# What does f(x,y) in the 2D CT image account for?

1. Mar 21, 2015

What does f(x,y) in the 2D CT image account for?
If for density, then which material is considered as a reference for that density?
If for atomic number, then again which is the reference substance?

2. Mar 21, 2015

### Choppy

If I understand the question, it's called a Houndsfield unit, or sometimes just a CT number.

The values is a CT image correspond to the difference in the mean value of the linear attenuation coefficient through the small volume at that point in the reconstructed image from that of water, normalized to that of water (or water minus air) and then multiplied by a scaling factor of 1000. So values of -1000 are expected for air. 0 for water. Soft tissues are in the range of -200 to +200 or so. Bone gets into the +500 to +1200 range and the really high density metals are a few thousand and can go off the scale depending on how many bits you have for storage and the specific details of your system - such as the spectrum of energy you're using.

The CT numbers can generally be mapped on a one-to-one basis to properties such as mass density, relative electron density, or effective atomic number.

3. Mar 22, 2015

What I read is that the attenuation coefficient based on Compton effect is proportioned to ρA/E, where ρ is the density, A is the effective atomic number and E is the energy. So if we have 2 tissues of different ρ and A, say ρ1, A1 and ρ2, A2 at a known E, will be their attenuation coefficients and consequently their Hounsfield Units be the same if ρ1A12A2?

Last edited: Mar 22, 2015
4. Mar 22, 2015

### Choppy

With CT the dominant interaction is the photoelectric effect because most CT machines use energies in the ballpark of 60-120 kVp. In megavoltage CT, the Compton process becomes more important, but you tend to give up contrast at the higher energies. PE has a strong dependence on atomic number Z to the 4 or 5, which translates into large attenuation differences for different media.

5. Mar 22, 2015

1) So, what is the relation between PE and energy? Compton and energy?
2) Will a composite attenuation of 2 materials of different Z&ρ be linearly dependent at 2 different energies? For example, lets expose 2 pixels containing 2 different materials to 2 beams of different energies one at a time, will In I0/I, at high E which is equal to μ1Δx+μ2Δx, be linearly dependent on, In I0/I, at low E? This is very important question for me. (I is the penetrated intensity, I0 is the incident intensity, μ is the linear attenuation coefficient).

Last edited: Mar 22, 2015
6. Mar 22, 2015

### Choppy

1. You should be able to look this up online. Generally speaking the photoelectric cross section varies as E-3. Compton is a fairly complex function of energy - look up the Klein-Nishina formulation. To a rough approximation for energies above about 10 MeV it goes at 1/E though.

2. I'm not sure I completely understand the question. One thing that's important to remember with CT is that you're not dealing with single energy photons (at least, not the conventional CT image sense). You're dealing with a spectrum of energies. You can change the spectra by changing the voltage on your x-ray tube, or changing the filtration. You can try to extract more information (a greater degree of contrast) from an image by scanning the same thing with different spectra and looking at the differences - maybe this is what you're getting at.

For further reading you might want to look up "Radon transform" and "filtered back projection" to see how one goes from linear attenuation to an image of Houndsfield units.

7. Mar 23, 2015

But lets assume that we expose only 2 pixels using 2 monochromatic photons one at a time in such a way that the photon traverses the first pixel and passes to the second one, then is received by sensor after it.

{source}---------->[pixel 1]------>[pixel 2]----------> {sensor}

Can we then be able to know the density of the 2 pixels by solving the 2 projection equations with the densities of 2 pixels as unknown? Will be those 2 equations be linearly dependent and then can not be solved?

8. Mar 24, 2015

### Choppy

Well, if you're only dealing with two individual photons then there really isn't much you can extract from an experiment like that.

But I think I see what you're getting at. You mean two different monochromatic sources that travel through two separate materials that yield two different signals in a detector - given that, is it possible to extract the linear attenuation coefficient information (and therefor a material description) from each separate material?

It's an interesting question. I'm not sure that you can (although I can't prove it off the top of my head). I think the issue is: is there a way of knowing which material you went through first?

In CT, the basic idea is that you have a single source that moves around the object in question. Along any ray line, you get a signal that results from the integration of the linear attenuation coefficients - the Radon transform. You then recover the information in each voxel when you project that information backwards and filter out the 1/r dependence, but this requires the use of multiple angles.

9. Mar 25, 2015