$\mathrm{Consider\; the\; unity\; negative\; feedback\; control\; system\; with}G\left(s\right)=\frac{K}{s(s+7)(s+11)}.$

$\text{The value of gain}K(K>0)\text{at which the given system will remain marginally stable is ----.}$

$\text{(Answer in integer)}$

$\mathrm{}$

The correct answer is 1386.

Step-by-Step Explanation:

The provided block diagram illustrates a standard closed-loop unity negative feedback control system with input $X\left(s\right)$, a comparator showing positive feedback for the input and negative feedback for the feedback path, and output $C\left(s\right)$. While the block diagram shows a reference forward transfer function of $\frac{7}{s^2+3s+2}$, the problem specifies the open-loop transfer function of the system under consideration as:
$G\left(s\right)=\frac{K}{s(s+7)(s+11)}$
with unity negative feedback, meaning $H\left(s\right)=1$.

To analyze the stability of the closed-loop system, we first determine its characteristic equation:
$1+G\left(s\right)H\left(s\right)=0$
Substituting $G\left(s\right)$ and $H\left(s\right)=1$ into the equation:
$1+\frac{K}{s(s+7)(s+11)}=0$
Multiplying both sides by the denominator:
$s(s+7)(s+11)+K=0$
Expanding the polynomial terms:
$s(s^2+18s+77)+K=0$
$s^3+18s^2+77s+K=0$

Next, we construct the Routh-Hurwitz array to determine the range of $K$ for stability:
- Row $s^3$: 1, 77
- Row $s^2$: 18, K
- Row $s^1$: $\frac{(18\times 77)-K}{18}$
- Row $s^0$: K

For the system to be marginally stable, all elements in the first column of the Routh array must be positive, except that a row must become zero to yield roots on the imaginary axis. Since $K>0$, marginal stability occurs when the row corresponding to $s^1$ becomes zero:
$\frac{18\times 77-K}{18}=0$
Solving for $K$:
$18\times 77-K=0$
$K=18\times 77$
$K=1386$

Thus, the value of gain $K$ at which the system is marginally stable is 1386.