GATE EC 2023 Set 1 — Question 52

Given:
Characteristic impedance $Z_0 = 50\ \Omega$
Voltage Standing Wave Ratio (VSWR) $S = 2.0$
Distance between successive minima $\frac{\lambda}{2} = 30\ \text{cm} \Rightarrow \lambda = 60\ \text{cm}$.
Distance of the first minimum from the load $d_{min1} = 10\ \text{cm}$.At a voltage minimum, the impedance is purely resistive and equal to:
$R_{min} = \frac{Z_0}{S} = \frac{50}{2} = 25\ \Omega$At a voltage maximum, the impedance is purely resistive and equal to:
$R_{max} = Z_0 \cdot S = 50 \cdot 2 = 100\ \Omega$We want to find an equivalent line of length $l_m$ terminated in $R_m$ such that its input impedance is $Z_L$.Case 1: $R_m = 25\ \Omega$ (Voltage Minimum)
If the termination $R_m$ is a voltage minimum, then $l_m$ is the distance from a minimum to the load position moving towards the generator.
In the original line, the minimum is at $d = 10\ \text{cm}$ from the load. To reach the load position from this minimum moving towards the generator, we must travel a distance:
$l_m = \frac{\lambda}{2} - d_{min1} = 30 - 10 = 20\ \text{cm}$.
Thus, $R_m = 25\ \Omega, l_m = 20\ \text{cm}$ is a valid equivalent. (Option B is correct)Case 2: $R_m = 100\ \Omega$ (Voltage Maximum)
If the termination $R_m$ is a voltage maximum, then $l_m$ is the distance from a maximum to the load position moving towards the generator.
Voltage maxima are located at $d_{max} = d_{min} \pm \frac{\lambda}{4} = 10 \pm 15\ \text{cm}$.
This gives $d_{max1} = 25\ \text{cm}$ and $d_{max2} = -5\ \text{cm}$.
A maximum at $d = -5\ \text{cm}$ means it is $5\ \text{cm}$ "behind" the load. Moving from this maximum towards the generator by $5\ \text{cm}$ reaches the load position ($d=0$).
Thus, $l_m = 5\ \text{cm}$ for $R_m = 100\ \Omega$.
So, $R_m = 100\ \Omega, l_m = 5\ \text{cm}$ is also a valid equivalent. (Option C is correct)