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Unformatted text preview: ME375 Handouts Root Locus
• Motivation
• Sketching Root Locus
• Examples School of Mechanical Engineering
Purdue University ME375 Root Locus  1 1 ME375 Handouts Servo Table Example
DC Motor Position Control
The block diagram for position control of the servo table is given by:
by:
Position
Controller θD 0.09 θDV KP θV Plant Gp(s)
Ei 168
s(0.3s + 1) θ 0.09 See how the closedloop poles move as proportional gain KP varies from 0 to ∞.
closedFind closedloop characteristic equation:
closed School of Mechanical Engineering
Purdue University ME375 Root Locus  2 2 ME375 Handouts Servo Table Example (cont.) School of Mechanical Engineering
Purdue University ME375 Root Locus  3 3 ME375 Handouts Motivation – Example 1
Revisit the DC motor positioning system with proportional control.
control.
corresponding block diagram is:
is: Its Plant Gp(s)
R(s) + Controller
KP − U(s) Y(s)
27.6
s( s + 57.5) Sketch the closedloop poles as the controller gain KP varies from 0 to ∞.
closedFind closedloop characteristic equation:
closed s 2 + 57.5 s + 27.6 K P = 0 School of Mechanical Engineering
Purdue University ME375 Root Locus  4 4 ME375 Handouts Example 1
Formulate an expression for the roots of the characteristic equation:
equation: Find the roots for KP = 0 and KP → ∞: Find KP when the roots are repeated.
repeated. School of Mechanical Engineering
Purdue University ME375 Root Locus  5 5 ME375 Handouts Example 1
Sketch the root locus:
Img. Axis
30 20 10 Real Axis 0
60 50 40 30 20 10 0
10 20 30 School of Mechanical Engineering
Purdue University ME375 Root Locus  6 6 ME375 Handouts Example 2
Using the same plant as in Example 1, try a different controller choice:
choice:
Plant Gp(s)
R(s) + Controller
Kd(s + 80) − U(s) Y(s)
27.6
s( s + 57.5) Sketch the root locus of the closedloop poles as the controller gain Kd varies from
closed0 to ∞.
Find closedloop characteristic equation:
closed School of Mechanical Engineering
Purdue University ME375 Root Locus  7 7 ME375 Handouts Example 2
Formulate an expression for the roots of the characteristic equation:
equation: Find the roots for Kd = 0 and Kd → ∞: Find repeated roots.
roots. School of Mechanical Engineering
Purdue University ME375 Root Locus  8 8 ME375 Handouts Example 2
Repeated roots (cont.): School of Mechanical Engineering
Purdue University ME375 Root Locus  9 9 ME375 Handouts Example 2
Sketch the root locus: Imag Axis
40
30
20
10
Real Axis 0
10
20
30
40
140 120 100 80 60 40 20 School of Mechanical Engineering
Purdue University 0
ME375 Root Locus  10 10 ME375 Handouts ClosedClosedLoop Characteristic Roots (CL Poles)
Reference
Input
R(s) + Error − E(s) Control
Input KP U(s) Output Gp(s) Y(s) Plant H(s)
The closedloop transfer function GCL(s) is:
closed Img. jω GCL ( s ) =
The closedloop characteristic equation is:
closed Real − jω
School of Mechanical Engineering
Purdue University ME375 Root Locus  11 11 ME375 Handouts Definitions
• Root Locus
Root Locus plotting is the method of determining the roots of the following
equation on the complex plane when the parameter K varies from 0 → ∞:
1 + K ⋅ G OL ( s ) = 0 or 1+ K ⋅ N (s)
=0
D (s) where N(s) and D(s) are known polynomials in factorized form:
N ( s ) = ( s − z1 )( s − z 2 ) L ( s − z N z )
D ( s ) = ( s − p1 )( s − p2 ) L ( s − p N P ) The NZ roots of the polynomial N(s) , z1 , z2 , …, zNz , are called the finite openopenloop zeros.
The NP roots of the polynomial D(s) , p1 , p2 , …, pNp , are called the finite
openopenloop poles.
School of Mechanical Engineering
Purdue University ME375 Root Locus  12 12 ME375 Handouts Root Locus
Methods of obtaining root locus:
• Given a value of K, numerically solve the 1 + K GOL(s) = 0 equation for a set
of roots. Repeat this for a set of K values and plot the corresponding roots on
roots.
the complex plane. (This is what we did in the last inclass exercise.)
plane.
in• Use MATLAB. In MATLAB use the commands rlocus and rlocfind.
rlocus
rlocfind
You
You can use online help to find the usage for these commands.
on1 + K P ⋅ 0 .03 ⋅ 16
=0
s ( 0 .0174 s + 1) ⇒ 1 + KP ⋅ 0 . 48
=0
0 .0174 s 2 + s >> op_num=[0.48];
>> op_den=[0.0174 1 0];
>> Gol=tf(op_num,op_den);
>> rlocus(Gol)
rlocus(Gol)
>> [K, poles]=rlocfind(Gol); • Apply root locus sketching rules to obtain an approximate root locus plot.
approximate
School of Mechanical Engineering
Purdue University ME375 Root Locus  13 13 ME375 Handouts Root Locus Sketching Rules
Reading material: Palm book 11.111.2
material:
11. 11.
Motivation: In the design of feedback controllers, a qualitative understanding is needed on
the influence of control parameters on the output performance of the controlled system.
We have seen earlier the role played by the poles of the closedloop transfer functions
closed(CLFTs) on the transient and steadystate response of the output response. The influence
steadyof control parameters on the placement of the CLTF poles in the complex plane can been
control parameters on the placement of the CLTF poles in the complex plane can been
seen by directly solving the characteristic equation of the CLTF. However, this process
does not allow for a qualitative understanding.
Here, we will use a set of rules for sketching the socalled root locus plots based on
soinformation known about the zeros and poles of the openloop system. From this, we can
enbetter understand the trends of parameter influence on response and use this in the design
of control systems. School of Mechanical Engineering
Purdue University ME375 Root Locus  14 14 ME375 Handouts Rules for sketching root locus
1. The number of branches in the root locus plot equals the number of openloop poles;
openpoles;
2. The locus starts (K=0) at an openloop pole, and as K → ∞ , m branches ends at open
(K=0
openloop zeros, and nm branches will end at infinity and asymptotically approach
straight lines. (n: number of openloop poles; m: number of open loop zeros)
lines. (n:
openpoles;
3. The locus is symmetric about the real axis.
axis.
4. The locus, on the real axis, includes all segments left of an odd number of open loop
poles and zeros.
zeros.
K → ∞ , nm branches will approach straight lines (asymptotes). These straight lines
5. As
intersect the real axis at σ 0 with an angle θ k , where:
where:
σ0 = ∑ pi − ∑ zi
n−m , ( pi : OL poles; zi : OL zeros.) θ k = (2k − 1) 180o
, k = 0,1,..., n − m − 1
n−m 6. A locus leaves (“break away”) or enters (“breaks in”) the real axis at values of K for
which repeated roots exist in the closed loop characteristic equation. These points
and values of K at these points satisfy:
satisfy:
dK
=0
ds
School of Mechanical Engineering
Purdue University ME375 Root Locus  15 15 ME375 Handouts Rules for sketching root locus
7. The departure angle for an OL pole (or the arrival angle for an OL zero) can be
calculated from:
rom:
φl ,dep = ∑ψ i − ∑ φi − 180o
i ≠l ψ l ,arr = ∑ φi − ∑ψ i + 180o
i ≠l where φi is the angle formed by the positive direction of the real axis and the vector
connecting another pole to the pole(or zero) of interest; ψ i is the angle formed by
interest;
the positive direction of the real axis and the vector connecting another zero to the
pole(or zero) of interest.
interest.
8. If the locus passes through the imaginary axis (boundary between stable and unstable
responses), the crossing point and corresponding values of K can be found by:
 Replacing s by jω in the closedloop characteristic equation, and
closed Setting the real and imaginary parts to zero and solving for K and jω
School of Mechanical Engineering
Purdue University ME375 Root Locus  16 16 ME375 Handouts Steps for sketching root locus
Step 1. Determine the closedloop transfer function;
closedfunction;
Step 2. Express the characteristic equation for the closedloop system in the standard form:
closedorm:
1+ K N (s)
=0
D( s) Step 3. Determine the openloop poles and zeros using, respectively: D(s)=0 and N(s)=0.
openrespectively: D(s)=0
N(s)=0
Mark the openloop poles with an “x” and zeros with an “o” in the complex plane.
openplane.
Step 4. Apply Rule 4.
Step 5. Find the number of asymptotes and the corresponding values of σ 0 and θ k . (Rule 5)
Step 6. If necessary, find the breakaway and breakin points. (Rule 6)
breakbreak points.
Step 7. If necessary, find the departure and arrival angles. (Rule 7)
angles.
Step 8. Sketch the root locus.
locus. School of Mechanical Engineering
Purdue University ME375 Root Locus  17 17 ME375 Handouts Example 3
A feedback control system is proposed. The corresponding block diagram is:
proposed.
is:
Plant Gp(s) Controller
R(s) +
− K
(s + 4) U(s) 1
s(s + 2) Y(s) Sketch the root locus of the closedloop poles as the controller gain K varies from
closed0 to ∞.
Find closedloop characteristic equation:
closed School of Mechanical Engineering
Purdue University ME375 Root Locus  18 18 ME375 Handouts Example 3
Step 1: Formulate the (closedloop) characteristic equation into the standard form for
(closedsketching root locus:
ocus: Step 2: Find the openloop zeros, zi , and the openloop poles, pi :
openopen School of Mechanical Engineering
Purdue University ME375 Root Locus  19 19 ME375 Handouts Example 3
Step 3: Determine locations of repeated roots, if any.
any. School of Mechanical Engineering
Purdue University ME375 Root Locus  20 20 ME375 Handouts Example 3
Step 4: Determine the imaginary axis crossings, if any. School of Mechanical Engineering
Purdue University ME375 Root Locus  21 21 ME375 Handouts Example 3
Step 5: Use the information from Steps 14 to sketch the root locus.
1Imag Axis 4
3
2
1
Real Axis 0
1
2
3
4
6 5 4 3 2 1 School of Mechanical Engineering
Purdue University 0
ME375 Root Locus  22 22 ME375 Handouts Example 4
The transfer function for a simple springmassdashpot oscillator is known to be:
springmassG( s) = 1
Ms + Cs + K
2 Using the root locus rules, sketch the root locus for this system:
a. in terms of the stiffness parameter K with M = 2 and C = 8 .
with
b. in terms of the damping parameter C with M = 2 and K = 8 .
with School of Mechanical Engineering
Purdue University ME375 Root Locus  23 23 ME375 Handouts Example 4
a. Root locus for varying K School of Mechanical Engineering
Purdue University ME375 Root Locus  24 24 ME375 Handouts Example 4
b. Root locus for varying C School of Mechanical Engineering
Purdue University ME375 Root Locus  25 25 ME375 Handouts Example 4
b. Root locus for varying C School of Mechanical Engineering
Purdue University ME375 Root Locus  26 26 ME375 Handouts Revisit PID Control Examples
Disturbance
D(s)
Reference
Input
R(s) Controller + Error − E(s) Control
Input GC (s) Plant GP (s) + + 4
( 2 s + 1)(0.5s + 1) U (s ) Output
Y(s) Sensor H (s) 1 (A) Proportional (P) control: GC (s) = KP
Proportional GYR ( s ) = 4K P
(2 s + 1)(0.5s + 1) + 4K P (B) Add Integral action, ⇒ ProportionalPlusIntegral (PI) control:
Add
ProportionalPlusGC ( s ) = K P + K I 1 KPs + KI
=
s
s GYR ( s ) = 4( KPs + KI )
s(2 s + 1)(0.5s + 1) + 4 ( K P s + K I ) (C) Add Derivative action, ⇒ ProportionalPlusIntegralPlus Derivative (PID) control:
Add
ProportionalPlusIntegralK s2 + K P s + K I
1
GC ( s ) = K P + K I + K D s = D
s
s GYR ( s ) = School of Mechanical Engineering
Purdue University 4( KDs2 + KPs + KI ) s (2s + 1)(0.5s + 1) + 4 ( K D s 2 + K P s + K I ) ME375 Root Locus  27 27 ME375 Handouts Effect of Proportional Control
Closed Loop Characteristic Equation: s 2 + 2.5s + (1 + 4K P ) = 0 Rewrite the characteristic equation:
OpenOpenloop poles and zeros: 5
4 Repeated roots: Imaginary part of s 3
2
1
0
1
2
3
4
5
8 7 6 5 4 3 2 1
Real part of s
School of Mechanical Engineering
Purdue University 0 1 2 ME375 Root Locus  28 28 ME375 Handouts Effect of Proportional Control
Root Locus for K P
Locus for
1.5 Imaginary Axis 1
0.5
0
0.5
1
1.5
2.5 2 1.5 1
0.5
Real Axis School of Mechanical Engineering
Purdue University 0 0.5
ME375 Root Locus  29 29 ME375 Handouts Effect of PI Control
Closed Loop Characteristic Equation: s 3 + 2.5s 2 + (1 + 4K P ) s + 4 K I = 0
Rewrite the characteristic equation:
OpenOpenloop poles and zeros: 5
4 Repeated roots:
roots Imaginary axis crossings: Imaginary part of s 3
2
1
0
1
2
3
4
5
8 7 6 5 4 3 2 1
Real part of s
School of Mechanical Engineering
Purdue University 0 1 2 ME375 Root Locus  30 30 ME375 Handouts Effect of PI Control
Locus or
RootRoot LocusforKK I
Locus ffor I
Locus for Imaginary A xis 30 20 Imaginary Axis 30 20 10 10 0 0 10 10 20
30
30 20 30
30 20 20 10 10
0
0
Real Axis 10 10 20 20 30 Real Axis
School of Mechanical Engineering
Purdue University ME375 Root Locus  31 31 ME375 Handouts Effect of PID Control
Closed Loop Characteristic Equation: s 3 + ( 2.5 + 4 K D ) s 2 + (1 + 4K P ) s + 4 K I = 0 Rewrite the characteristic equation:
OpenOpenloop poles and zeros: 5
4 Repeated roots:
Repeated roots Imaginary axis crossings: Imaginary part of s 3
2
1
0
1
2
3
4
5
8 7 6 5 4 3 2 1
Real part of s
School of Mechanical Engineering
Purdue University 0 1 2 ME375 Root Locus  32 32 ME375 Handouts Effect of PID Control
Root Locus for K D
Locus for Imaginary Axis 5 0 5 8 6 4
Real Axis 2 School of Mechanical Engineering
Purdue University 0
ME375 Root Locus  33 33 ME375 Handouts Controller Design using Root Locus
Objective:
Design a system that has zero stead state error for step inputs with %OS < 10%
dy
%OS 10%
and TS (2%) < 6 [sec].
(2%)
[sec].
Disturbance
D(s)
Reference
Input Controller + R(s) Error Control
Input GC (s) − E(s) Plant GP (s) + U (s ) + Output 4
( 2 s + 1)(0.5s + 1) Y(s) Sensor H (s) 1 GC ( s ) = K P + K I 1 KPs + KI
=
s
s Closed loop transfer functions:
4
⎛ KPs + KI ⎞
⎜
⎟⋅
4( KPs + KI )
GC ( s ) ⋅ GP ( s )
s
⎝
⎠ (2s + 1)(0.5s + 1) =
=
GYR ( s ) =
4
1 + GC ( s ) ⋅ GP ( s ) ⋅ H ( s )
s (2 s + 1)(0.5s + 1) + 4 ( K P s + K I )
⎛ K s + KI ⎞
1+ ⎜ P
⎟⋅
s
⎝
⎠ (2 s + 1)(0.5s + 1)
School of Mechanical Engineering
Purdue University ME375 Root Locus  34 34 ME375 Handouts Design PI Control
Closed Loop Characteristic Equation:
Loop Characteristic Equation s (2 s + 1)(0.5s + 1) + 4 ( K P s + K I ) = 0 let
let ζ = 0.707 and ζωn = 1
and ζω s 3 + 2.5s 2 + (1 + 4K P ) s + 4 K I = 0 School of Mechanical Engineering
Purdue University ME375 Root Locus  35 35 ME375 Handouts Step Response with PI Control
Step Response
1.4
1.2 Amplitude 1
0.8
0.6
0.4
0.2
0
0 2 4
6
Time (sec) School of Mechanical Engineering
Purdue University 8 10
ME375 Root Locus  36 36 ME375 Handouts PID Control Design
Objective:
Design a system that has zero stead state error for step inputs with %OS < 10%
dy
%OS 10%
and TS (2%) < 6 [sec].
(2%)
[sec].
Disturbance
D(s)
Reference
Input Controller + R(s) Error − E(s) Control
Input GC (s) Plant GP (s) + U (s ) + 4
( 2 s + 1)(0.5s + 1) Output
Y(s) Sensor H (s) 1 GC ( s ) = K P + K I K s2 + K P s + K I
1
+ KDs = D
s
s Closed loop transfer functions:
K Ds2 + KP s + K I
4
⋅
4( KDs2 + KPs + KI )
GC ( s ) ⋅ GP ( s )
s
(2 s + 1)(0.5s + 1)
GYR ( s ) =
=
=
K s2 + K Ps + K I
4
1 + GC ( s ) ⋅ GP ( s ) ⋅ H ( s )
s (2s + 1)(0.5s + 1) + 4 ( K D s 2 + K P s + K I )
1+ D
⋅
s
(2s + 1)(0.5s + 1)
School of Mechanical Engineering
Purdue University ME375 Root Locus  37 37 ME375 Handouts Design PID Control
Closed Loop Characteristic Equation:
Loop Characteristic Equation s (2 s + 1)(0.5s + 1) + 4 ( K D s + K P s + K I ) = 0 let
let ζ = 0.707, ζωn = 1, and a = 5
ζω
and 2 s 3 + ( 2.5 + 4 K D ) s 2 + (1 + 4K P ) s + 4 K I = 0 School of Mechanical Engineering
Purdue University ME375 Root Locus  38 38 ME375 Handouts Step Response with PID Control
Step Response
1.4
1.2 Amplitude 1
0.8
0.6
0.4
0.2
0
0 2 4
6
Time (sec) School of Mechanical Engineering
Purdue University
8 10
ME375 Root Locus  39 39 ME375 Handouts Additional Examples School of Mechanical Engineering
Purdue University ME375 Root Locus  40 40 ...
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 Fall '10
 Meckle
 Mechanical Engineering

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