The loop transfer function of a negative feedback system isGs Hs - Ks-11ss-2s-8The value of K - for which the system is marginally stable is

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The loop transfer function of a negative feedback system is
$G (s) H (s) = \frac{K (s + 11)}{s (s + 2) (s + 8)}$
The value of $K$ , for which the system is marginally stable is

160

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Solution

The correct option is A 160
Characteristic equation $q (s)$ for the given open loop system will be

$q (s) = s^{3} + 10 s^{2} + 16 s + K s + 11 K = 0$

Using R-H criteria,

$\begin{matrix} s^{3} & 1 & 16 + K s^{2} & 10 & 11 K s^{1} & \frac{10 (16 + K) - 11 K}{10} & 0 s^{0} & 11 k & 0 \end{matrix}$

For system to be marginally stable

$\frac{10 (16 + K) - 11 K}{10} = 0$

$160 + 10 K - 11 K = 0$

$K = 160$

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The loop transfer function of a negative feedback system is G(s)H(s)=K(s+11)s(s+2)(s+8) The value of K , for which the system is marginally stable is 160

The correct option is A 160Characteristic equation q(s) for the given open loop system will be q(s)=s3+10s2+16s+Ks+11K=0 Using R-H criteria, s3116+Ks21011Ks110(16+K)−11K100s011k0 For system to be marginally stable 10(16+K)−11K10=0 160+10K−11K=0 K=160

The loop transfer function of a negative feedback system is
$G (s) H (s) = \frac{K (s + 11)}{s (s + 2) (s + 8)}$
The value of $K$ , for which the system is marginally stable is

160