Question

The break away point of the root locus on the real axis for a closed-loop system with a loop gain
$G\left(s\right)H\left(s\right)=\frac{K(s+10)}{(s+2)(s+5)}lies$

• A. between - 2 and origin
• B. between -2 and - 5
• C. between - 10 and $-\infty$
• D. at $-\infty$

Answer 1

The correct option is B between -2 and - 5

The singularities are at,

$s=-2.-5.-10$

diagram
so, the break away point can lie only in the range between $(-5,-2)(-\infty ,-10)$
$1+\frac{K(s+10)}{(s+2)(s+5)}=0$
$\therefore k=\frac{-(s+2)(s+5)}{(s+10)}$
$\frac{dK}{ds}=\left\{\frac{(s+10)[s+2+s+5]-(s+2)(sS+5)}{(s+10{)}^2}\right\}=0$
$\therefore (s+10)(2s+7)=(s+2)(s+5)$
$2s^2+7s+20s+70=s^2+7s+10$
$\therefore s^2+20s+60=0$
$s=\frac{-20\pm \sqrt{400-240}}{2}$
$=\frac{-20\pm 12.64}{2}=-3.68,-16.32$

$\therefore$ the breakaway point is - 3.68 as the 16.32 do not lie in the set interval.
$\therefore$ the breakaway point lies in the interval -2 and -5.