Good evening guys,
I wanna ask you a question regarding the analysis of a function in the domain of frequencies (Fourier). I have two vectors: one containing 7700 values for pressure, and the other one containing 7700 values (same number) for time.
For example, I call the firt vector "a" and the second one "b". With the command "figure(1),plot(a,b)" I obtain the curve in the domain of time.
How can I do to plot this curve in the domain of frequency, to make Fourier transform?
I've read about the function "fft", but I've not understood very well how it can be used...can anyone help me?
Thanks in advance for your attention!
fft returns spectrum as complex numbers. In order to analyze it you have to use its absolute value or phase. In general, it should look like this (let's assume that t is vector containing time and y is the one with actual signal, N is the number of samples):
fY = fft(y) / (N/2) % scale it to amplitude, typically by N/2
amp_fY = abs(fY)
phs_fY = angle(fY)
Additionally, it would be nice to have FFT with known frequency resolution. For that, you need sampling period/frequency. Let's call that frequency fs:
fs = 1/(t(1) - t(0))
and the vector of frequencies for FFT (F)
should be:
F = (0:fs/N:(N-1)*fs/N)
and finally plots:
plot(F, amp_fY)
% or plot(F, phs_fy) according to what you need
I you can use stem instead of plot to get some other type of chart.
Note that the DC component (the average value) will be doubled on the plot.
Hope it helps
Related
I have a time varying signal (a), which I take the fft of. I need to multiply a frequency dependent weighting factor by the fft's y axis value; however if I do:
xdft = fft(a);
xdft = xdft(1:length(x)/2+1); % only retaining the positive frequencies
freq = Fs*(0:(L/2))/L;
And plot(freq,xdft) I get a peak fft value (y axis)of ~2000 at the correct frequency of the signal. But original signal peak value (amplitude) was ~46. I need to know how the numbers relate so I can weight the fft values.
You forgot to divide by the DFT length. Take a look at this example.
Consider that the output of fft is complex. So if you want to plot the real power spectral density you should multiply the output by its complex conjugate like this:
Pyy = xdft.*conj(xdft)/L;
Edit:
For the amplitude spectrum you should do something like this:
xdft=abs(xdft/L); % same as sqrt(xdft.*conj(xdft))/L
Y=xdft(1:L/2+1); % copy half of data since the other half is redundant
Y(2:end-1) = 2*Y(2:end-1); % correct the amplitudes
Edit 2:
Just wanted to point to a really great book (the best in my opinion) which explains how fourier series work (and much more) in a really easy and understandable way.
Alan V. Oppenheim, Alan S. Willsky, and S. Hamid Nawab. 1996. Signals & Systems (2nd Ed.). Prentice-Hall, Inc., Upper Saddle River, NJ, USA.
I am trying to use the ifft function in MATLAB on some experimental data, but I don't get the expected results.
I have frequency data of a logarithmic sine sweep excitation, therefore I know the amplitude [g's], the frequency [Hz] and the phase (which is 0 since the point is a piloting point).
I tried to feed it directly to the ifft function, but I get a complex number as a result (and I expected a real result since it is a time signal). I thought the problem could be that the signal is not symmetric, therefore I computed the symmetric part in this way (in a 'for' loop)
x(i) = conj(x(mod(N-i+1,N)+1))
and I added it at the end of the amplitude vector.
new_amp = [amplitude x];
In this way the new amplitude vector is symmetric, but now I also doubled the dimension of that vector and this means I have to double the dimension of the frequency vector also.
Anyway, I fed the new amplitude vector to the ifft but still I don't get the logarithmic sine sweep, although this time the output is real as expected.
To compute the time [s] for the plot I used the following formula:
t = 60*3.33*log10(f/f(1))/(sweep rate)
What am I doing wrong?
Thank you in advance
If you want to create identical time domain signal from specified frequency values you should take into account lots of details. It seems to me very complicated problem and I think it need very strength background on the mathematics behind it.
But I think you may work on some details to get more acceptable result:
1- Time vector should be equally spaced based on sampling from frequency steps and maximum.
t = 0:1/fs:N/fs;
where: *N* is the length of signal in frequency domain, and *fs* is twice the
highest frequency in frequency domain.
2- You should have some sort of logarithmic phases on the frequency bins I think.
3- Your signal in frequency domain must be even to have real signal in time domain.
I hope this could help, even for someone to improve it.
I want to ask some questions related to the last question of mine so I don't want to post in another thread. My question contains a code, I therefore can't post it as a comment. So I have to edit my old question into a new one. Please take a look and help. Thank you.
I'm new to FFT and DSP and I want to ask you some questions about calculating FFT in Matlab. The following code is from Matlab help, I just removed the noise.
Can I choose the length of signal L different from NFFT?
I'm not sure if I used window correctly. But when I use window (hanning in the following code), I can't get the exact values of amplitudes?
When L and NFFT get different values, then the values of amplitudes were different too. How can I get the exact value of amplitude of input signal? (in the following code, I used a already known signal to check if the code work correctly. But in case, I got the signal from a sensor and I dont know ahead its amplitude, how can I check?)
I thank you very much and look forward to hearing from you :)
Fs = 1000; % Sampling frequency
T = 1/Fs; % Sample time
L = 512; % Length of signal
NFFT=1024; % number of fft points
t = (0:L-1)*T; % Time vector
x = 0.7*sin(2*pi*50*t) + sin(2*pi*120*t); input signal
X = fft(hann(L).*x', NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(X(1:NFFT/2+1))) % Plot single-sided amplitude spectrum.
L is the number of samples in your input signal. If L < NFFT then the difference is zero-padded.
I would recommend you do some reading on the effect of zero-padding on FFTs. Typically it is best to use L = NFFT as this will give you the best representation of your data.
An excepted answer on the use of zero-padding and FFTs is given here:
https://dsp.stackexchange.com/questions/741/why-should-i-zero-pad-a-signal-before-taking-the-fourier-transform
In your experiment you are seeing different amplitudes because you will have different amount of spectral leakage with each different L.
You need to apply a window function prior to the FFT to get consistent results with frequency components that have non-integral number of periods within your sampling window.
You might also want to consider using periodogram instead of using the FFT directly - it takes care of window functions and a lot of the other housekeeping for you.
I have a set of data that is periodic (but not sinusoidal). I have a set of time values in one vector and a set of amplitudes in a second vector. I'd like to quickly approximate the period of the function. Any suggestions?
Specifically, here's my current code. I'd like to approximate the period of the vector x(:,2) against the vector t. Ultimately, I'd like to do this for lots of initial conditions and calculate the period of each and plot the result.
function xdot = f (x,t)
xdot(1) =x(2);
xdot(2) =-sin(x(1));
endfunction
x0=[1;1.75]; #eventually, I'd like to try lots of values for x0(2)
t = linspace (0, 50, 200);
x = lsode ("f", x0, t)
plot(x(:,1),x(:,2));
Thank you!
John
Take a look at the auto correlation function.
From Wikipedia
Autocorrelation is the
cross-correlation of a signal with
itself. Informally, it is the
similarity between observations as a
function of the time separation
between them. It is a mathematical
tool for finding repeating patterns,
such as the presence of a periodic
signal which has been buried under
noise, or identifying the missing
fundamental frequency in a signal
implied by its harmonic frequencies.
It is often used in signal processing
for analyzing functions or series of
values, such as time domain signals.
Paul Bourke has a description of how to calculate the autocorrelation function effectively based on the fast fourier transform (link).
The Discrete Fourier Transform can give you the periodicity. A longer time window gives you more frequency resolution so I changed your t definition to t = linspace(0, 500, 2000).
time domain http://img402.imageshack.us/img402/8775/timedomain.png (here's a link to the plot, it looks better on the hosting site).
You could do:
h = hann(length(x), 'periodic'); %# use a Hann window to reduce leakage
y = fft(x .* [h h]); %# window each time signal and calculate FFT
df = 1/t(end); %# if t is in seconds, df is in Hz
ym = abs(y(1:(length(y)/2), :)); %# we just want amplitude of 0..pi frequency components
semilogy(((1:length(ym))-1)*df, ym);
frequency domain http://img406.imageshack.us/img406/2696/freqdomain.png Plot link.
Looking at the graph, the first peak is at around 0.06 Hz, corresponding to the 16 second period seen in plot(t,x).
This isn't computationally that fast though. The FFT is N*log(N) operations.
I have 4 matrices of data F1,F2,O1,O2. All are neural signals collected at 1ms for a second. F1 and O1 were collected at the same time as with F2 and O2. I need to find how the data collected differes between the 2 trials and also compare the components of each trial (F1 and O1) to each other to notice and differences in respones. I'm new to MATLAB but I believe I need to use the fft function. I'm just not sure where to start, any help would be greatly appeciated.
Based on your sampling rate (1000 times per second), you will only be able to analyze signals which have frequency of up to 500 hz. Any neural signal components which have higher components will appear as signals of lower components (unless your device has highpass filter etc..)
The command for fft from Matlab Help is:
Y = fft(signal, n)
Signal is either F1 or O1 or F2 or O2 and should be a vector 1000 long. n determines how many samples your FFT have. This is essentially how finely you will split up the frequency values between 0 hz and 1000 hz (your sampling rate). For example, if you choose n =256, your Y will be a 256 long vector with a measure corresponding to the frequencies (0*1000/256 hz, 1*1000/256 hz, ... 255*1000/256 hz).
Y will be a vector of complex values. Often, you want to see the strength or power of the signal. You can use 'abs()' to find the magnitude. myPSD = abs(Y). Next because your signals are real signals, their fft's are symmetric about half the sampling rate (500hz). Thus, you should only look at the first half. Here is a code snippet to look at the first half.
Y = fft(signal, n); % choose your n
myPSD = abs(Y);
myHalfPSD = myPSD(1:ceil(n/2))
myFreqValues = [0:1000/n:500-1000/n] % both myHalfPSD and myFreqValues should be n/2 long
plot(myFreqValues, myHalfPSD)
Typically, PSD is displayed on a log scale or even decibal. So you might add a line.
Y = fft(signal, n); % choose your n
myPSD = abs(Y);
myHalfPSD = myPSD(1:ceil(n/2))
myHalfDBPSD = 20*log(myHalfPSD)
myFreqValues = [0:1000/n:500-1000/n] % both myHalfPSD and myFreqValues should be n/2 long
plot(myFreqValues, myHalfDBPSD)
If you want to plot all 4 graphs at once you might want to use something like
subplot(4,1,1), subplot(4,1,2) etc..
Hope that helps,
Chuan
If you're trying to compare the frequency spectrums of two separate acquisitions, then fft is the tool you want to use.
Mathworks has pretty good documentation on how to use the fft function, and you can probably cut and paste your data right into the example code they provide.
If you want to plot the data on the same axes, you can use the hold on command, plot different line colors (Ex: plot(x,y,'r') will plot a red line, 'b' a blue one, etc. - see lineseries properties) and include a legend to label the plots.