Midtermsolution

Midtermsolution - ECE 201A Midterm Solution 1. E = ( ax +...

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ECE 201A Midterm Solution 1. () ( ) 00 - ˆˆ +2 jy xz Eaj a e aj a e ββ =+ r (a) - 0 1 zx Ha j a e a j a e η ⎡⎤ =− + + ⎣⎦ r (b) and fields propagating in +y direction are E r H r 0 - a e β r and 0 - 0 1 j a e + r . Hence the time average power density is () () () 2 0 ˆ 1 1 1 ˆ Re 1 1 W/m 2 y y av z z x y a a j a e a j a e a ηη −+ + ⎛⎞ + × = + = ⎜⎟ ⎝⎠ r * 11 Re Re 22 x SE H = rr (c) E r and fields propagating in -y direction are H r 0 2 Ea j a e r and 0 0 1 2 j a e r . Hence the time average power density is 2 0 1 1 1 5 ˆ ˆ 2 2 Re 4 1 W/m 2 2 y av x z z x y y S a ja e a ja e a a × = = r * Re Re E H = (d) Net time average power density is 2 15 3 1= - W / m yy av av av y y SSS a a +− =+= rrr . (e) The part propagating in +y direction has two mutually orthogonal components with equal amplitude and 90 out of phase. So by definition this part is circularly polarized. The part propagating in - y direction has two mutually orthogonal components with unequal amplitude and 90 out of phase. So this part is elliptically polarized.
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2. μ 0 εεε = r1 0 ε r = 4 0 = r 0 d 0 0 Air z x The equivalent transmission line representing this problem is d Z 1 β 1 Z 2 2 Z 0 0 z 1 2 Each uniform region is represented by a uniform piece of a transmission line. The characteristic impedance of each transmission line is equal to the wave impedance in +z direction in each uniform region. The propagation constant of each transmission line is equal to the z component of the k vector in each uniform region. Since the last transmission line is infinitely long we see its characteristic impedance looking into it, i.e., to the right of junction 1. Therefore we can replace it with a load impedance which is equal to its characteristic impedance. Then d Z 1 1 Z 0 0 Z 2 Z in z 2 1 in Z is the input impedance seen at the input of the second transmission line, i.e., to the right of junction 2. Then the reflection coefficient at junction 2 is
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tan 0 tan 0 r in i in EZ Z Z Γ= = + Since the incidence is normal 0 ii i Z μ η ε == and 0 2 i i k π βω μ ε λ = . For no reflection back to the free space or . 0 0 in ZZ = 211 10 12 1 tan tan in Zj Z d Z Z d β + + 1 2 12 1 02 1 tan tan Z Z d Z Z Z d += + Since all the quantities are real we have to equate the real and imaginary parts of this equation, which yields Z Z = and 1 2 21 tan tan Z dZ Z d = . The first equality yields Z Z = , which is the trivial solution. The other one yields 1 2 Z = which is the desired solution. Obviously we cannot have both conditions at the same time. A moments reflection shows that if 1 tan d →∞ then imaginary parts of the denominator and the numerator in the in Z expression overwhelms the real parts and we obtain the desired solution. For 1 tan d 11 /2 dk d π . This means () 1 2 2 dm λ =+ or 1 4 , where 0,1. ... m = . Clearly 0 1 r 1 ε = , where 0 is the free space wavelength.
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Midtermsolution - ECE 201A Midterm Solution 1. E = ( ax +...

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