Questions about converting a signal from frequency domain to time domain

[amigo_pp]

as we know, according IFFT, we can convert a signal from frequency domain to time domain, the problem is , if the result of IFFT is complex, then what's the time-domain signal, the real part of IFFT or something else? And what's the physical meanings of the imaginary part of a time-domain signal?
i am really puzzled!

 

[shebo]

Re: for help about IFFT!

i can answer the second question the imaginary part means the phase so any signal has an imaginary part this mean there is a phase difference

 

[amigo_pp]

Re: for help about IFFT!

but phase is information in frequency-domain, isn't it?
so i think for a time-domain signal expressed in complex, it is just for computational convenience.
does anyone agree with me?

 

[safwatonline]

for help about IFFT!

i think we take the real part, but not sure,
i dont see a direct meaning for a complex value in time

 

[LouisSheffield]

for help about IFFT!

A real (i.e. measurable) signal will have a complex Fourier trasnform (or series, if one looks at it that way). The inverse Fourier transform of such a complex spectrum will again be real only.

 

[amigo_pp]

Re: for help about IFFT!

A real (i.e. measurable) signal will have a complex Fourier trasnform (or series, if one looks at it that way). The inverse Fourier transform of such a complex spectrum will again be real only.

of course it is.
but the case is that after a real signal is transformed into frequency-domain by FT,another function(always not linear) has effect on it and we get a new function in fre-domain, so we cann't conclude the IFT of this new function is a real signal, is not that?

 

[LouisSheffield]

for help about IFFT!

After a Fourier transform, any function applied to the frequency domain result, which leaves a non-real IFFT cannot be physically realized.

I have run across MANY instances where the time origin of the original signal has been the culprit of what you my have run across.
(i.e. a textbook FFT integrates from 0 to N-1, due solely to the algorithm, but in most cases it would have been preferred to be integrated from (1-N)/2 to (N-1)/2)

Do you have a concrete function (including nonlinear ones) in particular that illustrates your point?
(by the way, the question you asked is deceptively fascinating)

Added after 8 minutes:

(I'll try to remember tomorrow at work)
I have a reference that talks about the baseband (complex) representation of real signals.
That allows you to consider a wider class of signals.
I rarely work with real signals (in radar) any more.
The first thing to do is make them complex.

 

[safwatonline]

for help about IFFT!

hello Lousi,
can u upload this book

 

[amigo_pp]

Re: for help about IFFT!

to Louis
perhaps i just thought the problem as a pure mathematic one. In fact ,i have realized this point.

by the way, i also want to get more information about the book you mentioned, it will be best if you can upload it!

 

[LouisSheffield]

for help about IFFT!

(coming soon) - it's just a section from powerpoint course notes ...

Separately, I suggest you'll both be rewarded by looking into the Fast Hartley transform - in it's framing, both the time and frequency domain are purely real, since the transform kernel itself is purely real.
And algorithms exist which are every bit as fast (order of magnitude) as the standard FFT/IFFT.

Added after 1 hours 3 minutes:

https://mathworld.wolfram.com/HartleyTransform.html

 

[wolfi64]

for help about IFFT!

FFT is special (fast) implementation of DFT. In DFT you get a symmetric spectrum (conjugate complex). As long as you make sure, that the symmetric part (-fs/2...0) is the conjugated complex part of frequencies between 0 and fs/2, you will always get a real time-domain signal via IFFT. This is because all complex numbers add up with conjugate complex to real numbers.