K
2) 1+ =0
s (s +1)(2 s+1)
num=[1];
>> den=[2 3 1 0];
>> HsGs=tf(num, den)
Transfer function:
1
-----------------
2 s^3 + 3 s^2 + s
>> rlocus(HsGs)
>>
TRANSFER FUNCTION INFO
For the open loop transfer function, G(s)H(s):
n=3 poles at s=0, -1, -1/2
m=0
q= n-m
So there exists q= 3 zeroes as s goes to infinity (q = n-m = 3 – 0 = 3)
We can rewrite the open loop transfer function as G(s) H(s)= N(s)/D(s) where
N(s) is the numerator polynomial, and D(s) is the denominator polynomial.
N(s) = 1
D(s) = 2 s 3+ 3 s 2+ s
Characteristic Equation is 1 + KG(s) H(s) = 1 + KN(s)/D(s) = 1 + K¿)]
ROOT LOCUS SYMMETRY
The locus is symmetric about the real axis.
NUMBER OF BRANCHES
The open loop transfer function, G(s) H(s), has 3 poles, therefore the locus has 3
branches. Each branch is display in a different color.
START/END POINTS
Root locus starts (K = 0) at poles of open loop transfer function, G(s) H(s). These
are shown by an “x” on the diagram below.
LOCUS ON REAL AXIS
Root locus exists on real axis between: 0, -0.5 and -1 and negative infinity
because on the real axis, we have 3 poles at s = 0, -1, -1/2 and we have no zeros.
ASYMPTOTES AS │S │GOES TO INFINITY
In the open loop transfer function, G(s) H(s), we have n = 3 finite poles, and m =
o finite zeros, therefore we have q = n – m = 3 zeros at infinity.
180 ° 180 °
Angle of asymptote as s goes to infinity = ± =± =± 6 0°
q 3
There exist three poles at s=-1/2, -1, 0; so, the sum of poles= -3/2
There exist zero zeroes; so, the sum of zeroes = 0
Intersect of asymptote is at ((sum of poles)-(sum of zeros))/q = -1/2
BREAK-OUT AND IN POINTS ON REAL AXIS
Break Out (or Break In) points occur where N(s) D’(s) - N’(s) D(s) = 0. This
polynomial has 2 root at s = -0.2113 and -0.7887
From these 2 root, there exists 2 real root at s = -0.2113 and -0.7887
These roots are all on the locus (k > 0).
N(s) and D(s) are numerator and denominator polynomials of G(s) H(s),
and the tick mark, ‘ ,denotes differentiation.
N(s) = 1
N’(s) = 0
D(s) = 2 s 3+ 3 s 2+ s
D’(s) = 6 s2 +6 s+1
N(s) D’(s) = (1) (6 s2 +6 s) = 6 s2 +6 s+1
N’(s) D(s) = ¿) (0) = 0
N(s) D’(s) - N’(s) D(s) = (6 s¿¿ 2+6 s+1)−( 0 )=6 s 2 +6 s +1 ¿
ANGLE OF DEPARTURE
No complex poles in loop gain, so no angles of departure.
ANGLE OF ARRIVAL
No complex zeros in loop gain, so no angles of arrival.
CROSS IMAGINARY AXIS
Locus crosses imaginary axis at value of K = 8.
Locus Crossing Axis
CHANGING K CHANGES CLOSED LOOP POLES
Characteristic Equation is 1 + KG(s)H(s) = 0, or 1 + KN(s)/D(s) =0,
or D(s) + K N(s) = 2 s 3+ 3 s 2+ s+ ¿ K(1) = 0
So, by choosing K we determine the characteristic equation whose roots
are the closed loop poles. For example, with K = -0.096 and 0.096, then
the characteristic equation is
6 s2 +6 s+1 = 0
s = -0.2113 and -0.7887
K = -1/G(s)
K =- 1/2 s 3+ 3 s 2+ s
K=-1¿ 2(−0.2113 )3 +3 (−0.2113)2 +(−0.2113)
K = -0.096
K=-1¿ 2(−0.7887)3 +3(−0.7887)2+(−0.7887)
K= 0.096
D(s) + K N(s) = 2 s 3+ 3 s 2+ s-0.096(1) = 0
=2 s 3+ 3 s 2+ s−0.096
This equation has 3 roots at s = 0.077, -0.7772 and -0.8
D(s) + K N(s) = 2 s 3+ 3 s 2 s+¿ 0.096(1) = 0
This equation has 3 roots at s = -1.077, -0.2 and -0.223
Root at K =±0.096
CHOOSE POLE LOCATION AND FIND K
Characteristic Equation is 1 + KG(s)H(s) = 0, or 1 + KN(s)/D(s) =0, or
K = - D(s)/ N(s) = - ¿)/1
We can pick a value of s on the locus, and find K = - D(s)/ N(s).
We chooses s = 0.172-1.05j then
K = 2(0.172−1.05 j)3 +3(0.172−1.05 j)2 + 1
K = -4.174-4.729×10−3 j
Design for poles at s = -0.2113 and -0.7887
