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.
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
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'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.
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).
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.