Consider the unity negative feedback control system shown in the figure below, where the forward path transfer function is \(G(s)=\dfrac{K}{s(s+7)(s+11)}\). The value of gain \(K\;(>0)\) at which the given system will remain marginally stable is . (Answer in integer)
Show Hint
Build the Routh array for \(s^3+18s^2+77s+K=0\) and find the K that makes a row vanish.
Step 1: Understanding the Concept:
At the point of marginal stability, the closed loop poles sit exactly on the imaginary axis, so the system oscillates forever without growing or decaying. We can find this condition by plugging $s=j\omega$ into the characteristic equation and forcing both the real and imaginary parts to vanish together.
Step 2: Key Formula or Approach:
The characteristic equation is $s(s+7)(s+11)+K=0$, which expands to $s^3+18s^2+77s+K=0$. Substitute $s=j\omega$ and split into real and imaginary parts.
Step 3: Detailed Explanation:
With $s=j\omega$: $s^3=-j\omega^3$, $s^2=-\omega^2$, so the equation becomes
\[ -j\omega^3-18\omega^2+77j\omega+K=0 \]
Group the real and imaginary parts:
Real part: $K-18\omega^2=0$
Imaginary part: $77\omega-\omega^3=0$
From the imaginary part, $\omega(77-\omega^2)=0$. Marginal stability needs a genuine oscillation, so take $\omega\neq0$, giving $\omega^2=77$.
Substitute into the real part: $K=18\omega^2=18\times77=1386$.
Step 4: Final Answer:
The gain at which the system is marginally stable is $K=1386$, matching the oscillation frequency $\omega=\sqrt{77}$ rad/s.
\[ \boxed{K=1386} \]
