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c2
c0
m2 e2
⋅
:=
Gamma12
eta2
eta1
−
eta2
eta1
+
:=
Gamma23
eta3
eta2
−
eta3
eta2
+
:=
Tau12
2 eta2
⋅
eta2
eta1
+
:=
Tau23
2 eta3
⋅
eta2
eta3
+
:=
Gamma21
Gamma12
−
:=
Tau21
2 eta1
⋅
eta2
eta1
+
:=
FrontFace_echo
Gamma12 Einc
⋅
:=
Second_echo
Einc Tau12
⋅
Gamma23
⋅
Tau21
⋅
:=
First_thru
Einc Tau12
⋅
Tau23
⋅
:=
Second_thru
Einc Tau12
⋅
Gamma23
⋅
Gamma21
⋅
Tau23
⋅
:=
delayrefl
2
d2
c2
⋅
:=
delaytrans
2
d2
c2
⋅
:=
from 30 MHz to 30 GHz
i
0 1000
..
:=
freq
i
30 10
6
⋅
i1
+
()
⋅
:=
om
i
2
π
⋅
freq
i
⋅
:=
Gamma2Ray
i
FrontFace_echo
Second_echo exp i om
i
⋅
delayrefl
⋅
⋅
+
:=
Tau2Ray
i
First_thru
Second_thru exp i om
i
⋅
delayrefl
⋅
⋅
+
:=
20
10
0
20 log Gamma2Ray
i
⋅
20 log Tau2Ray
i
⋅
11.22
Homework 6 solutions
Problem 6.1
Using the tworay rule to calculate the reflection coefficient of a three material problem
with the same medium on either side.
This is the halfwave window radome problem
e0
8.854 10
12
−
⋅
:=
m0
4
π
⋅
10
7
−
⋅
:=
The properties of space
eta0
m0
e0
:=
c0
1
m0 e0
⋅
:=
e1
1
:=
e2
2
6
+
:=
e3
1
:=
m1
1
:=
m2
1
:=
m3
1
:=
Einc
1
:=
d2
0.25 0.0254
⋅
:=
in
meters
First calculate the medium impedances
eta1
eta0
m1
e1
⋅
:=
eta2
eta0
m2
e2
⋅
:=
eta3
eta0
m3
e3
⋅
:=
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5
10
15
20
25
30
30
freq
i
10
9
The halfwave window condition:
Lambda2
d2 2
⋅
:=
fwindow
c2
Lambda2
:=
fwindow
1.119
10
10
×
=
Note the Transmission maximum and reflection minimum occur at the halfwave window frequencies and
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 Spring '09
 diaz

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