Lossy Medium

EE142 Dr. 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 V = IR  l  El = ()JA ρ  A  A  E r r E = Jρ resisitivity r 1 r J = E V ρ r r J = σE conductivity

 Low resistivity => “conductor ” ~<10 -5 Ω- cm (  T )  High resistivity => “insulator ” ~>10 10 Ω- cm  Intermediate resistivity => “semiconductor ” typical ~10 -3 to 10 5 Ω- cm (  eEg/kT )  unit of conductivity = S/m = Siemens/meter = mho/m = (Ω- m) -1 Lossy Medium - Dr. Ray Kwok

EM Wave through medium r ε∇ ⋅E = ρf r r r ∂H ∇ ⋅ E = 0 ∇× E = −µ (homogeneous, r r r ∂t linear, isotropic) ∇ × E =r− jωµ H ∇ ⋅H = 0 r ρ ≈ 0, J ≠ 0 ∇ ⋅ H = 0 r r ∂E r r r ∇× H = J + ε ∇ × H = J + jωε E f ∂t f

r r  r r  σ  ∇× H = ()σ + jωε E = jωε + E = jωε cE  jω σ ε ≡ ε − j ≡ ε'−jε" finite σ means complex ε c ω Lossy Medium - Dr. Ray Kwok

Loss Tangent ε" σ tan δ ≡ = ε' ωε

good conductor σ >> ωε good insulator σ << ωε

Low tan δ → low

the smaller, the better !!! Lossy Medium - Dr. Ray Kwok Example A sinusoidal E-field with 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. σ σ σ tan δ = .0 001 = = = −9 ωε ωε oεr 9 10  ()2 ⋅π 10  ()5.2  36 π  σ = 39.1 ⋅10 −4 S/m

The average power dissipated per unit volume is P 1 r r 1 1 ave = J ⋅ E = σE2 = ()39.1 ⋅10 −4 ()250 2 V 2 2 2 P ave 2 = 34.4 W/m Note: V 2 1 V2 1 (El) 1 P = = = σE2 ()lA ave 2 R 2 ρl / A 2 Lossy Medium - Dr. Ray Kwok

Wave Equation r r ∂H ∇ × E = −µ ∂rt r ∂E ∇ × H = εc ∂tr r 2 2 ∂ E ∇ E = µε plane wave equation still holds with modification of ε c ∂t 2 r r r (j ωt−k c )r jωt −γr allow k be complex since ε is )t,r(E = Eoe ≡ Eoe e

γ ≡ jk c = jω µε c ≡ α + jβ constant attenuation constant Lossy Medium - Dr. Ray Kwok

Complex Propagation Constant

 ε"  σ  γ = jω µε = jω µε 1− j  = jω µε 1− j  = jω µε ()1− jtan δ = α + jβ c  ε'   ωε 

Ther r r −γr −αr − jβr )r(E = Eoe = Eoe e

attenuation e-αr Lossy Medium - Dr. Ray Kwok

dB scale

power intensity ratio in log scale, not a unit !!       I P V > 0 gain dB( ) =10 log   =10 log   = 20 log    Io   Po   Vo  < 0 loss

sound intensity power

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 Lossy Medium - Dr. Ray Kwok

dBm & dBW  P  dBW ≡10 log   1W  become real units  P  dBm ≡10 log   1mW 

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 Lossy Medium - Dr. Ray Kwok

Example How much electricity generated by the solar cell? Isotropic What if a 40 W bulb is used? 200 W bulb? 100 W 100 100 W Intensity = power/area = = = 96.7 4πR 2 4π()1 2 m 2

1 m Power generated in solar cell  W  =  96.7 ()100 cm 2 ()40 % = 31 8. mW  m2   .0 0318 W  solar cell In terms of dB =10 log   = −35 dB 10 x 10 cm 2  100 W  40% efficiency system “gain”  P  40 W bulb? − 35 =10 log    40  Power of electricity generated = 12.6 mW  P  200 W bulb? − 35 =10 log    200  Power of electricity generated = 63.2 mW Lossy Medium - Dr. Ray Kwok

Attenuation

r r r For example, if the )r(E = E e−γr = E e−αre− jβr o o r intensity going through a medium )r(E attenuates at a rate of 0.4 dB/m, A )[r( dB ] = 20 log r = 20 log e−αr what is α ? )0(E -0.4 dB = -8.686 α (1 m) log c )b( α = 0.4/8.686 = 0.046 /m log a )b( = log c )a( Note: nepers (np) is not a real unit. − 20 αr A )[r( dB ] = = − .8 686 αr similar to !!! ln( 10 ) Note also α is a positive number for attenuation.

ααα[dB/m] = 8.686 ααα[np/m] Attenuation term k

How to express e-αr term??

y θθθ r x kˆ = xˆ sin θ + yˆ cos θ

2 2 e−αr = e−α x( sin θ+ y cos θ) = e−α x + y Same ? 2 2  x   y  x + y 2 2 xsin θ + ycos θ = x  + y  = = x + y Yes, Q.E.D.  r   r  x 2 + y2 r Can think of : α ≡ αkˆ r r α ⋅r = α()xsin θ + y cos θ

r r 2 2 e−α⋅r = e−α ( xsin θ + ycos θ ) = e−α x + y Lossy Medium - Dr. Ray Kwok

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

 ε"  σ  γ = jω µε 1− j  = jω µε 1− j  = jω µε ()1− jtan δ  ε'   ωε  n !n n(n − )1 1( + )x n = ∑ x n−k =1+ nx + x 2 +..... k=0 n(!k − )!k 2 1( + )x n ≈1+ nx for small x  ε" 2/1  1 ε" γ = jω µε 1− j  ≈ jω µε 1− j  ≡ α + jβ  ε'   2 ε'  ω µε ε" ω µε σ σ µ α = = = small 2 ε' 2 ωε 2 ε β = ω µε ≈ ω v/ µ µ µ  ε"− 2/1 µ  j ε" µ ηc ≡ = = 1− j  ≈ 1+  ≈ εc ε 1( − jε /" ε )' ε  ε'  ε  2 ε'  ε Lossy Medium - Dr. Ray Kwok

Good Conductor (σ>> ωε )  σ  σ γ = jω µε 1− j  ≈ jω − j µε  ωε  ωε 1− j − j = e− jπ 2/ = e− jπ 4/ = 2 1− j µσ µσω γ = jω = j( + )1 ≡ α + jβ 2 ω 2 µσω α = β = 2 µ µ µωε 1+ j µω µω α ηc ≡ = ≈ j ≈ = ()1+ j = ()1+ j εc ε 1( − jσ / ωε ) εσ 2 σ 2σ σ Lossy Medium - Dr. Ray Kwok

Skin Depth δδδ 1 2 δ ≡ = = δ (NOT loss tangent δδδ !!!!!) α µσω s r r −αr − jβr )r(E = Eoe e At r = δ, |E| decreases to 1/e (or 63% drop).

r )r(E A )[r( dB ] = 20 log r = 20 log e−αr = − .8 686 αr )0(E

At r = δ, |E| decreases by -8.7 dB. At r = 2 δ, |E| decreases by -17.3 dB…. Lossy Medium - Dr. Ray Kwok General Material

γ = jω µε c = jω µ ε'( − jε )" ≡ α + jβ 2 2 2 2 2 2 2 2 2  ω µε " ω µε 'tan δ γ = −ω µ ε'( −jε )" = α − β + j2 αβ β =   =  2α  2()1+ tan 2 δ −1 − ω2µε '= α2 −β2 real ω2µε 'tan 2 δ ()1+ tan 2 δ +1 jω2µε "= 2j αβ imaginary β2 = 2 2 2 2(1+ tan δ −1)(1+ tan δ +1)  ω2µε " 2 2 2   2 2 2 2 α = β − ω µε '=   − ω µε ' 2 ω µε 'tan δ()1+ tan δ +1  2α  β = 2 2()1+ tan δ −1 4 2 2 2 2 4α + 4α ω µε '−()ω µε " = 0 ω2µε ' β2 = ()1+ tan 2 δ +1 2 2 − 4ω2µε '± ()()4ω2µε ' +16 ω2µε " 2 α2 = 8 µ µ ω2µε ' η = = α2 = ()−1± 1+ tan 2 δ c ε ε' ()1− jtan δ 2 c 2 µ − 2/1 2 ω µε ' 2 () α = (1+ tan δ −1) ηc = 1− jtan δ 2 ε' Lossy Medium - Dr. Ray Kwok Summary Lossy Medium - Dr. Ray Kwok

Example - The skin depth of a certain nonmagnetic conducting material is 2 µm at 5 GHz. Determine the 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 = (2 π)(5 x 10 9) (2 x 10 -6) = 6.28 x 10 4 m/s

r )r(E A )[r( dB ] = 20 log r = 20 log e−αr = − .8 686 αr )0(E A )[r( dB ] = − .8 686 /r δ = − .8 686 10( )2/ = −43 4. dB in just 5 skin depth. (((−(−−− 43 dB = 1 / 20,000 !!!) Only surface current on conductors. Lossy Medium - Dr. Ray Kwok

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 ?

o Teflon: εr = 2.08, tan δ = 0.0004 at 25 C assuming frequency independence. σ (a) tan δ = ωε 10 −9  6   −8 S/m σ = ωε oεr tan δ = ()2 ⋅π 10  (08.2 )(.0 0004 = )6.4 ⋅10  36 π  −8 σ µ ση o ()6.4 ⋅10 ()377 −6 α = = = = 04.6 ⋅10 np/m 2 ε 2 εr 2 08.2 A dB( ) = − .8 686 αr = − .8 686 ()04.6 ⋅10 −6 ()100 = − .0 005 dB 10 −9  (b) 10   −4 σ = ωε oεr tan δ = ()2 ⋅π 10  (08.2 )(.0 0004 = )6.4 ⋅10 S/m  36 π  −4 σ µ ση ()6.4 ⋅10 ()377 −2 α = = o = = 04.6 ⋅10 np/m 2 ε 2 εr 2 08.2 A dB( ) = − .8 686 αr = − .8 686 ()04.6 ⋅10 −2 ()100 = −50 dB works well at low freq (TV to antenna) but not so well at high freq. !! Lossy Medium - Dr. Ray Kwok

Example – In a nonmagnetic, lossy, dielectricr medium, a 300-MHz plane wave is characterized by the magnetic field phasorH = (xˆ − 4j zˆ)e−2ye− 9j y A/m. Obtain time-domain expressions for the electric and magnetic field vectors. What is the state of this wave? r z r −2y (j ωt−9 )y H( t,r )= ℜe{(xˆ − 4j zˆ)e e } + r r t = 0 H()t,r = ˆex −2y cos( ωt − 9 )y + ˆ e4z −2y sin( ωt − 9 )y y, k t = 0 x

α = 2, β = 9 − ω2µε '= α2 − β2 LHEP ω2µε "= 2αβ ε" 2αβ )(2(2 )9 = = = .0 468 = tan δ ε' β2 − α2 92 − 22 ε' β2 − α2 77 c2 77 3( ⋅10 8 )2 εr = = 2 = 2 = 6 2 = 95.1 εo ω µoεo ω 2( ⋅π 300 ⋅10 )

µ − 2/1 ηo − 2/1 377 − 2/1 ηc = ()1− jtan δ = ()1− jtan δ = ()1− .0j 468 ε' εr 95.1 o 22.0j ηc = 257 ∠12 5. = 257 e r E = zˆη e−2y cos( ωt − 9 )y − xˆ 4η e−2y sin( ωt − 9 )y r c c E = zˆ257 e−2y cos( ωt − 9y + 22.0 ) − xˆ1028 e−2y sin( ωt − 9y + 22.0 ) Lossy Medium - Dr. Ray Kwok

Exercise (how to write the attenuation?)

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: ~ sin( ωt + 30 x −15 z)

ε = 4( − j 02.0 )ε o

r r Write H (r,t) = ? Lossy Medium - Dr. Ray Kwok

Surface resistance for conductors

−αz − jβz − 1( + /z)j δ J = σE = Joe e = Joe J r r ∞ − 1( + /z)j δ Jo wδ Jowδ − jπ 4/ I = ∫ J ⋅ ad = Jow∫ e dz = = e 0 ()1+ j 2

w L J w δ/√2 V = E L = o L on surface o σ J o V 1+ j L L Z = = ≡ Z similar to R & ρ I σδ w s w 1+ j Z = surface impedance ( Ω) s σδ Lossy Medium - Dr. Ray Kwok

Homework

1. 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 -3 (b) moist soil with εr = 2.5 and σ = 10 S/m.

2. Calculations concerning the electromagnetic effect of currents in a good conductor usually neglect the displacement current even at microwave frequencies. 7 (a) Assuming εr = 1 and σ = 5.7 x 10 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.