Root Locus Analysis
Paula Raica
Department of Automation
Dorobantilor Str., room C21, tel: 0264 - 401267
Baritiu Str., room C14, tel: 0264 - 202368
email: [email protected]
http://rrg.utcluj.ro/st
Technical University of Cluj-Napoca
Root Locus Analysis
Introduction
The transient response of a closed-loop system is closely
related to the location of the closed-loop poles.
Variable loop gain ⇒ the location of the poles depends on the
value of the loop gain chosen.
Discuss the design problems that involve the selection of a
particular parameter value such that the transient response
characteristics are satisfactory.
If the gain adjustment alone does not yield a desired result,
addition of a compensator to the system will become
necessary.
W.R. Evans: the root-locus method, the roots of the
characteristic equation are plotted for all values of a system
parameter.
Root Locus Analysis
Root-locus method
The values of s that make transfer function around the loop equal
−1 must satisfy the characteristic equation of the system.
R(s)
C(s)
G(s)
k
H(s)
open loop: Hd (s) = kG (s)H(s), closed loop
kG (s)
C (s)
=
R(s)
1 + kG (s)H(s)
Characteristic eq: 1 + kG (s)H(s) = 0 or kG (s)H(s) = −1
Root Locus Analysis
Root-locus
kG (s)H(s) is a ratio of polynomials in s, complex quantity.
| kG (s)H(s) | ∠kG (s)H(s) = −1 + j0
Angle condition: ∠kG (s)H(s) = ±180o (2k + 1), k = 0, 1, 2, ...
Magnitude condition: |kG (s)H(s)| = 1
The values of s that fulfill the angle and magnitude conditions
are the closed-loop system poles.
A plot of points of the complex plane satisfying the angle
condition alone is the root-locus.
The closed-loop poles corresponding to a given value of the
gain can be determined from the magnitude condition.
Root Locus Analysis
Example
kG (s)H(s) =
k(s + z1 )
(s + p1 )(s + p2 )(s + p3 )(s + p4 )
∠kG (s)H(s) = ϕ1 − θ1 − θ2 − θ3 − θ4 ,
Test point
s
A4
-p4
jw
A2
q2
-p2
B1
q4
|G (s)H(s)| =
A1
j1
-z1
q1
-p1
A3
q3
-p3
Root Locus Analysis
s
kB1
A1 A2 A3 A4
Root locus - second order system
G (s)H(s) =
k
;
s(s + 2)
C (s)
k
= 2
R(s)
s + 2s + k
We wish to find the locus of the poles of this system as k is varied
from zero to infinity.
Root Locus Analysis
Root locus - second order system
s1 = −1 +
s2 = −1 −
√
√
1−k
1−k
k ∈ [0, 1]: overdamped (real
poles)
k > 1: underdamped (complex
poles)
Root Locus Analysis
Root locus - second order system
The angle condition:
∠
k
s(s + 2)
= −∠s − ∠(s + 2)
= ±180o (2k + 1),
k = 0, 1, 2, ..
Point P: The complex quantities
s and s + 1 have angles θ1 and
θ2 respectively, and magnitudes
|s1 | and |s + 2|.
θ1 + θ2 = 180o
Root Locus Analysis
Root locus - second order system
k is determined from the magnitude condition for a pair of
closed-loop poles: s1,2 = −1 ±
j1:
k
1 = |G (s)H(s)| = s(s + 2) s1,2
or
k = |s(s + 1)|s=−1+j1 = 2
Root Locus Analysis
Root locus procedure
Write the characteristic equation in the form:
1 + kP(s) = 0.
Factor P(s) in terms of np poles and nz zeros
Qnz
(s + zi )
1 + k Qni =1
=0
p
i =1 (s + pj )
Locate the open-loop poles and zeros in the s-plane: x - poles,
o - zeros.
Determine the number of separate loci SL. SL = np , when
np ≥ nz , np = number of finite poles, nz = number of finite
zeros.
Root Locus Analysis
Root locus procedure
Determine the number of separate loci SL. SL = np , when
np ≥ nz , np = number of poles, nz = number of zeros.
Locate the segments of the real axis that are root loci:
1
2
Locus begins at a pole and ends at a zero (or infinity)
Locus lies to the left of an odd number of poles and zeros
The root loci are symmetrical with respect to the real axis.
The loci proceed to the zeros at infinity along asymptotes
centered at σA and with angles ΦA .
P
P
(pj ) − (zi )
2q + 1
σA =
, ΦA =
· 180o , q=0,1,...(np −nz −1)
np − nz
np − nz
Root Locus Analysis
Root locus procedure
From Routh-Hurwitz criterion ⇒ the intersection with the
imaginary axis (if it does so).
Determine the breakaway point on the real axis (if any)
1
2
3
1
Set k = − P(s)
= p(s), (from 1 + kP(s) = 0)
Obtain dp(s)/ds = 0
Determine roots of (b) or use graphical method to find
maximum of p(s).
Determine the angle of locus departure from complex poles
and the angle of locus arrival at complex zeros, using the
phase criterion
∠P(s) = ±180o (2k + 1), at s = pj or zi .
Root Locus Analysis
Root locus procedure
Determine the root locations that satisfy the phase criterion
∠P(s) = ±180o (2k + 1) at a root location sx
Determine the parameter value kx at a specific root sx
Qnp
j=1 | s + pj |
kx = Qnz
|s=sx
i =1 | s + zi |
Root Locus Analysis
Root locus example
Let us sketch the root-locus plot and determine the value of k so
that the damping factor ζ of a pair of dominant complex-conjugate
closed-loop poles is 0.5 for a closed-loop system with the
open-loop transfer function:
G (s)H(s) =
k
s(s + 1)(s + 2)
The characteristic equation is written as:
1 + kG (s) = 0, or 1 +
k
=0
s(s + 1)(s + 2)
The open-loop transfer function has no zeros: nz = 0, and three
poles, p1 = 0, p2 = −1, p3 = −2: np = 3.
Root Locus Analysis
Root locus example
Locate the open-loop poles of the open-loop transfer function with
symbol x.
Locus has np = 3 branches.
Locate the segments of the real axis that are root loci: on the
negative real axis, the locus exists between p1 = 0 and p2 = −1,
and from p3 = −2 to −∞.
As the locus begins at pole and ends at a zero (or infinity),
between 0 and -1 we must find a breakaway point.
The root locus is symmetrical with respect to the horizontal real
axis.
Root Locus Analysis
Root locus example
jw
k=6
1.4j
s
60o
-2
-1
0
s=-0.42
-1.4j
Root Locus Analysis
Root locus example
The locus proceed to the zeros at infinity along asymptotes
centered at σA and with angles ΦA .
P
P
(pj ) − (zi )
0−1−2
σA =
= −1
=
np − nz
3
ΦA =
or
2q + 1
· 180o , q = 0, 1, 2
np − nz
ΦA = 60o , 180o , 300o
Root Locus Analysis
Root locus example
Routh-Hurwitz criterion:
s 3 + 3s 2 + 2s + k = 0
Routh array:
s3
s2
s1
s0
:
1
2
:
3
k
: (6 − k)/3
:
k
⇒ k = 6:
s 3 + 3s 2 + 2s + 6 = 0 or (s + 3)(s 2 + 2) = 0
√
s1,2 = ±j 2
Root Locus Analysis
Root locus example
The breakaway point:
p ′ (s) = 0 where p(s) = k = −1/G (s) = −s(s + 1)(s + 2)
dp(s)
= −(3s 2 + 6s + 2) = 0 ⇒ s1 = −0.4226, s2 = −1.5774
ds
The breakaway point must lie between 0 and -1, ⇒ s1 = −0.4226
Root Locus Analysis
Root locus example
jw
k=6
1.4j
s
60o
-2
-1
0
s=-0.42
-1.4j
Root Locus Analysis
Root locus example
Remember. The complex poles of a second-order transfer function:
jw
wd = wn 1 - z 2
p1,2 = −ζωn ± jωn
wn
s
q
-s=zwn
cosθ =
0
p
ζωn
=ζ
ωn
θ = arccosζ
Root Locus Analysis
1 − ζ2
Root locus example
Closed-loop poles with ζ = 0.5 lie on the lines passing through the
origin and making the angles ±arccosζ = ±arccos0.5 = ±60o with
the negative real axis.
jw
s1,2 = −0.3337 ± j0.5780
k=1.03
k=1.03
60o
s
-2
-1
0
The value of k that yields such
poles is found from the magnitude condition:
k = |s(s+1)(s+2)|s1,2 = 1.0383
The third pole is found at s =
−2.3326.
Root Locus Analysis
Root locus
Constructing the root locus when a variable parameter does
not appear as a multiplying factor.
R(s)
G(s)
C(s)
H(s)
G (s) =
20
,
s(s + 1)(s + 4)
H(s) = 1 + ks
The open-loop transfer function is then:
G (s)H(s) =
20(1 + ks)
s(s + 1)(s + 4)
Root Locus Analysis
Root locus
The characteristic equation:
1+G (s)H(s) = 1+
20(1 + ks)
= 0, s 3 +5s 2 +4s+20+20ks = 0
s(s + 1)(s + 4)
By defining 20k = K and dividing by the sum of terms that do not
contain k, we get:
1+
s3
+
Ks
=0
+ 4s + 20
5s 2
Sketch the root locus of the system given from the new
characteristic equation.
Root Locus Analysis
Root locus
jw
j2
s
-5
-2.5
0
-j2
Root Locus Analysis