Sample Mid-Term 2 Fall 2009 Engr 297C Dr. Ray Kwok 1. An unpolarized but focused beam of light (λ = 700 nm) is shined onto a 2-cm thick piece of glass (n = 1.43) at an angle of 55o from normal. Calculate the relative power (in dB, relative to the incident power) of the first 2 reflections (R1 & R2) and the first 2 transmission (T1 & T2) as shown in figure. n1 sin θi = n 2 sin θt (1)sin (55o ) = (1.43)sin θt R2 = ? θ t = 35o T2 = ? θ t + θi = 35 + 55 = 90o R1 = ? C T1 = ? D A (55o is the Brewster angle) !!! ┴ polarization at A, n1 cos θi − n 2 cos θ t cos 55o − 1.43 cos 35o Γ⊥ = = = −0.34 o o n1 cos θi + n 2 cos θ t cos 55 + 1.43 cos 35 B τ⊥ = 1 + Γ⊥ = 0.66 ┴ polarization at B, C, D… ┴ polarization only 2 Γ⊥ = 2 R 1 = ΓA = − 0.34 = 0.116 2 2 R 2 = τA ΓB τC = (0.66 )(0.34 )(1.34 ) = 0.090 2 2 T1 = τA τB = (0.66 )(1.34 ) = 0.782 2 2 T2 = τA ΓBΓC τD = (0.66)(0.34 ) (1.34 ) = 0.0105 2 n1 cos θi − n 2 cos θ t = +0.34 n1 cos θi + n 2 cos θt τ⊥ = 1 + Γ⊥ = 1.34 1. (continue) R2 = ? T2 = ? R1 = ? C A D B // polarization at A, Γ// = n1 cos θ t − n 2 cos θi =0 n1 cos θt + n 2 cos θi τ// = η2 (1 − Γ// ) = n1 (1 − Γ// ) = n1 = 1 = 0.70 η1 n2 n 2 1.43 T1 = ? // polarization at B, C, D… Γ// = τ// = n1 cos θt − n 2 cos θi =0 n1 cos θt + n 2 cos θi n1 (1 − Γ// ) = n1 = 1.43 n2 n2 2 // polarization only: R 1 = ΓA = 0 2 T1 = τ A τB 2 R 2 = 0 = T2 1 = (1.34 ) = 1 1.34 1. (continue) R2 = ? T2 = ? R1 = ? C A D T1 = ? Altogether, R1 = ½ (0.116) = 0.058 (┴ only) R 1 = 10 log(0.058) = −12.4dB T1 = ½ (0.782 + 1) = 0.891 T1 = 10 log(0.891) = −0.50dB B R2 = ½ (0.090) = 0.045 (┴ only) R 2 = 10 log(0.045) = −13.5dB T2 = ½ (0.0105) = 0.0053 (┴ only) T2 = 10 log(0.0053) = −22.8dB Note: R1 + R2 + T1 + T2 = 0.058 + 0.891 + 0.045 + 0.0053 = 0.9993 accounts for almost all the input power already. 2. A submarine is trying to communicate with a helicopter above water as shown in figure. Sea water is nonmagnetic and has a relative εr = 72, and σ = 4 S/m. (a) What is the minimum angle of entrance φ (see figure) such that communication with air is possible? (b) There are 2 transmitters on board, one at HF (3 MHz) and the other at mmwave (120 GHz). Which transmitter would deliver a stronger signal to the helicopter? Explain. (a) n w sin θc = n air sin (90o ) z φ 1 1 = 72 8.5 sin θc = θc = 6.77 o y φ min = 90 − 6.77 = 83.23o (b) At 3 MHz, σ 4 = = 333 ωε 2π(3 ⋅106 )(72 )(10 −9 / 36π ) ( )( ) ωµσ 2π 3 ⋅106 4π ⋅10 −7 (4 ) = = 6.88 good conductor α = 2 2 σ 4 At 120 GHz, = = 0.0083 ωε 2π 120 ⋅10 9 (72 ) 10 −9 / 36π ( low loss dielectric α= ) ( ) σ µ σ ηo 4 377 = = = 88.9 2 ε 2 ε r 2 72 HF transmitter, lower loss 2. (c) If a 3 MHz RHCP signal is transmitted at an angle φ = 90o (normal incidence) with the E-field amplitude = 5 µV/m right below the surface, write the full expressions for the incident E & H fields. z (c) φ y ω = 2πf = 1.88 ⋅ 107 α = β = 6.88 ηc = (1 + j) α 6.88 = (1 + j) = (1 + j)1.72 = 2.43∠45o σ 4 r E i = 5e − αz [xˆ cos(ωt − 6.88z ) + yˆ sin (ωt − 6.88z )] r E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z µV/m r 5 H i = e − 6.88 z yˆ cos 1.88 ⋅ 107 t − 6.88z − xˆ sin 1.88 ⋅ 107 t − 6.88z ηc r H i = 2.06e − 6.88 z yˆ cos 1.88 ⋅ 107 t − 6.88z − π / 4 − xˆ sin 1.88 ⋅ 107 t − 6.88z − π / 4 µA/m [ ( [ ) ( [ ) ( )] ( )] ( ) ( Note: z < 0, so the lower the z, the higher the amplitude (closer to the source) )] 2. (d) Write the expression for the reflected E-field. What’s its polarization state? (e) Write the expression for transmitted H-field. (d) Γ = η2 − η1 = (1 + j)1.72 − 377 = j1.72 − 375 = 0.991∠179.5o ≈ −0.991 (1 + j)1.72 + 377 η2 + η1 j1.72 + 379 r E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z r E r = −4.96e 6.88 z xˆ cos 1.88 ⋅ 107 t + 6.88z + yˆ sin 1.88 ⋅ 107 t + 6.88z µV/m [ ( ) [ )] ( ( ) )] ( LHCP (e) τ = 1 + Γ = 1 − 0.991 = 0.009 ( ) ω 2π 3 ⋅ 106 k air = = = 0.02π = 0.0628 8 c 3 ⋅ 10 r E i = 5e − 6.88 z xˆ cos 1.88 ⋅ 107 t − 6.88z + yˆ sin 1.88 ⋅ 107 t − 6.88z r E t = 0.045 xˆ cos 1.88 ⋅ 107 t − 0.0628z + yˆ sin 1.88 ⋅ 107 t − 0.0628z r 0.045 Ht = yˆ cos 1.88 ⋅ 107 t − 0.0628z − xˆ sin 1.88 ⋅ 107 t − 0.0628z ηo r H t = 0.12 yˆ cos 1.88 ⋅ 107 t − 0.0628z − xˆ sin 1.88 ⋅ 107 t − 0.0628z nA/m [ ( [ ( [ [ ) ) ( ( ( ) ) )] ( ( ( )] )] )] 2. (f) If the submarine is at a depth of 100 m, what is the attenuation (in dB) of the signal by the time it gets to the helicopter (assuming air is a lossless medium)? (ignore the reflection “mismatch loss” for now.) (f) A (dB) = −8.686αr = −8.686(6.88)(100) = −5976dB This demonstrated the difficulty of underwater communications. By the same token, it’s difficult to detect a submarine from air. 3. Find Zin, Γin, return loss at input port, and VSWR on the 50Ω line. 100 Ω 0.1λ 50 Ω λ/4 50 Ω system Zin ZL = 150 – j 75 Ω βl = π / 4 Z2 Zin 75 Ω, λ/8 50 Ω λ/4 Z1 ( ) Z1 = (2 − j) + j = 2 = 0.5 − j0.5 ZL + j tan β l = 1 + jZL tan βl 1 + j(2 − j) 2 + j2 Z1 = 37.5 − j37.5Ω o Z 2 = Zop = − jZo cot β l = − j100 cot 36 = − j138Ω Zo2 50 2 Zin = = = 2.9 + j13.6Ω Z1 + Z 2 37.5 − j37.5 − j138 Γin = Zin − Zg 2.9 + j13.6 − 50 = = 0.897∠149o Zin + Zg 2.9 + j13.6 + 50 RL = −20 log(ρ) = −20 log(0.897) = 0.94dB VSWR = 1 + ρ 1 + Γ12 1 + 0.897 = = = 18.4 1 − ρ 1 − Γ12 1 − 0.897

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