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Lossy Medium 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 = jωε + 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 dielectric loss the smaller, the better !!! Lossy Medium - Dr. Ray Kwok Example 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. 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 r r P 1 1 1 2 ave = J ⋅ E = σE2 = ()39.1 ⋅10 −4 ()250 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 phase constant attenuation constant propagation 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 phasor 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 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 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 electric field )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 nepers/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 radians !!! 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 2 2 2 2()1+ tan δ −1 − ω µε '= α −β 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 2 4α4 + 4α2ω2µε '− ω2µε " = 0 2 () 2 ω µε ' 2 β = ()1+ tan δ +1 2 2 2 2 2 2 2 − 4ω µε '± ()()4ω µε ' +16 ω µε " α = 8 2 µ µ 2 ω µε ' 2 ηc = = ' α = ()−1± 1+ tan δ εc ε ()1− jtan δ 2 2 µ − 2/1 2 ω µε ' 2 η = ()1− jtan δ α = ( 1+ tan δ −1) c 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 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 = (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 2( . 08 )(.0 0004 ) = 6.4 ⋅10 36 π −8 σ µ ση o ()6.4 ⋅10 ()377 −6 α = = = 6= .
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