EE480 Digital Control Systems

Kunio Takaya

Electrical and Computer Engineering

University of Saskatchewan

January 14, 2008

** Go to full-screen mode now by hitting CTRL-L

1

Contents

1 System Components 10

2 System Description 12

3 Open-Loop Control 16

4 Closed-Loop Control 18

5 Compensator Design by Root Locus Method 29

6 Assessment of the Designed Control System 37

7 Compensator Design by Frequency Response Method 42

2

8 Step Response 54

9 Assignment No.1 60

3

Rµν −1

2Rδµν =

8πG

c4Tµν

Here Tµν is tensor of energy momentum.

black blue

red magenta

green cyan

yellow

4

University of Saskatchewan, Electrical Engineering

EE 480.3 Digital Control Systems

January 2008, Kunio Takaya

Textbook: Charles L. Phillips and H. Troy Nagles Jr. “Digital

Control System, Analysis and Design” Third Edition, Prentice

Hall, 1995 ISBN 0-13-309832-x

Marks: Midterm Exam: 30%, Final Exam 55%, and Assignments

15%

1. Review of Linear Continuous-Time Control Systems.

• Control for a single arm of robot manipulators

• Analog and digital compensator design

5

2. Discrete Time Systems and the z-Transform (Chapter 2).

• System description for discrete systems

• Computational aspects of the z-Transform (difference

equation)

3. Sampling and Reconstruction (Chapter 3).

• Sampling and periodicity in spectrum

• Impulse invariant systems

• Step invariant systems (inclusion of a zero order hold)

4. Open Loop Discrete-Time Systems. (Chapter 4, 11)

• Direct conversion from the s-domain to the z-domain

• Bilinear transformation and frequency prewarping

5. Closed Loop Discrete-Time Systems. (Chapter 5)

• Simplifications of complex closed-loop systems mixed with

analog and digital blocks.

6

6. Discrete Controller Design I. (Chapter 2, 4, 5 and 9)

• Review of continuous-time state-equations

• Discrete state-equations

• Pole assignment method

7. Discrete Controller Design II. (Chapter 7, 8)

• Root locus in the z-plane

• Constraints on root loci due to specifications

8. Discrete Controller Design III. (Chapter 8)

• Phase-lead, phase-lag, and phase-lead-lag compensators

• PID controllers

• Controllers for robot manipulators

7

Blank Page

8

Control for Planar Robot Arms

9

Fig. 1 Schematic Diagram of a Planar Robot Arm

1 System Components

The specifications of the servo-motor (embedded direct drive DC

motor) are:

10

1. General: High initial torque, linear torque DC motor

2. Input Voltage: ±28V maximum

3. Output Torque: not specified

4. Angular Velocity: 3,600 rpm at 24V

5. Output Power: 1/20 HP at 24V

6. Torque Linearity: less than 1% within ±28V

7. Input Current: 4A at 24V

8. Built in Gear: Internally coupled by a down gear of 60:1

The power supply to be used is a voltage source for which the

output voltage is controlled by the input voltage.

11

1. Maximum Current: 5A

2. Input Voltage: ±28V

3. Voltage Gain: unity

4. Input Impedance: 100KΩ

5. Settling Time: 10ms for 24V

The mechanical specifications are known only in terms of the time

constant associated with each arm. The particular arm of interest

is quoted as 2 seconds.

12

2 System Description

The basic equation describing the dynamics of the arm is a second

order differential equation,

Jθ + fθ = τ

where J is the inertia of the arm, f is the friction coefficient, θ is

the angle of rotation, and τ is the applied torque. From this

differential equation, the transfer function can be derived as

G0(s) =θ(s)

T(s)=

1

s(Js+ f)=

1/f

s(Jf s+ 1)

Since J/f = T , mechanical time constant, and the torque is a

linear function of the input voltage Ea, i.e. τ = AEa, the transfer

13

function from the input voltage Ea to the angle θ now becomes,

G(s) =θ(s)

Ea(s)=

Af

s(Jf s+ 1)=

K

s(Ts+ 1).

The transfer function from the input voltage Ea to the angular

velocity of the arm ω is also given by

Gω(s) =Ω(s)

Ea(s)=

K

Ts+ 1.

If this arm is vertically placed instead of horizontally, the arm is

affected by the gravitational field. In this case, the system’s

equation needs to be modified by taking an additional torque into

account. Assuming the distance from the pivot (centre of arm

rotation) to the centre of the gravity being equal to `, the

differential equation is,

Jθ + fθ = τ +Mg` cos(θ).

14

Since this equation involves θ, θ and cos(θ) in a differential

equation, this system is no longer linear.

From the given specifications, we know that the motor rotates at

3,600 rpm when the input voltage is 24 V. However, this motor has

a built-in gear (down-gear) of 60:1 in its gear ratio. The speed of

rotation at the output shaft is, therefore, 60 rpm which is 1 rps

(revolutions per second or cycles/s). We decide here to use rps

(cycles/s) in our system model. Using the final value theorem of

the Laplace transform, we now determine two important

parameters, the gain K and time constant T .

Ωss = lims→0

sK

Ts+ 1· 24

s= 24K = 1rps

Thus, K = 1/24. Time constant T is given as 2 seconds. To check

the efficiency of the robot arm, calculate the input and output

15

power. The input power is 24V×4A=96 watts. The output power

at the input voltage of 24V is 1/20 HP, which is 120 × 746 = 37.3

watts. The efficiency is calculated to be 0.3885.

3 Open-Loop Control

This particular one arm of the robot manipulator can do angular

positioning without negative feedback. By simply applying a fixed

voltage for a certain time period, the arm rotates and come to a

desired position. This is open-loop control. Let us determine how

long a voltage of 24V has to be applied to change the arm position

by 90 or 1/4 of a revolution. This on-off control is described by

θ(s) =1/24

s(2s+ 1)· V0(

1

s− e−sT

s),

16

where V0 is the input voltage and and T is the duration of time

that V0 is applied.

L−1V0

24

1

s2(2s+ 1)=V0

48(2t− 4 + 4 e−0.5t)

yields

θ(t) =V0

48f(2t−4u(t)+4 e−0.5t)−(2(t−T )−4u(t−T )+4 e−0.5(t−T ))g.

Thus, the final value at t =∞ is

limt→∞

θ(t) =V0

48(2T ) =

1

4revolutions

If we apply V0 = 24V, T is, therefore, 0.25 seconds. Remember this

value, to turn the arm 90 the required time is 0.25 seconds at 24V.

17

4 Closed-Loop Control

The same task as considered previously as open-loop control can be

accomplished by closed-loop control that provides negative

feedback. The open-loop control cannot correct the set position of

90 in the event that the angle is disturbed. Whereas, the feedback

has ability to detect the error caused by the disturbance and

correct the error. This added functionality is achieved at the

expense of response time as well as higher power requirements. We

will investigate how the performance of the system is altered by the

closed loop configuration with a negative feedback loop.

18

We need to use a position sensor to detect the angular position of

the robot arm. The simplest position detector is a potentiometer.

We supply ±12V between two end terminals and measure the

voltage of the brush relative to the ground. Since one revolution

gives a voltage change of 24V, the sensor gain is 24V/revolution.

The transfer function is

Ep(s) = 24θ(s).

19

Fig. 2 Block diagram of the closed loop system

20

Incorporating two potentiometers, one to set a reference angle and

the other to measure the output angular position, the block

diagram of the closed-loop control system is illustrated in Fig. 2.

After simplification, we obtain the block diagram as shown at the

bottom of Fig. 2.

We now examine the characteristics of the closed-loop system in

terms of the commonly used indices such as damping ratio, time

constant, settling time, phase margin etc. According to Fig. 2, the

open-loop (forward) transfer function is

G(s) =0.5

s(s+ 0.5).

The feedback transfer function is H(s) = 1. The closed-loop

21

transfer function is given by

Gc(s) =G(s)

1 +G(s)H(s)=

0.5

s2 + 0.5s+ 0.5.

From the characteristic equation s2 + 0.5s+ 0.5 = 0, the roots

representing the closed-loop poles are found to be

s = −1

4± j√

7

4= −0.25± j0.6614.

The root locus plot for the open-loop transfer function G(s) varying

the loop gain can be plotted easily with a Matlab statement,

rlocus([0.5],[1, 0.5, 0]); The calculated poles can be verified

in the root locus plot. The loop gain is unity in this case.

22

Fig. 3 The root locus plot for G(s)

23

Referring to the standard 2nd order transfer function,

G2(s) =ω2n

s2 + 2ζωns+ ω2n

,

the poles of this system are located at

s = −ζωn ± jωn√

1− ζ2.

This equation defines damping ratio ζ and damped natural

frequency ωd as

ζ = cos θ and ωd = ωn√

1− ζ2.

24

Fig. 4 The poles of underdamped 2nd order systems

25

Percent overshoot for underdamped systems is given by

Mp = e−( ζ√

1−ζ2)π.

Time constant is

T =1

ζωn.

Other indices often used to assess the performance of control

system are settling time ts

ts ' 4

ζωn= 4T for 2% criterion

ts ' 3

ζωn= 3T for 5% criterion

and phase margin φm calculated approximately by

ζ ' 0.01φm (in degrees).

26

The calculated indices for the unity gain feedback closed-loop

system are summarized in the table below.

ζ 0.3536

ωn 0.7071 rad/s 0.1125 Hz

ωd 0.6614 rad/s 0.1052 Hz

Mp ' 30%

T 4 sec.

ts (2%) 16 sec.

ts (5%) 12 sec.

φm 35.36

27

Comparing these results, it can be concluded that the straight

negative feedback has made the system’s response slower and very

underdamped. The time constant of open loop control is 2 seconds.

Rough calculation of the settling time for the open loop,

e−ts2 = 0.02 (2% criterion)

yields ts = 7.82 seconds. This settling time is also less than that of

the closed loop control.

28

5 Compensator Design by Root Locus

Method

Feedback control can do much better job than what we have seen

in our simple trial of closing the loop with an arbitrary choice of

the loop gain.

• We now consider if it is possible to reduce system’s time

constant down to 1 second (twice as fast as the open loop case)

and keep the damping ration at about ζ = 0.6.

• The new time constant T = 1/ζωn and the damping ratio

ζ = cosφ = 0.6 determine desired pole locations. They are at

s = −1± j1.33 compared with s = −0.25± j0.66 of the trial

case as shown in Fig. 5.

29

Fig. 5 New poles (compensated) and poles of the trial case (uncompensated)

30

Passive circuits of a first order phase-lead and a phase-lag

compensator are shown in Fig. 6. The transfer function of either

compensator circuit will result in the form,

Gp(s) = Kp1 + Ts

1 + αTs.

Fig. 6 Passive Circuits of a phase-lead and a phase-lag compensator

31

For the phase-lead compensator, 0 < α < 1, whereas the phase-lag

compensator takes α > 1. Besides the gain Kp, the difference is its

mutual positions of the pole and the zero. The zero is on the

right-hand-side of the pole for the phase-lead compensator. In the

phase-lag compensator, the pole is on the right-hand-side of the

zero. A phase-lead compensator Gp+(s) and a phase-lag

compensator Gp−(s),

Gp+(s) =s+ 0.7

s+ 1.3and Gp−(s) =

s+ 1.3

s+ 0.7.

are individually incorporated in the closed loop system having a

transfer forward transfer function,

G(s) =0.5

s(s+ 0.5)

to demonstrate their general effects on the root locus. The root loci

were drawn with the Matlab commands,

rloci(poly([-0.7]),poly([0,-0.5,-1.3])) and

32

rloci(poly([-1.3]),poly([0,-0.5,-0.7])) As seen in Fig. 7, the zero of the

compensator attracts the root locus towards the zero whereas the

pole repels the root locus away form the pole.

Fig. 7 Effects of phase-lead and phase-lag compensators on root locus

33

Referring to Fig. 5, the root locus needs to be moved from the

uncompensated poles towards the left to come near the new pole

locations. This means that we must use a phase-lead compensator.

The root locus must pass through the new pole location of

s = −1± j1.33 with a certain gain. Usually adjusting the zero and

pole of the compensator is accomplished by try-and-error. For our

simple second order system, pole-zero cancellation would be

simpler than the tedious process of try-and-error. If we set the

phase-lead compensator to be

Gp+(s) = Kps+ 0.5

s+ 2,

The combined open-loop transfer function,

G(s) ·Gp+(s) = Kp0.5

s(s+ 2)

gives us a straight upright root locus at s = −1 as shown in Fig. 8.

34

Fig. 8 The root locus plot resulting from pole-zero cancellation

35

Letting the combined gain of jG(s) ·Gp+(s)j to be unity to satisfy

the condition of the root locus, we obtain the compensator gain

Kp = 5.55 from

jG(s) ·Gp+(s)js=−1±j1.33 = Kpj0.5

s(s+ 2)js=−1±j1.33 = 1.

Thus, our compensator is designed and it is,

Gp+(s) = 5.55s+ 0.5

s+ 2

36

6 Assessment of the Designed Control

System

The designed controller is improved the closed-loop system’s

performance as we have intended.

Uncompensated (Sys-I) Compensated (Sys-II)

ζ 0.3536 0.6

ωd 0.7071 rad/s 1.333 rad/s

T 4 sec. 1 sec.

Now, we investigate the power requirement which might have been

substantially increased because of the shortened time constant from

T = 4 down to T = 1. The direct input to the robot arm controller,

more specifically the input to the power supply, should be

37

compared for both of the uncompensated and the compensated

control system shown in Fig. 9, respectively.

Fig. 9 Input errors e1 and e2 in the uncompensated and compensated systems

38

When the input command is to turn the arm exactly one

revolution, the input errors E1(s) and E2(s) are:

E1(s) =1

1 +G(s)

1

s=

s+ 0.5

s2 + 0.5s+ 0.5

E2(s) =Gp+(s)

1 +G(s)Gp+(s)

1

s=

5.55(s+ 0.5)

s2 + 2s+ 2.775.

By using the initial value theorem to find the initial error, which

occurs at the onset of the command, we find:

e1(0) = lims→∞

sE1(s) = 1

e2(0) = lims→∞

sE2(s) = 5.55.

Recall that the input to G(s) is expressed in terms of angle θ

rather than the actual input voltage to the power supply, we would

expect the initial voltage of 5.55×24V to occur when the

compensator is incorporated. This input level far exceeds the

39

maximum input range of ±28V.

We can examine position offset using the final value theorem.

Position offset is the error that remains uncorrected for a step

input.

e1(∞) = lims→0

sE1(s) = 0

e2(∞) = lims→0

sE2(s) = 0

Since our system equation is of type I, i.e. G(s) has one integrator

1/s, zero position offset is expected. The velocity offset ess is

calculated from the velocity error coefficient Kv, i.e. ess = 1Kv

.

Kv = lims→0

sG(s) = 1, for Sys-I

Kv = lims→0

sGp+(s)G(s) = 1.39, for Sys-II

The velocity offset is the position error when input is a ramp signal

1/s2. ess = 1 for the uncompensated system Sys-I and ess = 0.72

40

for the compensated system Sys-II. Thus, Small improvement is

gained.

41

7 Compensator Design by Frequency

Response Method

The same first order compensator can be designed by using the

frequency response method. We still have to use a phase-lead

compensator to decrease the control system’s time constant. The

phase-lead compensator now has a transfer function,

Gc(s) =T1s+ 1

T2s+ 1.

where T1 > T2. DC gain is unity and its high frequency gain is

T1/T2. The frequency response of this phase-lead compensator is

shown in Fig. 10.

42

Fig. 10 Frequency Response of the Phase-lead Compensator

There are two corner frequencies defined by the amplitude

response. The lower corner frequency occurs at ω1 = 1/T1 and the

upper corner frequency is at ω2 = 1/T2. It can be proved that the

43

maximum leading phase occurs at ωm =√ω1ω2. From the

frequency response,

Gc(jω) =jω + ω1

jω + ω2,

the maximum phase obtained at ωm is given by

φm = tan−1 ωmω1− tan−1 ωm

ω2

= tan−1

√

ω2

ω1− tan−1

√

ω1

ω2

Letting α =ω2

ω1, we can express φm as

φm = tan−1 α− 1

2√α

or tanφm =α− 1

2√α.

Then, we can prove that

sinφm =α− 1

α+ 1.

44

This is the most important equation that determines the ratio

α =ω2

ω1for a desired phase compensation φm. Another important

equation to be used in designing the phase-lead compensator is

about the gain increase at a new cross-over frequency for 0dB.

Since the gain increase for high frequency, say ω =∞ is given by

Gain = 20 log10

T1

T2= 20 log10

ω2

ω1= 20 log10 α

the gain at the center frequency ωm should be

Gain(ωm) =1

220 log10 α = 10 log10 α

The design objective is to design a phase-lead compensator which

gives the overall control system a phase margin of φm = 60.

Usually, the loop gain must be determined according to error

criteria, separately from the phase-lead compensator. In our case,

we have found the loop gain of 5.55 was inadequate by the previous

45

design based on the root locus. Let us use the loop gain of 1 to

alleviate too high input voltage for the power supply.

The design procedure requires the Bode plot of the open-loop

system without a compensator. The following Matlab program

allows us to plot this Bode diagram.

% K

% DC Motor: G(s) = --------

% s(Ts+1)

%

clg, clc

K=1;

num=K;

den=[2,1,0];

w=logspace(-2,1,100);

jw=j*w;

H=polyval(num,jw)./polyval(den,jw);

mag=20*log10(abs(H));

phase=angle(H)*180/pi;

46

subplot(211),semilogx(w,mag,’w’),grid

title(’Magnitude Response’);

xlabel(’frequency in rad/s’),ylabel(’Magnitude in dB’);

subplot(212),semilogx(w,phase,’w’),grid

title(’Phase Response’);

xlabel(’frequency in rad/s’),ylabel(’Phase in degrees’);

47

Fig. 11 Frequency Response of the Phase-lead Compensator

48

From the obtained Bode plot, we can find that the magnitude

cross-over frequency is 0.45 rad/s and that the phase margin is

approximately 45. Therefore, we need to add an additional 15 to

meet the required phase margin of 60. Since a new cross-over

frequency needs to be chosen at a frequency higher than the

uncompensated system’s cross-over frequency 0.45 rad/s, we add 5

to account the further decrease in phase. Therefore, the required

phase compensation is φm = 20. Using

sinφm =α− 1

α+ 1,

we determine the corner frequency ratio α = ω2/ω1. We obtain

α = 2.03. Then, we must find the increase in gain introduced by

inserting the compensator.

Gain = 20 log10 α = 6.15 dB

Thus, the new cross-over frequency must occur when the

49

magnitude response of the Bode plot intersects -6.15dB line. The

new cross-over frequency is found to be approximately ωm = 1.0

rad/s. Then, we find two corner frequencies ω1 and ω2 from two

equations, α = ω2/ω1 and ωm =√ω1ω2. Since ωm = ω1

√α, we

find ω1 = 0.70. From ω2 = αω1, ω2 = 1.42. The designed

phase-lead compensator is

Gc(s) =T1s+ 1

T2s+ 1

=ω2

ω1

s+ ω1

s+ ω2

= αs+ ω1

s+ ω2

= 2.03s+ 0.70

s+ 1.42.

The frequency response of the compensated system by the

phase-lead compensator designed by using the frequency response

50

is calculated by the following Matlab program.

% 0.5

% DC Motor: G(s) = --------

% s(s+0.5)

% compensated by 2.03(s+0.7)/(s+1.42)

%

clg, clc

K=2.03;

num=K*0.5*poly([-0.7]);

den=poly([0,-0.5,-1.42]);

w=logspace(-2,1,100);

jw=j*w;

H=polyval(num,jw)./polyval(den,jw);

mag=20*log10(abs(H));

phase=angle(H)*180/pi;

subplot(211),semilogx(w,mag,’w’),grid

title(’Magnitude Response’);

xlabel(’frequency in rad/s’),ylabel(’Magnitude in dB’);

subplot(212),semilogx(w,phase,’w’),grid

title(’Phase Response’);

xlabel(’frequency in rad/s’),ylabel(’Phase in degrees’);

51

Fig. 12 Frequency Response of the compensated system

52

The obtained frequency response is shown in Fig. 12. The phase

margin appears to be close to 60.

53

8 Step Response

The characteristics of the three systems, uncompensated,

compensated with a controller designed using pole/zero

cancellation, and compensated with a phase-lead compensator

designed by the frequency response method, can be compared from

the step response view point. The unit step reponse can be

calculated and displayed by a single Matlab statement

step(num,den). The following Matlab program shows how to

combine a compensator transfer function, a system transfer

function, and a feedback transfer function, and generate one

combined closed-loop transfer function. This general Matlab

program to construct a closed-loop transfer funtion expects all

transfer functions to be represented by poles and zeros and their

gain. Using this program and the built-in function

step(num,den)., the unit step responses of the above three cases

54

are shown if Fig. 13.

% ----------------------------------------------------

% Finding Closed-Loop Step Response

% From Open-Loop Transfer Function

% K. Takaya Jan, 1998

% ----------------------------------------------------

% G(s)=A1*N1(s)/D1(s), H(s)=A2*N(s)/D2(s)

% Gc(s)=G(s)/1+G(s)H(s)

% =A1*N1(s)D2(s)/D1(s)D2(s)+A1*A2*N1(s)N2(s)

% ----------------------------------------------------

% Specify poles and zeros of G(s) and H(s),

% G(s)=0.5/s(s+0.5), H(s)=1

N1=[]; D1=[0,-0.5];

N2=[]; D2=[];

A1=0.5; A2=1;

% ---------------------(No Compensator)

Num=A1*poly([N1,D2]);

Den1=poly([D1,D2]);

Den2=poly([N1,N2]);

n1=size(Den1,2);

n2=size(Den2,2);

55

Den=Den1+A1*A2*[zeros(1,n1-n2),Den2];

T=[0:0.1:20];

figure(1); step(Num,Den,T);

% ---------------------(Pole/Zero Cancelled Compensator)

% Specify poles and zeros of G(s), Gc(s) and H(s),

% G(s)=0.5/s(s+0.5), Gc(s)=5.55(s+0.5)/(s+2), H(s)=1

N1=[]; D1=[0,-2.0];

N2=[]; D2=[];

A1=0.5*5.55; A2=1;

% ----------------------------------------------------

Num=A1*poly([N1,D2]);

Den1=poly([D1,D2]);

Den2=poly([N1,N2]);

n1=size(Den1,2);

n2=size(Den2,2);

Den=Den1+A1*A2*[zeros(1,n1-n2),Den2];

T=[0:0.1:20];

figure(2); step(Num,Den,T);

% ---------------------(Phase Margin 60 deg. Compensator)

% Specify poles and zeros of G(s), Gc(s) and H(s),

% G(s)=0.5/s(s+0.5), Gc(s)=2.03(s+0.7)/(s+1.42), H(s)=1

N1=[-0.7]; D1=[0,-0.5,-1.42];

N2=[]; D2=[];

A1=0.5*2.03; A2=1;

56

% ----------------------------------------------------

Num=A1*poly([N1,D2]);

Den1=poly([D1,D2]);

Den2=poly([N1,N2]);

n1=size(Den1,2);

n2=size(Den2,2);

Den=Den1+A1*A2*[zeros(1,n1-n2),Den2];

T=[0:0.1:20];

figure(3); step(Num,Den,T);

57

58

59

Fig. 13 Step Responses of (1) uncompensated system (top), (2) compensated

by pole/zero cancellation (left), and (3) by phase margin 60 (right).

60

9 Assignment No.1

1. Redesign the phase lead compensator without imposing

pole-zero cancellation. For a given zero at s = −1, determine

only the pole e.g. s = −p and an appropriate gain Kp of

Gp+(s) = Kps+ 1

s+ p

such that the time constant T = 1/ζωn = 1.5 sec. and the

damping ratio of ζ = cosφ = 0.6 are satisfied for the closed

loop system. Use the root locus method.

2. Redesign the phase lead compensator by the frequency

response method to satisfy the phase margin of 70.

61