Sample Mid-Term 2
EE142
Dr. Ray Kwok
Sample Mid-Term 2 - Dr. Ray Kwok
1.
An unpolarized but focused beam of light (λ = 700 nm) is shined onto a 2-cm thick
piece of glass (n = 1.43) at an angle of 55o from normal. Calculate the relative
power (in dB, relative to the incident power) of the first 2 reflections (R1 & R2) and
the first 2 transmission (T1 & T2) as shown in figure.
n1 sin θi = n 2 sin θt
(1)sin (55o ) = (1.43)sin θt
R2 = ?
T2 = ?
θ t = 35o
θ t + θi = 35 + 55 = 90o
R1 = ?
C
T1 = ?
D
A
(55o is the Brewster angle) !!!
┴ polarization at A,
n1 cos θi − n 2 cos θ t cos 55o − 1.43 cos 35o
Γ⊥ =
=
= −0.34
o
o
n1 cos θi + n 2 cos θ t cos 55 + 1.43 cos 35
B
τ⊥ = 1 + Γ⊥ = 0.66
┴ polarization at B, C, D…
┴ polarization only
2
Γ⊥ =
2
R 1 = ΓA = − 0.34 = 0.116
2
2
R 2 = τA ΓB τC = (0.66 )(0.34 )(1.34 ) = 0.090
2
2
T1 = τA τB = (0.66 )(1.34 ) = 0.782
2
2
T2 = τA ΓBΓC τD = (0.66)(0.34 ) (1.34 ) = 0.0105
2
n1 cos θi − n 2 cos θ t
= +0.34
n1 cos θi + n 2 cos θt
τ⊥ = 1 + Γ⊥ = 1.34
Sample Mid-Term 2 - Dr. Ray Kwok
1.
(continue)
R2 = ?
T2 = ?
R1 = ?
C
A
D
B
// polarization at A,
Γ// =
n1 cos θ t − n 2 cos θi
=0
n1 cos θt + n 2 cos θi
τ// =
η2
(1 − Γ// ) = n1 (1 − Γ// ) = n1 = 1 = 0.70
η1
n2
n 2 1.43
T1 = ?
// polarization at B, C, D… Γ// =
τ// =
n1 cos θt − n 2 cos θi
=0
n1 cos θt + n 2 cos θi
n1
(1 − Γ// ) = n1 = 1.43
n2
n2
2
// polarization only:
R 1 = ΓA = 0
2
T1 = τ A τB
2
R 2 = 0 = T2
 1 
=
(1.43) = 1
 1.43 
Sample Mid-Term 2 - Dr. Ray Kwok
1.
(continue)
R2 = ?
T2 = ?
R1 = ?
C
A
D
T1 = ?
Altogether, R1 = ½ (0.116 + 0) = 0.058 (┴ only)
R 1 = 10 log(0.058) = −12.4dB
T1 = ½ (0.782 + 1) = 0.891 (both pol.)
T1 = 10 log(0.891) = −0.50dB
B
R2 = ½ (0.090) = 0.045 (┴ only)
R 2 = 10 log(0.045) = −13.5dB
T2 = ½ (0.0105) = 0.0053 (┴ only)
T2 = 10 log(0.0053) = −22.8dB
Note: R1 + R2 + T1 + T2 = 0.058 + 0.891 + 0.045 + 0.0053 = 0.9993
accounts for almost all the input power already.
Sample Mid-Term 2 - Dr. Ray Kwok
2.
A submarine is trying to communicate with a helicopter above water as shown in
figure. Sea water is nonmagnetic and has a relative εr = 72, and σ = 4 S/m.
(a) What is the minimum angle of entrance φ (see figure) such that
communication with air is possible?
(b) There are 2 transmitters on board, one at HF (3 MHz) and the other at mmwave (120 GHz). Which transmitter would deliver a stronger signal to the
helicopter? Explain.
(a) n w sin θc = n air sin (90o )
sin θc =
z
φ
1
1
=
72 8.5
θc = 6.77 o
y
φ min = 90 − 6.77 = 83.23o
θc = 0.5 if use k
more accurate
(b) At 3 MHz,
σ
4
=
= 333
ωε 2π(3 ⋅106 )(72 )(10 −9 / 36π )
(
)(
)
ωµσ
2π 3 ⋅106 4π ⋅10 −7 (4 )
=
= 6.88
good conductor α =
2
2
σ
4
At 120 GHz,
=
= 0.0083
ωε 2π 120 ⋅10 9 (72 ) 10 −9 / 36π
(
low loss dielectric
α=
) (
)
σ µ σ ηo
4 377
=
=
= 88.9
2 ε 2 ε r 2 72
HF transmitter (3 MHz),
lower loss
Sample Mid-Term 2 - Dr. Ray Kwok
2. (c) If a 3 MHz RHCP signal is transmitted at an angle φ = 90o (normal incidence) with
the E-field amplitude = 5 µV/m right below the surface, write the full expressions
for the incident E & H fields.
(c)
z
φ
ω = 2πf = 1.88 ⋅ 107
α = β = 6.88
y
ηc = (1 + j)
α
6.88
= (1 + j)
= (1 + j)1.72 = 2.43∠45o
σ
4
r
E i = 5e − αz [xˆ cos(ωt − 6.88z ) + yˆ sin (ωt − 6.88z )]
r
E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z µV/m
r
5
H i = e − 6.88 z yˆ cos 1.88 ⋅ 107 t − 6.88z − xˆ sin 1.88 ⋅ 107 t − 6.88z
ηc
r
H i = 2.06e − 6.88 z yˆ cos 1.88 ⋅ 107 t − 6.88z − π / 4 − xˆ sin 1.88 ⋅ 107 t − 6.88z − π / 4 µA/m
[
(
[
)
(
[
)
(
)]
(
)]
(
)
(
Note: z < 0, so the lower the z, the higher the amplitude (closer to the source)
)]
Sample Mid-Term 2 - Dr. Ray Kwok
2. (d) Write the expression for the reflected E-field. What’s its polarization state?
(e) Write the expression for transmitted H-field.
(d) Γ = η2 − η1 = (1 + j)1.72 − 377 = j1.72 − 375 = 0.991∠179.5o ≈ −0.991
(1 + j)1.72 + 377
η2 + η1
j1.72 + 379
r
E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z
µV/m
r
6.88 z
7
7
E r = −4.96e
xˆ cos 1.88 ⋅ 10 t + 6.88z + yˆ sin 1.88 ⋅ 10 t + 6.88z
[
(
)
[
LHCP
)]
(
(
)
)]
(
Or with 0.5 deg phase shift if you keep it in Γ.
(e) τ = 1 + Γ = 1 − 0.991 = 0.009
(
)
ω 2π 3 ⋅ 106
k air = =
= 0.02π = 0.0628
8
c
3 ⋅ 10
r
E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z
r
E t = 0.045 xˆ cos 1.88 ⋅ 107 t − 0.0628z + yˆ sin 1.88 ⋅ 107 t − 0.0628z
r
0.045
Ht =
yˆ cos 1.88 ⋅ 107 t − 0.0628z − xˆ sin 1.88 ⋅ 107 t − 0.0628z
ηo
r
H t = 0.12 yˆ cos 1.88 ⋅ 107 t − 0.0628z − xˆ sin 1.88 ⋅ 107 t − 0.0628z nA/m
[
(
[
(
[
[
)
)
(
(
(
)
)
)]
(
(
(
)]
)]
)]
Sample Mid-Term 2 - Dr. Ray Kwok
2. (f) If the submarine is at a depth of 100 m, what is the attenuation (in dB) of the
signal by the time it gets to the helicopter (assuming air is a lossless medium)?
(ignore the reflection “mismatch loss” for now.)
(f) A (dB) = −8.686αr = −8.686(6.88)(100) = −5976dB
This demonstrated the difficulty of underwater communications.
By the same token, it’s difficult to detect a submarine from air.