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6.013/ESD.013J Electromagnetics and Applications, Fall 2005
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Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.013 Electromagnetics and Applications
Problem Set #11
Fall Term 2005
Issued: 11/29/2005
Due: 12/9/05
Suggested Reading Assignment: Staelin, Sections 6.1-6.4, 10.1, 10.2, 10.4
Final Exam: Wednesday, Dec. 21, 2005, 1:30-4:30pm.
Problem 11.1
A popular 1-MHz AM radio station in the middle of Kansas has a single transmitting antenna on a flat
prairie that radiates 100kW isotropically (equally in all directions) over the upper 2π steradians (i.e., this
station has no underground audience.) The matched input impedance (the radiation resistance Rr ) of this
antenna is ~70 ohms, and it is driven by V0 sin ωt volts at maximum power.
a) What is V0 [Volts] ?
b) What is the radiated intensity I [W / m 2 ] 50 kilometers from this antenna?
c) What is the maximum power Pr that can be received from this station by an antenna 50 km away
with an effective area A = 10 m2?
Problem 11.2
A short dipole antenna, 10 cm in length and aligned along the ẑ axis, is driven uniformly along its length
with a sinusoidal current of peak value 1 amp.
a) What is the electric field E (r , θ , t ) in the far field?
b) At what frequency would this antenna radiate 1 watt of power?
c) If a receiver with effective area A = 0.1 m2 needed 10-20 watts for successful reception, how far
away could it be and still receive signals from the 1 watt dipole? In what direction?
Problem 11.3
An antenna consists of two short dipoles, oriented along the z-axis and separated along the y-axis by a
distance a. They are driven in phase, each with a current I 0 and an effective length d eff , ( d eff
λ ), at
an angular frequency of ω. (Assume that each antenna radiates as it would in the absence of the other.)
a
a) What is the intensity of the radiation in the far field as a function of angle φ in the x-y plane?
b) For a = 2λ , at what angles φmax and φmin is the intensity a relative maximum or zero?
Page 1 of 7
Problem 11.4
A "turnstile" antenna consists of two short Hertzian dipoles driven at an angular frequency ω and oriented
at right angles to each other as shown in the figure below. One dipole, oriented along the x-axis is driven
with a current Iˆ1 = Iˆ0 xˆ and the other, oriented along the y-axis is driven with Iˆ2 = jIˆ0 yˆ . Both have the
same effective length d eff .
a) Find the complex amplitude of the total electric field on the +z axis in the far field. (Express your
answer in Cartesian coordinates with unit vectors xˆ , yˆ , and ẑ .)
b) Why is the result of part (a) called left-handed circular polarization (LHCP) for +z directed waves
along the +z axis?
c) What is the complex amplitude of the magnetic field on the +z axis in the far field?
d) What is the intensity of the radiation on the z axis in the far field?
Hint: S =
1 ⎡ ˆ ˆ ∗⎤
Re E × H ⎥
⎦
2 ⎢⎣
Problem 11.5
Sketch the far field radiation patterns in the x-y plane for each of the following short dipole antenna
arrays. The identical dipoles are directed in either the +z or -z directions, as indicated, and the
currents have equal amplitudes of ±1 . In part (b) one current has a phase of
π
2
so that its complex
amplitude is j. In each case find the angles φ corresponding to nulls ( φn ) and peaks ( φ p ). If the peaks
are unequal, also evaluate their relative values.
Page 2 of 7
Problem 11.6
Using the format of Problem 11.5 design two-dipole arrays that could produce the far field antenna gain
patterns illustrated below. The two dipoles have the same current amplitude but may differ in phase.
Find the spacing a between the two dipoles and their relative phase that results in the radiation patterns
shown in parts (a) - (c).
Page 3 of 7
6.013 Final Exam Formula Sheet
December 21, 2005
Cartesian Coordinates (x,y,z):
∇Ψ = xˆ ∂Ψ + yˆ ∂Ψ + ẑ ∂Ψ
∂x
∂y
∂z
∂A x ∂A y ∂A z
∇iA =
+
+
∂x
∂y
∂z
∂A y ⎞
⎛ ∂A
∂A ⎞ ⎛ ∂A y ∂A x ⎞
⎛ ∂A
+ yˆ ⎜ x − z ⎟ + ẑ ⎜
−
∇ × A = xˆ ⎜ z −
⎟
∂
∂
∂z
∂x
∂x
∂y ⎟⎠
y
z
⎝
⎠
⎝
⎠
⎝
2
2
2
∇2Ψ = ∂ Ψ + ∂ Ψ + ∂ Ψ
∂x 2 ∂y 2 ∂z 2
Cylindrical coordinates (r,φ,z):
∇Ψ = rˆ ∂Ψ + φˆ 1 ∂Ψ + ẑ ∂Ψ
∂r
r ∂φ
∂z
∂ ( rA r ) 1 ∂A φ ∂A z
∇iA = 1
+
+
r ∂r
r ∂φ
∂z
rˆ
r φ̂
ẑ
⎛ ∂ ( rA φ ) ∂A ⎞ 1
⎛ 1 ∂A z ∂A φ ⎞
A
∂
∂A
⎛
⎞
1
− r ⎟ = det ∂ ∂r ∂ ∂φ ∂ ∂z
−
+ φˆ ⎜ r − z ⎟ + zˆ ⎜
∇ × A = rˆ ⎜
∂z ⎟⎠
∂r ⎠
r ⎝ ∂r
∂φ ⎠ r
⎝ ∂z
⎝ r ∂φ
A r rA φ A z
( )
2
2
∇ 2 Ψ = 1 ∂ r ∂Ψ + 1 ∂ Ψ + ∂ Ψ
r ∂r ∂r
r 2 ∂φ2 ∂z 2
Spherical coordinates (r,θ,φ):
∇Ψ = rˆ ∂Ψ + θˆ 1 ∂Ψ + φˆ 1 ∂Ψ
r ∂θ
r sin θ ∂φ
∂r
(
)
∂A φ
∂ r 2Ar
∂ ( sin θA θ )
+ 1
+ 1
∇iA = 1
∂r
r sin θ
∂θ
r sin θ ∂φ
r2
⎛ 1 ∂A 1 ∂ ( rA φ ) ⎞
⎛ ∂ ( sin θA φ ) ∂A θ ⎞
1 ⎛ ∂ ( rA θ ) − ∂A r ⎞
r −
∇ × A = r̂ 1 ⎜
−
⎟
⎟ + φˆ ⎜
⎟ + θˆ ⎜
r sin θ ⎝
∂θ
∂φ ⎠
r ⎝ ∂r
∂θ ⎠
⎝ r sin θ ∂φ r ∂r ⎠
rˆ
r θˆ
r sin θ φˆ
1
=
det ∂ ∂r ∂ ∂θ
∂ ∂φ
r 2 sin θ
A r rA θ r sin θA φ
(
)
(
)
1
∂ sin θ ∂Ψ +
∂ 2Ψ
∇ 2 Ψ = 1 ∂ r 2 ∂Ψ + 1
∂r
∂θ
r 2 ∂r
r 2 sin θ ∂θ
r 2 sin 2 θ ∂φ2
Gauss’ Divergence Theorem:
∫ ∇iG dv = ∫ Ginˆ da
Vector Algebra:
∇ = x̂∂ ∂x + ŷ∂ ∂y + ẑ∂ ∂z
A • B = A x Bx + A y By + A z Bz
Stokes’ Theorem:
∇ • ( ∇× A ) = 0
∫A ( ∇ × G )in̂ da = ∫C Gid
∇× ( ∇× A ) = ∇ ( ∇ • A ) − ∇2 A
V
A
Page 4 of 7
Basic Equations for Electromagnetics and Applications
Fundamentals
f = q ( E + v × μ o H ) [ N ] (Force on point charge)
E1// − E 2 // = 0
∇ × E = −∂ B ∂t
H1// − H 2 // = J s × n̂
B1⊥ − B2 ⊥ = 0
d
∫ c E • ds = − dt ∫A B • da
n̂
nˆ • ( D1⊥ − D 2 ⊥ ) = ρs
∇ × H = J + ∂ D ∂t
1
2
0 = if σ = ∞
d
∫ c H • ds = ∫A J • da + dt ∫A D • da
∇ • D = ρ → ∫ D • da = ∫ ρdv
Electromagnetic Quasistatics
∇ • B = 0 → ∫ B • da = 0
E = −∇Φ ( r ) , Φ ( r ) = ∫V′ ( ρ ( r ) / 4πε | r ′ − r | ) dv′
A
V
A
E = electric field (Vm-1)
H = magnetic field (Am-1)
D = electric displacement (Cm-2)
B = magnetic flux density (T)
Tesla (T) = Weber m-2 = 10,000 gauss
ρ = charge density (Cm-3)
J = current density (Am-2)
−ρf
ε
C = Q/V = Aε/d [F]
L = Λ/I
i(t) = C dv(t)/dt
v(t) = L di(t)/dt = dΛ/dt
we = Cv2(t)/2; wm = Li2(t)/2
Lsolenoid = N2μA/W
τ = RC, τ = L/R
σ = conductivity (Siemens m-1)
Λ = ∫ B • da (per turn)
∇ • J = −∂ρ ∂t
∇2 Φ =
A
KCL : ∑ i Ii (t) = 0 at node
-1
J s = surface current density (Am )
KVL : ∑ i Vi (t) = 0 around loop
-2
ρs = surface charge density (Cm )
εo = 8.85 × 10
-12
Q = ω0 wT / Pdiss = ω0 / Δω
-1
Fm
ω0 = ( LC )
μo = 4π × 10-7 Hm-1
−0.5
c = (εoμo)-0.5 ≅ 3 × 108 ms-1
V 2 ( t ) / R = kT
e = -1.60 × 10-19 C
ηo ≅ 377 ohms = (μo/εo)0.5
Electromagnetic Waves
(∇
2
− με∂ ∂t
2
2
) E = 0 [Wave Eqn.]
( ∇ 2 − με∂ 2
∂t 2 ) E = 0 [Wave Eqn.]
Ey(z,t) = E+(z-ct) + E-(z+ct) = Re{Ey(z)ejωt}
( ∇ 2 + k 2 ) Eˆ = 0, Eˆ = Eˆ o e− jk i r
Hx(z,t) = ηo-1[E+(z-ct)-E-(z+ct)] [or(ωt-kz) or (t-z/c)]
k = ω(με)0.5 = ω/c = 2π/λ
∫A ( E × H ) • da + ( d dt ) ∫V ( ε E 2 + μ H 2 ) dv
= − ∫V E • J dv (Poynting Theorem)
kx2 + ky2 + kz2 = ko2 = ω2με
2
2
vp = ω/k, vg = (∂k/∂ω)-1
θ r = θi
sin θt / sin θi = ki / kt = ni / nt
Media and Boundaries
D = εo E + P
θ c = sin −1 ( nt / ni )
∇ • D = ρf , τ = ε σ
θ B = tan −1 ( ε t / ε i )
∇ • ε o E = ρf + ρ p
θ > θ c ⇒ Eˆ t = Eˆ iTe+α x − jkz z
∇ • P = −ρp , J = σE
k = k ′ − jk ′′
Γ = T −1
B = μH = μo ( H + M )
(
)
ε ( ω) = ε 1 − ωp 2 ω2 , ω p = ( Ne 2 mε )
ε eff = ε (1 − jσ / ωε )
0.5
(plasma)
0.5
for TM
TTE = 2 / (1 + [ηi cos θt / ηt cos θi ])
TTM = 2 / (1 + [ηt cos θt / ηi cos θi ])
Page 5 of 7
Skin depth δ = ( 2 / ωμσ )
0.5
[ m]
Radiating Waves
Wireless Communications and Radar
2
∇ 2 A − 12 ∂ A
= −μ J f
c ∂t 2
∇2Φ −
A = ∫V ′
Φ = ∫V ′
G(θ,φ) = Pr/(PR/4πr2)
ρf
1 ∂2Φ
=−
2
2
ε
c ∂t
PR = ∫4π Pr ( θ, φ, r ) r 2 sin θ dθdφ
μ J f ( t − rQP / c ) dV ′
Prec = Pr(θ,φ)Ae(θ,φ)
4π rQP
ρ f ( t − rQP / c ) dV ′
A e (θ, φ) = G(θ, φ)λ 2 4π
4πε rQP
E = −∇Φ −
∂A
, B = ∇× A
∂t
G (θ , φ ) = 1.5sin 2 θ (Hertzian Dipole)
(
)
ˆ (r ) = ∫ ′ ρˆ (r )e − jk⏐r ′ − r⏐ / 4πε r '− r dV ′
Φ
V
R r = PR i 2 (t)
ˆ
A(r)
= ∫V ' ( μ Jˆ ( r ) e − jk r '− r 4π r '− r ) dV '
jk x + jk y y
E ff ( θ ≅ 0 ) = ( je jkr λr ) ∫ E t (x, y)e x
dxdy
A
μˆ
ˆ 4πr ) e − jkr sin θ
Eˆ ffθ =
H = ( jηkId
ε ffφ
Ê z = ∑ i a i Ee
ˆ ( x, y, z ) e jωt ⎤
ˆ + ω2 μεΦ
ˆ = −ρˆ ε , Φ ( x, y, z , t ) = Re ⎡Φ
∇2Φ
⎣
⎦
ˆ + ω2 μεA
ˆ = −μ Jˆ , A ( x, y, z , t ) = Re ⎡ Aˆ ( x, y, z ) e jωt ⎤
∇2 A
⎢⎣
⎥⎦
Ebit ≥ ~4 × 10-20 [J]
− jkri
= (element factor)(array f)
Z12 = Z21 if reciprocity
At ωo , w e = w m
Forces, Motors, and Generators
J = σ ( E + v × B)
( 2 4) dv
2
ˆ 4 ) dv
= ∫V ( μ H
w e = ∫V ε Eˆ
wm
F = I × B [ Nm-1 ] (force per unit length)
Q n = ωn w Tn Pn = ωn 2α n
E = − v × B inside perfectly conducting wire (σ → ∞ )
f mnp = ( c 2 ) [ m a ] + [ n b ] + [ p d ]
2
2
-2
Max f/A = B /2μ, D /2ε [Nm ]
dw T
vi =
+ f dz
dt
dt
f = ma = d(mv)/dt
(
2
sn = jωn - αn
Acoustics
P = fv = Tω (Watts)
P = Po + p, U = U o + u
T = I dω/dt
∇p = −ρo ∂ u ∂t
I = ∑ i mi ri
∇ • u = − (1 γPo ) ∂p ∂t
2
FE = λ E ⎡⎣ Nm −1 ⎤⎦ Force per unit length on line charge λ
WM ( λ , x ) =
1 λ2
1 q2
; WE ( q, x ) =
2 L ( x)
2 C ( x)
fM (λ, x ) = −
∂WM
∂x
∂WE
f E ( q, x ) = −
∂x
q
λ
1
1 dL ( x )
d
= − λ 2 (1/ L ( x ) ) = I 2
2 dx
2
dx
d
1
1 dC ( x )
= − q 2 (1/ C ( x ) ) = v 2
2 dx
2
dx
)
2 0.5
2
( ∇ 2 − k 2 ∂ 2 ∂t 2 ) p = 0
2
k 2 = ω2 cs = ω2 ρo γPo
cs = v p = vg = ( γPo ρo )
0.5
or ( K ρo )
0.5
ηs = p/u = ρocs = (ρoγPo)0.5 gases
ηs = (ρoK)0.5 solids, liquids
Optical Communications
p, u ⊥ continuous at boundaries
Page 6 of 7
E = hf, photons or phonons
p = p+e-jkz + p-e+jkz
hf/c = momentum [kg ms-1]
uz = ηs-1(p+e-jkz – p-e+jkz)
dn 2 dt = − ⎡⎣ An 2 + B ( n 2 − n1 )⎤⎦
∫A up • da + ( d dt )∫V ( ρo
u
2
)
2 + p 2 2γPo dV
Transmission Lines
Time Domain
∂v(z,t)/∂z = -L∂i(z,t)/∂t
∂i(z,t)/∂z = -C∂v(z,t)/∂t
∂2v/∂z2 = LC ∂2v/∂t2
v(z,t) = V+(t – z/c) + V-(t + z/c)
i(z,t) = Yo[V+(t – z/c) – V-(t + z/c)]
c = (LC)-0.5 = (με)-0.5
Zo = Yo-1 = (L/C)0.5
ΓL = V-/V+ = (RL – Zo)/(RL + Zo)
Frequency Domain
ˆ
(d 2 /dz 2 +ω 2 LC)V(z)
=0
ˆ
ˆ e-jkz + V
ˆ e +jkz , v( z, t ) = Re ⎡Vˆ ( z )e jωt ⎤
V(z)
=V
+
⎣
⎦
ˆI(z) = Y [V
ˆ e-jkz - V
ˆ e +jkz ], i ( z, t ) = Re ⎡ ˆI( z )e jωt ⎤
0
+
⎣
⎦
k = 2π/λ = ω/c = ω(με)0.5
ˆ
ˆI(z) = Zo Zn (z)
Z(z) = V(z)
Zn (z) = [1 + Γ(z) ] [1 − Γ(z) ] = R n + jX n
Γ(z) = ( V− V+ ) e 2 jkz = [ Zn (z) − 1] [ Zn (z) + 1]
Z(z) = Zo ( ZL − jZo tan kz ) ( Zo − jZL tan kz )
VSWR = Vmax Vmin
Page 7 of 7