I'm trying to find the maximum frequency of a periodic signal in Matlab and as i know when you convert a periodic signal to the frequency spectrum you get only delta functions however i get a few curves between the produced delta functions. Here is the code :
t=[-0.02:10^-3:0.02];
s=5.*(1+cos(2*pi*10*t)).*cos(2*pi*100*t);
figure, subplot(211), plot(t,s);
y=fft(s);
subplot(212), plot(t,y);
Here is a code-snippet to help you understand how to get the frequency-spectrum using fft in matlab.
Things to remember are:
You need to decide on a sampling frequency, which should be high enough, as per the Nyquist Criterion (You need the number of samples, at least more than twice the highest frequency or else we will have aliasing). That means, fs in this example cannot be below 2 * 110. Better to have it even higher to see a have a better appearance of the signal.
For a real signal, what you want is the power-spectrum obtained as the square of the absolute of the output of the fft() function. The imaginary part, which contains the phase should contain nothing but noise. (I didn't plot the phase here, but you can do this to check for yourself.)
Finally, we need to use fftshift to shift the signal such that we get the mirrored spectrum around the zero-frequency.
The peaks would be at the correct frequencies. Now considering only the positive frequencies, as you can see, we have the largest peak at 100Hz and two further lobs around 100Hz +- 10Hz i.e. 90Hz and 110Hz.
Apparently, 110Hz is the highest frequency, in your example.
The code:
fs = 500; % sampling frequency - Should be high enough! Remember Nyquist!
t=[-.2:1/fs:.2];
s= 5.*(1+cos(2*pi*10*t)).*cos(2*pi*100*t);
figure, subplot(311), plot(t,s);
n = length(s);
y=fft(s);
f = (0:n-1)*(fs/n); % frequency range
power = abs(y).^2/n;
subplot(312), plot(f, power);
Y = fftshift(y);
fshift = (-n/2:n/2-1)*(fs/n); % zero-centered frequency range
powershift = abs(Y).^2/n;
subplot(313), plot(fshift, powershift);
The output plots:
The first plot is the signal in the time domain
The signal in the frequency domain
The shifted fft signal
Related
Assume I have a smooth function (represented as a vector):
x=0:0.1:1000;
y=sin(2*x);
and I want to find its periodicity - pi (or even its frequency -2 ) .
I have tried the following:
nfft=1024;
Y=fft(y,nfft);
Y=abs(Y(1:nfft/2));
plot(Y);
but obviously it doesn't work (the plot does not give me a peak at "2" ).
Will you please help me find a way to find the value "2"?
Thanks in advance
You have several issues here:
You are computing the fft of x when your actual signal is y
x should be in radians
You need to define a sampling rate and use that to determine the frequency values along the x axis
So once we correct all of these things, we get:
samplingRate = 1000; % Samples per period
nPeriods = 10;
nSamples = samplingRate * nPeriods;
x = linspace(0, 2*pi*nPeriods, nSamples);
y = sin(2*x);
F = fft(y);
amplitude = abs(F / nSamples);
f = samplingRate / nSamples*[0:(nSamples/2-1),-nSamples/2:-1];
plot(f, amplitude)
In general, you can't use an FFT alone to find the period of a periodic signal. That's because an FFT does sinusoidal basis decomposition (or basis transform), and lots of non-sinusoidal waveforms (signals that look absolutely nothing like a sinewave or single sinusoidal basis vector) can be repeated to form a periodic function, waveform, or signal. Thus, it's quite possible for the frequency of a periodic function or waveform to not show up at all in an FFT result (it's called the missing fundamental problem).
Only in the case of a close or near sinusoidal signal will an FFT reliably report the reciprocal of the period of that periodic function.
There are lots of pitch detection/estimation algorithms. You can use an FFT as a sub-component of some composite methods, including cepstrums or cepstral analysis, and Harmonic Product Spectrum pitch detection methods.
I'm having problems plotting the FFT of a wav file. I managed to plot the magnitude and phase spectrums of the signal, however I need to repeat this in range -fs/2:fs/2.
%read sound files
%'y' is the vector holding the original samples & 'fs' refers to the sampling frequency
[y,fs] = wavread('handel.wav');
ydft = fft(y); %fft to transform the original signal into frequency domain
n = length (y); %length of the original signal
% y has even length
ydft = ydft(1:length(y)/2+1);
% create a frequency vector
freq = 0:fs/length(y):fs/2;
shiftfreq = fftshift(freq);
%plot original signal in time domain;
figure;
plot ((1:n)/fs, y);
title('handel.wav in time domain');
xlabel ('second');
grid on;
% plot magnitude in frequency domain
figure;
plot(freq,abs(ydft));
title('handel.wav in frequency domain');
xlabel ('Hz');
ylabel('Magnitude');
grid on;
% plot phase in frequency domain
figure;
plot(freq,unwrap(angle(ydft)));
title ('handel.wav in frequency domain');
xlabel ('Hz');
ylabel ('Phase');
grid on;
What you are currently doing now is plotting the half spectrum, so from 0 <= f < fs/2 where fs is the sampling frequency of your signal, and so fs/2 is the Nyquist frequency. Take note that considering the half spectrum is only valid if the signal is real. This means that the negative spectra is symmetric to the positive spectra and so you don't really need to consider the negative spectra here.
However, you would like to plot the full spectrum of the magnitude and phase. Take note that when calculating the fft using MATLAB, it uses the Cooley-Tukey algorithm so when computing the N point FFT, half of result is for the frequencies from 0 Hz inclusive up to fs/2 Hz exclusive and the other half is for the frequencies from -fs/2 Hz inclusive up to 0 Hz exclusive.
As such, to plot the full spectrum, simply perform a fftshift on the full signal so that the right half and left half of the spectrum is swapped so that the 0 Hz frequency is located in the centre of the signal. Also, you must generate frequencies between -fs/2 to fs/2 to cover the full spectrum. Specifically, you need to generate N points linearly spaced between -fs/2 to fs/2. However, take note that the Nyquist frequency at fs/2 Hz is being excluded at the end, so you need to generate N+1 points between -fs/2 to fs/2 and remove the last point in order for the right step size between each frequency bin to be correct. The easiest way to generate this linear array of points is by using the linspace command where the start frequency is -fs/2, the ending frequency is fs/2 and you want N+1 points between this range and remove the last point:
freq = linspace(-fs/2, fs/2, n+1);
freq(end) = [];
As such, borrowing some parts of your code, this is what the modified code looks like to plot the full spectrum of the magnitude and phase:
%// Read in sound file
[y,fs] = wavread('handel.wav');
%// Take N-point FFT where N is the length of the signal
ydft = fft(y);
n = numel(y); %// Get N - length of signal
%// Create frequency vector - make sure you remove last point
freq = linspace(-fs/2, fs/2, n+1);
freq(end) = [];
%// Shift the spectrum
shiftSpectrum = fftshift(ydft);
%//plot original signal in time domain;
figure;
plot ((0:n-1)/fs, y); %// Note you should start from time = 0, not time = 1/fs
title('handel.wav in time domain');
xlabel ('second');
grid on;
%// plot magnitude in frequency domain
figure;
plot(freq,abs(shiftSpectrum));
title('handel.wav in frequency domain');
xlabel ('Hz');
ylabel('Magnitude');
grid on;
%// plot phase in frequency domain
figure;
plot(freq,unwrap(angle(shiftSpectrum)));
title('handel.wav in frequency domain');
xlabel('Hz');
ylabel('Phase');
grid on;
I don't have access to your handel.wav file, but I'll be using the one provided with MATLAB. You can load this in with load handel;. The sampling frequency is stored in a variable called Fs, so I had to do fs = Fs; before the code I wrote above could work. The sampling frequency for this particular file is 8192 Hz, and this is approximately a 9 second long file (numel(y) / fs = 8.9249 seconds). With that file, this is the magnitude and phase that I get:
For the discrete Fourier transform (DFT) as well as its fast implementations (FFTs), the frequencies are normalized with the sampling frequency fs, i.e., the original range -fs/2:fs/2 is changed to -pi:pi.
Besides, the DFT/FFT always starts with 0, and you can use fftshift() to shift the 0 frequency to the center. Therefore, after fftshift(), the range is -pi:pi, then, you can scale to -fs/2:fs/2.
look at the following Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect accuracy:
https://www.mathworks.com/matlabcentral/fileexchange/63965-amplitude-and-phase-spectra-of-a-signal--fourier-transform-
This program calculates amplitude and phase spectra of an input signal with acceptable accuracy especially in the calculation of phase spectrum.The code does three main jobs for calculation amplitude and phase spectra. First of all, it extends the input signal to infinity; because for calculation Fourier transform(FT) (fft function in Matlab), we consider our signal is periodic with an infinite wavelength, the code creates a super_signal by putting original signal next to itself until the length of super_signal is around 1000000 samples, why did I choose 1000000 samples? Actually, it is just based on try and error!! For most signals that I have tried, a supper signal with 1000000 samples has the best output.
Second, for calculating fft in Matlab you can choose different resolutions, the Mathwork document and help use NFFT=2^nextpow2(length(signal)), it definitely isn't enough for one that wants high accuracy output. Here, I choose the resolution of NFFT=100000 that works for most signals.
Third, the code filters result of FT by thresholding, it is very important step! For calculating phase spectrum, its result is very noisy because of floating rounding off error, it causes during calculation "arctan" even small rounding off error produces significant noise in the result of phase spectrum, for suppressing this kind of noise you can define a threshold value. It means if amplitude of specific frequency is less than predefined threshold value (you must define it) it put zero instead of it.
These three steps help to improve the result of amplitude and phase spectra significantly.
IF YOU USE THIS PROGRAM IN YOUR RESEARCH, PLEASE CITE THE FOLLOWING PAPER:
Afshin Aghayan, Priyank Jaiswal, and Hamid Reza Siahkoohi (2016). "Seismic denoising using the redundant lifting scheme." GEOPHYSICS, 81(3), V249-V260. https://doi.org/10.1190/geo2015-0601.1
I have generated the following time signal:
Now I want to perform a Discrete Fourier Transform by using the matlab command fft
Here is my code:
function [ xdft, omega ] = FastFourier( t, fs )
%% Inputs from other functions %%
[P_mean, x, u] = MyWay( t ) %From here comes my signal x(t)
%% FFT %%
xdft1 = fft(x); % Perform FFT
xdft2 = abs(xdft1); % Take the absolute value of the fft results
xdft = xdft2(1:length(x)/2+1); % FFT is symmetric, second half is not needed
freq = 0:fs/length(x):fs/2; % frequency axis
plot (freq(1:100),xdft(1:100));
end
And here is the plot that I get:
And what is puzzling to me is the y axis? Shouldn't the y axis represent the amplitudes of the frequency components? Is there a way to get the amplitudes of all the frequency components?
Thanks!
EDIT:
I have found that some people do the following:
n = size(x,2)/2; %The number of components and second half can be neglected again
xdft2 = abs(xdft1)/n;
This way I seem to get the amplitude spectrum, but why do I have to divide the absolute value by n?
FFT gives you a complex pair in each Frequency Bin. The first bin in the FFT is like the DC part of your signal (around 0 Hz), the second bin is Fs / N, where Fs is the sample rate and Nis the windowsize of the FFT, next bin is 2 * Fs / N and so on.
What you calc with the abs() of such a pair is the power contained in a bin.
you might also want to check this out: Understanding Matlab FFT example
Most (not all) FFT libraries preserve total energy (Parseval's theorem), which means that the magnitude has to get bigger for longer FFT windows (longer stationary waveform -> more energy). So you have to divide the result by N to get a more "natural" looking magnitude height of sinewaves in the spectrum.
If you want the amplitudes of the harmonics, then you need to plot real(xdft1) and imag(xdft1). Real(xdft1) gives you coefficients of all the cosine harmonics present in your signal, from -Fs/2 to +Fs/2, (we assume your Fs is large enough to cover all frequencies in the signal) and the imag(xdft) give the amplitudes of the sines.
What you are doing is giving you the magnitude of the signal, which is the RMS value of the total energy at a bin in both the real and imaginary frequency component.
Its often the item of most interest to people looking at a spectrum.
Basics of this: (https://www.youtube.com/watch?v=ZKNzMyS9Z6s&t=1629s)
Consider the following script that plots a sine wave.
t = 0:pi/100:2*pi;
y = sin(t);
plot(t,y)
grid on % Turn on grid lines for this plot
This gives me a plot of sine wave. I understand the sine wave that appears continuous, should actually be discrete (my PC cannot store infinite no. of samples of continuous signal), and the matlab plot function does some kind of interpolation to connect the dots.
So In fact I also used stem instead of plot to see the sampled values (on time axis) of sine.
Now my question is there must be some sampling frequency used here. How much is that?
The sampling interval is the time interval between two consecutive samples of your signal.
The sampling frequency means how much samples of your signal you have in a fixed time interval, and it is reciprocal to the sampling interval.
You declared:
t = 0:pi/100:2*pi;
So your sampling interval is π/100. This means that your sampling frequency is 100/π.
If you want exact units, you'll have to determine the time units for t. If t is in seconds, then your sampling frequency is 100/π Hz (1Hz = 1sec-1).
By the way, MATLAB's plot connects the sampling with straight lines, there is no additional interpolation involved.
Here is the scenario: using a spectrum analyzer i have the input values and the output values. the number of samples is 32000 and the sampling rate is 2000 samples/sec, and the input is a sine wave of 50 hz, the input is current and the output is pressure in psi.
How do i calculate the frequency response from this data using MATLAB,
using the FFT function in MATLAB.
i was able to generate a sine wave, that gives out the the magnitude and phase angles, here is the code that i used:
%FFT Analysis to calculate the frequency response for the raw data
%The FFT allows you to efficiently estimate component frequencies in data from a discrete set of values sampled at a fixed rate
% Sampling frequency(Hz)
Fs = 2000;
% Time vector of 16 second
t = 0:1/Fs:16-1;
% Create a sine wave of 50 Hz.
x = sin(2*pi*t*50);
% Use next highest power of 2 greater than or equal to length(x) to calculate FFT.
nfft = pow2(nextpow2(length(x)))
% Take fft, padding with zeros so that length(fftx) is equal to nfft
fftx = fft(x,nfft);
% Calculate the number of unique points
NumUniquePts = ceil((nfft+1)/2);
% FFT is symmetric, throw away second half
fftx = fftx(1:NumUniquePts);
% Take the magnitude of fft of x and scale the fft so that it is not a function of the length of x
mx = abs(fftx)/length(x);
% Take the square of the magnitude of fft of x.
mx = mx.^2;
% Since we dropped half the FFT, we multiply mx by 2 to keep the same energy.
% The DC component and Nyquist component, if it exists, are unique and should not be multiplied by 2.
if rem(nfft, 2) % odd nfft excludes Nyquist point
mx(2:end) = mx(2:end)*2;
else
mx(2:end -1) = mx(2:end -1)*2;
end
% This is an evenly spaced frequency vector with NumUniquePts points.
f = (0:NumUniquePts-1)*Fs/nfft;
% Generate the plot, title and labels.
subplot(211),plot(f,mx);
title('Power Spectrum of a 50Hz Sine Wave');
xlabel('Frequency (Hz)');
ylabel('Power');
% returns the phase angles, in radians, for each element of complex array fftx
phase = unwrap(angle(fftx));
PHA = phase*180/pi;
subplot(212),plot(f,PHA),title('frequency response');
xlabel('Frequency (Hz)')
ylabel('Phase (Degrees)')
grid on
i took the frequency response from the phase plot at 90 degree phase angle, is this the right way to calculate the frequency response?
how do i compare this response to the values that is obtained from the analyzer? this is a cross check to see if the analyzer logic makes sense or not.
Looks OK at first glance, but a couple of things you're missing:
you should apply a window function to the time domain data before the FFT, see e.g. http://en.wikipedia.org/wiki/Window_function for windowing in general and http://en.wikipedia.org/wiki/Hann_window for the most commonly used window function (Hann aka Hanning).
you probably want to plot log magnitude in dB rather than just raw magnitude
You should consider looking at the cpsd() function for calculating the Frequency response. The scaling and normalisation for various window functions is handled for you.
the Frequency reponse would then be
G = cpsd (output,input) / cpsd (input,input)
then take the angle() to obtain the phase difference between the input and the output.
Your code snippet does not mention what the input and output data sets are.