GATE EC 2016 Set 2 — Question 64

The input reflection coefficient magnitude $|\Gamma|$ is measured at the input of $T_1$. Since $T_1$ is a lossless transmission line with characteristic impedance $Z_0 = 50 \Omega$, the magnitude of the reflection coefficient at its input is equal to the magnitude of the reflection coefficient at the junction where $T_1$ meets the $50 \Omega$ shunt resistor and $T_2$. The junction consists of a $50 \Omega$ resistor in parallel with the input impedance of $T_2$. $T_2$ is an open-circuited lossless transmission line of length $L$ and $Z_0 = 50 \Omega$. Its input impedance is:
$$Z_{in2} = -j Z_0 \cot(\beta L) = -j 50 \cot(\beta L)$$ The total impedance at the junction $Z_J$ is:
$$Z_J = 50 \parallel Z_{in2} = \frac{50 \cdot (-j 50 \cot(\beta L))}{50 - j 50 \cot(\beta L)} = \frac{-j 50 \cot(\beta L)}{1 - j \cot(\beta L)}$$ The reflection coefficient $\Gamma_J$ relative to the $50 \Omega$ characteristic impedance of $T_1$ is:
$$\Gamma_J = \frac{Z_J - 50}{Z_J + 50} = \frac{\frac{-j 50 \cot(\beta L)}{1 - j \cot(\beta L)} - 50}{\frac{-j 50 \cot(\beta L)}{1 - j \cot(\beta L)} + 50} = \frac{-j \cot(\beta L) - (1 - j \cot(\beta L))}{-j \cot(\beta L) + (1 - j \cot(\beta L))} = \frac{-1}{1 - 2j \cot(\beta L)}$$ The magnitude is:
$$|\Gamma| = |\Gamma_J| = \frac{1}{\sqrt{1 + 4 \cot^2(\beta L)}}$$ From the given plot, $|\Gamma|$ reaches its maximum value of 1 at frequencies $f = 0.5, 1.5, 2.5, \dots$ GHz.
$|\Gamma| = 1$ when $\cot(\beta L) = 0$, which implies $\beta L = (2n+1)\frac{\pi}{2}$.
For the first peak at $f = 0.5$ GHz:
$$\beta L = \frac{2\pi f}{v_p} L = \frac{\pi}{2}$$ Given $v_p = 2 \times 10^8$ m/s and $f = 0.5 \times 10^9$ Hz:
$$\frac{2\pi (0.5 \times 10^9)}{2 \times 10^8} L = \frac{\pi}{2}$$ $$5\pi L = \frac{\pi}{2} \implies 5L = 0.5 \implies L = 0.1 \text{ m}$$Alternatively, the period of the plot is $\Delta f = 1$ GHz. The period of $|\Gamma|$ corresponds to a change in $\beta L$ of $\pi$:$$\frac{2\pi \Delta f}{v_p} L = \pi \implies \frac{2 \cdot 10^9}{2 \times 10^8} L = 1 \implies 10L = 1 \implies L = 0.1 \text{ m}$$