GATE 2018 EC Q.29: solution and answer

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

1. With no direct feed-through term ($D=0$), the transfer function is $G(s)=C(sI-A{)}^{-1}B=\frac{C\text{adj}(sI-A)B}{\det (sI-A)}$.

2. $sI-A=\left[\begin{array}{cc}s+4& 1.5\\ -4& s\end{array}\right]$, so $\det (sI-A)=s(s+4)-(1.5)(-4)=s^2+4s+6$. This is the characteristic polynomial. All four options have this denominator, so the numerator decides the answer.

3. To get the adjugate of a $2\times 2$ matrix, swap the diagonal entries and change the sign of the off-diagonal ones: $\text{adj}(sI-A)=\left[\begin{array}{cc}s& -1.5\\ 4& s+4\end{array}\right]$.

4. $\text{adj}(sI-A)B=\left[\begin{array}{cc}s& -1.5\\ 4& s+4\end{array}\right]\left[\begin{array}{c}2\\ 0\end{array}\right]=\left[\begin{array}{c}2s\\ 8\end{array}\right]$.

5. Numerator: $C\left[\begin{array}{c}2s\\ 8\end{array}\right]=1.5(2s)+0.625(8)=3s+5$.

6. $$G(s)=\frac{Y(s)}{U(s)}=\frac{3s+5}{s^2+4s+6}$$

7. Two quick checks. (i) The leading numerator coefficient of a strictly proper second-order $G(s)$ is $CB=1.5\times 2=3$. That rules out (C), which has 4, and (D), which has 6. (ii) DC gain: $G(0)=-C{A}^{-1}B$. Here $\det A=6$ and $A^{-1}=\frac{1}{6}\left[\begin{array}{cc}0& 1.5\\ -4& -4\end{array}\right]$, so $A^{-1}B=\left[\begin{array}{c}0\\ -\frac{4}{3}\end{array}\right]$ and $-C{A}^{-1}B=0.625\times \frac{4}{3}=\frac{5}{6}$. That matches $G(0)=5/6$ for (A) and rules out (B), whose $G(0)$ is negative. Only (A) passes both checks.