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Chapter 9: Control Systems: 

1 Chapter 9: Control Systems

Control System: 

2 Control System Control physical system’s output By setting physical system’s input Tracking E.g. Cruise control Thermostat control Disk drive control Aircraft altitude control Difficulty due to Disturbance: wind, road, tire, brake; opening/closing door… Human interface: feel good, feel right…

Tracking: 

3 Tracking

Open-Loop Control Systems: 

4 Open-Loop Control Systems Plant Physical system to be controlled Car, plane, disk, heater,… Actuator Device to control the plant Throttle, wing flap, disk motor,… Controller Designed product to control the plant

Open-Loop Control Systems: 

5 Open-Loop Control Systems Output The aspect of the physical system we are interested in Speed, disk location, temperature Reference The value we want to see at output Desired speed, desired location, desired temperature Disturbance Uncontrollable input to the plant imposed by environment Wind, bumping the disk drive, door opening

Other Characteristics of open loop: 

6 Other Characteristics of open loop Feed-forward control Delay in actual change of the output Controller doesn’t know how well thing goes Simple Best use for predictable systems

Close Loop Control Systems: 

7 Close Loop Control Systems Sensor Measure the plant output Error detector Detect Error Feedback control systems Minimize tracking error

Designing Open Loop Control System: 

8 Designing Open Loop Control System Develop a model of the plant Develop a controller Analyze the controller Consider Disturbance Determine Performance Example: Open Loop Cruise Control System

Model of the Plant: 

9 Model of the Plant May not be necessary Can be done through experimenting and tuning But, Can make it easier to design May be useful for deriving the controller Example: throttle that goes from 0 to 45 degree On flat surface at 50 mph, open the throttle to 40 degree Wait 1 “time unit” Measure the speed, let’s say 55 mph Then the following equation satisfy the above scenario v t+1 =0.7*v t +0.5*u t 55 = 0.7*50+0.5*40 IF the equation holds for all other scenario Then we have a model of the plant

Designing the Controller: 

10 Designing the Controller Assuming we want to use a simple linear function u t =F(r t )= P * r t r t is the desired speed Linear proportional controller v t+1 =0.7*v t +0.5*u t = 0.7*v t +0.5P*r t Let v t+1 =v t at steady state = v ss v ss =0.7*v ss +0.5P*r t At steady state, we want v ss =r t P=0.6 I.e. u t =0.6*r t

Analyzing the Controller: 

11 Analyzing the Controller Let v 0 =20mph, r 0 =50mph v t+1 =0.7*v t +0.5(0.6)*r t =0.7*v t +0.3*50= 0.7*v t +15 Throttle position is 0.6*50=30 degree

Considering the Disturbance: 

12 Considering the Disturbance Assume road grade can affect the speed From –5mph to +5 mph v t+1 =0.7*v t +10 v t+1 =0.7*v t +20

Determining Performance: 

13 Determining Performance V t+1 =0.7*v t +0.5P*r 0 -w 0 v 1 =0.7*v 0 +0.5P*r 0 -w 0 v 2 =0.7*(0.7*v 0 +0.5P*r 0 -w 0 ) +0.5P*r 0 -w 0 =0.7*0.7*v 0 +(0.7+1.0)*0.5P*r 0 -(0.7+1.0)w 0 v t =0.7 t *v 0 +(0.7 t-1 +0.7 t-2 +…+0.7+1.0)(0.5P*r 0 -w 0 ) Coefficient of v t determines rate of decay of v 0 >1 or <-1, v t will grow without bound <0, v t will oscillate

Designing Close Loop Control System: 

14 Designing Close Loop Control System

Stability: 

15 Stability u t = P * (r t -v t ) v t+1 = 0.7v t +0.5u t -w t = 0.7v t +0.5P*(r t -v t )-w =(0.7-0.5P)*v t +0.5P*r t -w t v t =(0.7-0.5P) t *v 0 +((0.7-0.5P) t-1 +(0.7-0.5P) t-2 +…+0.7-0.5P+1.0)(0.5P*r 0 -w 0 ) Stability constraint (I.e. convergence) requires |0.7-0.5P|<1 -1<0.7-0.5P<1 -0.6<P<3.4

Reducing effect of v0: 

16 Reducing effect of v 0 u t = P * (r t -v t ) v t+1 = 0.7v t +0.5u t -w t = 0.7v t +0.5P*(r t -v t )-w =(0.7-0.5P)*v t +0.5P*r t -w t v t =(0.7-0.5P) t *v 0 +((0.7-0.5P) t-1 +(0.7-0.5P) t-2 +…+0.7-0.5P+1.0)(0.5P*r 0 -w 0 ) To reduce the effect of initial condition 0.7-0.5P as small as possible P=1.4

Avoid Oscillation: 

17 Avoid Oscillation u t = P * (r t -v t ) v t+1 = 0.7v t +0.5u t -w t = 0.7v t +0.5P*(r t -v t )-w =(0.7-0.5P)*v t +0.5P*r t -w t v t =(0.7-0.5P) t *v 0 +((0.7-0.5P) t-1 +(0.7-0.5P) t-2 +…+0.7-0.5P+1.0)(0.5P*r 0 -w 0 ) To avoid oscillation 0.7-0.5P >=0 P<=1.4

Perfect Tracking: 

18 Perfect Tracking u t = P * (r t -v t ) v t+1 = 0.7v t +0.5u t -w t = 0.7v t +0.5P*(r t -v t )-w =(0.7-0.5P)*v t +0.5P*r t -w t v ss =(0.7-0.5P)*v ss +0.5P*r 0 -w 0 (1-0.7+0.5P)v ss =0.5P*r 0 -w 0 v ss =(0.5P/(0.3+0.5P)) * r 0 - (1.0/(0.3+0.5P)) * w o To make v ss as close to r 0 as possible P should be as large as possible

Close-Loop Design: 

19 Close-Loop Design u t = P * (r t -v t ) Finally, setting P=3.3 Stable, track well, some oscillation u t = 3.3 * (r t -v t )

Analyze the controller: 

20 Analyze the controller v 0 =20 mph, r 0 =50 mph, w=0 v t+1 = 0.7v t +0.5P*(r t -v t )-w = 0.7v t +0.5*3.3*(50-v t ) u t = P * (r t -v t ) = 3.3 * (50-v t ) But u t range from 0-45 Controller saturates

Analyze the controller: 

21 Analyze the controller v 0 =20 mph, r 0 =50 mph, w=0 v t+1 = 0.7v t +0.5*u t u t = 3.3 * (50-v t ) Saturate at 0, 45 Oscillation! “feel bad”

Analyze the controller: 

22 Analyze the controller Set P=1.0 to void oscillation Terrible SS performance

Analyzing the Controller: 

23 Analyzing the Controller

Minimize the effect of disturbance: 

24 Minimize the effect of disturbance v t+1 = 0.7v t +0.5*3.3*(r t -v t )-w w=-5 or +5 39.74 Close to 42.31 Better than 33 66 Cost SS error oscillation

General Control System: 

25 General Control System Objective Causing output to track a reference even in the presence of Measurement noise Model error Disturbances Metrics Stability Output remains bounded Performance How well an output tracks the reference Disturbance rejection Robustness Ability to tolerate modeling error of the plant

Performance (generally speaking): 

26 Performance (generally speaking) Rise time Time it takes form 10% to 90% Peak time Overshoot Percentage by which Peak exceed final value Settling time Time it takes to reach 1% of final value

Plant modeling is difficult: 

27 Plant modeling is difficult May need to be done first Plant is usually on continuous time Not discrete time E.g. car speed continuously react to throttle position, not at discrete interval Sampling period must be chosen carefully To make sure “nothing interesting” happen in between I.e. small enough Plant is usually non-linear E.g. shock absorber response may need to be 8 th order differential Iterative development of the plant model and controller Have a plant model that is “good enough”

Controller Design: P: 

28 Controller Design: P Proportional controller A controller that multiplies the tracking error by a constant u t = P * (r t -v t ) Close loop model with a linear plant E.g. v t+1 = (0.7-0.5P)*v t +0.5P*r t -w t P affects Transient response Stability, oscillation Steady state tacking As large as possible Disturbance rejection As large as possible

Controller Design: PD: 

29 Controller Design: PD Proportional and Derivative control u t = P * (r t -v t ) + D * ((r t -v t )-(r t-1 -v t-1 )) = P * e t + D * (e t -e t-1 ) Consider the size of error over time Intuitively Want to “push” more if the error is not reducing fast enough Want to “push” less if the error is reducing really fast

PD Controller: 

30 PD Controller Need to keep track of error derivative E.g. Cruise controller example v t+1 = 0.7v t +0.5u t -w t Let u t = P * e t + D * (e t -e t-1 ), e t =r t -v t v t+1 =0.7v t +0.5*(P*(r t -v t )+D*((r t -v t )-(r t-1 -v t-1 )))-w t v t+1 =(0.7-0.5*(P+D))*v t +0.5D*v t-1 +0.5*(P+D)*r t -0.5D*r t-1 -w t Assume reference input and distribance are constant, the steady-state speed is V ss =(0.5P/(1-0.7+0.5P)) * r Does not depend on D!!! P can be set for best tracking and disturbance control Then D set to control oscillation/overshoot/rate of convergence

PD Control Example: 

31 PD Control Example

PI Control: 

32 PI Control Proportional plus integral control u t =P*e t +I*(e 0 +e 1 +…+e t ) Sum up error over time Ensure reaching desired output, eventually v ss will not be reached until e ss =0 Use P to control disturbance Use I to ensure steady state convergence and convergence rate

PID Controller: 

33 PID Controller Combine Proportional, integral, and derivative control u t =P*e t +I*(e 0 +e 1 +…+e t )+D*(e t -e t-1 ) Available off-the shelf

Software Coding: 

34 Software Coding Main function loops forever, during each iteration Read plant output sensor May require A2D Read current desired reference input Call PidUpdate, to determine actuator value Set actuator value May require D2A

Software Coding (continue): 

35 Software Coding (continue) Pgain, Dgain, Igain are constants sensor_value_previous For D control error_sum For I control

Computation: 

36 Computation u t =P*e t +I*(e 0 +e 1 +…+e t )+D*(e t -e t-1 )

PID tuning: 

37 PID tuning Analytically deriving P, I, D may not be possible E.g. plant not is not available, or to costly to obtain Ad hoc method for getting “reasonable” P, I, D Start with a small P, I=D=0 Increase D, until seeing oscillation Reduce D a bit Increase P, until seeing oscillation Reduce D a bit Increase I, until seeing oscillation Iterate until can change anything without excessive oscillation

Practical Issues with Computer-Based Control: 

38 Practical Issues with Computer-Based Control Quantization Overflow Aliasing Computation Delay

Quantization & Overflow: 

39 Quantization & Overflow Quantization Can’t store 0.36 as 4-bit fractional number Can only store 0.75, 0.59, 0.25, 0.00, -0.25, -050,-0.75, -1.00 Choose 0.25 Result in quantization error of 0.11 Sources of quantization error Operations, e.g. 0.50*0.25=0.125 Can use more bits until input/output to the environment/memory A2D converters Overflow Can’t store 0.75+0.50 = 1.25 as 4-bit fractional number Solutions: Use fix-point representation/operations carefully Time-consuming Use floating-point co-processor Costly

Aliasing: 

40 Aliasing Quantization/overflow Due to discrete nature of computer data Aliasing Due to discrete nature of sampling

Aliasing Example: 

41 Aliasing Example Sampling at 2.5 Hz, period of 0.4, the following are indistinguishable y(t)=1.0*sin(6 π t), frequency 3 Hz y(t)=1.0*sin( π t), frequency of 0.5 Hz In fact, with sampling frequency of 2.5 Hz Can only correctly sample signal below Nyquist frequency 2.5/2 = 1.25 Hz

Computation Delay: 

42 Computation Delay Inherent delay in processing Actuation occurs later than expected Need to characterize implementation delay to make sure it is negligible Hardware delay is usually easy to characterize Synchronous design Software delay is harder to predict Should organize code carefully so delay is predictable and minimized Write software with predictable timing behavior (be like hardware) Time Trigger Architecture Synchronous Software Language

Benefit of Computer Control: 

43 Benefit of Computer Control Cost!!! Expensive to make analog control immune to Age, temperature, manufacturing error Computer control replace complex analog hardware with complex code Programmability!!! Computer Control can be “upgraded” Change in control mode, gain, are easy to do Computer Control can be adaptive to change in plant Due to age, temperature, …etc “future-proof” Easily adapt to change in standards,..etc