GATE 2018 EE Q.32: solution and answer

Official answer: ASource: Official answer key, GATE 2018 EE (IIT Guwahati)

1. Reduce one shunt branch. The inductor and the capacitor are in series, so $Z_2=j5+(-j1)=j4\Omega$. The series element in each section is $Z_1=j9\Omega$.

2. Use the self-similarity of an infinite ladder. Removing the first section leaves the same infinite ladder, so the impedance seen to the right of the first series arm is $Z_{eq}$ again. It sits in parallel with the first shunt branch. Hence $$Z_{eq}=Z_1+Z_2\parallel Z_{eq}=j9+\frac{j4Z_{eq}}{j4+Z_{eq}}$$

3. Form the quadratic. Multiply both sides by $(j4+Z_{eq})$: $$Z_{eq}(j4+Z_{eq})=j9(j4+Z_{eq})+j4Z_{eq}$$ Using $j9\cdot j4=-36$: $$\begin{array}{rl}{Z}_{eq}^2+j4Z_{eq}& =-36+j9Z_{eq}+j4Z_{eq}\\ Z_{eq}^2-j9Z_{eq}+36& =0\end{array}$$

4. Solve it. $$Z_{eq}=\frac{j9\pm \sqrt{(j9{)}^2-4(36)}}{2}=\frac{j9\pm \sqrt{-225}}{2}=\frac{j9\pm j15}{2}$$ This gives $Z_{eq}=j12\Omega$ or $Z_{eq}=-j3\Omega$.

5. Pick the physical root. At this frequency every element of a section is inductive: $j9$ in series and a net $j4$ in shunt. Two positive reactances combine to a positive reactance whether they are in series ($j{X}_1+j{X}_2$) or in parallel ($j\frac{X_1{X}_2}{X_1+X_2}$). So a ladder cut off after any number of sections is inductive, and so is its limit. That rules out $-j3\Omega$.

6. Check by iteration. Put $Z_{eq}=jX$, so $X=9+\frac{4X}{4+X}$. Start from the one-section value $X=9+4=13$ and repeat: $13\to 12.06\to 12.004\to \cdots \to 12$. This confirms $Z_{eq}=j12\Omega$. The iteration settles on $12$ and not on $-3$ because $f(X)=9+\frac{4X}{4+X}$ has slope $f^{\prime }(X)=\frac{16}{(4+X{)}^2}$. That slope is $\frac{1}{16}$ at $X=12$, so each step shrinks the error and the iteration converges there. At $X=-3$ the slope is $16$, so the error grows and $-3$ is never reached. Answer: (A) $j12\Omega$.