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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
Quiz 2, November 17, 2005
6.013 Formula Sheets attached.
Problem 1
Rs = 50Ω
i (t )
v ( t ) = 100 cos (π × 108 t )
+
_
λ1
2
λ2
= 1 meter
4
= 0.4 meter
jX s
Z 02 = 50Ω
Z 01 = 100Ω
Z L = 50 (1 + j ) Ω
A transmission line system incorporates two transmission lines with characteristic impedances of
Z 01 = 100Ω and Z 02 = 50Ω as illustrated above. A voltage source is applied at the left end,
v(t ) = 100 cos (π ×108 t ) . At this frequency, line 1 has length of
λ2
λ1
2
= 1 meter and line 2 has length
= 0.4 meter , where λ1 and λ2 are the wavelengths along each respective transmission line.
4
The two transmission lines are connected by a series reactance jX s and the end of line 2 is
of
loaded by impedance Z L = 50 (1 + j ) Ω . The voltage source is connected to line 1 through a
source resistance Rs = 50Ω .
a) What are the speeds c1 and c2 of electromagnetic waves on each line?
b) It is desired that X s be chosen so that the source current i ( t ) = I 0 cos (π ×108 t ) is in
phase with the voltage source. What is X s ?
c) For the value of X s in part (b), what is the peak amplitude I 0 of the source current i ( t ) ?
Note that the value of X s itself is not needed to answer this question or part ( d ) .
Problem 2
A parallel plate waveguide is to be designed so that only TEM modes can propagate in the
frequency range 0 < f < 2 GHz . The dielectric between the plates has a relative dielectric
constant of ε r = 9 and a magnetic permeability of free space μ0 .
a) What is the maximum allowed spacing d max between the parallel plate waveguide plates?
b) If the plate spacing is 2.1 cm, and f = 10 GHz, what TE n and TM n modes will
propagate?
Problem 3
Rs = 100Ω
Switch
opens at
t=0
+
100 Volts
_
Z 0 = 100Ω, T =
c
Z L = 300Ω
z
0
A transmission line of length
, characteristic impedance Z 0 = 100Ω , and one-way time of flight
T=
is connected at z = 0 to a 100 volt DC battery through a series source resistance
c
Rs = 100Ω and a switch . The z = end is loaded by a 300Ω resistor.
a) The switch at the z = 0 end has been closed for a very long time so that the system is in
the DC steady state. What are the values of the positive and negative traveling wave
voltage amplitudes V+ ( z − ct ) and V− ( z + ct ) ?
Part b, on the next page, to be handed in with your exam. Put your name at the top of the next
page.
Name:
b) With the system in the DC steady state, the switch is suddenly opened at time t = 0 .
Plot the positive and negative traveling wave voltage amplitudes, V+ ( z − ct ) and
i)
V− ( z + ct ) , as a function of z at time t = T .
2
V+ ( z − cT / 2 )
V− ( z + cT / 2 )
100
50
100
50
z
ii)
z
-50
-50
-100
-100
Plot the transmission line voltage v ( z , t ) as a function of z at time t = T
(
v z, t = T
2
2.
)
100
50
z
-50
-100
Please tear out this page and hand in with your exam. Don’t forget to put your name at the top of
this page.
6.013 Quiz 2 Formula Sheet
November 17, 2005
Cartesian Coordinates (x,y,z):
∇Ψ = xˆ ∂Ψ + yˆ ∂Ψ + zˆ ∂Ψ
∂x
∂y
∂z
∂A x ∂A y ∂A z
∇iA =
+
+
∂x
∂y
∂z
∂A y ⎞
⎛ ∂A
∂A ⎞ ⎛ ∂A y ∂A x ⎞
⎛ ∂A
∇ × A = xˆ ⎜ z −
+ yˆ ⎜ x − z ⎟ + zˆ ⎜
−
⎟
∂
∂
∂
∂
∂
∂y ⎟⎠
y
z
z
x
x
⎝
⎠
⎝
⎠
⎝
2
2
2
∇2Ψ = ∂ Ψ + ∂ Ψ + ∂ Ψ
∂x 2 ∂y 2 ∂z 2
Cylindrical coordinates (r,φ,z):
∇Ψ = r̂ ∂Ψ + φˆ 1 ∂Ψ + zˆ ∂Ψ
∂z
∂r
r ∂φ
∂ ( rA r ) 1 ∂A φ ∂A z
∇iA = 1
+
+
∂z
r ∂r
r ∂φ
rˆ
r φˆ
zˆ
⎛ ∂ ( rA φ ) ∂A ⎞ 1
⎛ 1 ∂A z ∂A φ ⎞
A
A
∂
∂
⎛
⎞
1
∇ × A = rˆ ⎜
−
+ φˆ ⎜ r − z ⎟ + zˆ ⎜
− r ⎟ = det ∂ ∂r ∂ ∂φ ∂ ∂z
r ⎝ ∂r
∂z ⎟⎠
∂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
∇iA =
+ 1
+ 1
∂r
r sin θ
∂θ
r sin θ ∂φ
r2
⎛ ∂ ( sin θA φ ) ∂A θ ⎞
⎛ 1 ∂A 1 ∂ ( rA φ ) ⎞
1 ⎛ ∂ ( rA θ ) − ∂A r ⎞
r −
∇ × A = rˆ 1 ⎜
−
⎟
⎟ + θˆ ⎜
⎟ + φˆ ⎜
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
2
r 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:
∫V ∇iG dv = ∫ A Ginˆ da
Stokes’ Theorem:
∫ ( ∇ × G )inˆ da =
A
∫ C G id
Vector Algebra:
∇ = xˆ ∂ ∂x + yˆ ∂ ∂y + zˆ ∂ ∂z
A • B = A x Bx + A y By + Az Bz
∇ • ( ∇× A ) = 0
∇× ( ∇× A ) = ∇ ( ∇ • A ) − ∇ 2 A
Basic Equations for Electromagnetics and Applications
f = q ( E + v × μo H ) [ N ]
E1// − E 2 // = 0
H1// − H 2 // = J s × n̂
∇ × E = −∂ B ∂t
B1⊥ − B2 ⊥ = 0
d
∫ c E • ds = − dt ∫A B • da
∇ × H = J + ∂ D ∂t
D1⊥ − D 2 ⊥ = ρs
Fundamentals
d
Electromagnetic Waves
( ∇ 2 − με∂ 2
∂t 2 ) E = 0 [Wave Eqn.]
( ∇ 2 + k 2 ) E = 0, E = E o e− jk ir
∇ • J = −∂ρ ∂t
k = ω(με)0.5 = ω/c = 2π/λ
kx2 + ky2 + kz2 = ko2 = ω2με
vp = ω/k, vg = (∂k/∂ω)-1
θr = θi
sin θt sin θi = k i k t = n i n t
= electric field (Vm-1)
= magnetic field (Am-1)
= electric displacement (Cm-2)
= magnetic flux density (T)
Tesla (T) = Weber m-2 = 10,000 gauss
θc = sin −1 ( n t n i )
ρ = charge density (Cm-3)
θB = tan −1 ( ε t εi )
J = current density (Am-2)
θ > θc ⇒ E t = E i Te +αx − jk z z
0.5
for TM
k = k '− jk ''
Γ = T −1
σ = conductivity (Siemens m-1)
-1
J s = surface current density (Am )
ρs = surface charge density (Cm-2)
T TE = 2 (1 + [ ηi cos θt ηt cos θi ])
εo ≈ 8.854 × 10-12 Fm-1
T TM = 2 (1 + [ ηt cos θt ηi cos θi ])
μo = 4π × 10-7 Hm-1
c = (εoμo)-0.5 ≅ 3 × 108 ms-1
e = -1.60 × 10-19 C
ηo ≅ 377 ohms = (μo/εo)0.5
( ∇ 2 − με∂ 2 ∂t 2 ) E = 0 [Wave Eqn.]
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
Ey(z,t) = E+(z-ct) + E-(z+ct) = Re{E y (z)e
jω t
}
Hx(z,t) = ηo [E+(z-ct)-E-(z+ct)] [or(ωt-kz) or (t-z/c)]
-1
∫ A ( E × H ) • da + ( d dt ) ∫V ( ε E 2 + μ H 2 ) dv
= − ∫V E • J dv (Poynting Theorem)
2
2
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
Media and Boundaries
D = εo E + P
∇ • D = ρf , τ = ε σ
(d2/dz2 + ω2LC)V(z) = 0
∇ • ε o E = ρf + ρ p
V(z) = V+e-jkz + V-e+jkz
∇ • P = −ρp , J = σE
I(z) = Yo[V+e-jkz – V-e+jkz]
B = μH = μo ( H + M )
k = 2π/λ = ω/c = ω(με)0.5
(
)
ε = ε o 1 − ωp ω2 , ωp = ( Ne 2 mεo )
2
1
2
0 = if σ = ∞
∫ c H • ds = ∫A J • da + dt ∫A D • da
∇ • D = ρ → ∫ D • da = ∫ ρdv
A
V
∇ • B = 0 → ∫ B • da = 0
A
E
H
D
B
n̂
0.5
(Plasma)
Z(z) = V(z) I(z) = Zo Zn (z)
ε eff = ε (1 − jσ ωε )
Zn (z) = [1 + Γ(z) ] [1 − Γ(z) ] = R n + jX n
skin depth δ = (2/ωμσ)0.5 [m]
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 = V max V min