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Medical Imaging Systems
Fall 2008 Exam 2
The Johns Hopkins University
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Medical Imaging Systems
Fall 2009 Exam 1 Solutions
The Johns Hopkins University
Problem 1 (SP, IQ) (30 points)
A medical imaging system has the following line spread function:
l(x) = cfw_
cos(x) |x| 1
2
0
o.w.
when the input is a vertical
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Medical Imaging Systems
Spring 2005 Exam 3
The Johns Hopkins University
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Medical Imaging Systems
Fall 2010 Exam 4
The Johns Hopkins University
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Medical Imaging Systems
Fall 2008 Exam 4 Solutions
The Johns Hopkins University
Problem 1 (PET,CT) (20 points)
(a) Suppose s0 represents the center of the circle.
R
(x(s ), y(s ); E)ds =
expcfw_
s0
s0
(x(s ), y(s ); E)ds =
expcfw_
R
N+
N
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Medical Imaging Systems
Fall 2010 Exam 4 Solutions
The Johns Hopkins University
Problem 1 (SP) (25 points)
(a) Answer:
f ( )f (x x0 )d
g(x) =
(b) Answer: Not linear. Given input: af1 (x) + bf2 (x), the output is
g (x) = (af1 (x) + bf2 (x)
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Medical Imaging Systems
Fall 2009 Exam 4 Solutions
The Johns Hopkins University
Problem 1 (SP,IQ,PR) (25 points)
Consider using an extended source in projection radiography.
2
2
(a) Suppose h1 (x, y) = e(x +y )/4 is the PSF of the source f
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Medical Imaging Systems
Spring 2006 Exam 4
The Johns Hopkins University
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Problem 1 (PR) (25 points)
(a) The radius of the image of the whole on the detector plane is
r =
a+b
a
r.
(b)
I0 = Id (0, 0) =
Is
,
4(a + b)2
by inverse square law (cf. eq. (5.4) in the textbook).
(c)
xd yd
,
M M
xd yd
= Is (xd , yd ) c
,
M M
xd yd
= Is c
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Medical Imaging Systems
Spring 2006 Exam 4 Solutions
The Johns Hopkins University
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Medical Imaging Systems
Spring 2007 Exam 4
The Johns Hopkins University
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Exam 4 Solution
Problem: 1
a. Id = Is cos3 exp x
cos
3
6
Id (3) = Id (3) = Is 35 = 0.716Is
1
3
6
Id (1.5) = Id (1.5) = Is 1.517 exp (1cm 6 )(3cm) = 0.0415Is
1.5
Id (0.75) = Id (0.75) =
17
1
3
Is 0.756 65 exp (1cm 6)(3cm)
0.75 65
1
3
= 0.0475Is
1
6
Id (0.6
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Medical Imaging Systems
Spring 2005 Exam 4
The Johns Hopkins University
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Medical Imaging Systems
Fall 2009 Exam 4
The Johns Hopkins University
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Medical Imaging Systems
Fall 2008 Exam 4
The Johns Hopkins University
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Medical Imaging Systems
Fall 2012 Exam 4
The Johns Hopkins University
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Medical Imaging Systems
Spring 2006 Exam 3 Solutions
The Johns Hopkins University
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Medical Imaging Systems
Spring 2006 Exam 3
The Johns Hopkins University
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Medical Imaging Systems
Fall 2012 Exam 4 Solutions
The Johns Hopkins University
Problem 1 (Projection Radiography) (25 points)
(a) Answer:
In the fracture
t = 0 e100.15
= 0 e1.5
= 0 0.2231
= 22.31mm2
Behind the bone
b = 0 e70.1530.3
= 0 e1
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Medical Imaging Systems
Fall 2013 Exam 4
The Johns Hopkins University
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Exam 4: Solutions
September 22, 2014
Problem 1 (SP) (25 points)
(a) fs (x) = [p(x) f (x)]s (x; x) where p(x) is the prole of the detector. In our case, p(x) = rect( x )
d
and x = d so:
fs (x) = [rect( x ) f (x)]s (x; d)
d
(b) The malfunction is equivalent
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Medical Imaging Systems
Fall 2013 Exam 4 Solutions
The Johns Hopkins University
Problem 1 (Projection Radiography) (25 points)
(a) Answer: The points x-y position will be magnied on the detector:
(xd , yd ) =
d d
x ,y
z z
= 200
x y
,
z z
c
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Medical Imaging Systems
Fall 2011 Exam 4
The Johns Hopkins University
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