Number of cycles before and after FFT

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Brad 2020-4-11

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评论： Star Strider 2020-4-18

Hi.

I have the signal y1 (fig1), take its fft, do some processing on it and take ifft to take it back to time domain. What I get is fig 2. I have two questions though:

1. Why the resultant signal is less dense than the original one (with less number of cycles)? Shouldn't they be similar?

2. Why the early indices of the resultant vector (y2), at t~ 0, have much large values? This happens sometimes in my analysis.

Thank you

Brad

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dpb 2020-4-11

"do some processing on it and take ifft to take it back to time domain."

Well, then the result could be anything -- it all depends on what "some processing" is.

What processing you did resulted in, well, the results you got. Some other processing would produce other results.

Brad 2020-4-11

Thank you for your response.

It involved a few simple multiplications and divisions. Do these two signal don't necessary have to be similar?

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Star Strider 2020-4-12

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Brad — You requested that I answer this so I’ll do my best.

It appears to me that ‘some processing’ is essentially some sort of lowpass filter. There are fewer cycles because the higher frequencies have been filtered out. My guess is that this is the single explanation for both of your observations.

The superposition of the higher and lower frequencies (and the absence of the higher frequencies after filtering) may account for the amplitude variations.

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dpb 2020-4-12

编辑：dpb 2020-4-13

在 MATLAB Online 中打开

"R function is not symmetric and ...useless if it is applied to fft result?"

You must make it symmetric around the zero frequency point and then apply to both negative and positive frequencies...

Or, if you're careful to put things in the right place/order, you can operation on the positive frequency side and then flip that result around the zero frequency. BUT the symmetry point is in the middle of the complex vector and goes both ways from there so have to be careful...also NB: the DC and Fmax components are unique in the vector -- there's only one of each of them so they aren't in the symmetry game...

Using the example at doc fft because had it still in memory owing to recent other Q? (with the exception that subtracted mean so DC component is almost identically zero as that had to do with other poster's problem), the figure for real()/imag() components looks like

and since L=1500, the vertical line is at point 751 (L/2+1).

>> arrayfun(@(i) fprintf('%d %12.5e + %12.5e\n',i,real(Y(i)),imag(Y(i))),L/2+1+[-5:5])
2.42442e-07 +  2.41797e-08
-7.05677e-09 +  1.68338e-08
1.43380e-07 +  5.22938e-08
-4.92976e-08 +  4.09230e-08
8.34723e-08 +  3.18481e-08
-1.14062e-07 +  0.00000e+00
8.34723e-08 + -3.18481e-08
-4.92976e-08 + -4.09230e-08
1.43380e-07 + -5.22938e-08
-7.05677e-09 + -1.68338e-08
2.42442e-07 + -2.41797e-08
>>

shows symmetry around the midpoint, while the two special components are:

Y(1)
ans =
   1.2904e-20
>> Y(end)
ans =
  -2.3242e-08 + 1.3694e-05i
>> 

Where the DC is real (0 imaginary component) and the Fmax component is unique in the array.

That's why in that example computation/normalization of the one-sided PSD, P1, the factor of 2 is applied to (2:end-1).

Star Strider 2020-4-18

As always, my pleasure! Thank you.

You can plot the fft however you wish. If you plot the entire fft (after doing fftshift), the frequency axis goes from the negative of the Nyquist frequency to the positive of the Nyquist frequency, with the same number of elements as in the fft result (the reason for using linspace to create it). Depending on the intended result, this can occasionally be useful. However for most purposes, plotting only the positive half of the fft gives all the desired informaiton.

The information in the fft result is essentially duplicated in the second half (or negative frequencies after doing fftshift), so there is no real point in plotting it. The only difference is that the second half (or negative) part is the complex conjugate for the values in the first (or positive) half, a distinction that the abs function obliterates. The normalisation is done by dividing it by the length of the original signal (as opposed to the increased length of the fft if zero-padded) . The energy is equally divided between the halves of the fft result, so doubling it in half the fft result approximates the amplitude of the original signal.

Brad 2020-4-18

1. What is the physical or mathematical reason for fftshift?

2. Is it always necessary to use fftshift? if not, when do we have to use it?

Thanks

Star Strider 2020-4-18

在 MATLAB Online 中打开

As always, my pleasure!

1. The Fourier transform is symmetrical by definition, going from

to ∞, and fftshift allows for that symmetry convention.

2. It is never necessary to use it. In some instances it is convenient to use it, for example to identify symmetry in the Fourier transform of a square wave:

syms f(t) t w 
f(t) = 1;
F(w) = int(f*exp(-1i*w*t), t, -1, 1)
figure
fplot(F, [-10*pi 10*pi])
grid

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