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Bottom-line: how to create a 90 degrees out of phase signal from the real part of the fourier transform of a 1D signal (i.e., fft of a line of an image). ? *(I *(I think the Fourier shift thm can help but mot exactly sure how yet)

In MRI the actual FID echo signal is acquired in quadrature (cf. http://mriquestions.com/real-v-imaginary.html ). Also each line of the k-space (=2D Fourier transform raw data in MRI jargon) is acquired sequentially, i.e., a sampled 1D (temporal) signal (in the receiver coil). Each sample is put in each x,y position in the 2D kspace (which correspond to spatial frequencies), which is the Fourier transformed to get a spatial (visually interpretable for humans) image.

That being said, I want to take any image, for the sake of the example, lets say the artificial 'cameraman' image of Matlab. We can do a DFT of this image and get a real and imaginary image. I want to make as if/simulate that this image was MRI acquired. Therefore, as I understand the description here http://mriquestions.com/real-v-imaginary.html , one could take the real part (but it doesnt matter if it was real, it could be the imaginary, but let's take the real one for the example), the dephase it w.r.t itself by 90 degrees. Then discard the imaginary part and replace it with this "dephased" version of the real part, since the real and imag parts are supposed to be the same but just dephased by 90 degrees.

The special thing is that this will result in a "phase" image e.g.: https://www.researchgate.net/figure/a-A-256-256-MRI-head-phase-image-b-its-corresponding-residue-distribution_fig6_232321363, in the spatial domain too. I thought about the Fourier Shift theorem: https://www.dsprelated.com/freebooks/sasp/Shift_Theorem.html ... but not sure how to use it...

ideas ?

SheppLogan
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  • I am having difficulty extracting what you question is specifically. Would you be able to add a first paragraph that bottom-lines your specific question? – Dan Boschen Feb 27 '20 at 17:20
  • Ok, I will try. Thanks for trying to answer – SheppLogan Feb 27 '20 at 17:22
  • So you want to create a quadrature of a given signal, but further instead of using the Hilbert Transform (which is one approach), you are looking for a way to do it from the real part of the Fourier Transform. Do I follow correctly? – Dan Boschen Feb 27 '20 at 17:33
  • I dont care which method is used to create the quadrature signal, as long as the I amd Q channels are dephased by 90 degrees (and the we take the real (or imaginary) doesn't matter, part of the Fourier transform of a 1D signal as the signal to work with if you see what i mean – SheppLogan Feb 27 '20 at 17:47
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    Then look into the Hilbert Transform. This is the typical way to convert signals that are over a wider frequency range to quadrature. – Dan Boschen Feb 27 '20 at 17:48
  • Or maybe more precisely I want to assume that e.g. the real part was measured and i want to create the other channel dephased by 90, which i will then use as the imagonary part (but that s just how i use it – SheppLogan Feb 27 '20 at 17:49
  • Please be more precise about it – SheppLogan Feb 27 '20 at 17:49
  • You can research it and ask any specific questions once you do. We’re more Q&A here and not a tutorial site. For example we typically wouldn’t answer “How do I do the Hilbert Transform” since you can easily research that online – Dan Boschen Feb 27 '20 at 17:51
  • Yes but maybe why is it relevant here?etc – SheppLogan Feb 27 '20 at 17:54
  • But thanks anyways – SheppLogan Feb 27 '20 at 17:55
  • If you measured the real part and want to determine a signal that is the quadrature version of it, then you can use the Hilbert Transform to so that. – Dan Boschen Feb 27 '20 at 18:02
  • And to your FT question as I answered below the imaginary part of the FT will be the Hilbert Transform of the Real part of your signal was causal. So if you have the fill FT you will already have what you are looking for – Dan Boschen Feb 27 '20 at 18:04
  • Thanks, sounds interesting, i will look at that. – SheppLogan Feb 27 '20 at 18:12
  • Was typing with fat fingers --- if you have the "full FFT" is what I meant. Good luck! – Dan Boschen Feb 27 '20 at 18:33

1 Answers1

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For a causal time domain signal, the Fourier Transform will be complex with the imaginary part as the Hilbert Transform of the real part.

This may be clearer by noting the following additional properties of the Fourier Transform:

  • The FT of a non-causal real even waveform ($f(t) = f(-t)$) will be all real.
  • The FT of a non-causal real odd waveform ($f(t) = -f(-t)$) will be all imaginary.
  • The FT of a non-causal imaginary odd waveforrm will be all real.
  • The FT of a non-causal imaginary even waveform will be all imaginary.

It is easy to prove all of the above by observing waveforms as rotating phasors of constant magnitude in time on a complex IQ plane, since the FT is the mangitude of each of those phasors at the position in frequency based on their rate of rotation. (In fact the Fourier Series Expansion is specifically to decompose any arbitrary single valued analytic function into such rotating phasors).

With those properties in mind, you can extend that to causal and anti-causal functions by adding or subtracting the cases listed above: Causal and anti-causal functions are the sum of an even and odd functions. Understanding this provides more insight into many other Fourier Transform properties.

As far as the first statement given about the interesting relationship between the real and imaginary components; if it is still not clear how this occurs it may be easier for some to swap the time and frequency domains and consider what occurs: In the time domain if you add the Hilbert Transform of a real signal as an imaginary component the result is a one-sided spectrum. A one-sided spectrum is the equivalent in the frequency domain to what a causal or anti-causal signal is in the time domain. Stated another way, a causal signal is a one-sided time-domain signal.

Consider the very simple case of a cosine which is a two sided spectrum with a positive negative frequency. This is also apparent in Euler's Identity where we see the two rotating phasors I was mentioning above, one rotating counter-clockwise (positive frequency) and one rotating clockwise (negative frequency):

$$cos(\omega t) = \frac{1}{2}e^{j\omega t} + \frac{1}{2}e^{-j\omega t}$$

Now consider the identity given as:

$$e^{j\omega t} = cos(\omega t) + j sin(\omega t)$$

$sin(\omega t)$ is the Hilbert Transform of $cos(\omega t)$, and $e^{j\omega t}$ is the single sided spectrum referred to earlier.

So we see that the following relationship holds:

$$f(t) = a(t) + j H\{a(t)\}$$

Where $f(t)$ is a function with a single-sided Fourier Transform, and $H\{\}$ is the Hilbert Transform.

Similarly by swapping time and frequency domain the following relationship will hold:

$$F(\omega) = A(\omega) + j H\{A(\omega)\}$$

Where here $F(\omega)$ is the Fourier Transform of a causal time domain function $f(t)$. The resulting $f(t)$ which has a single-sided Fourier Transform is referred to as the "analytic signal" and it also has the property that it will be minimum phase: it's Laplace transform will have all poles and zeros in the left half plane (and similarly for discrete-time systems the z-transform will have all poles and zeros inside the unit circle). A minimum phase system will have a single-sided frequency spectrum but not all single-sided frequency spectrums are minimum phase systems: We can cascade a minimum phase system with an all-pass filter that only modifies phase resulting in a mixed-phase or maximum-phase system, but since this only modifies phase in the spectrum and doesn't add new frequencies, the spectrum will still be single-sided.

For further details on this see:

FFTs of a complex signal - separating the real and imaginary parts

Dan Boschen
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    What is $A(\omega)$? – SheppLogan Feb 28 '20 at 20:11
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    That would be the real part of the Fourier Transform of f(t) – Dan Boschen Feb 28 '20 at 20:14
  • I dont really understand : I tried to replace each line of the fourier transform of the cameraman image in Matlab by the hilbert transform of this line. The problem is that the ifft of that yields a cameraman that is like folded onto itself on half of the image, the other part is black... any idea? – SheppLogan Feb 28 '20 at 20:40
  • It seems like the quadrature is more than just the real part with imaginary juste 90 degrees out of phase.. it looks like there is also a "conjugate " symmetry if I can say so...I mean that would be needed - when I looked at the true k space of MRI images – SheppLogan Feb 28 '20 at 20:41
  • I am not really familiar with image processing at all- so I can’t speak to the final result you are trying to achieve but am detailing how the real and imaginary components of the Fourier Transform are related for causal signals. I wouldn’t know much more specific to your details. – Dan Boschen Feb 28 '20 at 20:45
  • In summary what do you think of replacing each row of the Fourier transform by the hilbert transform of that row s real part.? – SheppLogan Feb 28 '20 at 20:46
  • But was trying to answer your “bottom line” question but perhaps that bottom line needs more detail as to what you are really trying to ask? – Dan Boschen Feb 28 '20 at 20:47
  • You did already give very interesting answers, but considering you were trying to answer the question s bottom line, does replacing each row of the Fourier transform by the hilbert transform of that row s real part , seem the right solution? – SheppLogan Feb 28 '20 at 20:49
  • What do you say about the last statemen – SheppLogan Feb 28 '20 at 21:11
  • The right solution for what? – Dan Boschen Feb 28 '20 at 23:22
  • For the question You were answering – SheppLogan Feb 29 '20 at 00:05
  • I guess I still need to ask more about your question: are you trying to create a 90 degree out of phase signal of the time domain signal (which is the Hilbert Transform) and somehow looking to use the Real part of the Fourier Transform to do that (if so why not just use the Hilbert Transform directly?) . I do not see how to create the quadrature time domain signal from just the real part of the Fourier Transform. – Dan Boschen Feb 29 '20 at 00:09
  • Well the problem is that as you can see here : http://mriquestions.com/real-v-imaginary.html the actual raw data acquired in MRI is not the spatial data, it is the Fourier transform. But the latter (each of its samples) is obtained by sampling points of a TEMPORAL signal (as strange as it may seem). And I consider the situation where 1 coil is used to mesure 1 time signal. This is considered as the "real" part of the DFT. And wrt previous website, the imaginary part is created as a that same measured signal BUT dephased by 90 degrees wrt itself. Then to get the image in spatial domain I – SheppLogan Feb 29 '20 at 13:39
  • do an inverse Fourier transform of this 2D matrix, where EACH ROW is a 1D signal with real and imaginary dephased in the way just mentioned above. (REM: In MRI each row of the 2D DFT is aquired in a sequential way, hence my idea of doing what i just mentioned). BTW thanks for trying to answer this complicated question, – SheppLogan Feb 29 '20 at 13:41
  • *REM: and so 1 sample of the real signal and 1 sample of the imaginary signal (obtained by dephasing) create 1 complex number for 1 frequency bin of the 2D k-space (Fourier) matrix – SheppLogan Feb 29 '20 at 13:43