1. CONTROL SYSTEMS
THEORY
A Graphical Tool
CHAPTER 7
STB 35103

2. Objectives

To learn the definition of a root locus.

To sketch a root locus.

To refine your sketch of a root locus.

To use the root locus to find the poles of a
closed-loop system.

To use the root locus to describe qualitatively the
changes in transient response and stability of a
system as a system parameter varied.

3. Introduction

What is root locus?


Root locus is graphical presentation of the closedloop poles as the parameter is varied.
Why do we need to use root locus?

We use root locus to analyze the transient
qualitatively. E.g. We can use root locus to analyze
qualitatively the effect of varying gain upon
percent overshoot, settling time and peak time.

We can also use root locus to check the stability of
our system qualitatively.

4. Introduction

What kind of system that uses root
locus?

Usually we use root locus to analyze the
feedback control system qualitatively.
K in the feedback
system is called a
gain. Gain is used to
vary the system in
order to get a
different output
response.

5. Introduction

What does a
root locus look
like?

The line with arrow
is the movement
path for closed
loop poles as the
gain is varied on
the s-plane.
Figure 1 – An example of root locus.
1 Pole
1 zero

6. Drawing the root locus

First step in drawing the root locus is to draw
the open-loop poles and zeros in the s-plane.

The most important thing that we must
understand in the movements of the poles is,
it is the closed-loop poles that move not the
open-loop poles.

Next slide shows an example of poles
movements when gain is varied.

7. Drawing the root locus

Given a feedback system for a video
camera that can automatically follow a
subject.
(a)Feedback system
(b)Closed loop transfer function of
the system, T(s)

8. Drawing the root locus

Based on the video camera feedback
system, the open-loop transfer function is
The modified unity feedback system
for an open-loop transfer function
can be displayed as
K2
G ( s ) = K1 ×
×1
s ( s + 10 )
=
K
s ( s + 10 )
K = K1 × K 2

9. Drawing the root locus

We are going to start drawing the root
locus for the video camera. The first step
is to draw the poles and zeros of the
system’s open-loop transient response.
*Fill up a table
with the
corresponding
K and pole
values

10. Drawing the root locus

Next step is to plot the poles values on the
s-plane by varying the gain, K, value.

11. Drawing the root locus

Join the poles with solid lines and you will
the shape of the locus (path)

12. Drawing the root locus

The process of drawing a root locus is time
consuming. If the system is complex, the
calculation will be much harder especially
if you use hand calculation.

An alternative approach is to sketch the
root locus instead of drawing the root
locus.

13. Sketching the root locus

In order to sketch the root locus we must
follow these five rules.
1.
Number of branches. We know that the closed loop
poles move when the gain, K, is varied. A branch is a
path where the a pole moves. The first rule is the
number of branches of the root locus equals to the
number of closed-loop poles.

14. Sketching the root locus
2.
Symmetry. The root locus is symmetrical
about the real axis.
imaginary axis
real axis

15. Sketching the root locus
3.
Real-axis segment. On the real-axis, for K
> 0 the root locus exists to the left of an odd
number of real-axis, finite open-loop poles
and/or finite open-loop zeros. Below is an
example of real-axis segment
Blue lines s-plane above are parts of the
real-axis where the root locus exists.

16. Sketching the root locus
4.
Starting and ending points. Where does
the root locus begin (zero gain) and end
(infinite gain)?
The root locus begins at the open-loop poles
and ends at open-loop zeros.
1 Pole
1 zero

17. Sketching the root locus
5. Behavior at infinity. A function can have finite
and infinite poles and zeros.



A function must have an equal number of poles
and zeros, finite and infinite.
If a function approaches infinity when s
approaches infinity – pole at infinity
If a function approaches zero when s
approaches infinity – zero at infinity

18. Sketching the root locus
5.
Behavior at infinity. The root locus
approaches straight lines as asymptotes as
the locus approaches infinity. Further, the
equation of the asymptotes is given by the
real-axis intercept, σa and angle θa as follows.
∑ finite poles − ∑ finite zeros
σa =
#finite poles − # finite zeros
θa =
( 2k + 1) π
#finite poles − # finite zeros
Where k = 0, ±1, ±2, ±3 and the angle is given in radians
with respect to the positive extension of the real axis

19. Sketching the root locus

We will sketch a root locus in Example 8.2

Problem: Sketch the root locus for the system
shown in figure below.

20. Sketching the root locus




It will greatly help our locus design if we know
the asymptotes for our locus.
Based on the open-loop transient response for
this system, the finite poles are at 0, -1, -2, and
-4 and the finite zero is at -3. So, there are 4
poles but only 1 zero. Hence there must be 3
infinite zeros, 3 asymptotes.
#Asymptotes = #finite poles – #finite zeros
Using the equation to calculate the asymptotes
crossing with the real axis,
( −1) + ( −2 ) + ( −4 ) − ( −3) 

 =−4
σa =
4 −1
3

21. Sketching the root locus

The angles of the lines that intersect at -4/3
given the equation for θa
θa =
( 2k + 1) π
# finite poles − #finite zeros
There are 3
=π /3
for k = 0
asymptotes so 3
=π
for k = 1
values of k=0,1,2
each representing
= 5π / 3
for k = 2
the angle of each
asymptote
= 7π / 3
for k = 3
As k continues to increase, the angle would begin to
repeat. The number of asymptotes equals the difference
between the number of finite poles and the number of
finite zeros.

22. Sketching the root locus

Root locus with the 3 asymptotes

23. Plotting and calibrating the root locus
sketch

Once we sketch the root locus using the
five rules discussed in the previous slides,
we may want to accurately locate points
on the root locus as well as find their
associated gain.

We might want to know the exact
coordinates of the root locus as it crosses
the radial line representing an overshoot
value.

24. Plotting and calibrating the root
locus sketch

Overshoot value can be represented by a
radial line on the s-plane.
Radial line
representing
overshoot value
on the s-plane

25. Plotting and calibrating the root
locus sketch

We learnt in Chapter 3, the value of zeta
on the s-plane is
ζ = cos θ

Given percent overshoot value we can
calculate the zeta value
ζ =
− ln ( %OS / 100 )
π 2 + ln 2 ( %OS / 100 )

26. Plotting and calibrating the root
locus sketch

Example

Draw a radial line on an S-plane that
represents 20% overshoot.

Solution

Overshoot is represented by zeta (damping ratio) on
the S-plane. So, the first step is to find the value of
zeta.
− ln ( %OS /100 )
ζ =
=
π 2 + ln 2 ( %OS /100 )
− ln(20 / 100)
π 2 + ln 2 ( 20 /100 )
= 0.456

27. Plotting and calibrating the root
locus sketch

Next step is to find the angle of the radial line.
ζ = cos θ
0.456 = cos θ
ζ
jω
θ
θ = cos −1 0.456
= 62.87o
σ

28. Plotting and calibrating the root
locus sketch

The point where our root locus intersect with the
n percent overshoot radial line is the point when
the gain value produces a transient response with
n percent overshoot.
Our root locus
intersect with the
radial line.
Meaning the gain
at the intersection
produces
transient
response with
zeta = 0.45

29. Plotting and calibrating the root
locus sketch

We know that when our root locus
intersect with the %OS radial line, the
gain during that intersection will produce
transient response with the same %OS.

Since we just sketch the root locus, we do
not know the exact coordinate of the
intersection between radial line and the
root locus.

30. Plotting and calibrating the root
locus sketch

A point on the radial line is on the root locus
if the angular sum (zero angle –pole angles)
in reference to the point on the radial line add
up to an odd multiple of 180,
̊

Odd multiple of 180 ̊  (2k+1)180 , k = 1,2,3, 4, ….
̊


180 ̊ , 540 , 900 , 1260 , …
̊
̊
̊
We must then calculate the value of gain. The
calculation for the gain by multiplying the
pole lengths drawn to that point and dividing
by the product of the zero length drawn to
that point.

31. Plotting and calibrating the root
locus sketch

Refer to the previous root locus that we have calculated,
we will find the exact coordinate as it crosses the radial line
representing 20% overshoot
ζ = 0.45

32. Plotting and calibrating the root
locus sketch

We can find the point on the radial line that crosses the
root locus by selecting a point with radius value then add
the angles of the zeros and subtract the angles of the poles
(θ =Σzero - Σpole =θ 2 – [θ 1 + θ 3 + θ 4 + θ 5]
Theory : Odd multiple of 180 ̊ 180 ̊ , 540 , 900 , 1260 …point on
̊
̊
̊
the radial line is on the locus

33. Plotting and calibrating the root
locus sketch

We will calculate the angle and magnitude
using the sine, cosine and tangent rules
b = a + c − 2ac cos ( β )
2
2
2

34. Plotting and calibrating the root
locus sketch

Let us take r = 0.747 and check if this
point on the radial line intersects with the
root locus

35. Plotting and calibrating the root
locus sketch

We will calculate the value of A, B, C, D, and E
and also θ1, θ2, θ3, θ4, and θ5

E length is equal to the radius r because of the
E
D
poles is at origin.
=
E = 0.747
θ5 = 180 − 63.256
= 116.744
D 2 = E 2 + 12 − 2 E (1) cos(63.256)
= 0.747 + 1 − 2(0.747) cos(63.25)
= 0.886
2
D = 0.886
= 0.941
sin θ 4
sin(θζ )
0.941sin(63.256)
0.941
= 45.141o
θ4 =
D
E
θζ
θ4
1
θ5

36. Plotting and calibrating the root
locus sketch

We will then calculate the length of C and θ3

You can either take the triangle E C or the
triangle D and C but the easiest is triangle E
and C.
C 2 = E 2 + 22 − 2 E (2) cos ( θζ )
C 2 = 0.747 2 + 22 − 2(0.747)(2) cos(63.256)
C = 1.793
E
C
=
sin θ3 sin(θζ )
0.747 sin(63.256)
1.793
= 21.842o
θ3 =
C
E
θζ
θ3
2
θ5

37. Plotting and calibrating the root
locus sketch

Calculate the length of B and θ2
B 2 = E 2 + 32 − 2 E (3) cos ( θζ )
B 2 = 0.747 2 + 32 − 2(0.747)(3) cos(63.256)
B = 2.746
E
B
=
sin θ 2 sin(θζ )
0.747 sin(63.256)
θ2 =
2.746
= 14.06o
B
E
θζ
θ2
3
θ5

38. Plotting and calibrating the root
locus sketch

Calculate the length of A and θ1
A2 = E 2 + 42 − 2 E (4) cos ( θζ )
A2 = 0.747 2 + 42 − 2(0.747)(4) cos(63.256)
A = 3.725
E
A
=
sin θ1 sin(θζ )
0.747 sin(63.256)
θ1 =
3.725
= 10.316o
E
A
θζ
θ1
4
θ5

39. Plotting and calibrating the root
locus sketch

We can calculate the gain, K, value where
root locus intersect the radial line.
∏ poles length
K=
∏ zeros length
K=
AC D E
B
3.725 ×1.793 × 0.941× 0.747
=
2.746
= 1.71
∏ → hasil darab

40. Plotting and calibrating the root
locus sketch

For root locus that do not have zeros, the
equation to find the gain is
K = ∏ poles length
∏ → hasil darab

41. Transient Response Design via Gain
Adjustment

In our previous example, the root locus
crossed the 0.45 damping ratio line with a
gain of 1.71.

Does this mean that the system will respond
with 20% overshoot, the equivalent to a
damping ratio of 0.45?

It must be emphasized that the formulas
describing percent overshoot, settling time,
and peak time were derived only for a
system with two closed-loop poles and
no closed-loop zeros.

42. Transient Response Design via Gain
Adjustment

If our system has additional poles and zeros,
we can still assume the system with only two
poles and no zero if it fulfills the requirement
below:

Higher order poles are much farther into the left
half of the s-plane measured from the jω-axis. (we
assume a factor of five times farther from jω-axis
than the dominant second-order pair. )

Closed-loop zeros near the closed-loop second
order pole pair are nearly canceled by the close
proximity of higher-order closed-loop poles.

43. Transient Response Design via Gain
Adjustment

Closed-loop zeros not canceled by the close
proximity of higher-order closed-loop poles are
far removed from the closed-loop second order
pole pair.
The best

44. Transient Response Design via Gain
Adjustment

Skill-Assessment Exercise 8.6
K
G(s) =
( S + 2)( S + 4)( S + 6)
a)
b)
c)
d)
Sketch the root locus
Using a second order approximation, design the
value of K to yield 10% overshoot for a unit step
input (Hint r = 3.431).
Estimate the settling time, peak time, and steadystate error for the value of K designed in (b)
Determine the validity of your second-order
approximation

45. Transient Response Design via Gain
Adjustment
If we look back at the equation for peak time, Tp and
settling time Ts.
4
Ts =
ζω n
Tp =
π
ωn 1 − ξ 2
Imaginary
Real

46. Transient Response Design via Gain
Adjustment

Solution

K=45.55
Ts=1.97

Tp=1.13



estep (∞) = 0.51
Kp=0.949,
Comparing this value to the real part of the
dominant pole, -2.028, we find that it is not
five times further. The second-order
approximation is not valid