Lecture Notes on Control Systems/D. Ghose/2012
13.5
127
Steps to obtain the Root Locus
Note: Different books give these set of steps differently. We will follow something that
is simple and works in most cases. You can take a look at D’Azzo and Houpis for a
complete set of steps.
Step 1: Put the closed-loop system in Evan’s form.
Gc (s) =
G1 (s)
1 + kG2 (s)
where, k is the parameter to be varied. We assume that k ∈ [0, ∞).
Step 2: Find the starting (k = 0) and the ending (k = ∞) points.
(a) Locate the open loop poles and zeros. Let there be n open loop poles and m
open loop zeros.
(b) Identify the number of loci. (n poles ⇒ n loci).
(c) Now the starting and ending points of the loci:
(i) As k → 0, the loci approach the open loop poles.
Proof: 1 + kG(s) = 0 ⇒ |kG(s)| = 1, for all s on the root locus.
⇒ G(s) → ∞ as k → 0.
⇒ If s0 is on the root locus as k → 0 then s0 must be a pole of G(s).
(ii) As k → ∞, m of the loci approach the open loop zeros, and the rest go
to ∞.
Proof: 1 + kG(s) = 0 ⇒ |kG(s)| = 1, for all s on the root locus.
⇒ G(s) → 0 as k → ∞.
m +b
m−1 +···+b s+b
(s)
m−1 s
1
0
Now, G(s) = N
= ssn +a
n−1 +···+a s+a
D(s)
n−1 s
1
0
and, m ≤ n.
So, for G(s) → 0
Either N (s) = 0 ⇒ s is a zero of G(s).
Or, D(s) → ∞ ⇒ s → ±∞.
Step 3: Find the real axis segments. That is, on the real axis, draw the root locus to the
left of the odd number of open loop poles and zeros.
Step 4: Find the asymptotes.
The n − m poles that do not approach the zeros go to ∞ along asymptotes.
(a) Find the asymptote angles.
φα =
π + 2απ
180o + α · 360o
=
, α = 0, 1, . . . , n − m − 1
n−m
n−m
Proof. As s → ∞,
Lecture Notes on Control Systems/D. Ghose/2012
kG(s) = k ·
128
k
sm
sm + · · · + b1 s + b0 ∼
= n−m
k
·
=
n
n
s + · · · + a1 s + a0
s
s
Now, 1 + kG(s) = 0 for all s on the root locus, so,
kG(s) ∼
=
k
sn−m
= −1
1
⇒ s = (−k) n−m
For k > 0, we have
−k = kej(π+α2π) , α = 0, 1, 2, . . .
So,
s = kej(π+α2π)
1
n−m
1
= k n−m ej (
π+α2π
n−m
) , α = 0, 1, 2, . . . , n − m − 1
Why does α go only up to n − m − 1? It is because the angles repeat after
this point.
So, as k → ∞, n − m poles go to ∞ along asymptotes with angles,
φα =
π + 2απ
180o + α360o
=
, α = 0, 1, . . . , n − m − 1
n−m
n−m
An Example:
Let
kG(s) = k
(s2
s+a
+ 2ζωn s + ωn2 )(s + b)2
Here, n = 4, m = 1.
As k → ∞, n − m = 3 of the closed loop poles will go to ∞. Only one of
them will approach the zero s = −a.
The asymptotes are given by,
180o + α360o
, α = 0, 1, 2
n−m
1800 1800 + 360o 1800 + 720o
=
,
,
3
3
3
= 60o , 180o , 300o
φα =
Lecture Notes on Control Systems/D. Ghose/2012
129
Figure 13.16: Open loop poles and zeros and the asymptotic centroid
Figure 13.17: Possible root loci
(b) Find the asymptotic centroid.
σ=
n
i=1
pi − m
i=1 zi
n−m
where, pi and zi are the open loop poles and zeros.
Proof.
sm + bm−1 sm−1 + · · · + b1 s + b0
sn + an−1 sn−1 + · · · + a1 s + a0
(s − z1 )(s − z2 ) · · · (s − zm )
= k·
(s − p1 )(s − p2 ) · · · (s − pn )
kG(s) = k ·
Note that,
bm−1 = −
m
k=1
zk
Lecture Notes on Control Systems/D. Ghose/2012
130
m
an−1 = −
pj
j=1
Try proving the above assertion yourself.
Now, let us do some long division (back to school!).
Insert
figure
6-eqn
here
Inse
se
er
rt
t th
the
e fig
f
igur
ig
ur
u
re 6
-eqn
-e
qn
n he
ere
e
Continuing in this way, we can obtain,
1
= −1
+ (an−1 − bm−1 )sn−m−1 + · · ·
⇒ sn−m + (an−1 − bm−1 )sn−m−1 + · · · + k = 0
kG(s) = k ·
sn−m
Let the roots of the polynomial be given by,
si = βi , i = 1, 2, . . . , n − m
Then,
(s − β1 )(s − β2 ) · · · (s − βn−m ) = s
n−m
−
n−m
βi sn−m−1 + · · · + k
i=1
Then,
an−1 − bm−1 = −
n−m
βi
i=1
But, note that −(an−1 − bm−1 ) =
But,
n−m
i=1
−an−1 =
βi is the sum of the poles going to ∞.
n
pj
j=1
is the sum of the open loop poles, and
−bm−1 =
m
j=1
zj
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131
is the sum of the open loop zeros.
Finally, the centroid is given by,
n−m
n
−(an−1 − bm−1 )
βi
pi − m
i=1 zi
σ =
=
= i=1
n−m
n − m
n−m
open loop poles −
open loop zeros
=
Number of poles − Number of zeros
i=1
Note that the centroid has to be real since complex conjugate poles or zeros
cancel the imaginary parts.
Also note that the whole proof depends on the fact that,
(s − f1 )(s − f2 ) · · · (s − fn ) = sn + gn−1 sn−1 + · · · + g1 s + g0
and
gn−1 = −
n
fi
i=1
An Example: Let the open loop plant transfer function be
G(s) =
1
s(τ s + 1)
With P-control, the closed-loop transfer function would be,
Gc (s) =
k
τ s2 + s + k
which has poles,
p1,2 =
−1 ±
√
1 − 4τ k
2τ
Asymptotic angles:
m = 0, n = 2, n − m = 2.
So there are two asymptotes.
180o + α360o
, α = 0, 1
n−m
1800 1800 + 360o
,
=
2
2
= 90o , 270o
φα = φα =
Asymptotic centroid:
Lecture Notes on Control Systems/D. Ghose/2012
Figure 13.18: The example problem
σ=
−1/τ
1
pi
=
=−
n−m
2
2τ
These are the same as that given in the figure.
Another Example: Let the open loop plant transfer function be,
G(s) =
1
(s + 4)2 + 16
With an integral control K(s) = k/s, the closed-loop transfer function is,
Gc (s) =
k
s
1+
1
(s+4)2 +16
k
1
·
s (s+4)2 +16
·
which is already in Evan’s form.
There are 3 open loop poles (n = 3) and no open loop zeros (m = 0).
p1 = 0, p2,3 = −4 ± j4.
So, the asymptotic angles are,
180o + α360o
, α = 0, 1, 2
n−m
1800 1800 + 360o 1800 + 720o
,
,
=
3
3
3
= 60o , 180o , 300o
φα = φα =
The centroids are,
σ=
0 + (−4 + j4) + (−4 − j4)
−8
pi
=
=
3
3
3
The sketch of the root locus is shown in the figure below.
132
Lecture Notes on Control Systems/D. Ghose/2012
133
Figure 13.19: Root locus of the example problem
Step 5: Find the departure and arrival angles.
(a) The departure angle θdi of the locus from pole pi is,
m
k=1
θz k −
n
j=1, j=i
θpj − θdi = 180o + l · 360o , l = 0, ±1, ±2, · · ·
where θzk is the angle made by a vector drawn from the zero zk to the pole
pi , and θpj is the angle made by a vector drawn from pole pj to pole pi . See
the figure given below for illustration.
Figure 13.20: Illustration for the departure angle
Proof: The departure angle expression is obtained directly from an application
of the angle criterion. Take a point si very close to pi .
Lecture Notes on Control Systems/D. Ghose/2012
134
Figure 13.21: Proof for the departure angle
Let the angle made by the vector from the pole pi to the point si be θpi . The
angle made by the vectors from the other poles and zeros to si is almost the
same as the angles made by vectors from other poles and zeros to pi (since si
and pi are close). Now apply the angle criterion.
m
k=1
θz k −
n
j=1, j=i
θpj − θpi = 180o + l · 360o , l = 0, ±1, ±2, · · ·
Now, solve for θpi = θdi .
An Example: Let
G(s) =
1
,
(s + 4)2 + 16
K(s) =
k
s
Then,
K(s)G(s) =
k
s[(s + 4)2 + 16]
Figure 13.22: Departure angle example
Departure angle for pole p2 ,
−(135o + 90o ) − θd2 = 180o ⇒ θd2 = −45o or 315o
Lecture Notes on Control Systems/D. Ghose/2012
135
(b) The arrival angle θai of the locus at zero zi is computed from,
m
k=1, k=i
θzk −
n
θpj + θai = 180o + l360o , l = 0, ±1, ±2, · · ·
j=1
where, θzk is the angle made by a vector drawn from zero zk to zero zi and θpj
is the angle made by a vector drawn from pole pj to zero zi . See the figure
below for illustration.
Figure 13.23: Illustration for the arrival angle
Proof: Similar to the previous proof.
An Example: Let
G(s) =
(s − 4)2 + 16
,
(s + 4)2 + 16
K(s) =
Then,
K(s)G(s) =
k[(s − 4)2 + 16]
s[(s + 4)2 + 16]
Note that
p1 = 0, p2,3 = −4 ± j4, z1,2 = 4 ± j4.
Figure 13.24: Arrival angle example
k
s
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136
The angle θd2 is determined from,
(135o + 180o ) − (135o + 90o ) − θd2 = 180o ⇒ θd2 = −90o or 270o
The arriving angle θa2 is,
90o − (45o + 45o ) + θa2 = 180o ⇒ θa2 = 180o
Step 6: Find the real axis breakaway points.
Real axis locus segments that meet always break away at ±90o . Generally, two
approaching loci always meet at a relative angle of 180o , and then break away
changing direction by ±90o . Breakaway point locations can be difficult to solve
for (in fact, this could be as difficult as solving for the exact root locus!).
As illustration, look at the example given for the asymptotic centroid computation.
Step 7: Use the Routh-Hurwitz criterion judiciously to identify the gain values for which
poles cross from stable to unstable region. But remember that this may not always
work, especially if there are poles already existing in the unstable region and you
are trying to find out the value of k for which some stable pole migrates to the
unstable region. Some times the auxiliary equations and its roots help in this.