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} _________.
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} _________.
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Solution and Explanation
\( P_{{incident}} = 10 \, {mW} \) 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.
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