written 7.7 years ago by |
G(S)H(S)=$\frac{K(S+4)(S+5)}{(S+3)(S+1)}$
Here, no. of poles P=2
No. of zeros Z=2
No. of branches ending at ∞,
N=P-Z=2-2
=0
Angle of Asymptoe
Θ=$\frac{(2q+1)180}{(P-Z)}$, q=0, 1,…….,(P-Z-1)
As P-Z=0, Angle of asymptote is not applicable
Centroid -> NA
The characteristic equaton is,
1+G(S)H(S)=0
$\frac{1+k(S^2+9S+20)}{(S^2+4S+3)}$=0
$\frac{k(S^2+9S+20)}{(S^2+4S+3)}$=-1
k=$\frac{-(S^2+4S+3)}{(S^2+9S+20)}$
$\frac{dk}{ds}$=0
-$\frac{(S^2+4S+3)(2S+9)-(S^2+9S+20)(2S+4)}{(S^2+9S+20)^2]}$=0
$(S^2+4S+3)(2S+9)-(S^2+9S+20)(2S+4)$=0
$2S^3+8S^2+6S+9S^2+36S+27-(2S^3+4S^2+18S^2+36S+40S+80)$=0
$(17S^2-22S^2)+42S-76S+27-80$=0
$-5S^2-34S-53$=0
$5S^2+34S+53$=0
S=$\frac{-3±\sqrt{(34^2-4 x S x 53)}}{(2 x S)}$
S=-4.79, -2.241
As both the values of ‘S’ lies on root locus. Both are valid break away points.
Point of intersection with imaginary axis,
$S^2$ (1+k) (20k+3)
$S^1$ (4+9k) 0
$S^0$ (20k+3) 0
(1+k) ≥ 0 k_max=-1
(4+9k) ≥ 0 k_max=(-4)/9
(20k+3) ≥ 0 k_max=(-3)/20
There is no point of intersection with imaginary axis.