Question

The open loop transfer function G(s) of a unity feedback control system is given as,

G(s) = $\frac{K(s+\frac{2}{3})}{s^2(s+2)}$

From the root locus, it can be inferred that when k tends to positive infinity

• A. Three roots with nearly equal real parts exist on the left half of the s-plane.
• B. One real root is found on the right half of the s-plane
• C. The root loci cross the j$\omega$ axis for a finite value of k : k $\ne$ 0.
• D. Three real roots are found on the right half of the s-plane.

Answer 1

The correct option is A Three roots with nearly equal real parts exist on the left half of the s-plane.

G(s) = $\frac{K(s+\frac{2}{3})}{s^2(s+2)}$

and H(s) = 1

Characteristic equation, 1 + G(s) H(s) = 0

$\Rightarrow$ 1+$\frac{K(s+\frac{2}{3})}{s^2(s+2)}=0$

$\Rightarrow$ $s^3+2s^2+K(s+\frac{2}{3})=0$

$\Rightarrow$ $s^3+2s^2+Ks+\frac{2k}{3}=0$

Routh Array:



As k>0, there is no sign change in the 1st column of routharray. So the system is stable and all the three roots lie on LHS of s-plane.

For k>0 (K$\ne$0), none of the row of Routh array becomes zero. So root loci does not cross the j$\omega$-axis.

Number of zero = Z = 1

Number of poles = P = 3

Number of branches terminating at infinity = P - Z = 3 - 1 = 2

Angle of asympototes = $\frac{(2k+1)\times {180}^0}{P-Z}$

= $\frac{(2k+1)\times {180}^0}{2}$

= (2k + 1) x ${90}^0$

= ${90}^0$ and ${270}^0$

Centeroid = $\frac{\sum poles-\sum zero}{P-Z}$

= $\frac{0+0-2-(-\frac{2}{3})}{2}=-\frac{2}{3}$



Since, all the three branches terminates at

$R_e\left(s\right)=-\frac{2}{3}$. So, all the three roots have nearly equal real part.