Download - University of Virginia Proportional Control Spring 2015 Jack Stankovic University of Virginia
University of Virginia
Proportional Control
Spring 2015
Jack StankovicUniversity of Virginia
University of Virginia
Outline
• Definition and Value of Proportional Control
• (Review) Desirable Properties of a Controller
• Analyzing Proportional Control– Root Locus
• P-Controller Design / Pole Placement Design
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Proportional Control
• Control Law– It quantifies how to set control input to the target system
• Proportional Control Law
u(k) = Kpe(k)
• Kp is a constant that is chosen when designing the proportional controller
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Kp G(z)R(z) E(z) Y(z)+
-
+ +U(z)
D(z)
Proportional Control
Feedback ControlUsing P Control
(set the control proportionalto the error)
Controller Target System
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43.0
47.0
)(
)()(
zzU
zYzG
Note: Simulation is in time domain
System youwant to control
Systemstarts atzero
Goal
Steady State GainG(1) = y(ss)/u(ss)
and if goal is y(ss) = 10=> u(ss) = 12
OPENLOOP
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Why is u(ss) = 12
• G(1) = y(ss)/u(ss)
• G(1) = .47/(1-.43) = .82
• y(ss)/u(ss) = 10/u(ss) = .82 (goal is 10)
• 12 = u(ss)
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Disturbance
Outputnot atdesired Value
y(ss) = 26.3
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Why is y(ss) = 26.3
• G(1) = y(ss)/[u(ss) + d(ss)]
• .82 = y(ss)/(12 + 20) * recall from before .47/(1-.43) = .82
• .82 * 32 = y(ss)
• 26.3 = y(ss)
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Let Kp = 2
Just guessTrial and Error
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Output
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Proportional Controller
Kp G(z)
R(z) E(z) T(z)+
-
+ +U(z)
D(z)
Feedback using Proportional Control in the presence of measurement noise
+ +
N(z)
Y(z)
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Simulation of a P Controller with disturbance and noise
G(z)=0.47/(z-0.43), and Kp=2
SteadyStateError
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Advantages of Proportional Control
•Does not require precise analytical model of the system being controlled
•Simple implementation
•Proper for applications with simple requirements (Overshoot, settling time, stability)
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Deficiency of Proportional Control
• May cause steady-state error to be nonzero
– Disturbance input is non zero– Noise input
• May cause oscillations
• Generally, one or more of the requirements for response time, overshoot, and oscillation may be impossible to fulfill at any proportional gain setting
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Review - Desirable Properties
of Controllers•SASO
•S: Stability•A: Accurate, steady-state error•S: Settling Time•O: Maximum Overshoot
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Desirable Properties of Controllers
• SASO Properties in the step response of a closed-loop system.
Settling time
Overshoot
Controlledvariable
Time
Reference
%
Steady StateTransient State
Steady state error
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Analyze Properties of Controllers
•Stability
–For any bounded input over any amount of time, the output will also be bounded. “BIBO”
– In P-Control, stability is assessed by determining if the poles of the closed-loop transfer function have a magnitude less than 1 (pole must be in the unit circle)
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Analyze Properties of Controllers
•Accuracy
– Quantified by the magnitude of the steady state control error.
– In P-Control, accuracy is assessed by computing the steady-state gain.
»There is a zero steady-state error if and only if this gain is 1.
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Analyze Properties of Controllers
•Accuracy
ssssk
ss yrkykre
)()(lim
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Analyze Properties of Controllers
•Accuracy
•Note also that steady-state error can be reduced by using a larger Kp.
))1(1)(1(
1
ifonly
0
HGK
e
p
ss
(see handoutfor derivation)
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Analyze Properties of Controllers
•Settling Time
– Settling time is a function of the closed-loop poles and is estimated using the following equation by employing the dominant pole approximation.
pks log
4
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Analyze Properties of Controllers
•Overshoot
– Overshoot is a property of the response to a step change in the reference input.
– Overshoot is also suggestive of oscillatory behavior since a large overshoot is typically followed by a large undershoot.
– Analyzed by an approximation
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Analyze Properties of Controllers
•Maximum Overshoot - Mp
•Mp is used to denote the maximum overshoot to a unit step input.
•Mp = |y(max) – y(ss)| / |y(ss)|
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Maximum Overshoot
•For the first order STABLE system–By definition Mp=0 if y(ss) > y(max)
–Mp=|p1| otherwise
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Analyze Properties of Controllers
•Second order STABLE system (Approx.)
• if real dominant pole p1>=0
• if real dominant pole p1<0• if dominant poles
0pM
1pM p /rM p
jrepp 21,
Note: Poles are in unit circle if stable
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Analyze Properties of Controllers
K(z) G(z)
R(z) E(z) T(z)+
-
+ +U(z)
D(z)
Closed-loop system
+ +
N(z)
Y(z)
H(z)
W(z)
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Analyze Properties of Controllers
•Closed-loop Transfer Functions
)()()(1
)()(
)(
)()(
zHzGzK
zKzG
zR
zYZFR
)()()(1
)(
)(
)()(
zHzGzK
zG
zD
zYZFD
)()()(1
1
)(
)()(
zHzGzKzN
zTZFN
)()()()()()()( zNzFzDzFzRzFZT NDR
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Method to analyze p-controller
• From the example, three transfer functions have the same poles, which are the solutions to the characteristic equation
• To visualize the relation between Kp and closed-loop poles, we introduce the root locus of the system
0)()()(1 zHzGzK
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Root Locus
• The root locus is a graphical procedure for determining the poles of a closed-loop system given the poles and zeros of a forward-loop system.
• Graphically, the locus is the set of paths in the complex plane traced by the closed-loop poles as the root locus gain is varied from zero to infinity.
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Root Locus
• In p-control, the root locus is the locations of all possible roots of the characteristic equation as Kp varies from zero to infinity
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• IBM Lotus Domino Server with measurement delay
K(z) G(z)
R(z) E(z)+
-
U(z)
P control with Measurement Delay
Y(z)
H(z) = Z-
2
W(z)
Example
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Example
• IBM Lotus Domino Server with measurement delay
• 43.0
47.0
)(
)()(
zzU
zYzG
2
1
)(
)()(
zzY
zWzH
Pole at .43
Two poles atzero
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43.0
47.0
)(
)()(
zzU
zYzG H(z) = 1/z*z
Example of Root Locus for 3 Pole SystemGiven Below
Two polesAt zero
PoleAt.43
1 + K(z)G(z)H(z)
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Analyze Properties of Controllers
•In P-Control, the closed-loop system is stable at the values of Kp that make all the poles lie inside the unit circle.
•Any system that has at least one zero at infinity will always become unstable for a large enough Kp.
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Root Locus/Matlab
•Specify transfer function>>G = tf(np,dp)
where np = numerator polynomial
dp = denominator polynomial
•Plot root locus figure of transfer function G>>rlocus(G)
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Design of Proportional Controllers• So far, we analyzed properties of a P
controller based on a specific value of Kp (like Kp=2) or found all values of Kp that make system stable.
• Instead, design the controller, choosing Kp, so that the resulting system has desirable properties. (SASO)
• Pole Placement Design:– Select controller parameters to get desired closed-loop
poles.
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Pole Placement Design• Step 1. Determine the required SASO
properties of the closed-loop system. • Step 2. Construct the closed-loop transfer
function as a function of the proportional gain KP.
• Step 3. Solve for the closed-loop poles in terms of KP.
• Step 4. Derive KP from the equation that modeled poles are equal to the desired poles. (Take intersection of solutions)
• Step 5. Check Robustness
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Example for Pole Placement Design
• For Proportional Control of the IBM Locus Domino Server
• Step 1. Decide Design Goals
• Stability
• ess<0.1
• ks<10
• MP<0.1
43.0
47.0
)(
)()(
zzU
zYZG
Note: May not be satisfied! (new control needed or relax requirements)
Recall: obtainedby system ID
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Pole Placement• Step 2. Closed loop transfer function.
P
PR Kz
K
zR
zYzF
47.043.0
47.0
)(
)()(
This is first order so one pole.
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Pole Placement
• Step 3. Consider 4 goals separately.
147.043.01 PKp
1.047.043.01
47.01)1(1
P
PRss K
KFe
1047.043.0log
4
P
s Kk
1.047.043.0 PP KM
Must be in unit circle
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Pole Placement
• Step 4. Solve for the conditions on Kp – one at a time
• Stability: -1.21 < KP < 3.0
147.043.01 PKp
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Step 4 continued
• Steady State error: ess< 0.1: KP > 10.9
• Recall – this is derived from the final value theorem and only valid if system is stable AND KP must be less than 3.0 to be stable BUT greater than 10.9 for this error
• Hence, this e(ss) cannot be satisfied
1.047.043.01
47.01)1(1
P
PRss K
KFe
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Step 4 continued
• Setting time requirement ks< 10• Depends on magnitude of largest closed
loop pole• With absolute value we get two
inequalities
-0.51 < KP < 2.3
1047.043.0log
4
P
s Kk
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Step 4 continued
• Overshoot requirement MP < 0.1: KP < 1.1
1.047.043.0 PP KM
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Step 4 continued
•Overall Solution:
– Intersection of equations from 1 to 4
– Bound on steady state error is not possible
– But if -0.51 < KP < 1.1 (meets other requirements)
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Step 5: Robustness of P-Control System
• A controller is robust if its behavior does not change much if there are errors in estimating parameters of the target system.
• For stability,
1 bKa P
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• Robustness of P-Control to errors in estimating the open-loop pole and gain of a first-order target system with a=0.43, b=0.47.
Step 5: Robustness
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Summary
•Basics of P Control•How to analyze properties of a P
controller (for a given Kp)•How to use pole placement design to
select the right parameter for the P controller to meet requirements
•Introduction to Root Locus
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Extra Slides
•These following slides were not presented in class
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Design of Proportional Controllers• Another example
• IBM Lotus Domino Server with a Moving-Average Filter
• The stochastic nature of resource consumption in computing systems typically results in substantial variability, especially for performance metrics such as response times, queue lengths, and utilizations.
• So a MOVING-AVERAGE FILTER is used to mitigate the effect of the stochastics.
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Moving-Average Filter• The input of the moving-average filter is the
“raw” signal and its output is a smoothed signal. • Y(k) is the unfiltered signal and w(k) is the
filtered signal.• w(k+1) = cw(k) + (1-c)y(k)• The filter has a transfer function
cz
c
zY
zWZH
1
)(
)()(
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Moving-Average Filter•w(k+1) = cw(k) + (1-c)y(k)
•C is a constant, 0 <= c < 1•If c = 0, the output has the same
time-domain characteristics as the input signal, although it is delayed by one time step.
•Value of c greater than 0 specifies the degree of smoothing.
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Moving Average Filter• Effect of a moving average filter on a sine wave
for different values of the filter parameter c.
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Design of Proportional Controllers
KpR(z) E(z) Y(z)+
-
U(z)
Moving Average Filter used in IBM Lotus Domino Server
az
b
Controller Lotus Server
W(z) cz
c
1
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Design of Proportional Controllers
•Two parameters Kp and c need to be set
•We plot the relations between Kp/c and desirable properties of controllers.
P
PR Kcczz
cK
zR
zYZF
)1(47.0))(43.0(
)1(47.0
)(
)()(
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Design of Proportional Controllers• Effect of a moving average filter constant c and
the proportional gain Kp
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Design of Proportional Controllers• From the plots
• c does not affect steady-state error,• A large kp decreases ess
• When c is small, ks increases when kp increases; when c is large, ks decreases as kp increases.
• When c is small, Mp increases when kp increases; when c is large, Mp decreases as kp increases.
• Finally, c = 0.95 and Kp = 2 is good!
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Limit Cycle
•How can CPU utilization be unbounded?– It must vary somewhere between zero and one (so
not unbounded)– Implies – can’t be unstable???
– For these types of physical systems where the “physics” of the system do not permit BIBO stability issues
» Instability is manifested as a limit cycle CPU utilization bounces between zero and
one