PIXEL Correlation
- Resolution
- Size
- In order to determine resolution you must first identify your area of acquisition or the size of the detector head
- If the imaging area is 400 by 400 mm and the matrix size is 64 x 64. Determine the size of a pixel
- This can be calculated as follows 400 mm/64 pixels = 6.25mm or 0.625 cm/pixel
- Hence, each pixel represents 0.625 cm of collected data
- If a photon deficient area or cold lesion is 2.0 cm in size then hypothetically you should be able to resolve it since your pixel size is smaller than the area deficient of counts
- However, if the photon deficient area was below 6.25 mm then the disease process is smaller what can be resolved
- Therefore, determining your matrix size becomes critical to resolving the photopenic area
- What happens if a 128 x 128 matrix is used?
- 400 mm/128 = 3.125mm/pixel
- At this point the ability to visualize the disease (cold or hot spot) increases
- Conclusion: Spatial resolution increases as the matrix size increases
- Other factors to consider regarding matrix size and its application to SPECT
- When you increase your matrix size you increase
- Need more counts to reduce noise
- Required storage space X4
- Increase the process time by a factor of 4
- Determining the matrix size
- The field of view (FOV) of a gamma camera is digitized into one of two matrix sizes
- 64 x64
- 128 x 128
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- Determine the best imaging resolution for the system
- Let's keep the same detector size 400 by 400 mm
- A LEHR collimator may be able to resolve an 8 mm object at a depth of 10 cm (this also means that the FWHM is 8-mm)
- Hypothetically it is suggested that you at least double the mm/pixel to the system’s resolving ability
- However, sampling resolution has determined that it should be three times the amount
- So if the FWHM is 8-mm, then the mm/pixel can be calculated as follows:
- If a 64 matrix is used then note the following calculation: 500 mm (detector)/128 (matrix) = 3.125 mm/pixel
- Sampling theory states that pixel size should be three times size of the object you wish to image
- Therefore,