GATE 2024 EC Q.58: solution and answer

Official answer: 200Source: Official answer key, GATE 2024 EC (IISc Bengaluru)

1. On a lossless line the reflection coefficient at distance $d$ from the load is $\Gamma (d)={\Gamma }_L{e}^{-j2\beta d}$. Its magnitude stays at $|\Gamma |=0.6$. Only its phase changes, falling by $2\beta d$, so over any stretch of $\lambda /2$ the phase passes through every angle from $0$ to $2\pi$.

2. The input impedance at distance $d$ is $$Z_{in}(d)=Z_0\frac{1+\Gamma (d)}{1-\Gamma (d)}\Rightarrow |Z_{in}(d)|=Z_0\frac{|1+\Gamma (d)|}{|1-\Gamma (d)|}$$

3. With $|\Gamma (d)|=0.6$ fixed, the triangle inequality gives $|1+\Gamma (d)|\le 1+|\Gamma |$ and $|1-\Gamma (d)|\ge 1-|\Gamma |$. Both bounds are reached at the same point, where $\angle \Gamma (d)=0$. Step 1 guarantees the line has such a point: it is a voltage maximum, and $Z_{in}$ is purely real there.

4. So the largest impedance magnitude is $Z_0$ times the VSWR: $$|Z_{in}{|}_{\max }=Z_0\frac{1+|\Gamma |}{1-|\Gamma |}=Z_0\cdot \text{VSWR}$$

5. Numerically, $\text{VSWR}=\frac{1+0.6}{1-0.6}=\frac{1.6}{0.4}=4$, so $|Z_{in}{|}_{\max }=50\times 4=200\Omega$.

6. Check: at a voltage minimum ($\angle \Gamma (d)=\pi$) the value is $Z_0/\text{VSWR}=50/4=12.5\Omega$. Along the line, $|Z_{in}|$ moves between $12.5\Omega$ and $200\Omega$ and repeats every $\lambda /2$. The phase of the unknown load only decides where these points fall, not how large they are.