Complex FFT then Inverse FFT MATLAB - matlab

I am using the FFT function in Matlab in an attempt to analyze the output of a Travelling Wave Laser Model.
The of the model is in the time domain in the form (real, imaginary), with the idea being to apply the FFT to the complex output, to obtain phase and amplitude information in the frequency domain:
%load time_domain field data
data = load('fft_data.asc');
% Calc total energy in the time domain
N = size(data,1);
dt = data(2,1) - data (1,1);
field_td = complex (data(:,4), data(:,5));
wavelength = 1550e-9;
df = 1/N/dt;
frequency = (1:N)*df;
dl = wavelength^2/3e8/N/dt;
lambda = -(1:N)*dl +wavelength + N*dl/2;
%Calc FFT
FT = fft(field_td);
FT = fftshift(FT);
counter=1;
phase=angle(FT);
amptry=abs(FT);
unwraptry=unwrap(phase);
Following the unwrapping, a best fit was applied to the phase in the region of interest, and then subtracted from the phase itself in an attempt to remove wavelength dependence of phase in the region of interest.
for i=1:N % correct phase and produce new IFFT input
bestfit(i)=1.679*(10^10)*lambda(i)-26160;
correctedphase(i)=unwraptry(i)-bestfit(i);
ReverseFFTinput(i)= complex(amptry(i)*cos(correctedphase(i)),amptry(i)*sin(correctedphase(i)));
end
Having performed the best fit manually, I now have the Inverse FFT input as shown above.
pleasework=ifft(ReverseFFTinput);
from which I can now extract the phase and amplitude information in the time domain:
newphasetime=angle(pleasework);
newamplitude=abs(pleasework);
However, although the output for the phase is greatly different compared to the input in the time domain
the amplitude of the corrected data seems to have varied little (if at all!),
despite the scaling of the phase. Physically speaking this does not seem correct, as my understanding is that removing wavelength dependence of phase should 'compress' the pulsed input i.e shorten pulse width but heighten peak.
My main question is whether I have failed to use the inverse FFT correctly, or the forward FFT or both, or is this something like a windowing or normalization issue?
Sorry for the long winded question! And thanks in advance.

You're actually seeing two effects.
First the expected one goes. You're talking about "removing wavelength dependence of phase". If you did exactly that - zeroed out the phase completely - you would actually get a slightly compressed peak.
What you actually do is that you add a linear function to the phase. This does not compress anything; it is a well-known transformation that is equivalent to shifting the peaks in time domain. Just a textbook property of the Fourier transform.
Then goes the unintended one. You convert the spectrum obtained with fft with fftshift for better display. Thus before using ifft to convert it back you need to apply ifftshift first. As you don't, the spectrum is effectively shifted in frequency domain. This results in your time domain phase being added a linear function of time, so the difference between the adjacent points which used to be near zero is now about pi.

Related

correct sampling points for FFT

I want to calculate the Fourier series of a signal using FFT on matlab. And I came across the following unexpected issue. Mere example
If I define a grid and then compute the fft as:
M=59;
x= deal(1*(0:M-1)/M);
y=3*cos(2*pi*x);
Yk=fftshift(fft2(y)/(M));
which gives me the exact analytic values expected: Yk(29)=1.5; Yk(31)=1.5; zeros anything else
but if I define the grid as, and repeat the fft calculation:
x=0:1/(M-1):1;
y=3*cos(2*pi*x);
Yk=fftshift(fft2(y)/(M));
got the Yk's values completely screwed up
This is an annoying issue since I have to analyse many signals data that was sampled as in the second method so the Yk's values will be wrong. Is there a way to workaround this? an option to tell something to the fft function about the way the signal was sampled. Have no way to resample the data in the correct way.
The main reason to avoid have spectral leaking, is that I do further operations with these Fourier terms individually Real and Imag parts. And the spectral leaking is messing the final results.
The second form of sampling includes one sample too many in the period of the cosine. This causes some spectral leaking, and adds a small shift to your signal (which leads to non-zero imaginary values). If you drop the last point, you'll cosine will again be sampled correctly, and you'll get rid of both of these effects. Your FFT will have one value less, I don't know if this will affect your analyses in any way.
x = 0:1/(M-1):1;
y = 3*cos(2*pi*x);
Yk = fftshift(fft2(y(1:end-1))/(M-1));
>> max(abs(imag(Yk)))
ans =
1.837610523517500e-16

Matlab fft on one period of sinewave returns phase of -pi/2. Why?

While trying to understand Fast Fourier Transform I encountered a problem with the phase. I have broken it down to the simple code below. Calculating one period of a 50Hz sinewave, and applying an fft algorithm:
fs = 1600;
dt = 1/fs;
L = 32;
t=(0:L-1)*dt;
signal = sin(t/0.02*2*pi);
Y = fft(signal);
myAmplitude = abs(Y)/L *2 ;
myAngle = angle(Y);
Amplitude_at_50Hz = myAmplitude(2);
Phase_at_50Hz = myAngle(2);
While the amplitude is ok, I don't understand the phase result. Why do I get -pi/2 ? As there is only one pure sinewave, I expected the phase to be 0. Either my math is wrong, or my use of Matlab, or both of them... (A homemade fft gives me the same result. So I guess I am stumbling over my math.)
There is a similar post here: MATLAB FFT Phase plot. However, the suggested 'unwrap' command doesn't solve my problem.
Thanks and best regards,
DanK
The default waveform for an FFT phase angle of zero is a cosine wave which starts and ends in the FFT window at 1.0 (not a sinewave which starts and ends in the FFT window at 0.0, or at its zero crossings.) This is because the common nomenclature is to call the cosine function components of the FFT basis vectors (the complex exponentials) the "real" components. The sine function basis components are called "imaginary", and thus infer a non-zero complex phase.
That is what it should be. If you used cosine, you would have found a phase of zero.
Ignoring numerical Fourier transforms for a moment and taking a good old Fourier transform of sin(x), which I am too lazy to walk through, we get a pair of purely imaginary deltas.
As for an intuitive reason, recall that a discrete Fourier transform is averaging a bunch of points along a curve in the complex plane while turning at the angular frequency of the bin you're computing and using the amplitude corresponding to the sample. If you sample a sine curve while turning at its own frequency, the shape you get is a circle centered on the imaginary axis (see below). The average of that is of course going to be right on the imaginary axis.
Plot made with wolfram alpha.
Fourier transform of a sine function such as A*sin((2*pi*f)*t) where f is the frequency will yield 2 impulses of magnitude A/2 in the frequency domain at +f and -f where the associated phases are -pi/2 and pi/2 respectively.
You can take a look at its proof here:
http://mathworld.wolfram.com/FourierTransformSine.html
So the code is working fine.

how to use ifft function in MATLAB with experimental data

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.

Fast fourier transform for deasonalizing data in MATLAB

I'm very much a novice at signal processing techniques, but I am trying to apply the fast fourier transform to a daily time series to remove the seasonality present in the data. The example I am working with is from here:
http://www.mathworks.com/help/signal/ug/frequency-domain-linear-regression.html
While I understand how to implement the code as it is written in the example, I am having a hard time adapting it to my specific application. What I am trying to do is create a preprocessing function which deseasonalizes the training data using similar code to the above example. Then, using the same estimated coefficients from the in-sample data, deseasonalize the out-of-sample data to preserve its independence from the in-sample data. Basically, once the coefficients are estimated, I will normalize each new data point using the same coefficients. I suspect this is akin to estimating a linear trend, then removing it from the in-sample data, and then using the same linear model on unseen data to detrend it i the same manner.
Obviously, when I estimate the fourier coefficients, the vector I get out is equal to the length of the in-sample data. The out-of-sample data is comprised of much fewer observations, so directly applying them is impossible.
Is this sort of analysis possible using this technique or am I going down a dead end road? How should I approach that using the code in the example above?
What you want to do is certainly possible, you are on the right track, but you seem to misunderstand a few points in the example. First, it is shown in the example that the technique is the equivalent of linear regression in the time domain, exploiting the FFT to perform in the frequency domain an operation with the same effect. Second, the trend that is removed is not linear, it is equal to a sum of sinusoids, which is why FFT is used to identify particular frequency components in a relatively tidy way.
In your case it seems you are interested in the residuals. The initial approach is therefore to proceed as in the example as follows:
(1) Perform a rough "detrending" by removing the DC component (the mean of the time-domain data)
(2) FFT and inspect the data, choose frequency channels that contain most of the signal.
You can then use those channels to generate a trend in the time domain and subtract that from the original data to obtain the residuals. You need not proceed by using IFFT, however. Instead you can explicitly sum over the cosine and sine components. You do this in a way similar to the last step of the example, which explains how to find the amplitudes via time-domain regression, but substituting the amplitudes obtained from the FFT.
The following code shows how you can do this:
tim = (time - time0)/timestep; % <-- acquisition times for your *new* data, normalized
NFpick = [2 7 13]; % <-- channels you picked to build the detrending baseline
% Compute the trend
mu = mean(ts);
tsdft = fft(ts-mu);
Nchannels = length(ts); % <-- size of time domain data
Mpick = 2*length(NFpick);
X(:,1:2:Mpick) = cos(2*pi*(NFpick-1)'/Nchannels*tim)';
X(:,2:2:Mpick) = sin(-2*pi*(NFpick-1)'/Nchannels*tim)';
% Generate beta vector "bet" containing scaled amplitudes from the spectrum
bet = 2*tsdft(NFpick)/Nchannels;
bet = reshape([real(bet) imag(bet)].', numel(bet)*2,1)
trend = X*bet + mu;
To remove the trend just do
detrended = dat - trend;
where dat is your new data acquired at times tim. Make sure you define the time origin consistently. In addition this assumes the data is real (not complex), as in the example linked to. You'll have to examine the code to make it work for complex data.

How to use inverse FFT on amplitude-frequency response?

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.