Question

A 50 \(\Omega\) lossless transmission line is terminated with a load \( Z_L = (50 - j75) \, \Omega.\) { If the average incident power on the line is 10 mW, then the average power delivered to the load
(in mW, rounded off to one decimal place) is} _________.

Hint:

To solve for the average power delivered to a load in a transmission line, first calculate the reflection coefficient \( \Gamma \) based on the load impedance and the characteristic impedance. Then, use the formula for power transfer, which incorporates the magnitude of the reflection coefficient to adjust the incident power. The formula \( P_{{load}} = \frac{P_{{incident}}}{1 + \left| \Gamma \right|^2} \) is crucial for this calculation.

Answer 1

We use the formula for the average power delivered to the load: \[ P_{{load}} = \frac{P_{{incident}}}{1 + \left| \Gamma \right|^2}, \] where:
\( P_{{incident}} = 10 \, {mW} \) is the incident power,
\( \Gamma \) is the reflection coefficient, which is given by: \[ \Gamma = \frac{Z_L - Z_0}{Z_L + Z_0}. \] Here, \( Z_0 = 50 \, \Omega \) is the characteristic impedance of the transmission line. Substituting \( Z_L = (50 - j75) \, \Omega \): \[ \Gamma = \frac{(50 - j75) - 50}{(50 - j75) + 50} = \frac{-j75}{100 - j75}. \] To simplify, multiply both the numerator and denominator by the complex conjugate of the denominator: \[ \Gamma = \frac{-j75(100 + j75)}{(100 - j75)(100 + j75)} = \frac{-j75(100 + j75)}{100^2 + 75^2} = \frac{-j7500 - 5625}{15625}. \] Now, calculate \( \left| \Gamma \right|^2 \): \[ \left| \Gamma \right|^2 = \frac{(-5625)^2 + (-7500)^2}{15625^2} = \frac{31890625 + 56250000}{244140625} = \frac{88140625}{244140625} = 0.36. \] Thus, the average power delivered to the load is: \[ P_{{load}} = \frac{10}{1 + 0.36} = \frac{10}{1.36} \approx 6.5 \, {mW}. \] Thus, the average power delivered to the load is 6.5 mW.