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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 ?

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  • $\begingroup$ 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? $\endgroup$ – Dan Boschen Feb 27 at 17:20
  • $\begingroup$ Ok, I will try. Thanks for trying to answer $\endgroup$ – Machupicchu Feb 27 at 17:22
  • $\begingroup$ 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? $\endgroup$ – Dan Boschen Feb 27 at 17:33
  • $\begingroup$ 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 $\endgroup$ – Machupicchu Feb 27 at 17:47
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    $\begingroup$ Then look into the Hilbert Transform. This is the typical way to convert signals that are over a wider frequency range to quadrature. $\endgroup$ – Dan Boschen Feb 27 at 17:48
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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 imaginary odd waveform ($f(t) = -f(-t)$) will be all imaginary.
  • The FT of a non-causal real odd waveforrm will be all imaginary.
  • The FT of a non-causal imaginary even waveform will be all real.

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).

For further details on this see:

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

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  • $\begingroup$ What is $A(\omega)$? $\endgroup$ – Machupicchu Feb 28 at 20:11
  • $\begingroup$ That would be the real part of the Fourier Transform of f(t) $\endgroup$ – Dan Boschen Feb 28 at 20:14
  • $\begingroup$ 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? $\endgroup$ – Machupicchu Feb 28 at 20:40
  • $\begingroup$ 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 $\endgroup$ – Machupicchu Feb 28 at 20:41
  • $\begingroup$ 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. $\endgroup$ – Dan Boschen Feb 28 at 20:45

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