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I. HOMEWORK #2 SOLUTIONS
1)
MTF
=

H
(
u
)

H
(0)
a)
H
1
(
u
) =
F{
h
1
(
x
)
}
Note:
F{
e

ax
2
}
=
p
π
a
e

π
2
u
2
a
H
1
(
u
) =
√
5
πe

5
π
2
u
2
MTF
=
√
5
πe

5
π
2
u
2
√
5
πe

5
π
2
0
2
=
e

5
π
2
u
2
b)
MTF
total
=
MTF
1
·
MTF
2
MTF
2
=
e

10
π
2
u
2
by the same methods used in part a.
MTF
total
=
e

5
π
2
u
2
·
e

10
π
2
u
2
=
e

15
π
2
u
2
c) The system blurring increases (image quality decreases) exponentially.
2) To solve for the FWHM, we need to ﬁnd the point where the exponential term
equals 1/2 and double it.
a)
h
1
(
x
) =
e

x
2
2
= 0
.
5

x
2
2
=
ln
(0
.
5)
x
2
= 2
ln
(2)
x
=
p
2
ln
(2)
FWHM
= 2
x
= 2
p
2
ln
(2)
h
2
(
x
) =
e

x
2
200
= 0
.
5

x
2
200
=
ln
(0
.
5)
x
2
= 200
ln
(2)
x
=
p
200
ln
(2)
FWHM
= 2
x
= 2
p
200
ln
(2)
b)
FWHM
total
=
p
FWHM
2
1
+
FWHM
2
2
FWHM
total
=
q
(2
p
2
ln
(2))
2
+ (2
p
200
ln
(2))
2
1
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View Full Document FWHM
total
=
p
8
ln
(2) + 800
ln
(2)
FWHM
total
=
p
808
ln
(2)
c) System two (
h
2
(
x
)) has the largest aﬀect on the overall FWHM.
d) System two (
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This note was uploaded on 10/30/2010 for the course MP 230 taught by Professor Macfall during the Fall '10 term at Duke.
 Fall '10
 MacFall

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