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.
Related
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.
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
Today I have stumbled upon a strange outcome in matlab. Lets say I have a sine wave such that
f = 1;
Fs = 2*f;
t = linspace(0,1,Fs);
x = sin(2*pi*f*t);
plot(x)
and the outcome is in the figure.
when I set,
f = 100
outcome is in the figure below,
What is the exact reason of this? It is the Nyquist sampling theorem, thus it should have generated the sine properly. Of course when I take Fs >> f I get better results and a very good sine shape. My explenation to myself is that Matlab was having hardtime with floating numbers but I am not so sure if this is true at all. Anyone have any suggestions?
In the first case you only generate 2 samples (the third input of linspace is number of samples), so it's hard to see anything.
In the second case you generate 200 samples from time 0 to 1 (including those two values). So the sampling period is 1/199, and the sampling frequency is 199, which is slightly below the Nyquist rate. So there is aliasing: you see the original signal of frequency 100 plus its alias at frequency 99.
In other words: the following code reproduces your second figure:
t = linspace(0,1,200);
x = .5*sin(2*pi*99*t) -.5*sin(2*pi*100*t);
plot(x)
The .5 and -.5 above stem from the fact that a sine wave can be decomposed as the sum of two spectral deltas at positive and negative frequencies, and the coefficients of those deltas have opposite signs.
The sum of those two sinusoids is equivalent to amplitude modulation, namely a sine of frequency 99.5 modulated by a sine of frequency 1/2. Since time spans from 0 to 1, the modulator signal (whose frequency is 1/2) only completes half a period. That's what you see in your second figure.
To avoid aliasing you need to increase sample rate above the Nyquist rate. Then, to recover the original signal from its samples you can use an ideal low pass filter with cutoff frequency Fs/2. In your case, however, since you are sampling below the Nyquist rate, you would not recover the signal at frequency 100, but rather its alias at frequency 99.
Had you sampled above the Nyquist rate, for example Fs = 201, the orignal signal could ideally be recovered from the samples.† But that would require an almost ideal low pass filter, with a very sharp transition between passband and stopband. Namely, the alias would now be at frequency 101 and should be rejected, whereas the desired signal would be at frequency 100 and should be passed.
To relax the filter requirements you need can sample well above the Nyquist rate. That way the aliases are further appart from the signal and the filter has an easier job separating signal from aliases.
† That doesn't mean the graph looks like your original signal (see SergV's answer); it only means that after ideal lowpass filtering it will.
Your problem is not related to the Nyquist theorem and aliasing. It is simple problem of graphic representation. You can change your code that frequency of sine will be lower Nyquist limit, but graph will be as strange as before:
t = linspace(0,1,Fs+2);
plot(sin(2*pi*f*t));
Result:
To explain problem I modify your code:
Fs=100;
f=12; %f << Fs
t=0:1/Fs:0.5; % step =1/Fs
t1=0:1/(10*Fs):0.5; % step=1/(10*Fs) for precise graphic representation
subplot (2, 1, 1);
plot(t,sin(2*pi*f*t),"-b",t,sin(2*pi*f*t),"*r");
subplot (2, 1, 2);
plot(t1,sin(2*pi*f*t1),"g",t,sin(2*pi*f*t),"r*");
See result:
Red star - values of sin(2*pi*f) with sampling rate of Fs.
Blue line - lines which connect red stars. It is usual data representation of function plot() - line interpolation between data points
Green curve - sin(2*pi*f)
Your eyes and brain can easily understand that these graphs represent the sine
Change frequency to more high:
f=48; % 2*f < Fs !!!
See on blue lines and red stars. Your eyes and brain do not understand now that these graphs represent the same sine. But your "red stars" are actually valid value of sine. See on bottom graph.
Finally, there is the same graphics for sine with frequency f=50 (2*f = Fs):
P.S.
Nyquist-Shannon sampling theorem states for your case that if:
f < 2*Fs
You have infinite number of samples (red stars on our plots)
then you can reproduce values of function in any time (green curve on our plots). You must use sinc interpolation to do it.
copied from Matlab Help:
linspace
Generate linearly spaced vectors
Syntax
y = linspace(a,b)
y = linspace(a,b,n)
Description
The linspace function generates linearly spaced vectors. It is similar to the colon operator ":", but gives direct control over the number of points.
y = linspace(a,b) generates a row vector y of 100 points linearly spaced between and including a and b.
y = linspace(a,b,n) generates a row vector y of n points linearly spaced between and including a and b. For n < 2, linspace returns b.
Examples
Create a vector of 100 linearly spaced numbers from 1 to 500:
A = linspace(1,500);
Create a vector of 12 linearly spaced numbers from 1 to 36:
A = linspace(1,36,12);
linspace is not apparent for Nyquist interval, so you can use the common form:
t = 0:Ts:1;
or
t = 0:1/Fs:1;
and change the Fs values.
The first Figure is due to the approximation of '0': sin(0) and sin(2*pi). We can notice the range is in 10^(-16) level.
I wrote the function reconstruct_FFT that can recover critically sampled data even for short observation intervals if the input sequence of samples is periodic. It performs lowpass filtering in the frequency domain.
I am trying to create an application for calculating coefficients for a graphic equalizer FIR filter. I am doing some prototyping in Matlab but I have some problems.
I have started with the following Matlab code:
% binamps vector holds 2^13 = 8192 bins of desired amplitude values for frequencies in range 0.001 .. 22050 Hz (half of samplerate 44100 Hz)
% it looks just fine, when I use Matlab plot() function
% now I get ifft
n = size(binamps,1);
iff = ifft(binamps, n);
coeffs = real(iff); % throw away the imaginary part, because FIR module will not use it anyway
But when I do the fft() of the coefficients, I see that the frequencies are stretched 2 times and the ending of my AFR data is lost:
p = fft(coeffs, n); % take the fourier transform of coefficients for a test
nUniquePts = ceil((n+1)/2);
p = p(1:nUniquePts); % select just the first half since the second half
% is a mirror image of the first
p = abs(p); % take the absolute value, or the magnitude
p = p/n; % scale by the number of points so that
% the magnitude does not depend on the length
% of the signal or on its sampling frequency
p = p.^2; % square it to get the power
sampFreq = 44100;
freqArray = (0:nUniquePts-1) * (sampFreq / n); % create the frequency array
semilogx(freqArray, 10*log10(p))
axis([10, 30000 -Inf Inf])
xlabel('Frequency (Hz)')
ylabel('Power (dB)')
So I guess, I am using ifft wrong. Do I need to make my binamps vector twice as long and create a mirror in the second part of it? If it is the case, then is it just a Matlab's implementation of ifft or also other C/C++ FFT libraries (especially Ooura FFT) need mirrored data for inverse FFT?
Is there anything else I should know to get the FIR coefficients out of ifft?
Your frequency domain vector needs to be complex rather than just real, and it needs to be symmetric about the mid point in order to get a purely real time domain signal. Set the real parts to your desired magnitude values and set the imaginary parts to zero. The real parts need to have even symmetry such that A[N - i] = A[i] (A[0] and A[N / 2] are "special", being the DC and Nyquist components - just set these to zero.)
The above applies to any general purpose complex-to-complex FFT/IFFT, not just MATLAB's implementation.
Note that if you're trying to design a time domain filter with an arbitrary frequency response then you'll need to do some windowing in the frequency domain first. You might find this article helpful - it talks about arbitrary FIR filter design usign MATLAB, in particular fir2.
To get a real result, the input to any typical generic IFFT (not just Matlab's implementation) needs to be complex-conjugate-symmetric. So doing an IFFT with a given number of independent specification points will require an FFT at least twice as long (preferably even longer to allow for some transition to zero from the highest frequency cut-off).
Trying to get a real result by throwing away the "imaginary" portion of a complex result won't work, as you will be throwing away actual required information content the time-domain filter needs for the given frequency response input to the IFFT. However, if the original data is conjugate-symmetric, then the imaginary portion of the IFFT/FFT result will be (usually insignificant) rounding-error noise that can be thrown away.
Also, the DTFT of a finite frequency response will produce an infinitely long FIR. To get a finite length FIR, you will need to compromise the specification for your frequency response specification so that there is little energy left in the latter portion of the time-domain representation that has to be truncated from the FIR to make it realizable or finite. One common (but not necessary the best) way to do this is to window the FIR result produced by the IFFT, and, by trial-and-error, try different windows until you find a FIR filter for which an FFT produces a result "close enough" to your original frequency spec.
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.