HW3Solution

HW3Solution - ECE 201A Homework 3 solution 1 RegionI 0 0 i...

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ECE 201A Homework 3 solution 1. We can describe the problem with a transmission line equivalent circuit, which is The propagation constants of the equivalent transmission lines, 0 β and 1 , are the components of the k vector in z direction, i.e., 00 0 2 cos z i k π θ λ == and 11 0 2 cos z t kn The characteristics impedances of the equivalent transmission lines, 0 Z and 1 Z , are the wave impedances in z direction, i.e., 0 cos z i ZZ η and 0 0 z n since the wave is TE polarized. Since regions I and III are infinite in extend so are the equivalent transmission lines. Since we see the characteristic impedance looking into an infinitely long transmission line, the equivalent circuit becomes i t i H G i E G 0 ε 0 μ Region I 1 0 Region II 0 0 Region III d z x 10 r εε = r n = . . . 0 Z 0 Z 1 Z 0 0 1 d
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To find the power reflected back to region I, let’s find the reflection coefficient Γ at the first junction. 0 0 in in Z Z Z Z Γ= + , where 00 1 1 1 01 0 1 tan tan in in in Z ZZ j Z d Z j Z d β −+ = ++ . Then ( ) () 22 10 1 1 1 0 1 1 1 0 1 1 0 1 tan tan tan tan tan 2t a n jZ Z d d ZZ d d Z Zj d ββ +− = + Power reflection coefficient is 2 Γ , so ( ) 2 2 21 0 1 2 2 2 2 1 0 1 tan 4t a n d Z Z d For total transmission into region III 2 0 . Because if there is no reflected power to region I, all the power must be transmitted to region III, since all the media are lossless. This requires either Z Z = , which is the trivial solution or 1 dm π = , which makes 1 tan 0 d = . 1 0 2 cos t n dd m θπ λ == , which yields 0 2c o s t m d n θ = . t is found using the Snell’s law, hence 1 sin sin sin sin it t i n n θθ =⇒ = 2 2 2 sin sin cos 1 sin 1 i i tt n nn =− = Therefore sin 2 sin 2 ii mm d n n −− . . 0 Z 0 Z 1 Z 0 1 d in Z
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2. . () cos sin 00 ˆˆ i jk x k y jk r iz z Ea E e a E e θ + == G G i G Upon reflection of the incident plane wave of k vector i k G plane waves with the k vectors indicated in the figure above will be generated. These are the plane waves with k vectors d k G , u k G and r k G . So the overall distribution will be given as the superposition of four plane waves with k vectors i k G , d k G , u k G and r k G . The amplitudes of the E G fields are found using the reflection coefficient for a planar conductor. In this case the E G field is tangential to the conductors. So 1 Γ=− . Then the E G fields of the four plane waves are cos sin 0 ˆ x k y E e + = G ( ) cos sin 0 ˆ x k y dz EaE e −− =− G ( ) cos sin 0 ˆ x k y uz e G cos sin 0 ˆ x k y rz E e −+ = G Then the total E G field is cos sin sin cos sin sin 0 2sin sin 2sin sin ˆ jkx jky jky jkx jky jky z y y Ee e e e e e θθ ⎛⎞ ⎜⎟ ⎝⎠ G ±²²²³ ² ² ² ´± ² ² ²³ ² ² ²´ ( ) cos cos 0 2sin cos ˆ sin jkx jkx z x j k y e G ±²²²³ ² ² ²´ σ =∞ = ∞ y x i k G i k G r k G u k G r k G d k G
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() 0 ˆ 4s i n c o ss i n s i n z Ea E k x k y θ =− G At a finite y say at ya = and for 0 45 = () () 1 sin cos 2 θθ == and sin 2 kx EA ⎛⎞ = ⎜⎟ ⎝⎠ G , where 0 i n 2 ka AE = Period of the standing wave x Δ pattern is found using 2 2 x π λ Δ = , which yields 2 2 2 x Δ= = for 0 45 = . In general 2cos x . 0 90 = will result in the shortest period.
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HW3Solution - ECE 201A Homework 3 solution 1 RegionI 0 0 i...

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