GATE 2020 EC Q.30: solution and answer

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

1. Take the energy-storage variables as the states: the capacitor voltage $v$ and the inductor current $i$. Call the voltage of the left top node (the node shared by $i_1$, the $2\Omega$ resistor and the inductor's left end) $v_A$, measured from the bottom rail.

2. KCL at the left node: $i_1$ enters, while the $2\Omega$ current and the inductor current $i$ leave. $$i_1=\frac{v_A}{2}+i\Rightarrow v_A=2i_1-2i$$

3. KVL around the loop formed by the $2\Omega$ branch, the inductor and the capacitor gives the inductor voltage (positive at its left end, the direction of $i$) as $v_A-v$: $$0.5\frac{di}{dt}=v_A-v=2i_1-2i-v\Rightarrow \frac{di}{dt}=-2v-4i+4i_1$$

4. KCL at the right top node: $i$ and $i_2$ enter; the capacitor current $0.25\frac{dv}{dt}$ and the $1\Omega$ current $v/1$ leave. $$i+i_2=0.25\frac{dv}{dt}+v\Rightarrow \frac{dv}{dt}=-4v+4i+4i_2$$

5. Assemble in the order $[v,i{]}^T$ with inputs $[i_1,i_2{]}^T$: $$\frac{d}{dt}\left[\begin{array}{c}v\\ i\end{array}\right]=\left[\begin{array}{cc}-4& 4\\ -2& -4\end{array}\right]\left[\begin{array}{c}v\\ i\end{array}\right]+\left[\begin{array}{cc}0& 4\\ 4& 0\end{array}\right]\left[\begin{array}{c}{i}_1\\ i_2\end{array}\right]$$

6. Check against the options. $\dot{v}$ has no $i_1$ term ($B_{11}=0$) because $i_1$ reaches the capacitor node only through the inductor. $\dot{i}$ has no $i_2$ term ($B_{22}=0$) because $i_2$ is in parallel with the capacitor, so it changes $v$, not $i$ directly. The $B$ matrix $\left[\begin{array}{cc}0& 4\\ 4& 0\end{array}\right]$ appears in (A) and (B), but only (A) also has $A=\left[\begin{array}{cc}-4& 4\\ -2& -4\end{array}\right]$. The answer is (A).