lecture 22 poynting’s theorem and normal incidence
TRANSCRIPT
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EECS 117
Lecture 22: Poyntings Theorem and Normal Incidence
Prof. Niknejad
University of California, Berkeley
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EM Power Carried by Plane Wave
In a lossless medium, we have found that
Ex = E0 cos(t z)
Hy =E00
cos(t z)
where = and = /The Poynting vector S is easily calculated
S = EH = zE2
00
cos2(t z)
S = z
E2020 (1 + cos (2(t z)))
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Average Power of Plane Wave
If we average the Poynting vector over time, themagnitude is
Sav =
E2020
This simple equation is very useful for estimating theelectric field strength of a EM wave far from its source
(where it can be approximated as a plane wave)
The energy stored in the electric and magnetic fieldsare
we = 12|Ex|2 = 1
2E20 cos
2(t z)
wm =
1
2|Hy|2
=
1
2
E20
20 cos
2
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Plane Wave Resonance
Its now clear that
wm =1
2
E20
cos2(t
z) = we
In other words, the stored magnetic energy is equal tothe stored electric energy. In analogy with a LC circuit,
we say that the wave is in resonance
We can also show that
tV
(wm + we)dV = SS dS
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Example: Cell Phone Basestation
A cell phone base station transmits 10kW of power.Estimate the average electric field at a distance of 1mfrom the antenna.
Assuming that the medium around the antenna islossless, the energy transmitted by the source at anygiven location from the source must be given by
Pt =
Surf
S dS
where Surf is a surface covering the source of radiation.Since we do not know the antenna radiation pattern,lets assume an isotropic source (equal radiation in all
directions)
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Example Cont.
In that case, the average Poynting vector at a distance rfrom the source is given by
S= Pt4r2
= 104
4Wm2
This equation is simply derived by observing that the
surface area of a sphere of radius r is given by 4r2
Using S= 12E200
, we have
E0 =
20S=
2 377 10
4
4= 775
V
m
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Example: Cell Phone Handset
A cell phone handset transmits 1W of power. What isthe average electric field at a distance of 10cm from thehandset?
S=Pt
4r2=
1
4.12W
m2= 77
W
m2
We can see that the electric field near a handset is at amuch lower level.
Whats a safe level?
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Complex Poynting Theorem
Last lecture we derived the Poynting Theorem forgeneral electric/magnetic fields. In this lecture wed liketo derive the Poynting Theorem for time-harmonic fields.
We cant simply take our results from last lecture and
simply transform t j. This is because the Poyntingvector is a non-linear function of the fields.
Lets start from the beginning
E = jB
H = jD + J = (j + )E
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Complex Poynting Theorem (II)
Using our knowledge of circuit theory, P = V I, wecompute the following quantity
(EH
) = H
EE H
(EH) = H (jB)E (jD + J)
Applying the Divergence TheoremV
(EH)dV =S
(EH) dSS
(EH) dS = V
E JdV+V
j(E DH B)dV
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Complex Poynting Theorem (III)
Lets define eff = + , and = . Since most
materials are non-magnetic, we can ignore magneticlossesS
(EH)dS = V
EDdVjV
(H H E E) d
Notice that the first volume integral is a real numberwhereas the second volume integral is imaginary
SEH dS = 2 V PcdV
SEH
dS = 4 V(wm we)dV
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Average Complex Poynting Vector
Finally, we have computed the complex Poynting vectorwith the time dependence
S = 12EH + EHe2jt
Taking the average value, the complex exponential
vanishes, so that
Sav =1
2 (EH)
We have thus justified that the quantity S = EHrepresents the complex power stored in the field.
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Example: Submarine Communication
Consider a submarine at a depth of z = 100 m. Wewould like to communicate with this submarine using aVLF f = 3 kHz. The conductivity of sea water is
= 4 Sm1
, r = 81, and 1.
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Ocean Water Conductivity
Note that we are forced to use low frequencies due tothe conductivity of the ocean water. The lossconductive tangent
tan c =
105 1
Thus the ocean is a good conductor even at 3 kHzThe propagation loss and constant are thus equal
= = 2 0.2
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O W W P i
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Ocean Water Wave Propagation
The wavelength in seawater is much smaller than in air(0 = 100 km in air)
= 2 = 29 m
Thus the phase velocity of the wave is also much
smallervp = f 9 104 m/s
The skin-depth, or the depth at which the wave is
attenuated to about 37 % of its value, is given by
=1
= 4.6 m
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Ocean Water Fields
The wave impedance is complex with a phase of 45
|c|
=
8
102
c = ej45
Notice that c 0, the ocean water thus generates avery large magnetic field for wave propagation
H =
E0
c ez
cos(6 103
t z )Where is the angle of the complex wave impedance,
45
in this particular case
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Shi t S b i C i ti
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Ship to Submarine Communication
Now lets compute the required transmission power ifthe receiver at the depth of z = 100 m is capable ofreceiving a signal of at least 1 V/m
Side-note: the receiver sensitivity is set by the noisepower at the input of the receiver. If the signal is toosmall, its swamped by the noise.
E0ez Emin = 1 V/m
This requires E0 = 2.8 kV/m, and a corresponding
magnetic field of H0 = E0/c = 37 kA/m
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P ti V t i O W t
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Poynting Vector in Ocean Water
This is a very large amount of power to generate at thesource. The power density at the source is
Sav = 12(EH)
Sav =1
2(2.84
37 cos(45)) = 37 MW/m2
At a depth of 100 m, the power density drops toextremely small levels
Sav(100m) = 4.6 1012 MW/m2
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R fl ti f P f t C d t
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Reflections from a Perfect Conductor
Consider a plane wave incidentnormally onto a conducting surface
Ei = xEi0ej1z
Hi = yEi00
ej1z
The reflected wave (if any) has thefollowing form
Er = xEr0ej1z
Hr = yEr00
ej1z
Ei
Hi
Er
Hr
Si
Sr
=
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B d C diti t I t f
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Boundary Conditions at Interface
The conductor forces the tangential electric field tovanish at the surface z = 0
E(z = 0) = 0 = x(Ei0 + Er0)
This implies that the reflected wave has equal andopposite magnitude and phase
Er0 = Ei0This is similar to wave reflection from a transmission
line short-circuit load.
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T t l Fi ld
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Total Field
We can now write the total electric and magnetic field inregion 1
E(z) = xEi0(ej1z
ej1z
) = xEi0j2sin(1z)
H(z) = yEi00
(ej1z + ej1z) = yEi00
2cos(1z)
The net complex power carried by the wave
EH
= zE2i0
0 4j sin(1z) cos(1z)
is reactive. That means that the average power is zero
Sav = 12(EH) = 0
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Standing Wave
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Standing Wave
The reflected wave interferes with the incident wave tocreate a standing wave
E(z, t) = (E(z)ejt
) = (Ei0j2sin(1z)ejt
)
E(z, t) = 2Ei0 sin(1z)cos(t)
H(z, t) = 2Ei01cos(1z) sin(t)
Note that the E and H fields are in time quadrature (90
phase difference)The instantaneous power is given by
S=4E2
i01 sin(1z)cos(1z)
2 sin(21z)
cos(t)sin(t) 2 sin(2t)
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Standing Wave Power
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Standing Wave Power
The electric and magnetic powers are readily calculated
we =1
2
1
|E1
|2 = 21
|Ei0
|2 sin2(1z)cos
2(t)
wm =1
21|H1|2 = 21|Ei0|2 cos2(1z)sin2(t)
Note that the magnetic field at the boundary of theconductor is supported (or equivalently induces) asurface current
Js = nH = x2Ei01
A/m
If the material is a good conductor, but lossy, then thiscauses power loss at the conductor surface.
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Normal Incidence on a Dielectric
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Normal Incidence on a Dielectric
Ei
Hi
Er
Hr
Si
Sr
Ht
S
Et
t
Consider an incident wave onto a dielectric region.
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Normal Incidence on a Dielectric
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Normal Incidence on a Dielectric
We have the incident and possibly reflected waves
Ei = xEi0ej1z
Hi = yEi01
ej1z
Er = xEr0ej1z
Hr = yEr01
ej1z
But we must also allow the possibility of a transmittedwave into region 2
Et
= xEt0ej2z
Ht = yEt02
ej2z
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Dielectric Boundary Conditions
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Dielectric Boundary Conditions
At the interface of the two dielectrics, assuming nointerface charge, we have
Et1 = Et2
Ei0 + Er0 = Et0
Ht1 = Ht2
Ei0
1 Er0
1=
Et0
2We have met these equations before. The solution is
Er0 =
2
12 + 1Ei0
Et0 =22
2 + 1
Ei0
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Transmission Line Analogy
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Transmission Line Analogy
z = 0
1 21 2
These equations are identical to the case of theinterface of two transmission lines
The reflection and transmission coefficients are thusidentical
=Er0
Ei0=
2 12 + 1
=Et0Ei0
= 1 +
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Harder Example
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Harder Example
Consider three dielectricmaterials. Instead of solvingthe problem the longway,
lets use the transmission lineanalogy.
First solve the problem at the
interface of region 2 and 3.Region 3 acts like a load toregion 2. Now transform thisload impedance by the length
of region 2 to present anequivalent load to region 1.
1
1
2
2
3
3
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Glass Coating
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Glass Coating
A very practical example is the case of minimizingreflections for eyeglasses. Due to the impedancemismatch, light normally reflects at the interface of air
and glass. One method to reduce this reflection is tocoat the glass with a material to eliminate thereflections.
From our transmission line analogy, we know that thiscoating is acting like a quarter wave transmission linewith
=
0 glass
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