya) < minSignalLen)
ya = [ya, zeros(1, minSignalLen - length(ya))];
end
df = Fs / length(ya); %actual frequency domain spacing;
targetFreqShift = 2.32; %lets shift it 2.32Hz
nSamplesShift = round(targetFreqShift / df);
%2a create frequency domain
ya_fft = fft(ya);
mag = abs(ya_fft);
phase = unwrap(angle(ya_fft));
ya_newifft=ifft(mag.*exp(i*phase));
% ----- changes start here ----- %
shift = nSamplesShift; % shift amount
N = length(ya_fft); % number of points in the fft
mag1 = mag(2:N/2+1); % get positive freq. magnitude
phase1 = phase(2:N/2+1); % get positive freq. phases
mag2 = mag(N/2+2:end); % get negative freq. magnitude
phase2 = phase(N/2+2:end); % get negative freq. phases
% pad the positive frequency signals with 'shift' zeros on the left
% remove 'shift' components on the right
mag1s = [zeros(1,shift) , mag1(1:end-shift)];
phase1s = [zeros(1,shift) , phase1(1:end-shift)];
% pad the negative frequency signals with 'shift' zeros on the right
% remove 'shift' components on the left
mag2s = [mag2(shift+1:end), zeros(1,shift)];
phase2s = [phase2(shift+1:end), zeros(1,shift) ];
% recreate the frequency spectrum after the shift
% DC +ve freq. -ve freq.
magS = [mag(1) , mag1s , mag2s];
phaseS = [phase(1) , phase1s , phase2s];
x = magS.*cos(phaseS); % change from polar to rectangular
y = magS.*sin(phaseS);
yafft2 = x + i*y; % store signal as complex numbers
yaifft2 = real(ifft(yafft2)); % take inverse fft
%pull out the original 1s of signal
plot(t,ya(1:length(t)),'-r',t,yaifft2(1:length(t)),'-b');
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
The final signal is a little over 4Hz which is what we expect. There is some distortion visible from the interpolation, but that should be minimized with a longer signal with a smother frequency domain representation.
Now that I've gone through all of that you may be wondering if there is an easier way. Fortunately for us, there is. We can take advantage of the hilbert transform and fourier transform properties to achieve a frequency shift without ever worrying about Fs or tolerance levels or bin spacing. Namely we know that a time shift leads to a phase shift in the Fourier domain. Well time and frequency are duals so a frequency shift leads to a complex exponential multiplication in the time domain. We don't want to just do a bulk shift of all frequencies because that that will ruin our symmetry in Fourier space leading to a complex time series. So we use the hilbert transform to get the analytic signal which is composed of only the positive frequencies, shift that, and then reconstruct our time series assuming a symmetric Fourier representation.
Fs = 44100;
t=linspace(0,1,Fs);
FShift = 2.3 %shift our frequency up by 2.3Hz
%1a create signal
ya = .5*sin(2*pi*2*t);
yaHil = hilbert(ya); %get the hilbert transform
yaShiftedHil = yaHil.*exp(1i*2*pi*FShift*t);
yaShifted = real(yaShiftedHil);
figure
plot(t,ya,'-r',t,yaShifted,'-b')
legend('Original signal (ya) ','New frequency signal (yaifft2) ')
Band-limited interpolation using a windowed-Sinc interpolation kernel can be used to change sample rate by arbitrary ratios. Changing the sample rate changes the frequency content of the signal, relative to the sample rate, by the inverse ratio.
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')