REVIEW OF UPW BASICS
Example : x̂-polarized UPW traveling in +zˆ direction
E(z) = x̂Eo e− jkz
E = xˆEo cos(ωt − kz)
H = yˆ
Eo
cos(ωt − kz)
ηo
Eo − jkz
H(z) = ŷ e
ηo
E × H : ẑ direction of propagation
c=
1
= 3 x108 m / s
μo εo
ω(rads / s) = 2πf
x
k(rads / m) =
μo
2π ω
=
ηo =
λ c
ε0
direction of propagation
E ( z,0 )
z
y
H ( z,0 )
λ=
2π
k
λ=
c
f
wavelength
L4-1
HOW DO WAVES CONVEY
POWER, ENERGY?
Recall:
E [V/m] i J [A/m2 ] = Pd [W/m3 ]
But E ⊥ H
Manipulate Ampere’s law to get E • J
∂B
∂D
For
symmetry,
compute
H
⋅
(
∇
×
E
=
−
)
E ⋅ (∇ × H = J +
)
∂t
∂t
H ⋅ ( ∇ × E ) − E ⋅ ( ∇ × H) = −H ⋅ (
Vector Identity
∇ ⋅ ( E × H) = − H ⋅
∂B
∂D
) − E ⋅ (J + ⋅ )
∂t
∂t
∂B
∂D
−E⋅ J −E⋅
(W/m3 )
∂t
∂t
This is Poynting’s Theorem
What does it mean?
L4-2
POYNTING THEOREM
Poynting’s Theorem: ∇ ⋅ (E × H ) = − H ⋅
∂B
∂D
−E⋅
−E⋅J
∂t
∂t
B = μH
∇ ⋅ ( E × H) = −
Poynting
vector,
S [W/m2]
D = εE
d ⎛1
d ⎛1
2⎞
2⎞
3
μ
H
−
ε
E
⎜
⎟
⎜
⎟ − (E ⋅ J ) (W/m )
dt ⎝ 2
⎠ dt ⎝ 2
⎠
Stored electric Power density
energy density, dissipated/m3,
Wd
We
Stored magnetic
energy density,
Wm
Energy is conserved!
Note :
d ⎛1
d|H|
dB
2⎞
μ
|
H
|
=
μ
|
H
|
=
H
i
⎜
⎟
dt ⎝ 2
dt
dt
⎠
⎛ volts amps watts ⎞
⋅
=
Poynting Vector S = E × H (W/m ) ⎜
m
m
m2 ⎟⎠
⎝
E
2
S
H
L4-3
INTEGRAL POYNTING THEOREM
ˆ
nda
Use:
∫S A ⋅ n̂ da = ∫V ∇ ⋅ A dv
dv
dv
Gauss’s Theorem (not Gauss’s Law)
Therefore:
∫S(E×H) ⋅ n̂ da
dv
dv
=
dv
∫V ∇⋅ (E×H) dv
2⎞
2⎞
⎡ d ⎛1
⎤
d ⎛1
= ∫ ⎢ − ⎜ μ H ⎟ − ⎜ ε E ⎟ − E ⋅ J ⎥ dv
V ⎣ dt ⎝ 2
⎠ dt ⎝ 2
⎠
⎦
2 1
2⎞
d ⎛1
E
×
H
⋅
n̂
da
=
−
ε
E
+
μ
H
dv − ∫ E ⋅ J dv
⎟
∫S
∫V dt ⎜⎝ 2
V
2
⎠
Power emerging = released stored energy - dissipation [W]
(
(
)
)
The Poynting vector
S = E × H gives both the magnitude
of the power density (intensity) and the direction of its flow.
L4-4
UNIFORM PLANE WAVE EXAMPLE
E = xEo cos ( ωt − kz )
⎛E ⎞
H = y ⎜ o ⎟ cos ( ωt − kz )
⎝ ηo ⎠
1
We = εoEo2 cos2 ( ωt − kz )
2
1 μo 2
2
cos
Wm =
E
( ωt − kz)
o
2 ηo2
⎛ E2 ⎞
S(t) = E × H = ẑ⎜ o ⎟ cos2 ( ωt − kz ) (W/m2 )
⎜ ηo ⎟
⎝ ⎠
S ( z at t = 0)
z
0
Eo2
1 Eo2
2
=
S = zˆ
cos ( ωt − kz ) ⇒ ⟨S⟩ = zˆ
2 ηo
η0
I ( θ, φ,r ) [w/m2 ]
The time average ⟨S(r,θ, φ⟩ is “intensity” [W/m2]
L4-5
COMPLEX NOTATION – POYNTING VECTOR
Defining a meaningful S and relating it to S is not obvious.
Let’s work backwards to find the time average ⟨S⟩ and then S
S(t) = E × H = Re ⎡E ⋅ e jωt ⎤ × Re ⎡H ⋅ e jωt ⎤
⎣
⎦
⎣
⎦
= [Er cos(ωt) − Ei sin(ωt)] × [Hr cos(ωt) − Hi sin(ωt)]
(
) (
)
1
⎡ Er × Hr + Ei × Hi ⎤
⎦
2⎣
1
1
= Re E × H∗ [ = R e {(Er + jEi ) × (Hr − jHi )}]
2
2
⇒ ⟨S(t)⟩ =
)
(
S (by definition
Thus, we can define ⟨S⟩ =
Recall: E = Er + jEi
(
1
Re E × H∗
2
H = Hr + jHi
)
and
S = E × H∗
e jωt = cos ωt + jsin ωt
L4-6
UPW REFLECTED BY PERFECT CONDUCTOR
ˆ − cos(ωt + kz)
E = x̂E+ cos(ωt − kz) + xE
Forward plus
reflected wave
= 0 at z = 0 (perfect conductor)
⇒ E− = −E+ (Solving for unknown reflection)
⇒
E = x̂2E+ sinωt ⋅ sinkz
Standing waves, oscillate without moving
(recall: cosα − cos β = −2 sin
α+β
isin
2
α−β
)
2
We [J/m3] = 2 εE2 sin2ωt sin2kz
E = 0 every half cycle (ωt = 0, π, etc.)
(Where does the energy go?)
ωt = -π/2 ωt = π/2
x
Perfect
Conductor
E
ωt = 0
Never any We here
z
z=0
L4-7
STANDING WAVE EXAMPLE - CONTINUED
E = x̂ [E+ cos(ωt − kz) + E− cos(ωt + kz)]
E+
E−
H = ŷ [
cos(ωt − kz) −
y cos(ωt + kz)]
ηo
ηo
(It’s in the H field!)
Note
E− = −E+ ⇒
H = ŷ
2E+
cos ωt ⋅ coskz
ηo
2
⎛ 2E ⎞
Wm [J/m3 ] = 1 μo ⎜ + ⎟ cos2ωt cos2kz = 2εE+2 cos2ωt cos2kz
2 ⎝ ηo ⎠
= 0 when ωt = π/2, 3π/2, etc.)
E2+
S = E × H = ẑ 4
cos ωt sinωt ⋅ coskz sinkz
ηo
y
ωt = π/2 ωt = π ωt = 0
2
E+
ˆ
H
=z
⋅ sin 2kz ⋅ sin 2ωt
ηo
1
⇒ ⟨S⟩ = 0 = Re E×H∗
2
(
)
S
z=0
z
L4-8
MIT OpenCourseWare
http://ocw.mit.edu
6.013 Electromagnetics and Applications
Spring 2009
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