Answer:
The answer is below
Explanation:
a) To solve this problem, we are going to use the smith chart. After entering the value of Zo = 200 ohm and ZL = (50 − j25) Ω, this give us [tex]z_L\ and\ y_L[/tex], the intersection between [tex]y_L[/tex] and the SWR line gives:
[tex]y(d)=1.026-j1.54\\\\We\ are\ using\ the\ imaginary\ part\ to calculate\ the\ capacitance,hence:\\\\wC=1.54*Y_o\\\\C=\frac{1.54}{Z_o*w}=\frac{1.54}{200*2\pi*800*10^6}=1.53*10^{-12}\\ \\C=1.53*10^{-12}F[/tex]
b) Also, we get:
[tex]y(d)=1.0-j1.52\\\\We\ are\ using\ the\ imaginary\ part\ to\ calculate\ the\ inductance,hence:\\\\1/wL=1.52*Y_o\\\\L=\frac{Z_o}{1.52*w}=\frac{200}{1.52*2\pi*800*10^6}=2.6*10^{-8}\\ \\L=2.6*10^{-8}H[/tex]