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.
Top Questions on Flip-Flop
In the circuit shown below, the AND gate has a propagation delay of 1 ns. The edge-triggered flip-flops have a set-up time of 2 ns, a hold-time of 0 ns, and a clock-to-Q delay of 2 ns. The maximum clock frequency (in MHz, rounded off to the nearest integer) such that there are no setup violations is (answer in MHz).
- An ideal p-n junction germanium diode has a reverse saturation current of 10 \(\mu A\) at 300 K. The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is _________. (Consider the Boltzmann constant \( k_B = 1.38 \times 10^{-23} { J/K} \) and the charge of an electron \( e = 1.6 \times 10^{-19} { C} \).
In the circuit shown below, the AND gate has a propagation delay of 1 ns. The edge-triggered flip-flops have a set-up time of 2 ns, a hold-time of 0 ns, and a clock-to-Q delay of 2 ns. The maximum clock frequency (in MHz, rounded off to the nearest integer) such that there are no setup violations is (answer in MHz).
- An ideal p-n junction germanium diode has a reverse saturation current of 10 \(\mu A\) at 300 K. The voltage (in Volts, rounded off to two decimal places) to be applied across the junction to get a forward bias current of 100 mA at 300 K is _________. (Consider the Boltzmann constant \( k_B = 1.38 \times 10^{-23} { J/K} \) and the charge of an electron \( e = 1.6 \times 10^{-19} { C} \).
- 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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Eight students (P, Q, R, S, T, U, V, and W) are playing musical chairs. The figure indicates their order of position at the start of the game. They play the game by moving forward in a circle in the clockwise direction.
After the 1st round, the 4th student behind P leaves the game.
After the 2nd round, the 5th student behind Q leaves the game.
After the 3rd round, the 3rd student behind V leaves the game.
After the 4th round, the 4th student behind U leaves the game.
Who all are left in the game after the 4th round?
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