I am missing something in the computation of the spectrum of my signal using FFT on Matlab.
My code:
%% compute the spectrum of the data (data(t))
L = length(time); % length of the sample
NFFT = 2^(nextpow2(L)-1); % Next power of 2 from length of y
Y = fft(data,NFFT);%/NFFT;%L;
Fs = 1/(mean(time(2:end)-time(1:end-1))); % compute the sampling frequency
f = Fs/2*linspace(0,1,NFFT/2+1);
loglog(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of My Data')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
Would you be so kind as to tell me where I messed up?
I tried to check if the algorythm works using these two sampling of the same signal (same sampling frequency ; over two different time range 0-10 and 0-100):
fs=1000;
time10 = [0:1/fs:10];
time100 = [0:1/fs:100];
data10 = sin(2*pi*0.23 .*time10)+cos(2*pi*12 .*time10);
data100 = sin(2*pi*0.23 .*time100)+cos(2*pi*12 .*time100);
I guess the two spectrum should supperpose but they don't... As seen here: https://www.dropbox.com/s/wfols9o409pr94u/FFT_spectrum_StackOverflow.png?dl=0
https://www.dropbox.com/s/a8vmzwto6x4130w/FFT_spectrum_StackOverflow.fig?dl=0
Thanks
The good news is that there is nothing wrong with your computation of the spectrum by itself.
The problem is that by looking at samples of different lengths you are effectively looking at two different samples altogether.
In the time-domain, they can be seen as the result of a multiplication of an infinitely long sinusoidal with a rectangular window of different lengths.
In the frequency-domain, the spectrum of the infinitely long continuous-time sinusoidal signal gets convoluted with the spectrum of the rectangular windows. With different window length the corresponding spectrum of those windows have different width (narrower spectrum for longer rectangular windows). As a result, the spikes in the spectrum of the infinitely long sinusoidal signal would get spread over different bandwidths. This is exactly what you are seeing.
Related
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
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 am new to Matlab and speech processing as well. I want to find the fundamental frequency of speech signal to determine the gender of the speaker. I removed the silence from the signal by analysing it within 10 msec periods.
After that I got the fft using this code :
abs(fft(input_signal_without_silences))
My plot of both the speech signal and the fft of it is below:
Now, I want to find the fundamental frequency but I could not understand which steps do I need to do this. Or do I misunderstand this concept?
As far as I have learnt, there are some methods like autocorrelation,
Since I am not familiar to both speech processing and matlab, any help and advice is very much appreciated.
The fft() help can solve most parts of your problem. I can give a brief overview of things based on the content of the help file.
At the moment what you are plotting is the two sided, unnormalized fft coefficients, which don't tell much. Use the following to get a more user informed spectral analysis of the voice signal. Using the single sided spectram you would be able to find the dominant frequency which might be the fundamental frequency of the speech signal.
y = []; %whatever your signal
T = 1e-2; % Sample time, 10 ms
Fs = 1/T; % Sampling frequency
L = length(y); % Length of signal
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
The problem is that you have a plot of Amplitude vs Sample Number instead of a plot of Amplitude vs Frequency.In order to calculate the fundamental frequency you need to find the frequency that corresponds to the highest frequency.
Matlab returns frequencies from -fs/2 to fs/2 so the frequency at index n is
f = n * (fs/N) - (fs/2)
where f = frequency, fs = sampling frequency, N = number of points in FFT.
So basically all you need to do is get the index where the plot is highest and substitute it in the equation above to get an estimate of the fundamental frequency.Make sure n > N/2 so that your fundamental frequency is positive.
I have a small input signal of 60Hz sine wave from signal generator, which is corrupted with 50Hz mains supply frequency. I want to measure the amplitude of the 60Hz signal using FFT because it is very small to see in the oscilloscope.
The Matlab FFT code:
y = data;
Fs = 2048;
[r, L] = size(y);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1)))
But the FFT plot doesn't give a sharp peak at 50 and 60Hz. The plot looks like this:
The consecutive points have high and low amplitude alternatively which gives a saw-tooth like plot. Why is it so? Is the amplitude of 60Hz affected by this?
Probably there are two effects
If one measures a time window of a signal, this leads unavoidable to a phase gap between the start and endpoint of the signal modes. The FFT of a gap or a rectangular signal causes high frequency oscillations. These oscillations caused by border effects can be damped due to a window function, which smooths out the signal to the borders.
There is a discrete frequency spectra in DFT. If one measures a signal which does not match to any of these discrete modes, a lot more frequencies are necessary to reconstruct the original signal.
Your 50 Hz signal may not be a pure perfect sine wave. Any differences from a perfect sine-wave (such as clipping or distortion) is equivalent to a modulation that will produce sidebands in the spectrum.
Windowing a signal whose period is not an exact sub-multiple of the FFT length will also convolve windowing artifacts with that signal.
I know that there are a lot of similar questions to this, I am still unable to figure out the answer.
Let's say we have time signal in MATLAB:
t=0:1/44100:1
and a cosine signal with frequency 500Hz:
x=cos(2*pi*500*t);
Now, I am trying to plot the magnitude spectrum obtained using the fft command on signal x
FFT=abs(fft(x))
plot(FFT)
According to the theory, we should get two peaks in the plot, one at -500 Hz and the other at 500Hz.
What I don't understand is that I do get two peaks but I can't figure out at what frequencies these peaks are. I know there is a way to figure out the frequency using the FFT index, length of the input signal and the sampling frequency but I still can't calculate the frequency.
I know that there are methods to align the FFT plots so that the peaks lie at the index number of the frequency they represent by using the fftshift function, but what I want is to figure out the frequency using the the plot resulting from simply calling this function:
FFT=fft(x)
In this case, I already know that signal contains a cosine of 500Hz, but what if the signal that we want to get the FFT of is not known before time. How can we get the frequency values of the peaks in that sample using the output from the fft function?
You need to generate the frequency array yourself and plot your FFT result against it.
Like this:
function [Ycomp, fHz] = getFFT(data, Fs)
len = length(data);
NFFT = 2^nextpow2(len);
Ydouble = fft(data, NFFT)/len; % Double-sided FFT
Ycomp = Ydouble(1:NFFT/2+1); % Single-sided FFT, complex
fHz = Fs/2*linspace(0,1,NFFT/2+1); % Frequency array in Hertz.
semilogx(fHz, abs(Ycomp))
end
You will see peaks at 500 Hz and Fs - 500 Hz (i.e. 44100 - 500 = 43600 Hz in your particular case).
This is because the real-to-complex FFT output is complex conjugate symmetric - the top half of the spectrum is a "mirror image" of the bottom half when you are just looking at the magnitude and is therefore redundant.
Note that of plotting power spectra you can usually save yourself a lot of work by using MATLAB's periodogram function rather than dealing directly with all the details of FFT, window functions, plotting, etc.