Question

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)

Hint:

Build the Routh array for \(s^3+18s^2+77s+K=0\) and find the K that makes a row vanish.

Answer 1

Correct Answer: 1386

Step 1: Write the closed loop characteristic equation.
For unity negative feedback with forward path
\[ G(s) = \frac{K}{s(s+7)(s+11)} \] the characteristic equation is
\[ 1+G(s) = 0 \implies s(s+7)(s+11)+K = 0 \]

Step 2: Expand the polynomial.
\[ (s+7)(s+11) = s^2+18s+77 \] \[ s(s^2+18s+77) = s^3+18s^2+77s \] So the characteristic equation is
\[ s^3+18s^2+77s+K = 0 \]

Step 3: Build the Routh array.
The rows for \(s^3\), \(s^2\), \(s^1\), \(s^0\) are
\[ s^3:\;1,\;77 \qquad s^2:\;18,\;K \qquad s^1:\;\frac{18\times77-K}{18},\;0 \qquad s^0:\;K \]

Step 4: Apply the marginal stability condition.
A system is marginally stable when a full row of the Routh array becomes zero, placing a pair of roots exactly on the imaginary axis. This happens when the \(s^1\) row vanishes,
\[ 18\times77-K = 0 \] \[ 1386-K=0 \] \[ K = 1386 \]

Step 5: Verify with the auxiliary equation.
The auxiliary equation comes from the \(s^2\) row,
\[ 18s^2+K=0 \implies 18s^2+1386=0 \implies s^2=-77 \implies s=\pm j\sqrt{77} \] These are purely imaginary roots, which confirms sustained oscillation and marginal stability at this value of \(K\).

Step 6: Final answer.
\[ \boxed{K = 1386} \]