Q21GATE 2003MCQ

The roots of the closed loop characteristic equation of the system shown in figure are

📖 Explanation

1+ GH= 0 (characteristic equation), $\; \; \; 1\, +\, \frac{45}{\left ( s+15 \right )\left ( s+1 \right )}\, =\, 0$ $\; \; \; \; \; s^{2}+16s+60\, =\, 0$ $\therefore \; \; \; \left ( s+10 \right )\left ( s+6 \right )\, =\, 0$ $\therefore$ Roots are $\left ( -6,\, -10 \right ).$

Q22GATE 2003MCQ

The loop gain GH of a closed loop system is given by the following expression $\frac{K}{s(s+2)(s+4)}$ The value of K for which the system just becomes unstable is

🚀 Solve in Practice Mode

📖 Explanation

Characteristic equation = 1 + GH = 0 $\Rightarrow \; \; \; 1+\frac{K}{s\left ( s+2 \right )\left ( s+4 \right )}\, =\, 0$ Characteristic equation, $\Rightarrow \; \; \; s^{2}\, +\, 6s^{2}\, +\, 8s\, +\, K\, =\, 0$ The Routh-array is formed as follows: According to Routh-Hurwitz criterion, for a stable system there should be no charge of sign in the first column of routh array. So, $\; \; \; \frac{48-K}{6} \gt 0$ and $K\gt 0$ $\; \; \; \; \; 0 \lt K \lt 48$ Hence for K=48, the system just becomes unstable.