EMSP 6sep2022 sol

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Electromagnetics and Signal Processing for Spaceborne Applications – EM part September 6th, 2022 1 2 3 4 do not write above Problem 1 A source with voltage Vg = 10 V and internal impedance Zg = 50  is connected to a transmission line with characteristic impedance ZC = 75  which terminates on a load ZL = 50 The frequency is f = 300 MHz and the length of the line is l = 5.25 m. 1) Calculate the power absorbed by ZL. 2) Calculate the voltage at the load section (VAA) if ZL becomes a short circuit due to a fault in the load; express VAA in the time domain. Solution 1) As there is no match at the generator section and at the load section, let us calculate the reflection coefficient at AA: Γ𝐿 = 𝑍𝐿 − 𝑍𝐶 𝑍𝐿 + 𝑍𝐶 = −0.2 The reflection coefficient at section BB is: Γ𝐵𝐵 = Γ𝐿𝑒−2𝑗𝛽𝑙 = 0.25 As a result, the input impedance is: Z𝐵𝐵 = Z𝐶 1 + Γ𝐵𝐵 1 − Γ𝐵𝐵 = 112.5 Ω As the line length is a multiple of /4, the input impedance could have been easily calculated as: SURNAME AND NAME __________________________________________________________ ID NUMBER __________________________________________________________ SIGNATURE __________________________________________________________ B A A B Vg Zg ZL ZC l Z𝐵𝐵 = 𝑍𝐶 2 𝑍𝐿 The reflection coefficient at the generator section (left side) is: Γ𝑔 = 𝑍𝐵𝐵 − 𝑍𝑔 𝑍𝐵𝐵 + 𝑍𝑔 = 0.3846 Therefore, the power crossing section BB is: 𝑃𝐵𝐵 = 𝑃𝐿 = |𝑉𝑔| 2 8Re[𝑍𝑔](1 − |Γ𝑔| 2 ) = 0.213 W This is also the power absorbed by the load, as no other element in the circuit beyond section BB can absorb power. 2) If the load becomes a short circuit, there is no need to perform calculations: indeed, as ZL = 0 Ω  Γ𝐿 = −1. As a result, whatever the progressive wave reaching the load, it will be totally reflected with a change in the sign. Therefore, the total voltage at the section…