I want to analyze an audiodata (.wav with pcm, 32k as sampling rate) and create the psd of it with the axes Sxx (watts/hertz not db) and f (hertz).
So I would start by reading out the audiodata with:
[x,fs]=audioread('test.wav');
After this I'm having some problems because I dont really know how to proceed and also Matlab always tells me that psd functions won't be supported in the future and that I should use pwelch.. (also tried to build the autocorr and afterwards use fourier to get to the Sxx but it didn't work out really well)
So could anybody tell me how I can get from my vector x to a vector with the psdvalues in watts/hertz and plot it afterwards?
very grateful for every kind of help! :)
Update1: Yes I did read the documentation of pwelch but I'm afraid my english is too bad to understand it completly.
So if I use the psd documentation:
nfft = 2^nextpow2(length(x));
Pxx = abs(fft(x,nfft)).^2/length(x)/fs;
Hpsd = dspdata.psd(Pxx(1:length(Pxx)/2),'fs',fs);
plot(Hpsd)
I'm able to get the plot in db with the peak at the right frequency. (I dont know how dspdata.psd work though)
I tried out:
[Pyy,f]=pwelch(x,fs)
plot(Pyy)
this gives me a non db-scale but the peak is at the wrong frequency
Update 2:
First of all, thanks a lot for your detailed answer! At the moment I'm working on my matlabskills as well as my english language but all the specific technical terms give me a hard time..
When using your example of pwelch on a wav-data with a clear frequency of 1khz, the plot shows me the peak at round about 0.14, could it maybe still be a special-scaled x-axis?
If I try it this way:
[y,fs]=audioread('test.wav');
N=length(y);
bin_vals=0:N-1;
fax_Hz= bin_vals*fs/N;
N_2=ceil(N/2);
Y=fft(y);
pyy=Y.*conj(Y);
plot(fax_Hz(1:N_2),pyy(1:N_2))
the result seems right (is this way correct?), but I still need some time to search for a proper way to display the y-axis in W/Hz, since I dont know how the audiosignal was created.
Update 3:
http://s000.tinyupload.com/index.php?file_id=33803229773204653857
This wav file should have a dominant frequency at 1khz with a duration of 3 seconds and a sampling frequency of 44100Hz. (If I plot the data received from audioread the oscillation seems reasonable)
with
[y,fs]=audioread('1khz.wav');
[pyy,f]=pwelch(y,fs);
plot(f,pyy)
I get a peak at 0.14 on the x-axis.
if I use
[y,fs]=audioread('1khz.wav');
[pyy,f]=pwelch(y,[],[],[],fs);
plot(f,pyy)
instead, the peak is at the 1000. Is this way right? And how could I interpret the difference scaling on the y-axis? (pwelch vs. square of abs)
I also wanted to ask if it is possible to get a flat psd of awgn in matlab? (since you just have finite elements I don't know to get there)
Thanks again for your detailed support!
Update 4
#A.Donda
So I have a new Problem for which I think it is probably necessary to go a bit more into detail. So my plan is basically to do the following:
Read and Audiodata ([y,fs]) and generate white Noise with a certain SNR ([n,fs])
Generate a Filter H which shapes the PSD(y) similiar to the PSD(n)
Generate an inverse Filter G=H^(-1) which reverts the effect of H.
My problem is that with using pwelch, the resulting vectorlength of pyy is way smaller than the vectorlength of y. Since my Filter is determined by P=sqrt(pnn/pyy), I can't multiply fft(y)*H and therefore get no results.
Do you know any help for this Problem?
Or is there a way to go back from a PSD (Welch estimated) to a normal signal (like an inverse function for pwelch)?
In the example you have from the psd documentation, you compute a psd estimate yourself, then put it into a dspdata.psd container and plot it. What dspdata.psd data does here for you is basically compute the frequency axis and provide it to the plot command, nothing more. You get a plot of the spectral density estimate, but that's the one you compute yourself using fft, which is the simplest and worst psd estimate you can get, a so-called periodogram.
Your use of pwelch is almost correct, you just forgot to use the frequency axis information in your plot.
[Pyy,f]=pwelch(x,fs)
plot(f,Pyy)
should give you the peak at the correct frequency.
Your use of pwelch is almost correct, but you have to give the sampling frequency as the 5th argument, and then use the frequency axis information in your plot.
[Pyy,f]=pwelch(y,[],[],[],fs);
plot(f,Pyy)
should give you the peak at the correct frequency.
What pwelch gives you is the spectral density of the signal over Hz. Correct axis labels would therefore be
xlabel('frequency (Hz)')
ylabel('psd (1/Hz)')
The signal you give pwelch is a pure sequence of numbers without physical dimensions. By specifying the sampling rate, the time axis gets a physical unit, s, therefore the resulting frequency is in Hz and the density is in 1/Hz. But still your time series values have no physical dimension, and therefore the density cannot be related to something like W. Has your audiosignal been obtained by a calibrated A/D converter? If yes, you should be able to relate your data to a physical dimension and units, but that's a nontrivial step.
On a personal note, I'd really advise you to brush up on your English, because using software, especially programming interfaces, without properly understanding the documentation is a recipe for disaster.
Related
I am trying to plot the spectrogram of my time domain signal given:
N=5000;
phi = (rand(1,N)-0.5)*pi;
a = tan((0.5.*phi));
i = 2.*a./(1-a.^2);
plot(i);
spectrogram(i,100,1,100,1e3);
The problem is I don't understand the parameters and what values should be given. These values that I am using, I referred to MATLAB's online documentation of spectrogram. I am new to MATLAB, and I am just not getting the idea. Any help will be greatly appreciated!
Before we actually go into what that MATLAB command does, you probably want to know what a spectrogram is. That way you'll get more meaning into how each parameter works.
A spectrogram is a visual representation of the Short-Time Fourier Transform. Think of this as taking chunks of an input signal and applying a local Fourier Transform on each chunk. Each chunk has a specified width and you apply a Fourier Transform to this chunk. You should take note that each chunk has an associated frequency distribution. For each chunk that is centred at a specific time point in your time signal, you get a bunch of frequency components. The collection of all of these frequency components at each chunk and plotted all together is what is essentially a spectrogram.
The spectrogram is a 2D visual heat map where the horizontal axis represents the time of the signal and the vertical axis represents the frequency axis. What is visualized is an image where darker colours means that for a particular time point and a particular frequency, the lower in magnitude the frequency component is, the darker the colour. Similarly, the higher in magnitude the frequency component is, the lighter the colour.
Here's one perfect example of a spectrogram:
Source: Wikipedia
Therefore, for each time point, we see a distribution of frequency components. Think of each column as the frequency decomposition of a chunk centred at this time point. For each column, we see a varying spectrum of colours. The darker the colour is, the lower the magnitude component at that frequency is and vice-versa.
So!... now you're armed with that, let's go into how MATLAB works in terms of the function and its parameters. The way you are calling spectrogram conforms to this version of the function:
spectrogram(x,window,noverlap,nfft,fs)
Let's go through each parameter one by one so you can get a greater understanding of what each does:
x - This is the input time-domain signal you wish to find the spectrogram of. It can't get much simpler than that. In your case, the signal you want to find the spectrogram of is defined in the following code:
N=5000;
phi = (rand(1,N)-0.5)*pi;
a = tan((0.5.*phi));
i = 2.*a./(1-a.^2);
Here, i is the signal you want to find the spectrogram of.
window - If you recall, we decompose the image into chunks, and each chunk has a specified width. window defines the width of each chunk in terms of samples. As this is a discrete-time signal, you know that this signal was sampled with a particular sampling frequency and sampling period. You can determine how large the window is in terms of samples by:
window_samples = window_time/Ts
Ts is the sampling time of your signal. Setting the window size is actually very empirical and requires a lot of experimentation. Basically, the larger the window size, the better frequency resolution you get as you're capturing more of the frequencies, but the time localization is poor. Similarly, the smaller the window size, the better localization you have in time, but you don't get that great of a frequency decomposition. I don't have any suggestions here on what the most optimal size is... which is why wavelets are preferred when it comes to time-frequency decomposition. For each "chunk", the chunks get decomposed into smaller chunks of a dynamic width so you get a mixture of good time and frequency localization.
noverlap - Another way to ensure good frequency localization is that the chunks are overlapping. A proper spectrogram ensures that each chunk has a certain number of samples that are overlapping for each chunk and noverlap defines how many samples are overlapped in each window. The default is 50% of the width of each chunk.
nfft - You are essentially taking the FFT of each chunk. nfft tells you how many FFT points are desired to be computed per chunk. The default number of points is the largest of either 256, or floor(log2(N)) where N is the length of the signal. nfft also gives a measure of how fine-grained the frequency resolution will be. A higher number of FFT points would give higher frequency resolution and thus showing fine-grained details along the frequency axis of the spectrogram if visualised.
fs - The sampling frequency of your signal. The default is 1 Hz, but you can override this to whatever the sampling frequency your signal is at.
Therefore, what you should probably take out of this is that I can't really tell you how to set the parameters. It all depends on what signal you have, but hopefully the above explanation will give you a better idea of how to set the parameters.
Good luck!
I am trying to measure the PSD of a stochastic process in matlab, but I am not sure how to do it. I have posted the exact same question here, but I thought I might have more luck here.
The stochastic process describes wind speed, and is represented by a vector of real numbers. Each entry corresponds to the wind speed in a point in space, measured in m/s. The points are 0.0005 m apart. How do I measure and plot the PSD? Let's call the vector V. My first idea was to use
[p, w] = pwelch(V);
loglog(w,p);
But is this correct? The thing is, that I'm given an analytical expression, which the PSD should (in theory) match. By plotting it with these two lines of code, it looks all wrong. Specifically it looks as though it could need a translation and a scaling. Other than that, the shapes of the two are similar.
UPDATE:
The image above actually doesn't depict the PSD obtained by using pwelch on a single vector, but rather the mean of the PSD of 200 vectors, since these vectors stems from numerical simulations. As suggested, I have tried scaling by 2*pi/0.0005. I saw that you can actually give this information to pwelch. So I tried using the code
[p, w] = pwelch(V,[],[],[],2*pi/0.0005);
loglog(w,p);
instead. As seen below, it looks much nicer. It is, however, still not perfect. Is that just something I should expect? Taking the squareroot is not the answer, by the way. But thanks for the suggestion. For one thing, it should follow Kolmogorov's -5/3 law, which it does now (it follows the green line, which has slope -5/3). The function I'm trying to match it with is the Shkarofsky spectral density function, which is the one-dimensional Fourier transform of the Shkarofsky correlation function. Is it not possible to mark up math, here on the site?
UPDATE 2:
I have tried using [p, w] = pwelch(V,[],[],[],1/0.0005); as I was suggested. But as you can seem it still doesn't quite match up. It's hard for me to explain exactly what I'm looking for. But what I would like (or, what I expected) is that the dip, of the computed and the analytical PSD happens at the same time, and falls off with the same speed. The data comes from simulations of turbulence. The analytical expression has been fitted to actual measurements of turbulence, wherein this dip is present as well. I'm no expert at all, but as far as I know the dip happens at the small length scales, since the energy is dissipated, due to viscosity of the air.
What about using the standard equation for obtaining a PSD? I'd would do this way:
Sxx(f) = (fft(x(t)).*conj(fft(x(t))))*(dt^2);
or
Sxx = fftshift(abs(fft(x(t))))*(dt^2);
Then, if you really need, you may think of applying a windowing criterium, such as
Hanning
Hamming
Welch
which will only somehow filter your PSD.
Presumably you need to rescale the frequency (wavenumber) to units of 1/m.
The frequency units from pwelch should be rescaled, since as the documentation explains:
W is the vector of normalized frequencies at which the PSD is
estimated. W has units of rad/sample.
Off the cuff my guess is that the scaling factor is
scale = 1/0.0005/(2*pi);
or 318.3 (m^-1).
As for the intensity, it looks like taking a square root might help. Perhaps your equation reports an intensity, not PSD?
Edit
As you point out, since the analytical and computed PSD have nearly identical slopes they appear to obey similar power laws up to 800 m^-1. I am not sure to what degree you require exponents or offsets to match to be satisfied with a specific model, and I am not familiar with this particular theory.
As for the apparent inconsistency at high wavenumbers, I would point out that you are entering the domain of very small numbers and therefore (1) floating point issues and (2) noise are probably lurking. The very nice looking dip in the computed PSD on the other hand appears very real but I have no explanation for it (maybe your noise is not white?).
You may want to look at this submission at matlab central as it may be useful.
Edit #2
After inspecting documentation of pwelch, it appears that you should pass 1/0.0005 (the sampling rate) and not 2*pi/0.0005. This should not affect the slope but will affect the intercept.
The dip in PSD in your simulation results looks similar to aliasing artifacts
that I have seen in my data when the original data were interpolated with a
low-order method. To make this clearer - say my original data was spaced at
0.002m, but in the course of cleaning up missing data, trying to save space, whatever,
I linearly interpolated those data onto a 0.005m spacing. The frequency response
of linear interpolation is not well-behaved, and will introduce peaks and valleys
at the high wavenumber end of your spectrum.
There are different conventions for spectral estimates - whether the wavenumber
units are 1/m, or radians/m. Single-sided spectra or double-sided spectra.
help pwelch
shows that the default settings return a one-sided spectrum, i.e. the bin for some
frequency ω will include the power density for both +ω and -ω.
You should double check that the idealized spectrum to which you are comparing
is also a one-sided spectrum. Otherwise, you'll need to half the values of your
one-sided spectrum to get values representative of the +ω side of a
two-sided spectrum.
I agree with Try Hard that it is the cyclic frequency (generally Hz, or in this case 1/m)
which should be specified to pwelch. That said, the returned frequency vector
from pwelch is also in those units. Analytical
spectral formulae are usually written in terms of angular frequency, so you'll
want to be sure that you evaluate it in terms of radians/m, but scale back to 1/m
for plotting.
I am wondering if I am using Fourier Transformation in MATLAB the right way. I want to have all the average amplitudes for frequencies in a song. For testing purposes I am using a free mp3 download of Beethovens "For Elise" which I converted to a 8 kHz mono wave file using Audacity.
My MATLAB code is as follows:
clear all % be careful
% load file
% Für Elise Recording by Valentina Lisitsa
% from http://www.forelise.com/recordings/valentina_lisitsa
% Converted to 8 kHz mono using Audacity
allSamples = wavread('fur_elise_valentina_lisitsa_8khz_mono.wav');
% apply windowing function
w = hanning(length(allSamples));
allSamples = allSamples.*w;
% FFT needs input of length 2^x
NFFT = 2^nextpow2(length(allSamples))
% Apply FFT
fftBuckets=fft(allSamples, NFFT);
fftBuckets=fftBuckets(1:(NFFT/2+1)); % because of symetric/mirrored values
% calculate single side amplitude spectrum,
% normalize by dividing by NFFT to get the
% popular way of displaying amplitudes
% in a range of 0 to 1
fftBuckets = (2*abs(fftBuckets))/NFFT;
% plot it: max possible frequency is 4000, because sampling rate of input
% is 8000 Hz
x = linspace(1,4000,length(fftBuckets));
bar(x,fftBuckets);
The output then looks like this:
Can somebody please tell me if my code is correct? I am especially wondering about the peaks around 0.
For normalizing, do I have to divide by NFFT or length(allSamples)?
For me this doesn't really look like a bar chart, but I guess this is due to the many values I am plotting?
Thanks for any hints!
Depends on your definition of "correct". This is doing what you intended, I think, but it's probably not very useful. I would suggest using a 2D spectrogram instead, as you'll get time-localized information on frequency content.
There is no one correct way of normalising FFT output; there are various different conventions (see e.g. the discussion here). The comment in your code says that you want a range of 0 to 1; if your input values are in the range -1 to 1, then dividing by number of bins will achieve that.
Well, exactly!
I would also recommend plotting the y-axis on a logarithmic scale (in decibels), as that's roughly how the human ear interprets loudness.
Two things that jump out at me:
I'm not sure why you are including the DC (index = 1) component in your plot. Not a big deal, but of course that bin contains no frequency data
I think that dividing by length(allSamples) is more likely to be correct than dividing by NFFT. The reason is that if you want the DC component to be equal to the mean of the input data, dividing by length(allSamples) is the right thing to do.
However, like Oli said, you can't really say what the "correct" normalization is until you know exactly what you are trying to calculate. I tend to use FFTs to estimate power spectra, so I want units like "DAC / rt-Hz", which would lead to a different normalization than if you wanted something like "DAC / Hz".
Ultimately there's no substitute for thinking about exacty what you want to get out of the FFT (including units), and working out for yourself what the correct normalization should be (starting from the definition of the FFT if necessary).
You should also be aware that MATLAB's fft has no requirement to use an array length that is a power of 2 (though doing so will presumably lead to the FFT running faster). Because zero-padding will introduce some ringing, you need to think about whether it is the right thing to do for your application.
Finally, if a periodogram / power spectrum is really what you want, MATLAB provides functions like periodogram, pwelch and others that may be helpful.
I want to remove noises from a recorded sound and make the fft of it finding fundamental frequencies of that sound, but I don't know how to remove those noises. I'm recording the sound of falling objects from different heights. I want to find the relation between the height and the maximum frequency of the recorded sound.
[y,fs]=wavread('100cmfreefall.wav');
ch1=y(:,1);
time=(1/44100)*length(ch1);
t=linspace(0,time,length(ch1));
L=length(ch1);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
Y1=log10(Y);
figure(1)
f = fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y1(1:NFFT/2+1))) ;
[b,a]=butter(10,3000/(44100/2),'high');
Y1=filtfilt(b,a,Y1);
% freqz(b,a)
figure(2)
plot(f,2*abs(Y1(1:NFFT/2+1))) ;
title('Single-Sided Amplitude Spectrum of y(t)');
xlabel('Frequency (Hz)');
ylabel('|Y(f)|')
xlim([0 50000])
% soundsc(ch1(1:100000),44100)
Saying that there is noise in your signal is very vague and doesn't convey much information at all. Some of the questions are:
Is the noise high frequency or low frequency?
Is it well separated from your signal's frequency band or is it mixed in?
Does the noise follow a statistical model? Can it be described as a stationary process?
Is the noise another deterministic interfering signal?
The approach you take will certainly depend on the answers to the above questions.
However, from the experiment setup that you described, my guess is that your noise is just a background noise, that in most cases, can be approximated to be white in nature. White noise refers to a statistical noise model that has a constant power at all frequencies.
The simplest approach will be to use a low pass filter or a band pass filter to retain only those frequencies that you are interested in (a quick look at the frequency spectrum should reveal this, if you do not know it already). In a previous answer of mine, to a related question on filtering using MATLAB, I provide examples of creating low-pass filters and common pitfalls. You can probably read through that and see if it helps you.
A simple example:
Consider a sinusoid with a frequency of 50 Hz, sampled at 1000 Hz. To that, I add Gaussian white noise such that the SNR is ~ -6dB. The original signal and the noisy signal can be seen in the top row of the figure below (only 50 samples are shown). As you can see, it almost looks as if there is no hope with the noisy signal as all structure seems to have been destroyed. However, taking an FFT, reveals the buried sinusoid (shown in the bottom row)
Filtering the noisy signal with a narrow band filter from 48 to 52 Hz, gives us a "cleaned" signal. There will of course be some loss in amplitude due to the noise. However, the signal has been retrieved from what looked like a lost cause at first.
How you proceed depends on your exact application. But I hope this helped you understand some of the basics of noise filtering.
EDIT
#Shabnam: It's been nearly 50 comments, and I really do not see you making any effort to understand or at the very least, try things on your own. You really should learn to read the documentation and learn the concepts and try it instead of running back for every single error. Anyway, please try the following (modified from your code) and show the output in the comments.
[y,fs]=wavread('100cmfreefall.wav');
ch1=y(:,1);
time=(1/fs)*length(ch1);
t=linspace(0,time,length(ch1));
L=length(ch1);
NFFT = 2^nextpow2(L);
f = fs/2*linspace(0,1,NFFT/2+1);
[b,a]=butter(10,3e3/(fs/2),'high');
y1=filtfilt(b,a,ch1);
figure(1)
subplot(2,1,1)
Y=fft(ch1,NFFT)/L;
plot(f,log10(abs(Y(1:NFFT/2+1))))
title('unfiltered')
subplot(2,1,2)
Y1=fft(y1,NFFT)/L;
plot(f,log10(abs(Y1(1:NFFT/2+1))))
title('filtered')
Answer to your question is highly dependent on the characteristics of what you call "noise" - its spectral distribution, the noise being stationary or not, the source of the noise (does it originate in the environment or the recording chain?).
If the noise is stationary, i.e its statistical characteristics do not change over time, you can try recording a few seconds (10-15 is a good initial guess) of noise only, preform FFT, and then subtract the value of the noise in FFT bin n from your measurement FFT bin n.
You can read some background here: http://en.wikipedia.org/wiki/Noise_reduction
We're trying to analyse flow around circular cylinder and we have a set of Cp values that we got from wind tunnel experiment. Initially, we started off with a sample frequency of 20 Hz and tried to find the frequency of vortex shedding using FFT in matlab. We got a frequency of around 7 Hz. Next, we did the same experiment, but the only thing we changed was the sampling frequency- from 20 Hz to 200 Hz. We got the frequency of the vortex shedding to be around 70 Hz (this is where the peak is located in the graph). The graph doesn't change regardless of the Cp data that we enter. The only time the peak differs is when we change the sample frequency. It seems like the increase in the frequency of vortex shedding is proportional to the sample frequency and this doesn't seem to make sense at all. Any help regarding establishing a relation between sample frequency and vortex shedding frequency would be greatly appreaciated.
The problem you are seeing is related to "data aliasing" due to limitations of the FFT being able to detect frequencies higher than the Nyquist Frequency (half-the sampling frequency).
With data aliasing, a peak in real frequency will be centered around (real frequency modulo Nyquist frequency). In your 20 Hz sampling (assuming 70 Hz is the true frequency, that results in zero frequency which means you're not seeing the real information. One thing that can help you with this is to use FFT "windowing".
Another problem that you may be experiencing is related to noisy data generation via single-FFT measurement. It's better to take lots of data, use windowing with overlap, and make sure you have at least 5 FFTs which you average to find your result. As Steven Lowe mentioned, you should also sample at faster rates if possible. I would recommend sampling at the fastest rate your instruments can sample.
Lastly, I would recommend that you read some excerpts from Numerical Recipes in C (<-- link):
Section 12.0 -- Introduction to FFT
Section 12.1 (Discusses data aliasing)
Section 13.4 (Discusses FFT windowing)
You don't need to read the C source code -- just the explanations. Numerical Recipes for C has excellent condensed information on the subject.
If you have any more questions, leave them in the comments. I'll try to do my best in answering them.
Good luck!
this is probably not a programming problem, it sounds like an experiment-measurement problem
i think the sampling frequency has to be at least twice the rate of the oscillation frequency, otherwise you get artifacts; this might explain the difference. Note that the ratio of the FFT frequency to the sampling frequency is 0.35 in both cases. Can you repeat the experiment with higher sampling rates? I'm thinking that if this is a narrow cylinder in a strong wind, it may be vibrating/oscillating faster than the sampling rate can detect..
i hope this helps - there's a 97.6% probability that i don't know what i'm talking about ;-)
If it's not an aliasing problem, it sounds like you could be plotting the frequency response on a normalised frequency scale, which will change with sample frequency. Here's an example of a reasonably good way to plot a frequency response of a signal in Matlab:
Fs = 100;
Tmax = 10;
time = 0:1/Fs:Tmax;
omega = 2*pi*10; % 10 Hz
signal = 10*sin(omega*time) + rand(1,Tmax*Fs+1);
Nfft = 2^8;
[Pxx,freq] = pwelch(signal,Nfft,[],[],Fs)
plot(freq,Pxx)
Note that the sample frequency must be explicitly passed to the pwelch command in order to output the “real” frequency data. Otherwise, when you change the sample frequency the bin where the resonance occurs will seem to shift, which is similar to the problem you describe.
Methinks you need to do some serious reading on digital signal processing before you can even begin to understand all the nuances of the DFT (FFT). If I was you, I'd get grounded in it first with this great book:
Discrete-Time Signal Processing
If you want more of a mathematical treatment that will really expand your abilities,
Fourier Analysis by Körner
Take a look at this related question. While it was originally asked about asked about VB the responses are generically about FFTs
I tried using the frequency response code as above but it seems that I dont have the appropriate toolbox in Matlab. Is there any way to do the same thing without using fft command? So far, this is what I have:
% FFT Algorithm
Fs = 200; % Sampling frequency
T = 1/Fs; % Sample time
L = 65536; % Length of signal
t = (0:L-1)*T; % Time vector
y = data1; % Your CP values go in this vector
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);
% Plot single-sided amplitude spectrum.
loglog(f,2*abs(Y(1:NFFT/2)))
title(' y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
I think there might be something wrong with the code I am using. I'm not sure what though.
A colleague of mine has written some nice GPL-licenced functions for spectral analysis:
http://www.mecheng.adelaide.edu.au/~pvl/octave/
(Update: this code is now part of one of the Octave modules:
http://octave.svn.sourceforge.net/viewvc/octave/trunk/octave-forge/main/signal/inst/.
But it might be tricky to extract just the pieces you need from there.)
They're written for both Matlab and Octave and serve mostly as a drop-in replacement for the analogous functions in the Signal Processing Toolbox. (So the code above should still work fine.)
It may help with your data analysis; better than rolling your own with fft and the like.