Lecture 22 Poynting’s Theorem and Normal Incidence

8/2/2019 Lecture 22 Poyntings Theorem and Normal Incidence

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EECS 117

Lecture 22: Poyntings Theorem and Normal Incidence

Prof. Niknejad

University of California, Berkeley

Universit of California Berkele EECS 117 Lecture 22 . 1/


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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)))



3/29

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

(t z)Universit of California Berkele EECS 117 Lecture 22 . 3/


4/29

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



5/29

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



7/29

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?



8/29

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



10/29

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



11/29


12/29

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.



13/29

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


O W W P i


15/29

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



16/29

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


Shi t S b i C i ti


17/29

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


P ti V t i O W t


18/29

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


R fl ti f P f t C d t


19/29

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

=


B d C diti t I t f


20/29

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.


T t l Fi ld


21/29

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


Standing Wave


22/29

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)


Standing Wave Power


23/29

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.


Normal Incidence on a Dielectric


24/29


Ei

Hi

Er

Hr

Si

Sr

Ht

S

Et

t

Consider an incident wave onto a dielectric region.




25/29


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


Dielectric Boundary Conditions


26/29

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


Transmission Line Analogy


27/29

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 +


Harder Example


28/29

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


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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Lecture 22 Poynting’s Theorem and Normal Incidence

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