Lossy Medium - San Jose State University€¦ · Lossy Medium - Dr. Ray Kwok Example – In a...

Lossy Medium

EE142Dr. Ray Kwok

•reference:

Fundamentals of Engineering Electromagnetics, David K. Cheng (Addison-Wesley)

Electromagnetics for Engineers, Fawwaz T. Ulaby (Prentice Hall)

Lossy Medium - Dr. Ray Kwok

Ohm’s Law

A

V

E

( )

EJ

E1

J

JE

AJAE

IRV

rr

rr

rr

ll

σ=

ρ=

ρ=

ρ=

=

resisitivity

conductivity

Low resistivity => “conductor” ~<10-5 Ω- cm ( T )

High resistivity => “insulator” ~>1010 Ω- cm

Intermediate resistivity => “semiconductor”

typical ~10-3 to 105 Ω- cm ( eEg/kT )

unit of conductivity = S/m = Siemens/meter = mho/m = (Ω- m)-1


EM Wave through medium

t

EJH

0Ht

HE

E

f

f

∂

∂ε+=×∇

=⋅∇∂

∂µ−=×∇

ρ=⋅∇ε

rrr

r

rr

r

EjJH

0H

HjE

0E

f

rrr

r

rr

r

ωε+=×∇

=⋅∇

ωµ−=×∇

=⋅∇(homogeneous,

linear, isotropic)

ρ ≈ 0, J ≠ 0

( )

"j'j

EjEj

jEjH

c

c

ε−ε≡ω

σ−ε≡ε

ωε=

ω

σ+εω=ωε+σ=×∇

rrrr

finite σ means complex ε


Loss Tangent

ωε

σ=

ε

ε≡δ

'

"tan

good conductor σ >> ωεgood insulator σ << ωε

Low tanδ → low dielectric loss

the smaller, the better !!!


Example

( ) ( )

4

99ro

1039.1

5.236

10102

001.0tan

−

−

⋅=σ

π⋅π

σ=

εωε

σ=

ωε

σ==δ

( )( )

34.4V

P

2501039.12

1E

2

1EJ

2

1

V

P

ave

242ave

=

⋅=σ=⋅= −rr

A sinusoidal E-field with amplitude of 250 V/m and frequency 1 GHz exists in a

lossy dielectric medium that has a εr = 2.5 and loss tangent of 0.001. Find the average power dissipated in the medium per cubic meter.

S/m

W/m2

The average power dissipated per unit volume is

( ) ( )AE2

1

A/

E

2

1

R

V

2

1P 2

22

ave ll

lσ=

ρ==

Note:


Wave Equation

2

2

c

2

c

t

EE

t

EH

t

HE

∂

∂µε=∇

∂

∂ε=×∇

∂

∂µ−=×∇

rr

rr

rr

plane wave equation still holds with modification of ε

β+α≡µεω=≡γ

≡= γ−ω−ω

jjjk

eeEeE)t,r(E

cc

rtj

o

)rkt(j

oc

rrrallow k be complex since ε is

propagation constant

attenuation constant

phase constant


Complex Propagation Constant

( ) β+α=δ−µεω=

ωε

σ−µεω=

ε

ε−µεω=µεω=γ jtanj1jj1j

'

"j1jj c

rjr

o

r

o eeEeE)r(E β−α−γ− ==rrr

The phasor

attenuation e-αr


dB scale

power intensity ratio in log scale, not a unit !!

=

=

=

ooo V

Vlog20

P

Plog10

I

Ilog10)dB(

sound intensity power voltage

10 log(2) ≈ 3, 3 dB = double

10 log(1/2)≈ -3, -3 dB = half

10 log(10) = 10, 10 dB = 10x

10 log(100) = 20, 20 dB = 100x

10 log(0.1) = -10, -10 dB = 1/10

What is 6 dB? −9 dB? 7 dB? −44 dB?

4x 1/8 5x 4 x 10-5

> 0 gain

< 0 loss


dBm & dBW

≡

≡

mW1

Plog10dBm

W1

Plog10dBW

0 dBm = 1 mW

30 dBW = 1 kW

-30 dBm = 1 µW

What is 40 dBW? -7 dBm? -26 dBm? 21 dBm?

10 kW 0.2 mW 2.5 µW 1/8 W

become real units


ExampleIsotropic100 W

1 m

solar cell10 x 10 cm2

40% efficiency

How much electricity generated by the solar cell?

What if a 40 W bulb is used? 200 W bulb?

Intensity = power/area( ) 222 m

W96.7

14

100

R4

100=

π=

π=

Power generated in solar cell

( )( ) mW8.31%40cm100m

W96.7 2

2=

=

In terms of dB dB35W100

W0318.0log10 −=

=

200 W bulb?

Power of electricity generated = 63.2 mW

=−

200

Plog1035

40 W bulb?

Power of electricity generated = 12.6 mW

=−

40

Plog1035

system “gain”


Attenuation

r686.8)10ln(

r20]dB)[r(A

)a(log

)b(log)b(log

elog20)0(E

)r(Elog20]dB)[r(A

eeEeE)r(E

c

ca

r

rjr

o

r

o

α−=α−

=

=

==

==

α−

β−α−γ−

r

r

rrr For example, if the electric field

intensity going through a medium

attenuates at a rate of 0.4 dB/m,

what is α ?

-0.4 dB = -8.686 α (1 m) α = 0.4/8.686 = 0.046 nepers/m

Note: nepers (np) is not a real unit.

similar to radians !!!

Note also α is a positive number for attenuation.

αααα[dB/m] = 8.686 αααα[np/m]

Attenuation term

22

22

22

yx)cosysinx(r

yxyx

yx

r

yy

r

xxcosysinx

eee

cosysinxk

22

+=+

+=

+

=θ+θ

==

θ+θ=

+α−θ+θα−α−

k

y

x

rθθθθ

Same ?

Yes, Q.E.D.

How to express e-αr term??

Can think of :

( )22

)cossin(

cossin

ˆ

yxyxreee

yxr

k

+−+−⋅− ==

+=⋅

≡

αθθαα

θθαα

αα

rr

rr

r


Low-loss dielectric (ε”<<ε’) or (σ<<ωε)

( )

ε

µ≈

ε

ε+

ε

µ≈

ε

ε−

ε

µ=

εε−ε

µ=

ε

µ≡η

ω≈µεω=β

ε

µσ=

ωε

σµεω=

ε

εµεω=α

β+α≡

ε

ε−µεω≈

ε

ε−µεω=γ

+≈+

+−

++=−

=+

δ−µεω=

ωε

σ−µεω=

ε

ε−µεω=γ

−

−

=

∑

'

"

2

j1

'

"j1

)'/"j1(

v/

22'

"

2

j'

"j

2

11j

'

"j1j

nx1)x1(

.....x2

)1n(nnx1x

)!kn(!k

!n)x1(

tanj1jj1j'

"j1j

2/1

c

c

2/1

n

2knn

0k

n

for small x

small


Good Conductor (σ>>ωε)

( ) ( )σ

α+=

σ

µω+=

σ

µω+≈

εσ

µωε≈

ωεσ−ε

µ=

ε

µ≡η

µσω=β=α

β+α≡µσω

+=ω

µσ−ω=γ

−===−

µεωε

σ−ω≈

ωε

σ−µεω=γ

π−π−

j12

j12

j1j

)/j1(

2

j2

)1j(2

j1j

2

j1eej

jjj1j

c

c

4/j2/j


Skin Depth δδδδ

s

21δ=

µσω=

α≡δ

rjr

o eeE)r(E β−α−=rr

At r = δ, |E| decreases to 1/e (or 63% drop).

r686.8elog20)0(E

)r(Elog20]dB)[r(A r α−=== α−

r

r

At r = δ, |E| decreases by -8.7 dB.

At r = 2δ, |E| decreases by -17.3 dB….

(NOT loss tangent δδδδ !!!!!)


General Material

( )

( ) ( )

( )

( )1tan12

'

tan112

'

8

"16'4'4

0"'44

'2

"'

2j"j

'

j2)"j'(

j)"j'(jj

22

2

22

2

22222

2

22224

2

22

222

2

222

2222

c

−δ+µεω

=α

δ+±−µεω

=α

µεω+µεω±µεω−=α

=µεω−µεωα+α

µεω−

α

µεω=µεω−β=α

αβ=µεω

β−α=µεω−

αβ+β−α=ε−εµω−=γ

β+α≡ε−εµω=µεω=γ

( )

( )( )( )

( )( )

( )1tan12

'

1tan12

1tan1tan'

1tan1

1tan1

1tan12

tan'

1tan12

tan'

2

"

22

2

2

2222

2

2

2

222

2

222

22

+δ+µεω

=β

−δ+

+δ+δµεω=β

+δ+

+δ+

−δ+

δµεω=β

−δ+

δµεω=

α

µεω=β

( )

( ) 2/1

c

'

c

c

tanj1'

tanj1

−δ−

ε

µ=η

δ−ε

µ=

ε

µ=η

real

imaginary


Summary


Example - The skin depth of a certain nonmagnetic conducting

material is 2µm at 5 GHz. Determine the phase velocity in the material.

What is the attenuation (in dB) when the wave penetrates 10 µm into the material?

phase velocity v = ω/β

for conductor, α = β = 1/δ

v = ωδ = (2π)(5 x 109) (2 x 10-6) = 6.28 x 104 m/s

dB4.43)2/10(686.8/r686.8]dB)[r(A

r686.8elog20)0(E

)r(Elog20]dB)[r(A r

−=−=δ−=

α−=== α−r

r

in just 5 skin depth.

((((−−−− 43 dB = 1 / 20,000 !!!)

Only surface current on conductors.


Example – (a) Calculate the dielectric loss (in dB) of an EM

wave propagating through 100 m of teflon at 1 MHz. (b) at 10 GHz ?

Teflon: εr = 2.08, tanδ = 0.0004 at 25oC assuming frequency independence.

( ) ( )( )

( )( )

( )( ) 005.01001004.6686.8r686.8)dB(A

1004.608.22

377106.4

22

106.40004.008.236

10102tan

tan

6

68

r

o

89

6

ro

−=⋅−=α−=

⋅=⋅

=ε

ση=

ε

µσ=α

⋅=

π⋅π=δεωε=σ

ωε

σ=δ

−

−−

−−

(a)

S/m

np/m

dB

Coaxial cable works well at low freq (TV to antenna) but not so well at high freq. !!

( ) ( )( )

( )( )

( )( ) 501001004.6686.8r686.8)dB(A

1004.608.22

377106.4

22

106.40004.008.236

10102tan

2

24

r

o

49

10

ro

−=⋅−=α−=

⋅=⋅

=ε

ση=

ε

µσ=α

⋅=

π⋅π=δεωε=σ

−

−−

−−

S/m

np/m

dB

(b)


Example – In a nonmagnetic, lossy, dielectric medium, a 300-MHz plane

wave is characterized by the magnetic field phasor A/m.Obtain time-domain expressions for the electric and magnetic field vectors.What is the polarization state of this wave?

α = 2, β = 9

( ) y9jy2 eez4jxH −−−=r

( ) ( ) ( )

22.0jo

c

2/12/1

r

o2/1

c

26

28

2

2

oo

2

22

o

r

2222

2

222

e2575.12257

468.0j195.1

377tanj1tanj1

'

95.1)103002(

)103(77c77'

tan468.029

)9)(2(22

'

"

2"

'

=∠=η

−=δ−ε

η=δ−

ε

µ=η

=⋅⋅π

⋅=

ω=

εµω

α−β=

ε

ε=ε

δ==−

=α−β

αβ=

ε

ε

αβ=µεω

β−α=µεω−

−−−

( ) ( ) ( ) )y9tsin(e4z)y9tcos(ext,rH

eez4jxet,rH

y2y2

)y9t(jy2

−ω+−ω=

−ℜ=

−−

−ω−

rr

rr

)22.0y9tsin(e1028x)22.0y9tcos(e257zE

)y9tsin(e4x)y9tcos(ezE

y2y2

y2

c

y2

c

+−ω−+−ω=

−ωη−−ωη=

−−

−−

r

r

x

z

y, k t = 0

t = 0+

LHEP


Exercise (how to write the attenuation?)

oj

zxt

εε

ω

)02.04(

)1530sin(~

−=

−+

Given Eo at the origin has a amplitude of 1 V/m along the y-axis in a

non-magnetic medium, with the propagation given by:

?),( =trHrr

Write


Surface resistance for conductors

σδ

+=

≡σδ

+==

σ==

j1Z

w

LZ

w

Lj1

I

VZ

LJ

LEV

s

s

oo

J

L

w

Jo

δ/√2w

( )4/joo

0

/z)j1(

o

/z)j1(

o

zjz

o

e2

wJ

j1

wJdzewJadJI

eJeeJEJ

π−∞

δ+−

δ+−β−α−

δ=

+

δ==⋅=

==σ=

∫∫rr

on surface

surface impedance (Ω)

similar to R & ρ


Homework1. Determine the frequency at which a time-harmonic electric field intensity

causes a conduction current density and a displacement current density

of equal magnitude in

(a) seawater with εr = 72 and σ = 4 S/m, and

(b) moist soil with εr = 2.5 and σ = 10-3 S/m.

2. Calculations concerning the electromagnetic effect of currents in a good

conductor usually neglect the displacement current even at microwave

frequencies.

(a) Assuming εr = 1 and σ = 5.7 x 107 S/m for copper, compare the magnitude of the displacement current density with that of the conduction

current density at 100 GHz.

(b) Write the differential equation in phasor form for magnetic field

intensity H in a source-free good conductor.

Lossy Medium - San Jose State University€¦ · Lossy Medium - Dr. Ray Kwok Example – In a...

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