digital control - lecture 3homes.et.aau.dk/yang/de5/dc/mm3.pdf · digital controller design lead...

Outline

Digital ControlLecture 3

Lecture 3 Digital Control

Outline

Outline

1 Digital Controller Design

2 Lead Compensator for Antenna - Design Example


What have we talked about in MM2?MM2?

• Sampling rate selection

• Equivalents between continuous & digital Systemsdigital Systems

Digital Controller DesignLead Compensator for Antenna - Design Example

ExercisesEmulation Method for Digital Control

Outline

1 Digital Controller DesignEmulation Method for Digital Control

2 Lead Compensator for Antenna - Design ExampleEffect of Sample TimesAccommodation for Sampling DelayEffect of Sampling Method




Digital Controller Design

Digital controller can be obtained using:

Emulation, which finds the discrete equivalent of a continuouscontroller

Direct discrete design (next lecture)




Frequency Issues

Continuous Systems

For a minimum-phase transfer function, the phase is uniquelydetermined by the magnitude curve:

∠G (jω) ≈ n × 90◦

where n is the slope of G (jω) in units of decade of amplitude

Discrete Systems

The amplitude and phase relationship is lost!The prediction of stability from the amplitude curve alone forminimum-phase systems is lostIt is typically necessary to determine both magnitude and phase fordiscrete systems




Emulation Method

1 A continuous controller is designed

2 Sample time is selected

3 Discrete equivalent is computed

4 Evaluation of design



Exercises

Effect of Sample TimesAccommodation for Sampling DelayEffect of Sampling Method

Outline

1 Digital Controller DesignEmulation Method for Digital Control

2 Lead Compensator for Antenna - Design ExampleEffect of Sample TimesAccommodation for Sampling DelayEffect of Sampling Method



Exercises


Case Study: Antenna Control

General System Model:

J θ̈ + B θ̇ = Tc + Td

Discarding the disturbances Td gives the transfer function:

Θ(s)

U(s)=

1

s(

s

a+ 1

)

where a = B

J= 0.1 and u(t) = Tc (t)

B.

Design Specifications:

Overshoot to a step input less than 16% (PM ≈ 55)

Settling time to 1% in less than 10s

Tracking error to ramp of slope 0.01 radsec less than 0.01rad

Sampling time to give at least 10 samples in a rise-time



Exercises


Lead Compensator Design for Antenna (FC pp. 375)

Step 1

Design the low frequency gain K with respect to the steady-stateerror specificationAntenna system case: K = 1

Step 2

Determine the needed phase lead

sys=tf(1,[10 1 0]);

margin(sys)

PM=18 at ω = 0.308



Exercises



Step 3

Using lead contribution of φmax = 45 should result in PM=63which is 8 more than needed.

Step 4

Determine:

α =1 − sinφmax

1 + sinφmax

=1 − sin 45

1 + sin 45= 0.1716

Step 5

T =1

ωmax

√α

=1

ωn

2

√α

=2

0.92√

α= 5.248

Giving a zero in s = − 1T

= −0.19 and a pole ins = − 1

αT= −1.11.



Exercises



Step 6

Draw the compensated frequency response, check PMUsing the formulation:

D(s) =Ts + 1

αTs + 1

we use:

sysD=tf([5.3 1],[0.9 1])

sysC=sys*sysD

margin(sysC)

step(feedback(sysC,1))



Exercises



−80

−60

−40

−20

0

20

40

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−180

−135

−90

Pha

se (

deg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 56.3 deg (at 0.505 rad/sec)

Frequency (rad/sec)

Figure: Frequency response

Step Response

Time (sec)

Am

plitu

de

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4

System: untitled1Time (sec): 5.65Amplitude: 1.16


Figure: Step response



Exercises



Step 7

Step 7: Iterate on the design until all specifications are met

sysD=tf([10 1],[1 1])

sysC=sys*sysD

margin(sysC)

sysCL=feedback(sysC,1)

step(sysCL)



Exercises



−80

−60

−40

−20

0

20

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)


Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

System: sysCLTime (sec): 3.63Amplitude: 1.16





Exercises


Digital Lead Compensator for Antenna - Fast Sampling

Continuous lead controller

D(s) =10s + 1

s + 1

Digitization - Fast Sample Rate

sysc=tf(1,[10 1 0]);

lead=tf([10 1],[1 1]);

syslead=sysc*lead;

Ts=1/20;

leadd1=c2d(lead,Ts,’zoh’);

sysd=c2d(sysc,Ts,’zoh’);

syscld=feedback(sysd*leadd1,1);

step(syscld)



Exercises


Digital Lead Compensator for Antenna - Fast Sampling

Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

System: syscldfastTime (sec): 3.65Amplitude: 1.18

System: syscldfastTime (sec): 8.75Amplitude: 0.99



Exercises


Digital Lead Compensator for Antenna - Slow Sampling


D(s) =10s + 1

s + 1

Digitization - Slow Sample Rate

sysc=tf(1,[10 1 0]);

lead=tf([10 1],[1 1]);

syslead=sysc*lead;

Ts=1/2;




step(syscld)



Exercises



Step Response

Time (sec)

Am

plitu

de

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4

System: syscldslowTime (sec): 3.35Amplitude: 1.35

System: syscldslowTime (sec): 13.5Amplitude: 0.99



Exercises


Effect of Sample Time on Step Response

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plitu

de

ContinuousFast samplingSlow sampling



Exercises


Effect of Sample Time on Frequency Response

−150

−100

−50

0

50

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−270

−180

−90

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

ContinuousFast samplingSlow sampling



Exercises


Effect of Sample Time on Pole Locations

Pole−Zero Map

Real Axis

Imag

inar

y A

xis

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

System: syscldslowPole : 0.737 + 0.422iDamping: 0.3Overshoot (%): 37.3Frequency (rad/sec): 1.09

System: syscldfastPole : 0.975 − 0.0433iDamping: 0.482Overshoot (%): 17.7Frequency (rad/sec): 1.01

System: untitled1Pole : −0.5 + 0.866iDamping: 0.5Overshoot (%): 16.3Frequency (rad/sec): 1



Exercises


Incorporating Sampling Delay in System

Continuous System

G (s) =1

s(10s + 1)

Continuous System with Delay

Gd(s) =2/T

s + 2/T

1

s(10s + 1)



Exercises


Lead Compensator for System using Slow Sampling Rate

Inserting T = 1/2

Gd(s) =2/T

s + 2/T

1

s(10s + 1)

=4

s(s + 4)(10s + 1)



Exercises


Lead Compensator Design for Antenna

Step 1

Design the low frequency gain K with respect to the steady-stateerror specificationSteady-state unchanged from original system: K = 1

Step 2

Determine the needed phase lead

sys=tf(1,[10 41 4 0]);

margin(sys)

PM=14 at ω = 0.308



Exercises



Step 3

Using lead contribution of φmax = 50 should result in PM=64which is 9 more than needed.

Step 4

Determine:

α =1 − sinφmax

1 + sinφmax

=1 − sin 50

1 + sin 50= 0.1325

Step 5

T =1

ωmax

√α

=1

0.4√

(0.1325)= 6.869

Giving a zero in s = − 1T

= −0.1456 and a pole ins = − 1

αT= −1.099.



Exercises



Step 6

Draw the compensated frequency response, check PMUsing the formulation:

D(s) =Ts + 1

αTs + 1

we use:

sysD=tf([6.9 1],[0.9 1])

sysC=sys*sysD

margin(sysC)

step(feedback(sysC,1))



Exercises



−150

−100

−50

0

50

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−270

−225

−180

−135

−90

Pha

se (

deg)

Bode DiagramGm = 16.9 dB (at 2.05 rad/sec) , Pm = 48.6 deg (at 0.607 rad/sec)

Frequency (rad/sec)


Step Response

Time (sec)

Am

plitu

de

0 5 10 15 20 250

0.2

0.4

0.6

0.8

1

1.2

1.4System: untitled1Time (sec): 4.76Amplitude: 1.21





Exercises



Step 7

Step 7: Iterate on the design until all specifications are met

sysD=tf([7.5 1],[0.68 1])

sysC=sys*sysD

margin(sysC)

sysCL=feedback(sysC,1)

step(sysCL)



Exercises



−150

−100

−50

0

50

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−270

−180

−90

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)


Step Response

Time (sec)

Am

plitu

de

0 5 10 15 20 25 300

0.2

0.4

0.6

0.8

1

1.2

1.4


System: sysCLTime (sec): 7Amplitude: 1.01





Exercises




D(s) =7.5s + 1

0.68s + 1

Digitization - Slow Sample Rate

sysc=tf(1,[10 1 0]);

lead=tf([7.5 1],[0.68 1]);

syslead=sysc*lead;

Ts=1/2;




step(syscld)



Exercises


Digital Lead Compensator for Antenna - Comparison

−80

−60

−40

−20

0

20

Mag

nitu

de (

dB)

10−2

10−1

100

101

102

−270

−180

−90

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

ContinuousSlow samplingCompensated slow sampling


0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plitu

de

ContinuousSlow samplingCompensated slow sampling




Exercises


Discretization in Matlab

Matlab

sysd=c2d(sys,Ts,method)

method:

’zoh’: Zero order hold

’foh’: First order hold (academic)

’tustin’: Bilinear approximation (trapezoidal)

’prewarp’: Tustin with a specific frequency used for prewarp

’matched’: Matching continuous poles with discrete



Exercises


Discretization of Lead Compensator - Fast Sample Rate

−40

−30

−20

−10

0

10

Mag

nitu

de (

dB)

10−1

100

101

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

ZOHTustinMatched


0 1 2 3 4 5 6 7 80

0.2

0.4

0.6

0.8

1

1.2

Step Response

Time (sec)

Am

plitu

de

ZOHTustinMatched




Exercises


Discretization of Lead Compensator - Fast Sample Rate

0.8 0.82 0.84 0.86 0.88 0.9 0.92 0.94 0.96 0.98 1−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

0.10.20.30.4

0.50.6

0.70.8

0.9

Pole−Zero Map

Real Axis

Imag

inar

y A

xis

ZOHTustinMatched



Exercises


Discretization of Lead Compensator - Slow Sample Rate

−80

−60

−40

−20

0

20

Mag

nitu

de (

dB)

10−1

100

101

−270

−225

−180

−135

−90

−45

0

Pha

se (

deg)

Bode Diagram

Frequency (rad/sec)

ZOHTustinMatched


0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

Am

plitu

de

ZOHTustinMatched




Exercises


Discretization of Lead Compensator - Slow Sample Rate

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1−1

−0.8

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

1

0.7π/T

0.8π/T

0.9π/T

π/T

0.1π/T

0.2π/T

0.3π/T

0.4π/T0.5π/T

0.6π/T

0.7π/T

0.8π/T

0.9π/T

π/T

0.1

0.20.3

0.40.50.60.70.80.9

0.1π/T

0.2π/T

0.3π/T

0.4π/T0.5π/T

0.6π/T

Pole−Zero Map

Real Axis

Imag

inar

y A

xis

ZOHTustinMatched



Exercises


Some important things to remember

Discretization of compensator

Use the method suited for implementation in the system

Discrete equivalent of plant

Use method corresponding to implementation (usually ZOH)

Simulink can combine discrete compensator with continuousplant (digitization of plant not necessary)



Exercises

Book: Digital Control

Problem 7.4

Problem 7.5

Problem 7.7


digital control - lecture 3homes.et.aau.dk/yang/de5/dc/mm3.pdf · digital controller design lead...

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