A weird feature in the obtained function from calculating the inverse Fourier transform using ifft

Hey,

I am calculating the inverse Fourier transform of a function and obtaining the result in the time domain using the ifft function in matlab. When I perform the Fourier transform and plot the absolute value of the obtained function in time I see a little figure at the other side of the time axis like a tail for the function. I am wondering does it mean there is an issue in the inverse Fourier transform taken?

I have attached the pictures for the whole function I obtain in the time domain using the ifft function and the two chunks of the function at the two ends of the time axis. My question is about the small chunk on the right of the whole function's picture shown.

I appreciate your comments.

[Walter Roberson]

You are showing the result of ifft(), right? So why do you have negative time??

[Shaily_T]

Thanks @Walter Roberson! Yes, this is the result of ifft(). Indeed, I have plotted the absolute value of the ifft() result in time domain. The original function that I am taking the inverse fourier transform from using ifft() is centered at the zero frequency and has both negative and positive frequencies. Isn't that a reason to get negative time? I have attached the original function.

[Paul]

Can you post the data and the code that generates the plot in question?

[Shaily_T]

Thanks @Paul! I have attached the code and data all in the file "ifft".

Answer 1

[Matt J]

IFFTs are N-periodic, so if you shift a lobe far enough to the left or right of a length N sampling, one side of the lobe will start to wrap around circulantly to the other side.

[Walter Roberson]

But this is supposedly an ifft being shown in the time domain.

[Matt J]

Yes, the same remarks apply to IFFTs.

[Shaily_T]

Thanks @Matt J! So, do you mean the small chunk is just because of the IFFT inherent property? And my ifft() result seems reasonable? Could you please elaborate more on your point? I appreciate your time.

[Matt J]

Yes, that is what I mean.

[Shaily_T]

Thanks @Matt J! So, is the small chunk a physical answer? I mean if I write a code for the Fourier transform myself do I still see that small chunk? And should I consider it when I am employing the result of the inverse Fourier transform in other calculations and physical effects?

Also, I am wondering is there a way I can see what is the behaviour of the function in times after the time I see the small chunk? I mean to see if I have my time axis for longer times how that small chunk evolve? In my case to see it at times after t = 1.49e-05. Now it seems it is cut after certain time.

Sorry for asking such a long question. I really appreiate your comment.

[Image Analyst]

What is the expected shape of the output pulse? A spike? A rectangular pulse of some width? If it's supposed to be a narrow rectangular pulse and it's not looking like that, it could be that you don't have enough samples so that the finite window size of your total signal is causing the sinc function that gets convolved with it (because of a finite signal length) having a strong effect. You should be able to reduce it by using more samples. And what is being plotted? The real part, the imaginary part, the magnitude, or the phase? If your spectrum is not symmetric then your ifft will be hermitian and so the phase would, I think, shift your time signal away from the origin and give rise to the signal having different amounts of signal on the right and left side. Did you use fftshift or ifftshift?

[Matt J]

@Shaily_T To sample a time horizon of width T, your frequency sampling interval must satisfy .

[Shaily_T]

Thanks @Image Analyst! I would say if I have only the output on the left of the whole obtained function in time and not the small chunk on the right it is reasonable and as expected. However, I do have that small chunk on the right of the obtained function in time when I calculate the inverse Fourier transform using the ifft() function. No, I don't use ifftshift. And the plot is for the absolute value of the the inverse fourier transform result using ifft() versus time.

[Shaily_T]

@Matt J Thanks! So, I'll play around with the frequency sampling interval to obtain the desirable time horizon of width T to see what is the behaviour of the small chunk on the right of the whole function image.

Answer 2

[Paul]

Hi Shaily_T,

It looks like the purpose of the code is to linearly convolve an input signal with an impulse response via mulitiplication in the frequency domain. However, multiplying the DFT's of two signals is equivalent to a circular convolution. Linear convolution requires zero padding the input and the impulse response in the time domain before taking their DFT's.

Are h and FourierInputField developed from taking the FFT with zero padding?

Linear and Circular Convolution

[Shaily_T]

No, they are not. Only FourierInputField is the result of taking analytical fourier transform from a gaussian pulse in the time domain but I have done it analytically and not using fft function.

[Paul]

Can you provide any more insight into where the data comes from and what you're trying to do?

I'm curious about a couple of things

fnames = unzip('ifft.zip').';
s1 = load(fnames{2});
s2 = load(fnames{4});
figure
plot(subplot(411),s1.nu,real(s2.FourierInputField))
plot(subplot(412),s1.nu,imag(s2.FourierInputField))
plot(subplot(413),s1.nu,real(s1.h))
plot(subplot(414),s1.nu,imag(s1.h))

I assume nu is a frequency vector? Is there a reason why nu is not symmetric around zero?

FourierInputField is the FT of a Gaussian pulse. Is that pulse complex? If not why does FourierInputField have an imaginary component?

Finally, any more insight into h will be hepful. It kind of looks like it's supposed to be perodic. Is it? Why does it have the the sharp discontinuties back to (slightly different?) nonzero values on the left and right sides? Is that feature expected for the problem under consideration?

[Shaily_T]

Thanks for your response @Paul. I appreciate your time. Sorry for the delay. I was not feeling well yesterday.

I am trying to obtain the output pulse from the system. nu can also be symmetric but with the symmetric nu I again see that small chunk on the right of the whole ifft result.

No, the input is not complex but is not centered at t=0 so there is a t-t0 which results in the imaginary component.

I am honestly not sure if that features are supposed to be there. However, I don't think those features are the reason of the small chunk. I tried to use fftshift after employing ifft and then plot the result. If I do that I don't see that small chunk. but the result is a bit different. I am still not sure which of those answers are acceptable.

Because this is indeed another question I posted it here https://www.mathworks.com/matlabcentral/answers/1949413-the-difference-between-the-obtained-result-using-fftshift-when-employing-the-ifft-function

I really appreciate your comment on it.

[Paul]

I wanted to explore more, but it appears the that the ifft.zip file is no longer part of this thread. At least, I can't find it to attach to this comment. Also, weren't there four .png (or .jpg) files attached to the question? Those aren't there anymore either.

[Walter Roberson]

@Paul

The file is still attached at https://www.mathworks.com/matlabcentral/answers/1947528-a-weird-feature-in-the-obtained-function-from-calculating-the-inverse-fourier-transform-using-ifft#comment_2708168

[Paul]

When I click the paperclip and tab "Link from this thread" the only item in the list is the .png file attached to this comment.

[Walter Roberson]

Good point -- but the file is still there and can be downloaded. I just did that and re-uploaded it to this discussion.

[Shaily_T]

@Paul @Walter Roberson It is my fault. I deleted the .png files and the code as I have put the exact same images and file in my new question here https://www.mathworks.com/matlabcentral/answers/1949413-the-difference-between-the-obtained-result-using-fftshift-when-employing-the-ifft-function?s_tid=mlc_ans_email_ques. But it seems it was not an issue to have them in both questions. So, I reattached them. Sorry for the confusion.

[Paul]

Without knowing much about FourierInputField, h, or nu .....

Load and plot the data

fnames = unzip('ifft.zip');
load(fnames{2},'nu','h');
load(fnames{4},'FourierInputField');
figure
plot(subplot(411),nu,real(FourierInputField))
plot(subplot(412),nu,imag(FourierInputField))
plot(subplot(413),nu,real(h))
plot(subplot(414),nu,imag(h))

We see that the data is not symmetric around 0, so I'll chop the data off on the right to make the data symmetric around zero.

ii = find(nu > -nu(1));
FourierInputField(ii) = [];
nu(ii)                = [];
h(ii)                 = [];

Plot the truncated data.

figure
plot(subplot(411),nu,real(FourierInputField))
plot(subplot(412),nu,imag(FourierInputField))
plot(subplot(413),nu,real(h))
plot(subplot(414),nu,imag(h))

It was stated in this comment that FourierInputField are samples of the CTFT of a Gaussian pulse. If this is the case, FourierInputField should have linear phase

figure
plot(nu,unwrap(angle(FourierInputField))),grid

The phase looks linear, but it should go through zero at nu = 0. I think it doesn't in this plot because of unwrapping confusion, but it really does

figure
plot(nu,angle(FourierInputField)),grid
xlim([-.1 .1]*1e8)

Because FourierInputField is on both sides of zero, we first iffshift, then take the ifft, and then fftshift to get the time domain result centered on n = 0

TimeInputField = fftshift(ifft(ifftshift(FourierInputField)));

Now make a discrete-time vector. No element in the nu vector is equal to zero, which seems odd

min(abs(nu))

ans = 1.3526e+04

so we don't have the typical case where we'd have nu = n*Fs; Instead I'll approximate the time vector as

N = numel(nu)

N = 8683

Fs = mean(diff(nu))*N;
Ts = 1/Fs;
t = (-8682/2:8682/2)*Ts;

Plot TimeInputField

figure
plot(t,abs(TimeInputField))
xline(34*Ts)
xlim([0 50]*Ts)

It's basically a pulse centered on t = 34*Ts. If it's a Gaussian pulse, the phase should be zero

figure
plot(t,angle(TimeInputField))
xlim([0 50]*Ts)

The phase is zero for values of t where where the pulse amplitude is not insignificant.

Now compute the output, which is the circular convolution of the TimeInputField and Timeh, whatever that is. Even though h those sharp discontinuites and flat spots, they occur out on the tails of FourierInputField, so maybe those features of h don't matter.

y = fftshift(ifft(ifftshift(FourierInputField.*h)));
figure
plot(t,abs(y)),grid
xlim([-50 100]*Ts)

y does have some peaking in negative time. Those peaks whould show up on the far right of the result if we hadn't applied fftshift

ynoshift = (ifft(ifftshift(FourierInputField.*h)));
figure
plot((0:numel(ynoshift)-1)*Ts,abs(ynoshift))
xlim([numel(ynoshift)-50, numel(ynoshift)]*Ts)

As done in the code provided with the question, we can zero pad in the frequency domain to interpolate the time domain result. When plotting we have to scale Ts to account for the additional samples added in the frequency domain.

y = fftshift(ifft(ifftshift([zeros(1,4e4),FourierInputField, zeros(1,4e4)].*[zeros(1,4e4), h, zeros(1,4e4)])));
Tsinterp = Ts*N/numel(y);
figure
t = (-88682/2:88682/2)*Tsinterp;
plot(t,abs(y)),grid
xline(34*Ts)
xlim([-500 1000]*Tsinterp)