The Complete Root Locus
Introduction
The complete root locus (CRL) displays the locations of the closed loop poles of a single
input, single output (SISO) system as one real parameter k ∈(-∞,∞) is varied. Figure 1 illustrates
a unity feedback system having the transfer function
s +1
G (s ) =
(1)
s (s + 4 )(s 2 + 2 s + 2 )
in the forward path. Figure 2 shows the loci of the roots as the proportional gain k is varied. The
solid lines are the locations corresponding to positive gains while the dashed line corresponds to
negative gains. The locus for positive gains is called the root locus while the locus for negative
gains is called the complementary root locus. The two loci together are called the complete root
locus. Also shown in the figure is the MATLAB® code for generating the root locus.
+
s +1
s (s + 4 ) s 2 + 2 s + 2
(
k
R(s)
-
)
Y(s)
Figure 1: Unity Feedback System
Root Locus
10
8
6
Imaginary Axis
4
2
0
>> G=zpk([-1],[0 -4 -1+i -1-i],1)
Zero/pole/gain:
-2
(s+1)
----------------------4
s (s+4) (s2 + 2s + 2)
-6 >> rlocus(G,'-',-G,'--')
-8
-10
-15
-10
-5
0
5
Real Axis
Figure 2: Root Locus of System of Figure 1
10
Although it is easy and fast to generate plots such as Figure 2, it is essential to understand
how to draw the root locus because it provides insight into the influence a compensator
(controller) pole or zero has on the system performance.
Most undergraduate control texts cover the root locus construction for only positive k. The
presentation will cover both signs. The motivation for doing this is completeness, the insight
offered by observing the influence of the other sign, and the opportunity to check the correctness
of the plot construction by observing certain properties.
Contained herein are the details of drawing the root locus and the properties of the open loop
transfer function that provide the key tools that help in drawing the diagram.
What is the Root Locus?
The Correct Form
As previously stated, the root locus is a plot of the closed loop poles as a single parameter in
the open loop transfer function is varied. Let this parameter be k. Let the closed loop transfer
function be written as
P(s )
P(s )
D( s )
(2)
T (s) =
=
D( s ) + kN ( s ) 1 + k N ( s )
D( s)
where P(s), D(s), and N(s) are polynomials and the denominator has been arranged so that the
polynomial terms that contain the factor k are grouped together and denoted as the polynomial
N(s). The terms in the closed loop transfer function denominator that do not contain the factor k
are grouped together and denoted as D(s).
The denominator on the right side of Eq. (2) is said to be in the “correct form” for root locus
construction.
In general, let G(s) be the forward transfer function and H(s) be the feedback transfer
function in the SISO system illustrated in Figure 3. In order to develop Eq. (2), first develop the
transfer function of the loop in Figure 3 which is
+
R(s)
G(s)
Y(s)
H(s)
Figure 3: SISO System
Y (s )
G (s )
=
= T (s ) .
R (s ) 1 + G (s )H (s )
(3)
Once the transfer function is known and the numerator and denominator fractions have been
cleared, the factorization that is shown in Eq. (2) can be performed.
Example 1
A servo position control system with derivative feedback is shown in Figure 4. In Figure 4,
k is a variable gain on the derivative feedback and the constants a, b, and KP are all known.
Determining the closed loop transfer function for the system of Figure 4 and performing the
operations in Eq. (2) shows
a
a
KP
KPa
KPa
s (s + b ) + K P a
Y (s )
s (s + b )
=
=
=
=
.
K a
aK P
R(s )
s (s + b ) + K P a(1 + k ) s (s + b ) + K P a + kK P a
1 + (1 + k ) P
1+ k
s (s + b )
s (s + b ) + K P a
KP
+
R(s)
a
s (s + b )
KP
-
(4)
Y(s)
1+k s
Figure 4: Servo Position Control System
The denominator of the last term of Eq. (4) is now in the correct root locus form.
Root Locus Construction – Step 1
Put the transfer function into the correct form.
In examining the transfer function of Eq. (1) and the system shown in Figure 1, it is seen that
the correct form of the closed loop denominator is
N (s )
s +1
1+ k
=1+ k
.
(5)
D (s )
s (s + 4 )(s 2 + 2 s + 2 )
The Nature of the Root Locus
The closed loop poles are the roots of the equation
N (s ) D (s ) + kN (s )
1+ k
=
=0.
D (s )
D (s )
(6)
N (s )
is referred to as the
D(s )
open loop transfer function. The roots of N(s) are called open loop zeros of which there are m.
The roots of D(s) are called open loop poles of which there are n. It is assumed in the
presentation that there are more open loop poles than there are zeros, i.e. n > m. The quantities n
which is also known as the characteristic equation (CE). The ratio
and m are also the number of finite open loop poles and zeros, respectively. Because n > m, open
loop zeros can also occur as s becomes infinite giving rise to infinite open loop zeros. If the case
of m > n were considered, then there would be open loop poles as s becomes infinite and then
there would be infinite open loop poles. This is the reason behind distinguishing between finite
and infinite open loop poles and zeros. The constraint of n > m will be maintained and more will
be said about open loop zeros occurring as s becomes large when asymptotes are discussed.
Because n > m, the degree of the characteristic equation in Eq. (6) is n. A branch is the path
of a root as the gain k is varied from zero to ∞ (or -∞). For positive k there are n branches on the
root locus and for negative k there are also n branches on the complementary root locus
Each branch of the CRL has a remarkable property. It starts at an open loop pole when k is
zero and ends at an open loop zero when k is either ±∞. To see why this occurs, first write the
CE as
D(s ) + kN (s ) = 0 .
(7)
If k is zero, then the roots of the CE occur at the roots of D(s), i.e. the open loop poles. Now
write the CE as
D (s )
k = −
.
(8)
N (s )
As s approaches an open loop zero, larger and larger values of k are needed to satisfy the CE.
The branch of the locus then occurs between the open loop pole and the open loop zero. Because
there are more open loop poles than finite open loop zeros, some of the zeros satisfying Eq. (8)
occur at infinite values of s. More will be said about these zeros when asymptotes are covered.
The root locus is drawn in the complex plane. The sketch begins by drawing the complex
plane and entering the finite open loop poles and zeros.
It should be noted that the coefficients of the characteristic equation are all real which means
the roots are either real or complex conjugate pairs. This means that the root locus is symmetric
about the real axis of the complex plane. It should also be appreciated that if there is another line
of symmetry for the open loop poles and zeros, then the root locus is also symmetric about this
other line of symmetry.
Example 2
Consider the root locus plot of Figure 7. The open loop poles and zeros are symmetric with
respect to the vertical line passing through the point -2 on the real axis. The root locus plot is
symmetric about this line as well as the real axis. The MATLAB® code necessary to generate
the plot is also shown in Figure 7. It should be noted that the plot in Figure 2 of the CRL of the
system in Figure 1 is only symmetric about the real axis.
Root Locus Construction – Step 2
Include the finite open loop poles and zeros in a
sketch of the complex plane. Note the lines of
symmetry.
Figure 8 shows the finite open loop pole – zero plot of the system in Figure 1. The only line
of symmetry is the real axis.
Root Locus
8
6
4
Imaginary Axis
2
0
-2
>> G=zpk([-2],[-1+i -1-i, -3-i, -3+i],1)
Zero/pole/gain:
(s+2)
-----------------------------(s2 + 2s + 2) (s2 + 6s + 10)
>> rlocus(G,'-',-G,'--')
-4
-6
-8
-10
-8
-6
-4
-2
0
2
4
Real Axis
Figure 7: Root Locus Plot Showing Two Lines of Symmetry
jω
j
-4
σ
-1
-j
Figure 8: Plot of Open Loop Poles and Zeros
6
Magnitude Criterion and Angle Criterion
Before presenting the next step of the locus construction, there is a topic that needs to be
discussed. This is magnitude criterion and angle criterion. This topic provides an additional tool
for helping in the locus construction.
Let so be some point on the CRL. Eq. (7) can be rewritten as
D (s o )
k=−
.
(9)
N (s o )
This equation needs to be satisfied in both magnitude and phase in order for the point so to be on
the locus. This requirement can be expressed in two parts as
D (s o )
k =
(10)
N (s o )
and
D (s o )
∠−k = ∠
(11)
N (s o )
where ∠ denotes the angle or phase of the complex number. Eq. (11) can be further divided as
D(so ) even (including zero ) multiple of π if k < 0
∠
=
.
(12)
odd multiple of π if k > 0
N (s o ) 
Eq. (10), called the magnitude criterion, is useful for determining the gain that locates a CE root
at a particular point on the CRL. Eq. (12), called the angle criterion, is used as a tool for the
construction of the CRL. Any point in the complex plane that satisfies Eq. (12) is a point on the
CRL.
Roots on the Real Axis
Information from the previous section can be used to demonstrate that the entire real axis is
on the CRL. This will also lead to third step in drawing the locus.
Consider the complex plane plot of Figure 9. From a given open loop transfer function, there
are real poles and zeros on the horizontal axis as well as a pair of complex conjugate poles. The
material from Graphical Transfer Function Evaluation will be used to show that any point on
the real axis satisfies Eq. (12). To begin the demonstration, pick a point on the positive real axis
and draw vectors from the finite open loop poles and zeros as shown in Figure 10. The vectors
in Figure 10 were not drawn on top of one another so that one could be distinguished from one
another. Recall that the phase of the transfer function at a point is the sum of the angle of the
vectors drawn from the transfer function zeros less the sum of the angles of the vectors drawn
from the transfer function poles.
Before this evaluation is developed further, observe the vectors drawn from the two complex
conjugate poles. Regardless of the real axis location of the point where the phase is desired, the
angles of the vectors from the two complex conjugate poles add to zero. Complex conjugate
poles and zeros do not contribute to the phase of the open loop transfer function when evaluated
on the real axis.
Excluding the vectors from the complex conjugate poles, it is seen that all of the vectors have
zero angle making the phase of the open loop transfer function zero and indicating that k is
negative. Also, notice that the same result for the phase would be obtained regardless of where
to evaluate the phase of the transfer function as long as the point is anywhere on the positive real
axis to the right of the origin. Thus, the positive real axis to the right of the origin is on the
complementary root locus.
jω
σ
Figure 9: Pole-Zero Plot
jω
σ
Figure 10: Determining the Phase of the Given Open Loop Transfer Function on the Positive
Real Axis
Going through the same operation in Figure 11(a) as was done in Figure 10, it is seen that the
phase of the open loop transfer function on the interval between origin and the pole to the left of
the origin is -180o. All of the vectors are zero degrees except the one drawn from the origin.
The two complex conjugate poles were eliminated from consideration. The angle of the vector
drawn from the zero less the sum of the angles drawn from the poles is 0 – 180 o = -180o.
Because this result is true for any point on the interval between the origin and the pole to the left
of the origin, this interval is on the root locus.
On the interval between the zero and the pole to its right, the angle from the zero is 0o as
shown in Figure 11(b). The sum of the angles from the poles to the point on the interval is 360o.
The phase of any point on this interval is 0o – 360o or zero which indicates that all points on this
interval are part of the complementary root locus.
In Figure 11(c), any point on the interval between the zero and the repeated pole, it is seen
that the angle from the zero to the point is 180o and the sum of the angles from the poles to this
point is 360o. The difference is 180o – 360o or -180o. This result indicates that the entire interval
is on the root locus.
Finally, in Figure 11(d), the point of interest is located to the left of the repeated poles. The
angle from the zero to the point is 180o and the sum of the angles from the poles to this point is
720o. The difference is 180o – 720o or -540o or -180o. Thus, the interval to the left of the
repeated poles is on the root locus. It is seen that the immediate intervals to the left and right of
the repeated poles are both root locus.
jω
σ (a)
jω
σ (b)
jω
σ (c)
jω
σ (d)
Figure 11: Determining the Locus on the Real Axis
The end result is shown in Figure 12. The intervals of the real axis on the root locus are
indicated with solid lines and the intervals that are on the complementary root locus are shown as
dashed lines.
jω
σ
Figure 12: Real Axis Parts of the Complete Root Locus
There is an easy way to distinguish which intervals are on the root locus and which intervals
are on the complementary root locus. On any real axis interval, if the number of real poles and
zeros to the right is even, then the interval is on the complementary root locus, otherwise it is on
the root locus. Note that for the application of this rule, the number zero is considered even.
Root Locus Construction – Step 3
Perform the “Roots on the Real Axis”
Determining the roots on the real axis for the open loop system of Figure 1 produces the
result on Figure 13.
jω
j
-4
σ
-1
-j
Figure 13: Root Locus of System in Figure 1 after Steps 1 – 3
Asymptotes and Asymptote Angles
Recall that there are more open loop poles (n) than open loop zeros (m) and also recall that as
the gain k increases in magnitude, each locus moves from an open loop pole to a open loop zero.
That this is true can be easily appreciated for the m open loop poles that approach the m open
loop zeros, but how the remaining n – m poles approach open loop zeros is at this point
perplexing. Notice that the degree of the open loop transfer function denominator is larger than
that of the numerator. As s becomes large, the magnitude to the open loop transfer function
evaluated at s decreases and as s grows without bound, the transfer function magnitude
approaches zero. This brief scenario indicates that there are open loop zeros at infinity. Not any
value of s at infinity will do because both the magnitude and angle criteria must be satisfied.
It will now be demonstrated that there are n – m zeros for infinite s on the root locus as well
as n – m zeros for infinite s on the complementary root locus. Taking the limit as s becomes
large of the characteristic equation provides
 sm 

N (s ) 
lim1 + k
 = 1 + k lim n  = 1 + k lim(s m−n ) = 0.
s →∞
s →∞ s
s →∞
D (s ) 

 
(13)
In taking the limit, it is seen that as sm-n becomes smaller and smaller, the magnitude of k must
grow in order to satisfy the characteristic polynomial. This situation affirms that as k increases
in magnitude, the locus approaches an open loop zero. For a large magnitude of s, Eq. (13) can
be solved for s as
s = (− k )
1 /( n −m )
(14)
which has two sets of solutions depending upon the sign of k. If k is positive, then the n – m
angles of s satisfying Eq. (14) are
∠s = (2i − 1)
π
n−m
, i = 0,1, 2, ..., m − n − 1 .
(15)
If k is negative, then the n – m angles of s satisfying Eq. (14) are
∠s = (2i )
π
n−m
, i = 0,1, 2, ..., m − n − 1 .
(16)
The solutions of Eq. (14) appear as lines having slopes given by Eqs. (15) and (16).
Eqs. (15) and (16) provide only slopes and, in order to use these slopes, a point is required.
A short example will demonstrate the validity of Eqs. (15) and (16).
Example 3
Determine the roots of the characteristic equation
s+4
1+ k 4
=0
3
s + 9 s + 3s 2 + 6 s + 12
for large magnitudes of k having either sign and plot the roots in the complex plane. Sketch in
lines having slopes determined by Eqs. (15) and (16) and compare.
The MATLAB script file of Figure 14 was used in this calculation. The plot is shown in
Figure 15. The asymptotes of the root locus are drawn in solid lines while the asymptotes of the
complementary root locus are drawn as dashed lines. The dots in Figure 15 are the actual roots
of the characteristic equation. In examining Figure 15, the fact that the lines are asymptotes can
be appreciated from the difference between the calculated roots and the asymptote lines. By
choosing larger values of k in the MATLAB script file, the difference between the roots and
asymptotes will diminish.
clear all
N
= [0 0 0 1 4];
D
= [1 9 3 6 12];
%
%
%
%
%
%
%
%
%
Numerator Polynomial
Denominator Polynomial
Same length polynomials are easy to add
Degree of numerator
Degree of denominator
Values of k
Length of the array ks
Use Figure 1
Put hold on so many items can be plotted
m
= 1;
n
= 4;
ks
= logspace(3,5,20);
nl
= length(ks);
figure(1);
hold on;
for i = 1:nl
pk = roots(D+ks(i)*N);
% Roots for positive k
nk = roots(D-ks(i)*N);
% Roots for negative k
plot(pk,'k.');
% Plot roots for positive k as dots
plot(nk,'k.');
% Plot roots for negative k as dots
end
sigmaa = (sum(roots(D))-(-4))/(n-m); % Asymptote intersection point
L
= 50;
% Length of asymptotes
for i=0:n-m-1
angle = (2*i-1)*pi/(n-m);
% Asymptote angle - positive k
X = [sigmaa L*cos(angle)];
% X coordinates
Y = [
0 L*sin(angle)];
% Y coordinates
plot(X,Y);
% Plot line
angle = (2*i)*pi/(n-m);
% Asymptote angle - negative k
X = [sigmaa L*cos(angle)];
% X coordinates
Y = [
0 L*sin(angle)];
% Y coordinates
plot(X,Y,'k--');
% Plot line
end
grid;
title('Example of Asymptotes')
xlabel('Real s');
ylabel('Imaginary s')
Figure 14: Code for Asymptote Angle Formula Verification
Eqs. (15) and (16) provide the slopes for the asymptotes, but these two equations do not
provide any information regarding where these lines intersect one or more of the complex plane
axes. It will now be demonstrated that there is a point on the real axis that all asymptote lines
intersect and this point is determined from a knowledge of the open loop poles and zeros. To
start this demonstration, write Eq. (7) as
D( s )
= −k
(17)
N (s)
where the polynomials are given by
D ( s ) = s + a1s
n
n −1
+ a2 s
n−2
n
+ ⋯ + an −1s + an = ∏ (s + pi )
(18)
i =1
and
m
N ( s ) = s m + b1s m−1 + b2 s m−2 + ⋯ + bm −1s + bm = ∏ (s + zi )
(19)
i =1
and where the ai and bi are the polynomial coefficients while the open loop poles are located at
points –pi and the open loop zeros are located at –zi. By dividing N(s) into D(s), the quotient
becomes
D (s ) n−m
n−m
(20)
= s + (a1 − b1 ) s n −m−1 + lower order terms ≈ (s − σ a ) = − k
N (s )
where σa is given by
−a +b
σa = 1 1
(21)
n−m
and where it is seen for large s, Eq.(20) is the same as Eq. (14). By multiplying out the
monomial term on the right of Eq. (20), it is seen that result matches the two highest power terms
of the quotient. The right side of Eq. (20) can be rearranged as
1
1+ k
=0
(22)
(s − σ a )n−m
which corresponds to a system consisting of n-m repeated poles at the point σa. Eq.(22) can be
solved for s and evaluated for any value of k, even zero. In Example 3, the evaluation of σa leads
to a value of -5/3 and n-m is 3. Plotting the values of s that satisfy Eq. (22) for Example 3 in the
complex produces the straight lines of Figure 15, i.e. the asymptotes. The quantity σa is called
the asymptote intersection point. This quantity was used in drawing the lines from Example 3
shown in Figure 15.
Example of Asymptotes
50
40
30
Imaginary s
20
10
0
-10
-20
-30
-40
-50
-50
-40
-30
-20
-10
0
Real s
10
20
30
40
50
Figure 15: Plot of Roots and Asymptotes
The asymptote intersection point can be readily determined from knowing the poles and
zeros of the open loop transfer function. To see how this arises, refer to Eq. (18). In multiplying
out the polynomial, it is seen that
n
a1 = ∑ pi
(23)
i =1
and from Eq.(19), it can be shown that
m
b1 = ∑ zi .
(24)
i =1
Because a1 is the negative of the pole sum, Eq. (23) can be rewritten as
n
a1 = −∑ open loop poles
(25)
i =1
and b1 can be written as
m
b1 = −∑ open loop zeros .
(26)
i =1
Substituting Eqs. (25) and (26) into Eq. (21) shows that the asymptote intersection point is
σa =
n
m
i =1
i =1
∑ open loop poles − ∑ open loop zeros
.
(27)
n−m
Root Locus Construction – Step 4
Sketch the root locus and complementary root
locus asymptotes.
For the open loop system specified in Eq. (1), the open loop zero occurs at s = -1 and the open
loop poles occur at s = 0, -4, and -1 ± j1. From Eq. (27) the asymptote intersection point is
n
m
∑ open loop poles − ∑ open loop zeros
0 − 4 − 1 + j1 − 1 − j1 − (−1)
5.
=−
n−m
4 −1
3
From Eq. (15), the angles of the asymptotes on the root locus are
∠s = -60o, +60o, and +180o.
From Eq. (16), the angles of the asymptotes on the complementary root locus are
∠s = 0o, +120o, and +240o.
Combining the asymptote lines with the results of Figure 13 shows the sketch of Figure 16.
σa =
i =1
i =1
=
Breakaway Points
Breakaway points occur in the complex plane where the characteristic equation has multiple or
repeated roots. Breakaway points are significant in that they are the only way a locus can either
enter or leave the real axis. The breakaway points offer useful information concerning the locus
and the locus construction.
The way to find breakaway points is to note that at a repeated root of a polynomial, both the
polynomial and the slope of the polynomial are zero. In order to locate the points of multiple
roots, it is sufficient to locate those points where the derivative of the characteristic equation with
respect to s vanishes. Applying this idea to Eq. (7) shows
d
(D(s ) + kN (s )) = 0 = d  D(s ) + k  = d  D(s )  = d  1 + N (s )  = d  N (s )  = 0 . (28)
ds
ds  N (s )
 ds  N (s )  ds  k D(s )  ds  D(s ) 
From Eq. (28), it can be seen that the points of multiple roots occur at the solutions to either
d  D (s ) 

=0
(29)
ds  N (s ) 
or
d  N (s ) 

 = 0.
(30)
ds  D (s ) 
That either Eqs. (29) or (30) provide the same results might seem disconcerting and a closer look
at Eq. (28) reveal the reasoning behind these two equations.
jω
j
σa
-4
σ
-1
-j
Figure 16: Root Locus Construction after Step 4
Step 6 Routh Tabulation
Step 7 Angles of Approach and Departure
Step 8 Sketch the locus