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The Root Locus Method
Chapter 7
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Introduction
The root locus concept
The root locus procedure
PID controllers
Outline
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The root locus is a powerful tool for design and analysis offeedback control systems. The transient performance of a feedback control system is directly
related to the location of the poles in the s-plane.
As we change one or more system parameters, the location ofpoles in s-plane changes.
Therefore, it is necessary to determine how the poles move in the
s-plane as the parameters are varied. Root locus (RL):
It is the path of the roots of the characteristic equationtraced out in the s-plane as a system parameter is changed.
For example: changes in the parameters of a controller
This is crucial because one can adjust the system response to getthe desired performance through proper selection of controllerparameters.
7.1: Introduction
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Root Locus Method:
A graphical method for sketching the locus of roots in the s-plane as a parameter is varied.
This method gives a qualitative information about thestability and performance of the system.
7.1: Introduction
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Root locus:
It is the path of the roots of the characteristic equationtraced out in the s-plane as a system parameter is changed.
Example:
7.2: The Root Locus Concept
0102 Ksssq
10
1
sssG
Kss
KsT
10
2
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Pole location as a function of gain
7.2: The Root Locus Concept
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Root locus:
As K varies the locationof the poles is changed.
Hence root locus showsthe changes in the
transient response as theparameter K is varied.
7.2: The Root Locus Concept
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Remarks:
For K25, the system has complex roots, and hence is: Underdamped
In underdamped case the real parts of the poles are alwaysthe same regardless of the value of the gain. This means that
the settling time remains the same. Note: for higher order systems, this analysis is not
practical.
7.2: The Root Locus Concept
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The performance of a control system is described by the
closed-loop transfer function:
The roots of the characteristic equation determine the modes
of response of the system. Example:
7.2: The Root Locus Concept
)(
)(
sD
sN
sR
sYsT
01 sKG
sKG
sKGsT
1
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Characteristic equation:
Where K is a variable parameter.
The characteristic roots (poles) of the system must satisfythis equation.
The characteristic equation is a function of
7.2: The Root Locus Concept
01 sKG
js
01 sKG
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Complex number algebra:
In polar form: Magnitude/absolute value
Angle/phase
7.2: The Root Locus Concept
0y0,xifNAN
0y0,xif
0y0,xif
0y0,xiftan
0y0,xiftan
0xiftan
2
2
x
y1
x
y1
x
y1
s
Imaginary
Real
planes
s22 yxs
x
y
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Complex function:
The function of a complex number is also a complex number
In case of control systems:
The magnitude
7.2: The Root Locus Concept
assF
j
n
j
i
m
i
ps
zssF
1
1
lengthspole
lengthszero
1
1
j
n
j
i
m
i
ps
zs
sF
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The phase or angle:
Example:
Find F(s) at the point s=-3+j4
7.2: The Root Locus Concept
n
j
j
m
i
i pszs
sF
11
anglespoleangleszero
2
1
ss
ssF
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KG(s) is a function of s and hence is a complex number
In polar form
Therefore:
7.2: The Root Locus Concept
1 sKG
01 jsKG
01 jsKGsKG
1sKG
,540,180)21(180360180 kksKG
,3,2,1,0 k
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Conclusion:
A pole of a system causes the angle of KG(s) (or simply G(s)since K is a scalar) to be an odd multiple of 180.
A pole of a system causes the absolute value of KG(s) to beequal to 1.
7.2: The Root Locus Concept
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An orderly process for locating the locus of roots as a
parameter varies. STEP 1: Locate the poles and zeros of P(s) in the s-plane
Write the characteristic equation as:
Then rearrange it in such away that the parameter of interest, K,appears as a multiplying factor:
Factors P(s) and write it in form of poles and zeros:
7.3: The Root Locus Procedure
01 sF
KsKP 0for01
01
1
1
j
n
j
i
m
i
ps
zs
K
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Locate the poles (-pi) and zeros (-zi) on the s-plane. By
convention, we use x to denote poles and o to denotezeros.
Rearrange the characteristic equation:
For K=0, the roots of the characteristic equation are thepoles of the P(s), since:
For K=, the roots of the characteristic equation are the
zeros of the P(s), since:
7.3: The Root Locus Procedure
011
i
m
i
j
n
j
zsKps
01
i
m
i
zs
01
j
n
j
ps
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Property of RL:
The RL of the characteristic equation 1+KP(s)=0 begins at thepoles of P(s) and ends at the zeros of P(s) as K increases fromzero to infinity. (verified)
STEP 2: Locate the segments of the real axis that are the root loci.
Property of RL: The root locus on the real axis always lies in a section of the real
axis to the left of an odd number of poles and zeros.
This property can be verified by the angle criterion.
A branch of the RL is the locus of one root when K varies.
The number of braches of RL is equal to the number ofpoles of P(s).
7.3: The Root Locus Procedure
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Property of RL:
The RL are symmetrical with respect to the horizontal realaxis because the complex roots appears as pairs of complexconjugate roots.
Complex conjugate roots:
Roots having same real parts but with imaginary parts equalin magnitude and opposite in sign.
7.3: The Root Locus Procedure
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Example:
characteristic equation of a control system:
7.3: The Root Locus Procedure
01)(1 2
4
1
2
1 1
ss
sKsF
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STEP 3: proceed the loci to the zeros at infinity along
the asymptotes. Asymptotes: it is the path that the root locus follows as the
parameter becomes very large and approaches infinity.
The number of asymptotes is equal to the number of poles
minus the number of zeros. Angle of the asymptote: the angle that the asymptote makes
with the real axis.
Real axis intercept: The asymptotes are centered at a point onthe real axis:
7.3: The Root Locus Procedure
1,2,1,0180
12
mnk
mn
k
A
mn
zp
mn
zerospoles
A
m
i
i
n
j
j
11
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Example
7.3: The Root Locus Procedure
242
111
sss
sKsF
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STEP 4: Determine the break away point on the real
axis (if any). The breakaway is a point on the real axis at which the root
locus leave the real axis.
To compute the break away point, we determine the
maximum of K. That is:
The roots will give the break away point.
The breakaway point can also be evaluated graphically.
7.3: The Root Locus Procedure
01 sKP
01 sPdsddsdK sPK
1
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STEP 5: Determine the angle of departure of the
locus from a pole. Angle of locus departure from a pole is the difference
between the net angle due to all other poles and zeros and thecriterion angle.
7.3: The Root Locus Procedure
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STEP 6: Determine where the locus crosses the
imaginary axis (if it does so), using the Routh-Hurwitzcriterion.
STEP 7: complete the sketch of the root loci.
7.3: The Root Locus Procedure
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Example 7.4:
Plot the RL for a system with following characteristicequation as K varies for K>0.
STEP1:
Zeros: this system has no finite zerosPoles at:
As the number of poles n=4, we have 4 separate loci.
The RL are symmetrical with respect to the real axis
7.3: The Root Locus Procedure
01 44444 jsjsssK
444441
jsjssssP
jj,, 44,4440
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STEP2:
A segment of the RL exists on the real axis between s=0 ands=-4.
STEP3:
Number of asymptotes=n-m=4-0=4
The angles of the asymptotes
The intercept/center of the asymptotes
7.3: The Root Locus Procedure
1,2,1,018012
mnk
mn
kA
mn
zp
A
m
i
i
n
j
i
11
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STEP4:
The breakaway point is evaluated through:
Or graphically, the breakaway point is between 0 and -4.
STEP5:
Angle of departure of RL from a pole:
STEP6:
The imaginary axis values of the loci.
STEP7:
Sketch the complete RL
7.3: The Root Locus Procedure
01 sPdsd
ds
dK
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7.3: The Root Locus Procedure
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PID (Proportional-Integral-Derivative) controller:
In industry, the most common controller that has so far beenimplemented is the simple proportional plus integral plusderivative (PID) controller.
The general form of the PID control (actuating) signal is:
7.6: Three-Term (PID) Controller
dt
tde
DIP KdtteKteKtu
Controller Process
Sensor
Measured output
_
+ sY sR
sE sU
sG c sG
p
1
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The transfer function of the PID Controller is:
If KP is the only non-zero term, the controller is said to be aproportional controller or P-control for short.
If KI is the only non-zero term, the controller is said to be anintegral controller or I-control for short.
If KP and KI are the only non-zero terms, the controller issaid to be aproportional plus integral controller or PI-
control for short.
If KP and KD are the only non-zero terms, the controller issaid to be aproportional plus derivative controller or PD-control for short.
7.6: Three-Term (PID) Controller
sKK Ds
K
PsE
sU I
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Objective:
Choose a suitable controller Gc(s) to make the closed-loopsystem:
1) stable;
2) the output y(t) of the plant be close to the reference signal
r(t) (i.e. error ess should be small)
7.6: Three-Term (PID) Controller
1
1
s
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Proportional control (P-control)
The controller transfer function is:
The system transfer function:
Stability: from R-H criterion, the system is stable if:
Steady-state error
7.6: Three-Term (PID) Controller
Pc
KsG
PP
Ks
K
sT 1
01
0
PK
P
ssK
e
1
1
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Proportional control (P-control)
The unit step response is:
For quick response (i.e. less settling time), the KP must be as
large as possible, for a given time constant .
7.6: Three-Term (PID) Controller
tK PK
P ety
1
1
planes
Imaginary
Real
PK
1
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Integral-control (I-control)
The problem with a P controller is that the steady-state error,ess, is not zero for a step input in case of type zero system.
Can we design a controller which makes ess = 0?
Yes. We need to introduce an integral term.
The output signal of an Integral controller has the form:
The controller transfer function is:
The system transfer function (2nd order system):
7.6: Three-Term (PID) Controller
dtteKtuI
s
K
c
IsG
I
I
Kss
KsT
2
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Integral-control (I-control)
Stability: from R-H criterion, the system is stable if:
Steady-state error in case of step input:
An integral controller eliminated the steady-state error of astep response .
7.6: Three-Term (PID) Controller
0
0
IK
0
lim10
s
sssG
Ae
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Integral-control (I-control)
Unit step response:
For a given time constant, KI can be adjusted to get thedesired transient response of the system
Example:
7.6: Three-Term (PID) Controller
22
2
12 2 nn
n
IK
IK
sssssT
n
IK
n 212
,
10
1
s
sGp
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PI-Control
A proportional plus integral (PI) controller has the transferfunction:
The task is to tune the control parameters KP and KI toachieve a better control.
By combining the advantages of P and I controllers, we havemore freedom of choice and can achieve better performancesince there are two parameters to tune.
7.6: Three-Term (PID) Controller
s
K
Pc
IKsG
1
1
s
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Summary
s
KsKP 2s
KsKP 3s
KsKP
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Summary
11
s
KsKP
11
ss
KsKP
11
1
ss
sK asKP
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Summary
11
21
ss
KsKP 11
21
sss
KsKP
111
21
sss
sK asKP
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Example 7.7: Automobile Velocity Control
A velocity control system for maintaining the relative velocitybetween the two vehicles.
The output Y(s) is the relative velocity between the twovehicles.
The input R(s) is the desired relative velocity between the two
vehicles. The transfer function of the process (automobile system) is:
Objective: To design a controller that can maintain a prescribed velocity
between the vehicles and maneuver the active vehicle.
7.7: Design Examples
82
1
ssp sG
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Example 7.7: Automobile Velocity Control continued
7.7: Design Examples
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Example 7.7: Automobile Velocity Control continued
Control Goal:
Maintain a prescribed velocity between the vehicles and maneuver theactive vehicle.
Variable to be controlled:
The relative velocity between the vehicles, denoted by y(t)
Design specifications:
DS1: zero steady state error to a step input.
DS2: steady state error due to a ramp input of less than 25% of theinput magnitude.
DS3: Percent overshoot less than 5% to a step input
DS4: settling time less than 1.5 seconds to a step input (using a 2%criterion to establish settling time)
7.7: Design Examples
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Example 7.7: Automobile Velocity Control continued
System configuration:
Obtain mathematical models:
7.7: Design Examples
Controller Process
_
+
sY sR sG
c sG
p
82
1
ssp sG
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