Find the frequency and gain K for which the Root locus crosses the imaginary axis. For what range of K is the system stable.

Mumbai University > Electronics Engineering > Sem 4 > Linear Control Systems

Topic: Stability Analysis in Time Domain

Difficulty : High

Marks : 10M

Given => G(s)H(s) = $\frac{k(s+3)}{s(s+1)(s+2)(s+4)}$

step 1=> obtain trial no of losi,

Here, p=poles=4

z= Zeros = 1

No of losi = p-z = 3

step 2=> Draw poles zeros poles are

s=0,-1,-2,-4

zeros are

s=-3

step 3=> Real axis losi=>

present between,

-$\infty$<$\sigma$<-4

-3<$\sigma$<-2

-1<$\sigma$<0</p>

step 4=> calculate number of asympototes and angle of asymptotes

a) Number of asymptotes = p-z = 4-1 = 3

b) Angle of asymptotes = $\beta$x = $\frac{(2x+1)180}{p-z}$ ; x=0,1,2

for x = 0 , $\beta$ = $60^o$

for x = 1 , $\beta$ = $180^o$

for x = 2 , $\beta$ = $300^o$

step 5=> centroid $\sigma_c$ = $\frac{sum of poles - sum of zeros}{p-z}$

$\sigma_c$ = $\frac{-1-2-4-(-3)}{3}$

$\sigma_c$ = -1.33

step 6=> Break away point=>

1 + G(s)H(s) = 0

1 + $\frac{k(s+3)}{s(s+1)(s+2)(s+4)}$ = 0

k = $\frac{-s^4 - 7s^3 - 14s^2 - 8s}{s+3}$

s(s+1)(s+2)(s+4)+k(s+3) = 0

$s^4 + 7s^3 + 14s^2 + 8s + k(s+3)$

$k(s+3) = -s^4 -7s^3 -14s^2 - 8s$

k = $\frac{-s^4 -7s^3 -14s^2 - 8s}{s+3}$

differentiate wrt s,

$\frac{dk}{ds}$ = $\frac{(s+3)(-4s^3-21s^2-28s-8)-(s^4-7s^3-14s^2-8s)}{(s+3)^2}$

but $\frac{dk}{ds}$ = 0

$0 = -3s^4 - 26s^3 -77s^2 -84s -24$

we get,

$S_1$ =-0.4349

$S_2$ = -1.6097

$S_3$ = -3.31 + 0.681 j

of these only,

$S_1$ = -0.4349 is valid

since $S_2$, $S_3$, $S_4$ do not lie on the root locus.

step 7=> Angle of departure or arrival

=> not required, since there is no complex poles or zeros.

Step 8=> Intersection with imaginary axis

=> 1+G(s)H(s) = 0

1 + $\frac{k(s+3)}{s(s+1)(s+2)(s+4)}$ = 0

s(s+1)(s+2)(s+4)+k(s+3) = 0

$s^4 + 7s^3 + 14s^2 + 8s + ks + k3$ = 0

$s^4 + 7s^3 + 14s^2 + (8+k)s + k3$ = 0

using routh array,

$\therefore \frac{(\frac{98-8-k}{7})(8+k)-21k}{\frac{98-(8+k)}{7}} = 0$

(90-k)(8+k) - 147k =0

720 + 90k + 8k - $k^2$ - 147k = 0

$K_1$ = 9.64 ; $K_2$ = -72.64

since, k can not be negative

$K_{max}$ = 9.64

Auxillary equation is,

$(\frac{98-(8+k_{max})}{7})s^2 + 3K_{max}$ = 0

$(\frac{98-9.64}{7})s^2 + (3 * 9.64) = 0$

11.48$s^2$ + 28.9 = 0

$s^2 = -2.517$

$\therefore = \pm 1.58j$

w = 1.58 rad/sec

Hence the root locus cuts the imaginary axis at $\pm$ 1.58 rad/sec