Soln_Homework6 - Homework 6 solutions Problem 6.1 Using the...

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c2 c0 m2 e2 := Gamma12 eta2 eta1 eta2 eta1 + := Gamma23 eta3 eta2 eta3 eta2 + := Tau12 2 eta2 eta2 eta1 + := Tau23 2 eta3 eta2 eta3 + := Gamma21 Gamma12 := Tau21 2 eta1 eta2 eta1 + := FrontFace_echo Gamma12 Einc := Second_echo Einc Tau12 Gamma23 Tau21 := First_thru Einc Tau12 Tau23 := Second_thru Einc Tau12 Gamma23 Gamma21 Tau23 := delayrefl 2 d2 c2 := delaytrans 2 d2 c2 := from 30 MHz to 30 GHz i 0 1000 .. := freq i 30 10 6 i 1 + ( ) := om i 2 π freq i := Gamma2Ray i FrontFace_echo Second_echo exp i om i delayrefl ( ) + := Tau2Ray i First_thru Second_thru exp i om i delayrefl ( ) + := 20 10 0 20 log Gamma2Ray i ( ) 20 log Tau2Ray i ( ) 11.22 Homework 6 solutions Problem 6.1 Using the two-ray rule to calculate the reflection coefficient of a three material problem with the same medium on either side. This is the halfwave window radome problem e0 8.854 10 12 := m0 4 π 10 7 := The properties of space eta0 m0 e0 := c0 1 m0 e0 := e1 1 := e2 2 6 + ( ) := e3 1 := m1 1 := m2 1 := m3 1 := Einc 1 := d2 0.25 0.0254 := in meters First calculate the medium impedances eta1 eta0 m1 e1 := eta2 eta0 m2 e2 := eta3 eta0 m3 e3 :=
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0 5 10 15 20 25 30 30 freq i 10 9 The halfwave window condition: Lambda2 d2 2 := fwindow c2 Lambda2 := fwindow 1.119 10 10 × = Note the Transmission maximum and reflection minimum occur at the halfwave window frequencies and
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