EMSP 18lug2022

Pagina 1

Electromagnetics and Signal Processing for Spaceborne Applications July 18th, 2022 1 2 3 4 do not write above Problem 1 A plane sinusoidal EM wave (f = 9 GHz) propagates from a medium with electric permittivity r1 = 4 (assume r = 1 for both media) into free space. The expression of the incident electric field is: 𝐸⃗ 𝑖(𝑧,𝑦)= −𝜇 𝑥 𝑒−𝑗√2 2 𝛽1𝑧𝑒𝑗√2 2 𝛽1𝑦 V/m 1) What is the polarization of the incident field (specify the details of the polarization)? 2) Determine the value of the electric field in A(z = 1 cm, y = 0 m). Solution 1) The wave polarization is linear, specifically a TE component (along -x). 2) The incidence angle can be derived, for example, from the y component of : 𝛽𝑦 = 𝛽1 sin(𝜃)= 𝛽1√2/2 → sin(𝜃)= √2/2 → 𝜃 = 45° To determine the transmitted wave, it is first necessary to calculate the refraction angle, which is: 𝜃2 = sin−1 (sin (𝜃)√ 𝜀𝑟1 𝜀𝑟2 ) = sin−1(√2)≈ sin−1(1.4142) SURNAME AND NAME ____________________________________________________ ID NUMBER ___________________________________________________________ SIGNATURE ___________________________________________________________ y z 2 1  A This is the sign of an evanescent wave: this wave is totally reflected, but the electric field will penetrate in the second medium. The expression of the transmitted field will be: 𝐸⃗ 𝑡(𝑧,𝑦)= −𝜇 𝑥(1+ Γ𝑇𝐸) 𝑒−𝑗𝛽2𝑧𝑧𝑒𝑗𝛽2𝑦𝑦 V/m As in A, y = 0 m → 𝐸⃗ 𝑡(𝑧,𝑦= 0 m)= −𝜇 𝑥Γ𝑇𝐸 𝑒−𝑗𝛽2𝑧𝑧 V/m Let us calculate 𝛽2𝑧: 𝛽2𝑧 = 𝛽2 cos(𝜃2)= 𝛽2√1− [sin(𝜃2)]2 = 𝛽0√1− [sin(𝜃2)]2 = 𝛽0√1 − √2 2 = 𝛽0√−1 = ±𝑗𝛽0 = − 𝑗𝛽0 The negative sign is chosen to obtain a physical solution: 𝑒−𝑗𝛽2𝑧𝑧 = 𝑒−𝑗(− 𝑗𝛽0)𝑧 = 𝑒−𝛽0𝑧 The reflection coefficient can be calculated as: 𝜂1 = 𝜂0 cos(𝜃)√𝜀𝑟1 = 266.6 Ω 𝜂2 = 𝜂0 cos(𝜃2)√𝜀𝑟2 = 𝜂0 − 𝑗= 𝑗377 Ω The choice of the negative sign in 𝜂2 is consistent with the one in 𝛽2𝑧. Γ = 𝜂2 − 𝜂2 𝜂2…