GATE EC 2024 Set 1 — Question 37

The closed-loop transfer function is given by $T(s) = \frac{G(s)}{1+G(s)}$.
The characteristic equation is $1+G(s) = 0$.
Substituting $G(s) = \frac{K}{(s+1)(s+2)(s+3)}$, we get:
$1 + \frac{K}{(s+1)(s+2)(s+3)} = 0$
$(s+1)(s+2)(s+3) + K = 0$
$s^3 + 6s^2 + 11s + 6 + K = 0$For the impulse response to decay faster than $e^{-t}$, all poles of the closed-loop system must have a real part less than $-1$. This means that if we shift the s-plane by $s' = s+1$ (or $s = s'-1$), all poles in the $s'$-plane must have negative real parts.
Let $s = p-1$. Substituting this into the characteristic equation:
$(p-1)^3 + 6(p-1)^2 + 11(p-1) + 6 + K = 0$
$(p^3 - 3p^2 + 3p - 1) + 6(p^2 - 2p + 1) + 11(p - 1) + 6 + K = 0$
$p^3 - 3p^2 + 3p - 1 + 6p^2 - 12p + 6 + 11p - 11 + 6 + K = 0$
$p^3 + (6-3)p^2 + (3-12+11)p + (-1+6-11+6+K) = 0$
$p^3 + 3p^2 + 2p + K = 0$Now, we apply the Routh-Hurwitz criterion to this polynomial in $p$ to ensure all roots have negative real parts:
Routh Array:
$p^3$: 1 2
$p^2$: 3 K
$p^1$: $\frac{3 \times 2 - 1 \times K}{3} = \frac{6-K}{3}$
$p^0$: KFor stability (all roots of $p$ to have negative real parts), all elements in the first column must be positive:Combining these conditions, we get $0 < K < 6$.Now, let's check the given options:
(A) $1 < K \leq 5$: This range is entirely within $0 < K < 6$.
(B) $7 < K \leq 21$: This range is outside $0 < K < 6$.
(C) $-4 \leq K \leq -1$: This range is outside $0 < K < 6$.
(D) $-24 \leq K \leq -6$: This range is outside $0 < K < 6$.Thus, the appropriate option is (A).The final answer is $\boxed{A}$