GATE 2020 EC Q.13: solution and answer

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

1. The Smith chart plots the reflection coefficient $\Gamma =\frac{z-1}{z+1}$ in the complex $\Gamma$-plane, where $z=Z/Z_0$ is the normalised impedance (with a real characteristic impedance $Z_0$). For $Z=jX$ the normalised value $z=jx$, with $x=X/Z_0$, is still purely imaginary, so the normalised resistance is $r=0$.

2. Substitute $z=jx$: $$\Gamma =\frac{jx-1}{jx+1}$$

3. Take the magnitude. The numerator and denominator have equal moduli: $$|\Gamma |=\frac{\sqrt{x^2+1}}{\sqrt{x^2+1}}=1\text{for every real }x$$

4. So every purely reactive load gives $|\Gamma |=1$: total reflection, because a reactance absorbs no power. The locus is the set of points with $|\Gamma |=1$, a circle of radius 1 centred at the origin, which is the outer boundary of the chart.

5. The whole circle is covered. $x=0$ (short circuit) gives $\Gamma =-1$, $x=+1$ gives $\Gamma =\frac{j-1}{j+1}=j$, $x=-1$ gives $\Gamma =-j$, and $x\to \pm \infty$ (open circuit) gives $\Gamma \to +1$. The phase is $\angle \Gamma =\pi -2{\tan }^{-1}x$. As $x$ runs from $-\infty$ to $\infty$, ${\tan }^{-1}x$ runs from $-\pi /2$ to $\pi /2$, so $\angle \Gamma$ sweeps the full range from $2\pi$ down to $0$.

6. Cross-check with the constant-$r$ circles: the circle for resistance $r$ has centre $\left(\frac{r}{1+r},0\right)$ and radius $\frac{1}{1+r}$. Putting $r=0$ gives centre $(0,0)$ and radius $1$, which is option (A).