M.Sc. in Medical Imaging
Nuclear Medicine
2014-15
Supplemental Examination
11:00 – 13:00 Friday 14th August 2015
Physics, Instrumentation and Computer
Technology
A total of 3 Questions MUST be answered
All parts of a chosen question must be answered
Each Question carries equal marks
Please use a separate answer book for each question.
Print the number of the question on each answer book.
Question 1
Image Processing can be used to alter “raw” image data in such a way as to highlight image
details that may not be easily visible in the original image. Discuss in detail the following
Image Processing techniques:
a] Histogram Equalisation
b] Maximum Intensity Projection [MIP]
[10 marks]
[10 marks]
Sample Answer and Marking Scheme
a] Histogram Equalisation
• Histogram Equalisation is a form of Contrast Enhancement [in fact it is Optimum
Contrast Enhancement. [1 mark]
• Definition of a image Histogram – “a graph showing the number of occurrences of
each pixel value in the image”. [1 mark]
• Figure showing a typical Histogram with proper labels [Frequency/No. Pixels] vs.
[Pixel Value]. [2 marks]
• In Histogram Equalisation Pixel Values are re-mapped to give a completely flat
histogram. [1 mark]
• Figure showing the flat Histogram of a processed image with proper labels. [2 marks]
• A major disadvantage is the computation time – a lot of calculation – but is often
combined with Region Of Interest [ROI] analysis to a specific area of the image.
[2 marks]
• In practice the resulting Histogram is never completely flat because of discrete
nature of the image data, but it is always flatter than the original image, giving
optimal contrast enhancement. [1 mark]
b] Maximum Intensity Projection [MIP]
• “Rendering” is a process of generating an image from a model i.e. by computer
programming. [1 mark]
• MIP is used in architecture, video games, simulators, movies, and MRI [an equivalent
technique called Minimum Intensity Projection is used in CT]. [1 mark]
• MIP is used to display 3D information on a 2D device i.e. computer monitor [1 mark]
• MIP is used in Magnetic Resonance Angiography [MRA] [for example] where there
usually is a stack of ‘slices’ representing a 3D volume in the body. [1 mark]
• With MIP, the computer simulates rays through the volume [slice by slice] for a
certain projection, and selects the highest pixel value on each ray for display.
[1 mark]
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Abdomen Slice
Simulated Rays for this Projection
[2 marks]
•
•
•
The resulting image resembles a conventional catheter angiographic image. [1 mark]
The projection is slightly changed, and the process repeated. [1 mark]
The resulting set of projection images are combined into a cine loop, giving the
observer a good impression of the 3D structure. [1 mark]
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Question 2
a) Radioactive decay is described by the following exponential decay equation:
Nt = N0 e-λt
where:
N0 is the initial number of nuclei,
Nt is the number of nuclei remaining after a time t, and
λ is the decay constant.
Derive this equation from first principles.
[10 marks]
b) At 9am a 99mTc source has an activity of 1200 MBq. What is its activity at 5 pm ?,
[Take the half life of Technetium to be 6 hours]
[10 marks]
Answer Plan
a]
dN/dt α N
dN/dt = - λN
[- means N is decreasing with time]
∫ dN/N = ∫ [-λ dt]
[2 marks]
[2 marks]
ln N = -λt + constant
at time t=0, initial number of nuclei is N0
hence constant = ln N0
ln N = -λt + ln N0
[2 marks]
ln N - ln N0 = -λt
N = N0 e-λt
[2 marks]
Graph of exponential decay with labels
[2 marks]
Page 4 of 9
b]
N = N0 e-λt
or, Activity[t] = Activity[0] e-λt
t1/2 = [ln [2] ] / λ
[2 marks]
or λ = [ln [2] ] / t1/2
λ = 0.693 / [6 hrs] = 0.116 hrs-1
[3 marks]
time interval 9am to 5pm => t = 8 hours
Activity[t=0] = 1200 MBq
[2 marks]
using Activity[t] = Activity[0] e-λt
Activity[t = 8 hours] = [1200MBq] exp[ - [0.116 hrs-1] [8 hrs]]
Activity[t = 8 hours] = [1200MBq] exp[- 0.928 ]
Activity[t = 8 hours] = [1200MBq] [0.395]
Activity after 8 hours is ~474 MBq
Page 5 of 9
[3 marks]
Question 3
a) In Gamma Camera Imaging, list the factors which may cause a reduction of image
quality and also detail the resultant effect on the image. [7 marks]
b) For the energy spectrum of photons incident on a gamma camera, outline some of
the mechanisms which cause the energy spectrum to be ‘smeared’ or spread out.
[5 marks]
c) Outline how additional extra peaks are formed in the energy spectrum. [3 marks]
d) Briefly describe how detected scattered photons can be rejected using pulse height
analysis. [5 marks]
Answer Plan
a) Factors which cause reduction of image quality in imaging [1 mark for each correct,
extra 0.5 mark for 3]:
• Attenuation of radiation in the patient
• Scatter of radiation in the patient
• Detection efficiency
• Spatial resolution
Resulting in… [1 mark for each correct, extra 0.5 mark for 3]:
• High image noise
• Poor resolution
• Low contrast
• Reconstruction artifacts and distortions
b) [1.5 mark for each, extra 0.5 for 3 correct, total of 5 marks]:
• Random variations in the number of light photons per kV of gamma ray
absorption
• Random variations in the number of light photons converted to photoelectrons
• Random variations in the number of secondary electrons emitted at each dynode
• Variations in the PMT kV
• Electronic Noise
c) [1 for each of the below, total of 3 marks]
• Backscatter from surrounding shielding material
• Iodine escape
• Lead Characteristic radiation
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d) [Total of 5 marks]
• Scattered photons have less energy than the original gamma ray
• Scattered photons will give less fluorescent light when they interact with the PMT
crystal
• The total sum of the electronic pulses from the PMTs will be less for scattered
photons
• It’s possible to reject scattered photons by rejecting pulses which do not have the
expected known isotope energy
• An appropriate energy window is set, only pulses which fall within that energy
window are accepted
Page 7 of 9
Question 4
a) In nuclear medicine imaging, explain in detail what is meant by the term “Aliasing”.
Why does it arise? What effect does it have on the image acquired? What steps can
we take to minimise its appearance in the image? [8 marks]
b) If we acquire an image with a field of view of 400mm, and a matrix of 128 x 128,
what is the minimum size object which can be resolved in the image? [4 marks]
c) Explain what an image histogram is and how it may be used to enhance the contrast
in an image with inherently low contrast. Illustrate your answer with an example
histogram of an image with low contrast and how this would be altered after
enhancing the contrast. [8 marks]
Answer Plan
a) Aliasing – due to using too low a sampling frequency. Relates to the spatial
frequencies present in the object which we wish to image using the gamma camera.
Should mention the Nyquist sampling theorem – state what this is, i.e. related to
sampling frequency used to acquire the image – must sample at a frequency at least
twice that of the highest frequency present in the objet to be imaged. Appears as
spurious low frequency features in the image – high spatial frequencies will not be
reproduced accurately in the image, i.e. results in a blurred image. Related to the
spatial resolution – increasing the resolution will decrease its appearance in the
image.
b) 400mm FOV, 128 matrix à pixels are 3.125mm. Nyquist states that the sampling
frequency must be at least twice the highest frequency present. Hence, the smallest
object which can be resolved in the image will be 3.125mm x 2 = 6.25mm
c) A histogram is a graph showing the distribution of pixel values in an image
It plots the number of pixels with different grey levels in the image. Should show an
example histogram of an image with low contrast [e.g. all pixels bunched towards
the centre]. [4 marks]
Enhance the contrast using the technique of windowing: apply a function like the
following, should explain how the function changes the histogram:
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[4 marks]
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