Tutorial - Transmission-reflection.pdf - Problem 8.9 The three regions shown in Fig P8.9 contain perfect dielectrics For a wave in medium 1 incident

# Tutorial - Transmission-reflection.pdf - Problem 8.9 The...

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Problem 8.9 The three regions shown in Fig. P8.9 contain perfect dielectrics. For a wave in medium 1, incident normally upon the boundary at z = d , what combination of ε r 2 and d produces no reflection? Express your answers in terms of ε r 1 , ε r 3 and the oscillation frequency of the wave, f . Medium 2 ε r 2 Medium 3 ε r 3 Medium 1 ε r 1 z = - d z = 0 z d Figure P8.9: Dielectric layers for Problems 8.9 to 8.11. Solution: By analogy with the transmission-line case, there will be no reflection at z = d if medium 2 acts as a quarter-wave transformer, which requires that d = λ 2 4 and η 2 = η 1 η 3 . The second condition may be rewritten as η 0 ε r 2 = bracketleftbigg η 0 ε r 1 η 0 ε r 3 bracketrightbigg 1 / 2 , or ε r 2 = ε r 1 ε r 3 , λ 2 = λ 0 ε r 2 = c f ε r 2 = c f ( ε r 1 ε r 3 ) 1 / 4 , and d = c 4 f ( ε r 1 ε r 3 ) 1 / 4 .

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Problem 8.10 For the configuration shown in Fig. P8.9, use transmission-line equations (or the Smith chart) to calculate the input impedance at z = d for ε r 1 = 1, ε r 2 = 9, ε r 3 = 4, d = 1 . 2 m, and f = 50 MHz. Also determine the fraction of the incident average power density reflected by the structure. Assume all media are lossless and nonmagnetic. Solution: In medium 2, λ = λ 0 ε r 2 = c f ε r 2 = 3 × 10 8 5 × 10 7 × 3 = 2 m . Hence, β 2 = 2 π λ 2 = π rad/m , β 2 d = 1 . 2 π rad . At z = d , the input impedance of a transmission line with load impedance Z L is given by Eq. (2.63) as Z in ( d ) = Z 0 parenleftbigg Z L + jZ 0 tan β 2 d Z 0 + jZ L tan β 2 d parenrightbigg . In the present case, Z 0 = η 2 = η 0 / ε r 2 = η 0 / 3 and Z L = η 3 = η 0 / ε r 3 = η 0 / 2, where η 0 = 120 π ( ) . Hence, Z in ( d ) = η 2 parenleftbigg η 3 + j η 2 tan β 2 d η 2 + j η 3 tan β 2 d parenrightbigg = η 0 3 parenleftBigg 1 2 + j ( 1 3 ) tan1 . 2 π 1 3 + j ( 1 2 ) tan1 . 2 π parenrightBigg = η 0 ( 0 . 35 j 0 . 14 ) .

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