AUTOMATIC CONTROL SYSTEMSprofsite.um.ac.ir/~karimpor/autocontrol/lec2_aut_con.pdf · •Regulatory...

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Ali Karimpour Associate Professor Ferdowsi University of Mashhad References: 1- Modern Control Technologies: Components and Systems, 2 nd Edition, by Kilian, Delmar Publication Co, ٢٠٠۵ AUTOMATIC CONTROL AUTOMATIC CONTROL SYSTEMS SYSTEMS

Transcript of AUTOMATIC CONTROL SYSTEMSprofsite.um.ac.ir/~karimpor/autocontrol/lec2_aut_con.pdf · •Regulatory...

Page 1: AUTOMATIC CONTROL SYSTEMSprofsite.um.ac.ir/~karimpor/autocontrol/lec2_aut_con.pdf · •Regulatory control despite the presence of disturbances •Supervisory control for SP b) Servomechanism

Ali KarimpourAssociate Professor

Ferdowsi University of Mashhad

References:1- Modern Control Technologies: Components and Systems, 2nd Edition, by Kilian, Delmar Publication Co, ٢٠٠۵

AUTOMATIC CONTROLAUTOMATIC CONTROLSYSTEMSSYSTEMS

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2

Lecture 2

Feedback Control PrincipalsFeedback Control PrincipalsTopics to be covered include: Introduction Performance Criteria On-off Controllers PID Controllers PIP Controllers Fuzzy Logic Controllers

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Introduction

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a) Heating systems

• Regulatory control despitethe presence of disturbances

• Supervisory control for SP

b) Servomechanism systems

• Tracking control

• More accurate model

c) Fuzzy logic control: A new and increasingly important type of control that does not use mathematical models but mimic the skill and experience of a human operator.

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Performance Criteria

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Performance criteria are various measurable parameters that indicate how good (or bad) the control system is.

• Transient (moving) parameters• Steady-state (not changing) parameters.

Transient response of a robot.

• Rise time (T)

• Overshoot and P.O.

• Settling time

• Steady-state error

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On-off Controllers

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Two-point control: (also called on–off control) is the simplest type of closed-loop control strategy.

Mostly suitable on slow-moving systems where it is acceptable for the controlled variable to move back-and-forth between the two limit points.

Reducing Tcyc?

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On-off Controllers

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Three-position control: is similar to two-point control, except in this case the controller has three states, such as forward–off–reverse

Typical Operation:

Strong Motors.

Weak Motors.

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PID Controllers

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We now consider more sophisticated control strategies that require “smart” controllers that use op-amps or a microprocessor.

• Proportional control

Explain the behavior if deg/2 VoltsK p

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PID Controllers

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Proportional control is • simple, • makes sense, and• the basis of most control systems, but • has one fundamental problem“steady-state error”.

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deg/2 VoltsK p

PID Controllers

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Example: Derive Dead bandof system if the motor need 6 volts to rotates (since of friction).

Clearly dead band is 6 deg.

Other sources of dead band?

Compensation methods?

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PID Controllers

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Example: Specify the controller required to position the robot arm to within 5 deg of the set point.

Total friction and gravity will beLess than 50 in. oz in the load side.

A 350 Deg. Pot is used.

Controller output is in A.

Answer:oz in.55 AoutputController 2.0

VAVAgainController /38.1/029.01oz in./

251oz/ in. 1

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PID Controllers

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Example: In the previous example Suppose set point is set by 0.87 VAnd the arm is at 0 deg.Specify the situation.

Answer:

87.0isError AoutputController 2.138.187.0 side) oz(load in.300side)oz(motor in.302.125 istorqueMotor

145.0029.02587.025 isatError

side) oz(load in.50side)oz(motor in.5)145.038.1(25 istorqueMotor

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PIP Controllers

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BiasOne way to deal with the gravity problem is to have the controller add in a constantvalue (to its output) that is just sufficient to support the weight.

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PIP Controllers

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Another Example of P controller

The flow sensor provides an output signal of 0-5 V, which correspond to 0-10 gal/min.

The flow valve is operated with a signal of 0-5 V, where 0 V corresponds to completelyclosed and 5 V is all the way open.

Design a 50% proportional band and set the controller value.

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PIP Controllers

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Another Example of P controller

Design set point for a 6 gal/min output.

Full opening of valve leads to 10 gal/min so : set point = 0.6*5v=3v.

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PIP Controllers

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Another Example of P controller

Exercise1: a) Draw block diagram of above system.b) Find the value of output flow in gal/minc) Set the value of set point such that output be exactly 6 gal/min

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PIP Controllers

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Bias

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deg/2 VoltsK p

PID Controllers

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PI Controller

How to remove steady state error?

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PID Controllers

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PI ControllerThe proportional feedback system is (KP = 10 in. · oz/deg) and has been modified to include integral feedback.

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PID Controllers

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PI Controller

Integral control may cause overshoot and oscillations.

One solution to the overshoot problem is to include derivative control. Derivative control“applies the brakes,” slowing the controlled variable just before it reaches its destination.

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PID Controllers

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Derivative Control

One solution to the overshoot problem is to include derivative control.

Derivative control “applies the brakes,” slowing the controlled variable just before it reaches its destination.

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PID Controllers

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PID Control

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PIP Controllers

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PI controller

Exercise2: a) Draw block diagram of above system with PI controller.b) Find the value of output flow in gal/minc) Find the output of Integral part of controller.

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Tuning of PID Controllers

Because of their widespread use in practice, we present below several methods for tuning PID controllers. Actually these methods are quite old and date back to the 1950’s. Nonetheless, they remain in widespread use today.In particular, we will study.

Ziegler-Nichols Oscillation Method Ziegler-Nichols Reaction Curve Method Cohen-Coon Reaction Curve Method Time domain design Frequency domain design

PIDتنظیم کنترلرهاي

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Ziegler-Nichols Reaction Curve Method(Open-Loop Case)

For open-loop tuning, we first find the plant parameters by applying a step input to the open-loop system.

The plant parameters K, TD and T1 are then found from the result of the step test as shown in Figure.

حالت حلقه باز نیکولز زیگلرطراحی

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PIDKTT19.0

Kp Ki Kd

2127.0

DKTT

PIDDKTT12.1

P

حالت حلقه باز نیکولز زیگلرطراحی

DKTT1

216.0

DKTT

KT16.0

Ziegler-Nichols Reaction Curve Method(Open-Loop Case)

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Numerical Example

Example4: Consider step response of an open-loop system as:

مثال عددي

sesGTTCK

DsT

D 20140)(sec20sec,5,40 :So 1

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PI 09.09.0 1 DKTT

Kp Ki Kd

0054.027.021

DKTT

PID 12.02.1 1 DKTT

P 1.01 DKT

T

012.06.021

DKTT

3.06.0 1 KT

sesGTTCK

DsT

D 20140)(sec20sec,5,40 :So 1

1.0)( sKP

ssKPI

0054.009.0)(

ss

sKPID 3.0012.012.0)(

Numerical Example مثال عددي

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Ziegler-Nichols Oscillation Method(Closed-loop)

This procedure is only valid for open loop stable plants and it is carried out through the following steps

Set the true plant under proportional control, with a very small gain.

Increase the gain until the loop starts oscillating. Note that linear oscillation is required and that it should be detected at the controller output.

Record the controller critical gain Kc and the oscillation period of the controller output, T.

Adjust the controller parameters according to Table

)حلقه بسته(نوسانی بروش نیکولز زیگلرطراحی

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PI cK45.0

Kp Ki Kd

TKc54.0

PID TKc2.1 TKc075.0cK6.0

P cK5.0

Ziegler-Nichols Oscillation Method(Closed-loop))حلقه بسته(نوسانی بروش نیکولز زیگلرطراحی

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Example5: Consider a plant with a model given by

Find the parameters of a PID controller using the Z-N oscillation method. Obtain a graph of the response to a unit step input reference.

30

Numerical Exampleمثال عددي

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Solution

Applying the procedure we find:Kc = 8 and ωc = 3. T=3.62

Hence, from Table, we have

The closed loop response to a unit step in the reference at t= 0 is shown in the next figure.

حل

17.2075.065.22.18.46.0 TKKTKKKK cdc

icp

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Response to step reference

0 5 10 150

0.5

1

1.5Step response for PID control

Time (sec)

Ampl

itude

پاسخ سیستم به پله

ss

sCPID 17.265.28.4)(

117.201.017.265.28.4)(

ss

ssCPID

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PID Controllers

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Stability

A stable system is one where the controlled variablewill always settle out at or near the set point.

An unstable system is one where, under some conditions, the controlled variable drifts awayfrom the set point or breaks into oscillations that get larger and larger until the system saturates on each side.

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PID Controllers

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Stability

Reason for instability

• Phase lag caused by dead time or backlash.

• Positive feedback.

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Digital PID Controller

or

Digital PID Controllers

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Digital PID Controller

or

Digital PID Controllers

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Digital PID Controllers

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Sampling rate

In a digital control system, sample rate is the number of times per second a controller reads in sensor data and produces a new output value.Shannon’s sampling theorem ?Aliasing

Digital PID Controllers

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Sampling rate

Digital PID Controllers

In practice, a sampling rate of at least ten times the highest frequency in the system is usually sufficient.In most systems, sampling is done once at the beginning of each pass through the program loop (that is, one sample for each iteration).

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Sampling rate

Digital PID Controllers

Example5: A microprocessor-based control system runs at a clock speed of 1 MHz.

• 55 instructions with an average execution time of 4 clocks/instruction

• 8-bit ADC with a 100-μs conversion time

What is the maximum sample rate?

Maximum sample rate is: 3.125 kHz

Is the microprocessor ok for this system?

Yes

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PIP Controllers

The set point has been defined as the place where you want the controlled variable to be.

In a dynamic system, such as a robot arm, the desired position is a moving target, in which case we are concerned with path control.

There are two ways to implement path control:

• Carrot-and-horse

• Feedforward, or PIP approach

A Proportional + Integral + Preview (PIP) controller is a system thatIncorporates information of the future path in its current output.

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PIP Controllers

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A Proportional + Integral + Preview (PIP) controller is a system thatIncorporates information of the future path in its current output.

Notice that the feed forward term,is proportional to the difference between where the controlled object is and whereit must be in the future.

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Fuzzy Logic Controllers

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Fuzzy logic, a relatively new concept in control theory, is simply the acceptance of principles that have existed since the beginning of time:

Real-world quantities are not usually“all or nothing” or “black and white” but something in-between

Time of Fajr?

Weather ?

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Fuzzy Logic Controllers

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Fuzzy logic controllers are modeled after the natural way people arrive at solutions:

• We apply different solution methodologies (rules), depending on the value of the stimulus.

• We frequently apply more than one of our “rules” at the same time to a single problem.• We accept a certain amount of imprecision, which allows us to arrive at workable solutions to problems.

Temperature is 20 degree.

Parking a car.

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Fuzzy Logic Controllers

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Fuzzy logic was first proposed by L. A. Zadeh working at Berkeley in 1965.

However, Japanese industry really embraced the idea and developed applications in the area of fuzzy logic control.

The Nissan system claims to cut fuel consumption by 12-17%.

Fuzzy logic-controlled washing machines adjust the amount of water, amount of detergent, and cycle time to how dirty and how many clothes are in the load.

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Fuzzy Logic Controllers

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Example 7: A One-Input System

• Rule 1: If the temperature is cool, then turn up the gas.• Rule 2: If the temperature is medium , then the gas is OK.• Rule 3: If the temperature is warm, then turn down the gas.

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Fuzzy Logic Controllers

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Example of a One-Input System

For example, if the temperature sensor reported 64°, the quantisizer would determine that 64° is 20% cool and 40% medium

Smallest Max Largest Max

Centroid of areaMean of max

Bisector of area

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Example of a One-Input System

For example, if the temperature sensor reported 64°, the quantisizer would determine that 64° is 20% cool and 40% medium

0.2*1+0.4*0.6=0.44

0.44/0.6=1

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Example8: A Two-Input System

• Rule 1: If T is cool and lowering, then increase the gas sharply.• Rule 2: If T is cool and steady, then increase the gas.• Rule 3: If T is cool and raising, then the gas is OK.• Rule 4: If T is medium and lowering, then increase the gas.• Rule 5: If T is medium and steady, then the gas is OK.• Rule 6: If T is medium and raising, then decrease the gas.

• Rule 7: If T is warm and lowering, then the gas is OK.• Rule 8: If T is warm and steady, then decrease the gas.• Rule 9: If T is warm and raising, then decrease the gas sharply.

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Find the output if T=64 degree and increasing 0.6 degree/min

• Rule 2: If T is cool and steady, then increase the gas.

• Rule 3: If T is cool and raising, then the gas is OK.

• Rule 5: If T is medium and steady, then the gas is OK.

• Rule 6: If T is medium and raising, then decrease the gas.

0.2 0.1 +2

0.2 0.5 0

0.4 0.1 0

0.4 0.5 -2

0.02 +2 = 0.04

0.1 0 = 0

0.04 0 = 0

0.2 -2 = -0.4

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Exercises

1- A thermostat is used to maintain a temperature of 87° F. An accurate recording of the room temperature is shown in Figure 11.42. The respective cut-in and cut-off points are currently 84° and 90°. What would you predict the cycle time would be if the cut-in and cut-off points were respectively changed to 86° and 88°?

2- A robot arm was commanded to go to a new position. Its response was recorded and isshown in following Figure. Determine the rise time, overshoot, settling time, and steady-state error of the response.

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Exercises

3-(a)Define Kp for shown system.(b) For output sensor of 2.5v and set Point of 3v, define valve input voltage(suppose Bias resistor=0).(c) If sensor steady state tarnsferfunction be 0.75 V.min/gal, set point resistor 500 ohm and Valve input voltege 3v, find the Bias Resistor?

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lecture 2

Ali Karimpour Apr 2013

Exercises

8- For the single-input fuzzy logic temperature controller specified in example 7, determine the defuzzified output for a temperature of 62°. Does your answer make sense?

9- For the two-input fuzzy logic temperature controller specified in Example 8, determine the defuzzified output for a temperature of 76° and ΔT of –0.3 deg/ min. Does your answer make sense?

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5- Draw a block diagram for an analog PID controller, indicating the function that eachblock performs.

4- Explain how the addition of integral feedback in a proportional control system eliminatessteady-state error.

6- What is the necessary condition that allows PIP control to be used?

7- What problem is solved and what new problem is created with the addition of integralfeedback?