How do two stripy k-space images reconstruct a white circle?

[BobP]

In this image the circle in the top left is the original. Next to it lies its K-space. We then see the result of running through k-space vertically and horizontally and then the combination of these two views.

I do not understand how the white circle can be reproduced. I though we would be adding pixel values in the two k-space data sets hence there should not be a black baground. This is obviously now what is happening. Please can someone explain how the two stripy images form the white circle. thanks

 

[Dale]

I believe that this is an image from Denis Hoa's website. You should probably give a link or a citation. It is an excellent resource.

On this slide there is only one picture of k-space, the one on the top row, middle column. The picture on the top row right column and bottom row middle column are NOT pictures of k-space. They don't really have a name, but they would be called something like x*ky space and kx*y space. You would never add them together.

 

[BobP]

I believe that this is an image from Denis Hoa's website. You should probably give a link or a citation. It is an excellent resource.

On this slide there is only one picture of k-space, the one on the top row, middle column. The picture on the top row right column and bottom row middle column are NOT pictures of k-space. They don't really have a name, but they would be called something like x*ky space and kx*y space. You would never add them together.

Absolutely. here is the link if anyone wants it
http://www.revisemri.com/tutorials/what_is_k_space/

What my question meant was how would you combine the "x*ky space and kx*y space" to produce the image? Thank you

 

[Dale]

What my question meant was how would you combine the "x*ky space and kx*y space" to produce the image? Thank you

You don't combine them.

So, the Fourier transform has the property that it is separable. In keeping with my above notation k-space is kx*ky space. If you Fourier transform in the X direction then you go from kx*ky space to x*ky space. If you then apply another Fourier transform, but in the y direction then you go from x*ky space to x*y space, which is the image.

 

[BobP]

You don't combine them.

So, the Fourier transform has the property that it is separable. In keeping with my above notation k-space is kx*ky space. If you Fourier transform in the X direction then you go from kx*ky space to x*ky space. If you then apply another Fourier transform, but in the y direction then you go from x*ky space to x*y space, which is the image.

OK so once we have decoded kx space --> x*ky space shouldn't we have a uniform signal at a particular x-value across the whole of y. My reasoning for this is that we know there is a y signal from ky but we don't know where it is. I don't understand why we see fancy patterns?
Thank you for your help by the way

 

[Dale]

Are you familiar with the sinc function and how it Fourier transforms into a boxcar function?

If you go the other way it may be easier to understand. Start from the image domain and Fourier transform each row to go to the kx*y domain.

If the row is all zeros then it Fourier transforms to all zeros. Otherwise the row looks like a boxcar function so it Fourier transforms to a sinc function.

Different rows will have different boxcar widths and therefore different sinc functions. The "fancy pattern" is just all of those sinc functions put together.

 

[BobP]

Are you familiar with the sinc function and how it Fourier transforms into a boxcar function?

If you go the other way it may be easier to understand. Start from the image domain and Fourier transform each row to go to the kx*y domain.

If the row is all zeros then it Fourier transforms to all zeros. Otherwise the row looks like a boxcar function so it Fourier transforms to a sinc function.

Different rows will have different boxcar widths and therefore different sinc functions. The "fancy pattern" is just all of those sinc functions put together.

Oh I see. So in any given row, the varying intensity on the figure displaying the kx*y domain arises because of the relative amplitude of the sinc function at the position in kx space?

 

[Dale]

Oh I see. So in any given row, the varying intensity on the figure displaying the kx*y domain arises because of the relative amplitude of the sinc function at the position in kx space?

Yes, relative amplitude and width.

 

[BobP]

Yes, relative amplitude and width.

Ah. thanks so much!

 

[Dale]

You are welcome. By the way, D M Higgins site is an excellent resource as is Denis Hoa's site
https://www.imaios.com/en/e-Courses/e-MRI/The-Physics-behind-it-all

If you are an engineer or physicist looking to do MRI research and development then the Haacke, Brown, Thompson, and Venkatesan book is great. If you are a radiologist or technologist looking to understand MR physics then the Dale, Brown, Semelka book is excellent.