I'm trying to get Fourier transform of a gaussian pulse. Here's a sample code I found on the Internet.
fs=80; %sampling frequency
sigma=180;
t=-0.5:10/fs:0.5; %time base
variance=sigma^2;
x=1/(sqrt(2*pi*variance))*(exp(-t.^2/(2*variance)));
subplot(2,1,1)
plot(t,x,'b');
title(['Gaussian Pulse \sigma=', num2str(sigma),'s']);
xlabel('Time(s)');
ylabel('Amplitude');
L=length(x);
NFFT = 1024;
X = fftshift(fft(x,NFFT));
Pxx=X.*conj(X)/(NFFT*NFFT); %computing power with proper scaling
f = fs*(-NFFT/2:NFFT/2-1)/NFFT; %Frequency Vector
subplot(2,1,2)
plot(f,abs(X)/(L),'r');
title('Magnitude of FFT');
xlabel('Frequency (Hz)')
ylabel('Magnitude |X(f)|');
xlim([-10 10])
I need an explanation. Assume that I'm trying to get Fourier transform of 100 femtosecond gaussian pulse. How can I calculate the sampling frequency, sigma and t variables.
http://www.gaussianwaves.com/2014/07/generating-basic-signals-gaussian-pulse-and-power-spectral-density-using-fft/
A gaussian pulse has infinite support so there's no such thing as a "100 femtosecond gaussian pulse", you can however have a gaussian pulse with a sigma of 100 femtoseconds which is probably what you want. For your sampling frequency, you probably want to choose something such that you actually capture interesting information about the pulse. Your sampling period hence be less than a sigma (the original code (not the one you posted) from the blog used sigma/8). As for t, if you want to capture most of the energy in the pulse, you need to go several sigmas to the left and right of 0 which is where the pulse is centered in time.
Related
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 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.
Here is my code for generating a triangular waveform in the time domain and for generating its corresponding fourier series/transform (I don't know whether its series or transform because matlab only has fourier transform function but since the signal is periodic, references say that the fourier counterpart must be called fourier series).
x = 0;
s = 50; % number of sinusoidal components
fs = 330; % hertz
dt = 1/fs; % differential time
t = [0:dt:4]; % seconds
const = 2 / (pi^2);
for k = 1:2:s,
x = x + (((-1)^((k - 1) / 2)) / (k^2)) * sin(4*pi*k*t);
end
x = const * x;
% amplitude = max(x) = 0.2477
% period = 0.5 seconds
f = linspace(-fs/2,fs/2,length(x));
xk = fftshift(fft(x));
figure;
subplot(3,1,1);
plot(t,x);
grid on;
xlabel('time(seconds)');
title('Time Domain');
subplot(3,1,2);
plot(f,abs(xk));
grid on;
xlabel('frequency(hertz)');
title('Magnitude Spectrum');
subplot(3,1,3);
plot(f,angle(xk));
grid on;
xlabel('frequency(hertz)');
title('Phase Spectrum');
And here is the generated plots for the time domain signal, magnitude spectrum, and phase spectrum.
link:
fs = 330hz
My problem is when I changed the sampling frequency (fs which is currently equal to 330 hz) to another value, the plots of the magnitude and phase spectra change.
Here is the plots of the magnitude and phase spectra when the sampling frequency is equal to 400 hz:
link:
fs = 400 hz
Can you explain why does this happen? And what can I do in order to get a constant plots for the magnitude and phase spectra given any sampling frequency?
I can't get your pictures to load over my proxy, but the spectrum of a FFT will be have a bigger "gap" in the middle at a higher sampling rate. A fundamental property of sampling is that it introduces copies of your original spectrum; you may have learned this if you studied the discrete-time Fourier transform. At a higher sampling rate, these copies are farther apart.
Additionally, your sampling points will be in different places at different sampling rates, so you may get different lobing behavior.
Incidentally, you are getting the discrete Fourier transform in Matlab -- you are giving it a finite sequence of discrete points, not a continuous, inifintely long signal.
If you want the plots to look the same, just make their x-axes match.
It because that the spectra by DFT/FFT are indeed the sampled and normalized version of the original analog spectra, therefore, with the sampling step changes, the sampling step in frequency domain also changes, thus the spectra lines that you see also change since the original spectra are not constant. Another factor may be the aliasing effects, since the analog spectra of triangular waveform are infinite in theory.
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 am trying to get the peak frequency of a musical note by using the FFT function that exists in MATLAB. I just copy-pasted the code for FFT of a mathematical function and replaced the function with the audio file.
Fs = 44100; % Sampling frequency
T = 1/Fs; % Sample time
L = 1000; % Length of signal
t = (0:L-1)*T; % Time vector
% Sum of a 50 Hz sinusoid and a 120 Hz sinusoid
y = wavread('c-note2.wav');
plot(Fs*t(1:50),y(1:50))
xlabel('time (milliseconds)')
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)|')
Here, instead of y=wavread('c-note2.wav'), we had something like y=0.15sin(5x)+0.32cos(50t)+rand(I) (To add noise to the signal).
Is what we are trying to do correct? Can we put a wavread instead of a mathematical signal?
From the graph obtained I want to get the peak frequency of the c-note and check whether it is matching with the actual frequency of a c-note, but I am getting absurd results.
The pitch of a musical note is very often different from the peak frequency returned by an FFT. Musical notes usually contain a ton of overtones, many often stronger than the pitch frequency, some possibly even slightly inharmonic in frequency. Search for pitch detection or estimation algorithms instead of just looking at the FFT spectrum.
Also, when using an FFT to look at the audio spectrum, the length of the FFT has to be longer than several periods of the lowest frequency of interest. Your FFT length appears to be much too short to resolve 50 Hz (20 mS period).