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# Art Report – Matlab Code

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**Question- How are the intensity values measured by a CT scanner processed to obtain projection data?**

The motivation behind a processed tomography obtaining is to measure x beam transmission through a patient for a large number of perspectives.

Diverse perspectives are accomplished in computed tomography principally by utilizing

- Detector with many detector components along the indicator circular segment (by and large 800-900 detector components),
- by turn of the x beam tube around the patient, taking around 1000 precise angular estimations.

Assigned values to the pixels in a ct scanned image are generally associated with the average linear attenuation coefficient µ (m-1) of that tissues that are present inside that pixel. linear attenuation coefficient µ have dependency on composition of material present, density of the material, its photon energy as mentioned in Beer’s law: –

I(x) =I0 e^{-µx}

where I(x) is the intensity of the attenuated X ray beam,

- I0 indicates unattenuated X ray beam,
- where x represents thickness

Beer’s law just describes the attenuation of the first beam and does not consider the intensity that is produced by scattered radiation.

For poly-energetic X beam bars Beer’s law ought to entirely be coordinated over all photon energies in the X beam range.

In the back projection strategies produced for CT reconstruction algorithm, this is commonly not actualized.

- Rather regularly a pragmatic solution is to assume where Beer’s law can be connected utilizing one value which is the average photon energy of the X ray spectrum.

This presumption makes errors in the reconstruction and it leads to the beam hardening artefact.

As a X beam bar is transmitted through the patient, distinctive tissues are experienced with various linear attenuation coefficients. The power of the attenuated X beam bar, transmitted a separation d, can be expressed as:

A CT images is composed of a matrix of pixels which represents the linear attenuation co-productive in the related volume components (voxels).

Matrix which have 4 columns and 4 rows represents the transmission value along one line. Every element in the matrix can on a basic level have an alternate estimation of the related linear attenuation coefficient. The equation for the attenuation can be expressed as:

From the above it tends to be seen that the essential information required for CT is the power of the lessened and unattenuated X beam pillar, individually I(d) and I0, and this can be estimated.

Reconstructing image procedures would then be able to be connected to determine he matrix of linear attenuation coefficients, which is the premise of the CT picture.

In CT the matrix of reproduced linear attenuation coefficients (µmaterial) is changed into a corresponding matrix of Hounsfield units (HUmaterial), where the HU scale is expressed with respect to the linear attenuation coefficient of water at room temperature (µwater):

It can be seen that

- HUwater = 0 as (µmaterial = µwater),
- HUair = -1000 as (µmaterial = 0)
- HU=1 is associated with 0.1% of the linear attenuation coefficient of water.

Typical values for body tissues.

The actual estimation of the Hounsfield unit relies upon the composition of the tissue or material, the cylinder voltage, and the temperature.

Hounsfield units are eight-bit grey scale which offers only 128 grey values.

4000+ HU The display is defined using

- Window level (WL) as CT number of mid-grey
- Window width (WW) as the number of HU from black -> w

The decision of WW and WL is directed by clinical need Optimal perception of the tissues of interest for the CT picture must be accomplished by choosing the most proper window width and window level. Distinctive settings of the WW and WL are utilized to picture for instance delicate tissue, lung tissue or bone.

Same image data at different WL and WW

marked an initial guess for the two values as a point (A) on the graph.

Starting at the point A on the graph above, draw a line that intersect either of the two blue lines at a right angle. This point of intersection is a solution (x,y) that satisfies one of the (projection) equations but not the other. Now draw a line from this point to intersect the other line at a right angle. Repeat this operation a few times, alternating between the lines.

**Does it matter where on the graph you start, or will this method always lead you to the correct solution?**

Yes, it is important process to obtain projection data.

**What is the smallest angular range required to produce an image that is free of significant artefacts?**

Artefacts can genuinely corrupt the nature of processed tomographic (CT) pictures, here and there to the point of making them symptomatically unusable. To enhance picture quality, it is important to know why artefacts occur happen and how they can be countered. CT artefacts start from a scope of sources. Material science based antiquities result from the physical procedures engaged with the procurement of CT information. Patient-based artefacts are generally cause of some factors such as patient movement or metallic materials presence on the patient body. Scanner-based artefacts result due to functions of scanner is not perfect, Helical and multi-section technique artefacts are generally the image reconstruction process. The outcomes at various angular range we have broken down which Resultant output picture is free of significant artefacts or not, when we entered precise angular range as 150 degrees, we get output picture of starting theory is free of artefacts yet beneath that precise angular range artefacts can be watched.

Just Presenting some outputs below angular range 150 degree and above it-

Above are the outputs at angular range 150 Degree, now let’s check the results below 150-degree range-

Above outputs are evaluated at 100-degree angular range, we can observe artefacts.

**Question-the optimum set of parameters for each of the three noise levels (none, low and high) i.e. the set of parameters that produces the best image.**

Best result we are getting for NOISE free image at given parameters below-

Figure Sinogram at Noise level At None

Figure Sinogram at Noise level At Low

Figure Sinogram at Noise Level At High

**Question-Comparison of all output images from Noise level none to High-**

We can observe best output image at different noise level and at different parameters. Let’s analyse the result-

For noise free image output images have best contrast level and smooth image among all other noise images output.

**For Noise level -low** output images we need to improve quality of images by applying some filters we have to include smoothness in the image at its best level, its contrast level has to be increased with the help of some image processing techniques.

**For Noise level- high** Output images we need to apply filter as we can see image is very much noise, we can apply Gaussian filter, as we can see the type of Noise is Gaussian.

**For Noise Level**-None Time taken to produce image was 3.067 approximate, we have made some changes in reconstruction parameter that is relaxation time previously it was 1 and this time we made it as 0.7 for the same best output images, now time decrease as 2.768, it is not completely 50% of the originally generated output image time, sometimes speed also depends on the processing speed of the system.

**For Noise level Low **-None Time taken to produce image was 3.469 approximate, we have made some changes in reconstruction parameter that is relaxation time previously it was 1 and this time we made it as 0.7 for the same best output images, Now time decrease as 2.898.

**For Noise level High**-None Time taken to produce image was 3.0625 approximate, we have made some changes in reconstruction parameter that is relaxation time previously it was 1 and this time we made it as 0.5 for the same best output images, now time decrease as 2.68.

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