GATE 2018 EC Q.10: solution and answer

Official answer: CSource: Official answer key, GATE 2018 EC (IIT Guwahati)

1. On a normalized impedance Smith chart, each point is a value of the reflection coefficient $\Gamma =\frac{z-1}{z+1}$, where $z=Z_L/Z_0=r+jx$. The chart centre is $\Gamma =0$, and the horizontal diameter is the real axis of $\Gamma$ (the line $x=0$), running from $\Gamma =-1$ on the left to $\Gamma =+1$ on the right.

2. Point $P$ (leftmost end of the real axis): $\Gamma =-1$. Solving $\frac{z-1}{z+1}=-1$ gives $z-1=-z-1$, so $z=0$ and $Z_L=0$: a short circuit. You can also read it off the chart: $P$ lies on the outer circle, which is the $r=0$ circle, and on the real axis, where $x=0$, so $z=0+j0$.

3. Point $Q$ (centre of the chart): $\Gamma =0$. Then $z-1=0$, so $z=1+j0$ and $Z_L=Z_0$: a matched load. The $r=1$ circle drawn in the figure passes through this point, and it lies on the real axis ($x=0$).

4. Point $R$ (rightmost end of the real axis): $\Gamma =+1$. The equation $\frac{z-1}{z+1}=1$ would need $z-1=z+1$, which no finite $z$ satisfies; $\Gamma \to 1$ only as $z\to \infty$, so $Z_L\to \infty$: an open circuit. On the chart, every constant-$r$ circle and every constant-$x$ arc (including the $r=1$ circle and the $x=\pm 1$ arcs drawn here) passes through this point, which is the limit $r\to \infty$.

5. So $P$ is a short circuit, $Q$ is a matched load and $R$ is an open circuit, which is option (C).