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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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6.013/ESD.013J — Electromagnetics and Applications
Fall 2005
Problem Set 11 - Solutions
Prof. Markus Zahn
MIT OpenCourseWare
Problem 11.1
A
Figure 1: Impedance model. (Image by MIT OpenCourseWare.)
V0
VRMS = √
2
2
�
VRMS
=⇒
VRMS = P Rr
Rr
�
=⇒ V0 = (2)(70)(105 ) = 3741.7 Volts (peak)
P =
B
Figure 2: Surface area of half-hemisphere. (Image by MIT OpenCourseWare.)
1
Problem Set 11
2πr2 I = P =⇒ I =
6.013, Fall 2005
P
100 × 103
=
= 6.366 × 10−6 Watts/m2
2
2πr
2π(50 × 103 )2
C
[Pr ]max = IA = (6.336 × 10−6 W/m2 )(10 m2 ) = 63.66 × 10−6 Watts
Problem 11.2
A
E(r, θ, t) in the far field limit
�
�
ηkId −jkr
jωt
Ê = Re j
e
sin θe
êθ
4πr
ηkId
=−
sin θ sin(ωt − kr) êθ
4πr
B
ω
c
=⇒ λ =
c
f
�
�
�2
π �� Idf ��
c
3Ptotal
Ptotal = η0 �
=⇒ f =
�
3 c
Id
πη0
�
3 × 108
3(1)
f=
= 150.99 × 106 Hz
(1)(0.1) π(377)
k=
C
� η � �� Id ��2
0 �
� sin2 θ [W/m2 ]
Ŝ = r̂
2 � 2λr �
First of all, the farthest you can go is when θ = ±π/2 because the power directed there is maximum since
sin2 θ = 1.
A|S|max = P∗
�
�
�
� η � �� Idf ��2
1
2 × 10−20
2(3 × 108 )
0 �
�
≈
A
= P∗ =⇒
2 � 2cr �
r
(377)(0.1) (0.1)(150.99 × 106 )
r ≈ 1.1 × 109 m
2
Problem Set 11
6.013, Fall 2005
Problem 11.3
Figure 3: Dipole configuration and spherical coordinate system. (Image by MIT OpenCourseWare.)
A
Intensity of radiation in the far field? This situation is similar to that developed in lecture, but the dipoles
are oriented on the y-axis rather than the x-axis.
For a single dipole, the field on the x, y-plane is
ˆ
ˆ θ = π , φ) = êθ η jk Ideff e−jkr
E(r,
2
4πr
For two dipoles, Iˆ1 = I0 and Iˆ2 = I0 ejψ , both with length deff
ˆθ,total = η jkI0 deff e−jkr1 + η jkI0 deff ejψ e−jkr2
E
4πr1
4πr2
a
sin φ
2
a
r2 ≈ r + sin φ
2
These small differences only matter for phase
r1 ≈ r −
Figure 4: Triangle details. (Image by MIT OpenCourseWare.)
3
Problem Set 11
6.013, Fall 2005
�
jkI0 deff � −jkr+jk a sin φ
a
2
e
+ e−jkr− 2 sin φ ejψ
4πr
�
ψ
jkI0 deff jψ/2 −jkr � j(k a sin φ− ψ )
a
2
=η
e
e
e 2
+ e−j(k 2 sin φ− 2 )
4πr
�
�
jkI0 deff jψ/2 −jkr
a
ψ
e
e
2 cos k sin φ −
=η
4πr
2
2
Êθ,total = η
For our case, ψ = 0
� a
�
jkI0 deff −jkr
e
2 cos k sin φ
4πr
2
�
�2
� a
�
1 |Êθ |2
kI0 deff
Intensity = �S� = êr
= êr 2η
cos2 k sin φ
2 η
4πr
2
Êθ,total = η
B
a
2π 2
�
�λ
=
= 2π
2
λ
� 2
�
�
�2
kI0 deff
�S� = êr 2η
cos2 (2π sin φ)
4πr
a = 2λ, k
φmax ?
cos2 (2π sin φmax ) = 1
π sin φmax = n�
π, n = 0, ±1, ±2, . . .
2�
n
sin φmax =
2
n=0
sin φmax = 0
φmax = 0, 180◦
n = ±1 sin φmax = ±1/2 φmax = ±30◦ , ±150◦
n = ±2 sin φmax = ±1
φmax = ±90◦
φmin ?
cos2 (2π sin φmin ) = 0
π
�
2�
π sin φmin = (2m − 1) , m = 0, ±1, ±2, . . .
2
1
sin φmin = (2m − 1)
4
m = 1, 0
sin φmin = ±1/4 φmin = ±14.48◦, ±165.52◦
m = 2, −1 sin φmin = ±3/4 φmin = ±48.59, ±131.41◦
4
Problem Set 11
6.013, Fall 2005
y
90 °
120 °
150 °
180 °
-1
1
60 °
30 °
0.5
-0.5
0.5
-0.5
210 °
240 °
1
0
x
330 °
300 °
-1
270 °
Figure 5: Plot of radiation pattern. (Image by MIT OpenCourseWare.)
Problem 11.4
A dipole in the êz -direction has an electric field in the far-field, in spherical coordinates, of
Ê = η
jkId −jkr
e
êθ sin θ
4πr
Figure 6: Dipole orientation. (Image by MIT OpenCourseWare.)
We have a dipole in the êx -direction. We can rotate the cartesian system such that we can use the solution
for the ẑ-directed dipole. If we transform the spherical solution back to cartesian coordinates correctly we
will have found our solution.
5
Problem Set 11
6.013, Fall 2005
Figure 7: Dipole with rotated coordinates. (Image by MIT OpenCourseWare.)
We are only interested in the z-axis: θ = π/2, φ = ±π/2
Ê = η
jIkd −jkr
e
êθ
4πr
On z-axis:
ex , r = |z |
êθ = −ˆ
ex η
Ê = −ˆ
jIkd −jk|z|
e
4π|z|
This dipole has current I = Iˆ0 and length d = deff :
jIˆ0 deff −jk|z|
ˆ = −ˆ
E
ex η
e
4π|z|
We also have a dipole in the êy -direction. We use the same method: The z-axis: θ = π/2, φ = ±π/2
ˆ = η jIkd e−jkr ˆ
E
eθ
4πr
Figure 8: Dipole with rotated coordinates. (Image by MIT OpenCourseWare.)
On z-axis, êθ = −êy , r = |z|
Ê = −ηˆ
ey
jIkd −jk|z|
e
4π|z|
6
Problem Set 11
6.013, Fall 2005
This dipole has I = jIˆ0 and d = deff .
Ê = ηêy
kIˆ0 deff −jk|z|
e
4π |z |
Total field is given by superposition:
Êtotal = (−jêx + êy )η
kIˆ0 deff −jk|z|
e
4π |z |
On the +z-axis, z > 0
Êtotal = (−jêx + êy )η
kIˆ0 deff −jkz
e
4πz
B
Polarization: As time advances, how does the direction and amplitude of the electric field change? For this,
we need to look at the real E-field, not just the complex amplitude:
�
�
ˆ0 deff
kI
Ê = Re{Êejωt } = Re (−jˆ
ex [cos(ωt − kz) + j sin(ωt − kz)] + êy [cos(ωt − kz) + j sin(ωt − kz)])η
4πz
Ê = (êx sin(ωt − kz) + êy cos(ωt − kz))η
kI0 deff
4πz
Let us look at one point in space, z = z1 , and see how the direction and magnitude of the E-field changes:
Only the direction of the field changes as time advances; the magnitude remains the same. Thus, it is
y
t�
k z1
Ω
x
t�
k z1
Π
+
Ω
2Ω
ωt − kz1 = 0
ωt − kz1 = π2
kdeff
→ Ê = êy η I04πz
1
I0 kdeff
→ Ê = êx η 4πz1
Figure 9: Time evolution of electric field. (Image by MIT OpenCourseWare.)
circularly polarized, since the field traces out a circle.
To determine whether the polarization is right-handed or left-handed, curl your fingers of both hands in
the direction of the path traced out by the field. If your right thumb points in the direction of propagation
(+z in this case), then the field is right-handed. If your left thumb points in the direction of propagation,
however, it is left-handed. In this case we have a left-handed circularly polarized wave.
7
Problem Set 11
6.013, Fall 2005
C
Find the magnetic field:
∂H
∂t
ˆ = −µjω H
ˆ
∇×E
∇ × E = −µ
�
� êx êy
�
�
∂
�∂
� ∂x ∂y
�
�E
x
Ey
−êx
êz
∂
∂z
Ez
�
�
�
�
�
�
�
�
Ê only varies with z =⇒ only has Ex , Ey components
∂Êy
∂Êx
ˆ
+ êy
= −µjω H
∂z
∂z
We assume that 1/z varies much slower than e−jkz , so we can treat 1/z as a constant:
∂Êy
= −jk Êy ,
∂z
−êx (−jk)η
∂Êx
= −jk Êx
∂z
kIˆ0 deff −jkz
kIˆ0 deff −jkz
ˆ
e
+ˆ
ey (−jk)(−j)η
e
= −µjω H
4πz
4πz
k 2 Iˆ0 deff −jkz
ˆ
(jˆ
ey )η
ex − ˆ
e
= −µjω H
4πz
�
k 2 Iˆ0 deff −jkz
µ ω 2� �
µεIˆ0 deff −jkz
Ĥ = (−êx − jêy )η
e
= (−êx − jêy )
e
µ�
µω4πz
ε �
ω4πz
√
ω µεIˆ0 deff −jkz
= −(êx + jêy )
e
4πz
kIˆ0 deff −jkz
e
= −(êx + jêy )
4πz
D
1
ˆ ∗}
Re{Ê × H
2 �
�
1
kI0 deff −jkz
kI0 deff jkz
= Re (−jêx + êy )η
e
× (−êx + jêy )
e
2
4πz
4πz
�
�
�2 �
1
kI0 deff
= Re (êz + êz )η
2
4πz
�
�2
kI0 deff
= êz η
4πz
�S� =
8
Problem Set 11
6.013, Fall 2005
Problem 11.5
A
Figure 10: Dipole configuration. (Image by MIT OpenCourseWare.)
In general:
jkI0 deff −jkr1
jkI0 ejψ deff −jkr2
e
+η
e
4πr1
4πr2
�
jkI0 deff jψ/2 � −jψ/2 jk a sin φ
a
=η
e
e
e 2
+ ejψ/2 e−jk 2 sin φ
4πr
�
�
jkI0 deff jψ/2
a
ψ
e
2 cos k sin φ −
=η
4πr
2
2
�
�
�
�
2
1 |Êθ |2
kI0 deff
a
ψ
�S� = êr
cos2 k sin φ −
= êr 2η
2 η
4πr
2
2
Eθ,total = η
a
2π λ
�
=�
= π, ψ = 0
2
2
λ
� �
�
�2
kI0 deff
�S� = êr 2η
cos2 (π sin φ)
4πr
Nulls:
a = λ, k
cos(π sin φ) = 0
π
�
π sin φ = (2n − 1) , n = 0, ±1, ±2, . . .
�
2
1
sin φ = (2n − 1)
2
1
n = 1, 0 sin φ = ±
φ = ±30◦ , ±150◦
2
Peaks:
cos2 (π sin φ) = 1
π sin φ = m�
π, m = 0, ±1, ±2, . . .
�
sin φ = m
m=0
sin φ = 0
φ = 0, 180◦
m = ±1 sin φ = ±1 φ = ±90◦
9
Problem Set 11
6.013, Fall 2005
y
90 °
1
120 °
150 °
180 °
-1
60 °
30 °
0.5
-0.5
0.5
-0.5
210 °
240 °
1
0
x
330 °
300 °
-1
270 °
Figure 11: Plot of radiation pattern. (Image by MIT OpenCourseWare.)
B
Figure 12: Dipole configuration. (Image by MIT OpenCourseWare.)
I2 = I1 ejψ
ψ=
π
2
a=
λ
4
ka
π
=
2
4
�S� = êr 2η
�
kI0 deff
4πr
�2
cos2
�π
4
sin φ −
π�
4
10
Problem Set 11
6.013, Fall 2005
Nulls:
cos
�π
π�
=0
4
4
π
π
π
�
�
�
sin φ − = (2m − 1) , m = 0, ±1, ±2, . . .
4
4
2
sin φ = 4m − 1
sin φ −
m=0
sin φ = −1
φ = −90◦
Peaks:
cos2
�π
n=0
π�
=1
4
4
π
π
π
�
�
�
sin φ − = (2m − 1) , m = 0, ±1, ±2, . . .
4
4
2
sin φ = 4m + 1
sin φ −
sin φ = 1
φ = 90◦
y
90 °
120 °
150 °
180 °
-1
1
60 °
30 °
0.5
-0.5
0.5
-0.5
210 °
240 °
1
0
x
330 °
300 °
-1
270 °
Figure 13: Plot of radiation pattern. (Image by MIT OpenCourseWare.)
11
Problem Set 11
6.013, Fall 2005
C
Figure 14: Dipole configuration. (Image by MIT OpenCourseWare.)
�N
�
π
jkηdeff � ˆ jkn a sin φ −jkr
Êθ (r, θ = , φ) =
IN e 2
e
,
2
4πr
−N
�
��
�
I2 = −I1 , I3 = I1
array factor
�
a
π
jkηdeff � jk a2 sin φ
, φ) =
I0 e
− 1 + e−jk 2 sin φ e−jkr
2
4πr
� a
�
jkηdeff I0 −jkr
=
e
[2 cos k sin φ − 1]
4πr
2
�
�2 �
�2
kI0 deff
a
�S� = êr η
2 cos(k sin φ) − 1
4πr
2
Êθ (r, θ =
a
2π 2
�
�λ
=
= 2π
2
2
λ
� �
�
�2
kI0 deff
�S� = êr η
[2 cos(2π sin φ) − 1]2
4πr
a = 2λ, k
Nulls:
2 cos(2π sin φ) = 1
1
cos(2π sin φ) =
2
π
5π 7π
2π sin φ = ± , ± , ± , . . .
3
3
3
1 5 �
7
�
sin φ = ± , ± , ± (larger than 1)
6 6 �6
φ = ±9.59◦, ±170.41◦, ±62.71◦, ±117.29◦
Peaks:
Largest Peaks?
12
Problem Set 11
6.013, Fall 2005
cos(2π sin φ) = −1
[2 cos(2π sin φ) − 1]2 = 9
2π sin φ = ±π, ±3π, ±5π, . . .
1 �
3
sin φ = ± , ±�
=⇒ φ = ±30◦ , ±150◦
2 �2
Smaller Peaks?
cos(2π sin φ) = 1
[2 cos(2π sin φ) − 1]2 = 1
2π sin φ = 0, ±2π, . . .
sin φ = 0, ±1
φ = 0◦ , 180◦ , ±90◦
What about cos(2π sin φ) = 0? Though [2 cos(2π sin φ) − 1]2 = 1 as well, when cos(2π sin φ) = 0, this is not
a peak as can be seen by taking the second derivative with respect to φ and evaluating it at that point.
y
90 °
120 °
150 °
180 °
-10
10
60 °
30 °
5
-5
5
-5
210 °
240 °
10
0
x
330 °
300 °
-10
270 °
Figure 15: Plot of radiation pattern. (Image by MIT OpenCourseWare.)
13
Problem Set 11
6.013, Fall 2005
Problem 11.6
A
Figure 16: Dipole configuration. (Image by MIT OpenCourseWare.)
Putting 2 identical dipoles 1/2 a wavelength apart means they will cancel along the x-axis. But since neither
is delayed with respect to each other, they add on the y-axis to a maximum.
�S� = êr 2η
�
kI0 deff
4πr
�2
cos2
�π
2
�
cos φ
B
Figure 17: Dipole configuration. (Image by MIT OpenCourseWare.)
This is the same pattern as in 11.3(b), but rotated.
�S� = êr 2η
�
kI0 deff
4πr
�2
cos2
�π
4
cos φ −
π�
4
14
Problem Set 11
6.013, Fall 2005
C
We need to come up with a maximum at φ = 0, but a minimum at φ = π. We have 2 dipoles of equal
amplitude, separated by a distance a.
�
�
jkI1 deff jψ/2 −jkr
a
ψ
E=η
e
e
2 cos k cos φ −
4πr
2
2
Figure 18: Dipole configuration. (Image by MIT OpenCourseWare.)
φ = 0 =⇒ must add to a peak
�
�
a ψ
2 cos k −
=2
2
2
a ψ
k − = 0, ±2π, ±4π, . . .
2
2
φ = π =⇒ must be a null
−k
a ψ
π
3π
− = ± ,± ,...
2
2
2
2
�
�
�
�
We want the solution with the fewest nulls and peaks, so let us take the lowest angles:
ka − ψ
+ − ka − ψ
−2ψ
=⇒ ψ
=0
=π
=π
= − π2
=⇒
We need a positive a, so we use
π
ψ = 3π
2 = −2
3π
ka−� 2 �= 0
λ
3λ
a = 3π
2
2π = 4
15
Problem Set 11
6.013, Fall 2005
Figure 19: Dipole configuration. (Image by MIT OpenCourseWare.)
�S� = êr 2η
�
kI0 deff
4πr
�2
2
cos
�
3π
π
cos φ +
4
4
�
16