413Presentation

EECE 413, Fall, 20061

Digital Controls Course Outline• Introduction to control systems; control system

examples; basic principle of feedback control’ brief review of control development history.

• Digital control systems design: Time domain specifications; Dynamic response to unit step and ramp functions. Stability of discrete time system; Digital PID design; Pole placement design.

• Digital Control system design: Frequency domain specifications; Gain and phase margins; Compensator design with bilinear transformations.

• Digital control system design through state space approach: State space description of discrete systems; State feedback design via pole placement; State estimator design; Controller design with state estimator.

EECE 413, Fall, 20062

What Is a Control System?

• Control system objective: To make the system output and desired reference as close as possible, i.e., to make the error as small as possible.

• Key issues:– How to describe the system to be controlled? (Modeling)– How to design the controller? (Control)

Controller System to be controlled

Desired

Performance:

REFERENCE

Difference:

ERROR

INPUTTo the System

Aircraft, missiles, cars, appliances, economic

systems, etc.

Information about the system:

OUTPUT

EECE 413, Fall, 20063

Simple System Model for Cruise Control

mux

mbxxmxbu

xb

xv

Mass m

acceleration

x displacement

Force uFriction force

By Newton’s Law of motion: f = ma, where f is the total force applied to an object with a mass m and a is the acceleration, we have

This is a 2nd order Ordinary Differential Equation (ODE) with respect to displacement x. It can be written as a 1st order ODE with respect to speed

muv

mbv Model of the cruise control system, u is input force, v is output.

EECE 413, Fall, 20064

Cruise Control System

Controller

REFERENCEINPUT OUTPUT

?

70 MPHu Speed v

muv

mbv

Control engineering is to design the controller in a way that the system meets desired performance specifications.

EECE 413, Fall, 20065

Basic Electrical Systems

v

i

RV = iR

Resistor

v(t)

i(t)

C

i = C dv/dt

Capacitor

v(t)

i(t)

L

v = L di/dt

Inductor

Kirchhoff’s Voltage Law (KVL)The sum of voltage drops around any closed loop in a circuit is 0.

v1

v2

v3

v5 v4

v1 + v2 + v3 + v4 + v5 = 0

Kirchhoff’s Current Law (KCL)The sum of currents entering/leaving a node/closed surface is 0.

i1 i2

i3

i5i4

i1 + i2 + i3 + i4 + i5 = 0

i1

i2

i3

i4i5

EECE 413, Fall, 20066

Modeling of a simple electrical system:

dtdvRCRiv o

R

dtdvCi o

0

0

io

oiRo

iRo

vdtdvRCvvvv

vvv

iooioo vvvRCvv

dtdvRC

To find out the relationship between the input (v i) and the output (vo) for the circuit:

vi

i

vo

RC

By KVL, we have

A dynamic model of the circuit

EECE 413, Fall, 20067

Dynamic AnalogiesPhysically dissimilar systems can have similar dynamic properties. For example, consider a spring-mass system.

MK

B

f(t)

v(t)

M = mass, B = friction constant, K = spring constant, f(t) = external force, and v(t) = velocity of the mass.

Summing the forces yields:

t

dvKtBvdttdvMtf )()()()(

Inertial forcefriction force

spring force

EECE 413, Fall, 20068

More Dynamic Analogies

t

dvLR

tvdttdvCti )(1)()()(

i(t)

iC(t) iR(t) iL(t)

C R L

Electric network

capacitanceresistance

inductance

Summing the currents (Kirchhoff’s Current law) yields:

EECE 413, Fall, 20069

Mechanical and Electrical systems are dynamically analogous

t

dvLR

tvdttdvCti )(1)()()(

t

dvKtBvdttdvMtf )()()()(

Thus, understanding one of these systems gives insights into the other.

EECE 413, Fall, 200610

Block Diagrams

KxxBxMdvKtBvdttdvMtf

t

)()()()(

dttdv )(

t

dv )(

A block diagram using integrators, adders, and gains:

f(t)+ 1/M

v(t)

B

K

+

- -

Conclusion: We have seen several types of descriptions of systems: spring-mass, RLC circuit, block diagram, and equations. All four descriptions define a system with the same dynamic properties. We will develop methods that characterize these systems efficiently and that abstract their critical dynamic properties.

Solve for x

KxxBtfM

x

KxxBtfxM

)(1)(

where )(&)( tvxdttdvx

for starting point

EECE 413, Fall, 200611

Voltage Control of the Simple Electrical System

Controller

REFERENCEINPUT OUTPUT

?

220 voltsvi vo

Control engineering is to design the controller in a way that the system meets desired performance specifications.

ioo vvvRC

vi

i

voR C

EECE 413, Fall, 200612

Ordinary Differential Equations (ODE)

tubtyatyaty 001

tubtyaty 00

Many real life problems can be modeled as an ODE of the following form:

This is called a 2nd order ODE as the highest order derivative in the equations is 2. The ODE is said to be homogeneous if u(t) = 0. Many systems can be modeled or approximated as a 1st order ODE.

An ODE is also called the time-domain model of the system, because it can be seen from the above equations that y(t) and u(t) are functions of time t. The key issue associated with ODE is: how to find its solution? That is: how to find an explicit expression for y(t) from the given equation(s)?

EECE 413, Fall, 200613

Laplace Transform and Inverse Laplace Transform

0 5 10-1

-0.5

0

0.5

1

mag

nitu

de

0.1 0.2 0.3 0.4 0.50

5

10

15

Freq

uenc

y m

agni

tude

of y

frequency (Hz)

0 5 10-2

-1

0

1

2

mag

nitu

de

0 0.5 10

5

10

15

Freq

uenc

y m

agni

tude

of y

frequency (Hz)

Let’s first examine the following time-domain functions:

Cosine function with a frequency f = 0.2 HzNote it has a period T = 1/f = 5 seconds

x(t) = cos(0.4*pi*t) + sin(0.8*pi*t)*cos(1.6*pi*t)What are the frequencies of this function?

The Laplace transform is a tool to convert a time-domain function into a frequency-domain function in which information about frequencies of the function can be captured. It is often much easier to solve problems in frequency-domain with the help of Laplace transform.

Signal plots

PSD plots

f = 0.2 Hz f = 0.2, 0.4, & 1.2 Hz

EECE 413, Fall, 200614

% l1_sinusoids.m%dt = 0.05; f1 = 0.2;f2 = 0.4;f3 = 0.8; ntime = 10;ti = [0:dt:ntime-dt];s1 = cos(f1*2*pi*ti);s2 = s1 + sin(f2*2*pi*ti).*cos(f3*2*pi*ti);figure(1) subplot(221)plot(ti',s1'),grid on % for 1 second trajectoryylabel('magnitude')

subplot(223)nfft = 512;Y = fft(s1',nfft);Pyy = Y.* conj(Y) / nfft;f = 1*(0:nfft/2)/nfft/dt;plot(f,Pyy(1:nfft/2+1)),grid onylabel('Frequency magnitude of y')xlabel('frequency (Hz)') subplot(222)plot(ti',s2'),grid on ylabel('magnitude') subplot(224)nfft = 512;Y = fft(s2',nfft);Pyy = Y.* conj(Y) / nfft;f = 1*(0:nfft/2)/nfft/dt;plot(f,Pyy(1:nfft/2+1)),grid onylabel('Frequency magnitude of y')xlabel('frequency (Hz)')

Matlab Code for Previous Sinusoid Plots

EECE 413, Fall, 200615

Laplace Transform

0

dtetftfLsF st

0Re

11101111 0

000

ssss

es

es

es

dtedtetfsF ststst

atetf

asasasas

eas

dtedteedtetfsF tastasstatst

Re

111011

0000

Given a time-domain function f(t), its Laplace transform is defined as follows:

Example 1: Find the Laplace transform of a constant function f(t) = 1.

Example 2: Find the Laplace transform of an exponential function

EECE 413, Fall, 200616

Inverse Laplace Transform

2

2

1

1

1

11

)()(

aste

ase

st

s

sFtf

at

at

22

22

22

22

cos

sin

cos

sin

)()(

basasbte

basbbte

assat

asaat

sFtf

at

at

Given a frequency-domain function F(s), the inverse Laplace transform is to convert it back to its original time domain function f(t).

Here are some useful Laplace and inverse Laplace transform pairs. (See Appendix B)

EECE 413, Fall, 200617

Laplace Transform Properties

sFasFatfLatfLatfatfaL 221122112211

ssFtfsLtfLdttdfL

sFstfLstfLdt

tfdL 222

2

sFs

tfLs

dfLt 11

0

1. Superposition:

2. Differentiation: Assume that f(0) = 0.

3. Integration:

EECE 413, Fall, 200618

Express ODE models using Laplace Transforms to get transfer functions

ubvvm

sG

bmssUsVsUsVbms

sUsbVsmsVuLvbLvmsLuLbvLvmLuLbvvmL

1

The mechanical system in the cruise-control problem can be represented by an ODE:

Taking Laplace transforms on both sides of the equation, we obtain

G(s) is called the transfer function of the system model.

EECE 413, Fall, 200619

Classical Feedback Control in Continuous Setting

)()()()()()( sUsGsYsUsYsG

)()()()()()()()()()( sYsRsDsGsEsDsGsUsGsY

)()(1)()(

)()()(

sDsGsDsG

sRsYsH

Examine block diagram of continuous time (CT) control system:

R(s)D(s)

E(s)G(s)

U(s) Y(s)+

-

Note that the objective of control system design is to match the output Y(s) to the reference R(s). So, we must compute the relationship between them.

We also have U(s) = D(s)E(s), and E(s) = R(s) – Y(s), and

)()()()()()(1)()()()()()()()()(

sRsDsGsYsDsGsYsDsGsRsDsGsUsGsY

Closed-loop transfer function from R to Y.

EECE 413, Fall, 200620

Simplified Block Diagram in Continuous Setting

)()(1)()(

)()(1)()(1)()()()()()(

sDsGsRsR

sDsGsDsGsRsHsRsYsRsE

)()(1)()()(sDsG

sDsGsH

The CT block diagram of the previous control system may be simplified as,

R(s) Y(s)

The control problem is to choose an appropriate D(s) such that H(s) will have desired properties. We have,

The problem is equivalent to finding an appropriate control law D(s) such that the resulting error function e(t) goes to zero as quick and smooth as possible, which is the same as saying that the output y(t) is tracking the given reference r(t) as quick and as smooth as possible. We will mainly focus on the cases when r(t) is either a step function or a ramp function.

r(t) y(t)

EECE 413, Fall, 200621

Two Approaches to Design a Digital ControllerThe 1st is to discretize the controller:

R(s)D(s)

E(s)G(s)

U(s) Y(s)+

-

R(s)D(z) G(s)

Y(s)+

-

ZOH

T

The 2nd is to discretize the plant and then apply digital control techniques for the controller:

R(z)D(z)

E(z)U(z)

Y(z)

+-

G(s)ZOH

T

R(z)D(z)

E(z)G(z)

U(z) Y(z)+

-

EECE 413, Fall, 200622

Classical Feedback Control in Discrete Setting

)()()()()()( zUzGzYzUzYzG

)()()()()()()()()()( zYzRzDzGzEzDzGzUzGzY

)()(1)()(

)()()(

zDzGzDzG

zRzYzH

Examine block diagram of discrete time (DT) control system:

R(z)D(z)

E(z)G(z)

U(z) Y(z)+

-

Note that the objective of control system design is to match the output Y(z) to the reference R(z). So, we must compute the relationship between them.

We also have U(z) = D(z)E(z), and E(z) = R(z) – Y(z), and

)()()()()()(1)()()()()()()()()(

zRzDzGzYzDzGzYzDzGzRzDzGzUzGzY

Closed-loop transfer function from R to Y.

EECE 413, Fall, 200623

Simplified Block Diagram in Discrete Setting

)()(1)()(

)()(1)()(1)()()()()()(

zDzGzRzR

zDzGzDzGzRzHzRzYzRzE

)()(1)()()(zDzG

zDzGzH

The DT block diagram of the previous control system may be simplified as,

R(z) Y(z)

The control problem is to choose an appropriate D(z) such that H(z) will have desired properties. We have,

The problem is equivalent to finding an appropriate control law D(z) such that the resulting error function e(k) goes to zero as quick and smooth as possible, which is the same as saying that the output y(k) is tracking the given reference r(k) as quick and as smooth as possible. We will mainly focus on the cases when r(k) is either a step function or a ramp function.

r(k) y(k)

EECE 413, Fall, 200624

Steady State Errors

)()()(1

1)()( sS

sGsDsRsE

)()(lim

,11

0sGsDK

Ke

sp

pss

In order to evaluate steady-state errors must evaluate the system error e:

S(s) is called the sensitivity. In the step input case to a stable system the final value theorem tells us:

Position error constant Kp

If Kp and ess are finite (no s in D(s)G(s)), the system is type 0.

)()(lim0

sGssDKsv

Kv is the velocity constant & if finite (one s in D(s)G(s)), the system is type 1.

)()(lim 2

0sGsDsK

sa Ka is the acceleration constant & if finite

(two s’ in D(s)G(s)), the system is type 2.

See Table 2.1, p. 23, for error vs. system type.

EECE 413, Fall, 200625

Steady State Errors - continuous

pss KDGsDsGs

sssEe

1

1)0()0(1

1)()(1

11lim)(lim)(00

Case 1: If r(t) is a unit step, R(s) = 1/s, & by the final value theorem of Laplace Transform

Obviously, we need the proportional error constant to be infinity, which implies G(s)D(s) has at least a factor of 1/s (type 1 system) in order to make e(inf) zero. If Kp is a finite scalar, the open-loop system G(s)D(s) is said to be a Type 0 system.

Here we need the velocity error constant to be infinity, which implies G(s)D(s) has at last a factor of 1/s2 (Type 2 system) in order to make e(inf) zero.

ProportionalError constant

Case 2: If r(t) is a unit ramp, R(s) = 1/s2, & by the final value theorem of Laplace Transform

vsss KsDssGsDsGs

sssEe 1)()(lim

1)()(1

11lim)(lim)(0

200

VelocityError constant

EECE 413, Fall, 200626

Steady State Errors - discrete

p

zz KDGzDzGzzzzEze

11

)1()1(11

)()(11

11lim)(1lim)(

11

Case 1: If r(k) is a unit step, R(z) = z/(z-1), & by the final value theorem of z-Transform

Obviously, we need the proportional error constant to be infinity, which implies G(z)D(z) has at least a factor of 1/(z-1) (type 1 system) in order to make e(inf) zero. If Kp is a finite scalar, the open-loop system G(z)D(z) is said to be a Type 0 system.

Here we need the velocity error constant to be infinity, which implies G(z)D(z) has at last a factor of 1/(z-1)2 (Type 2 system) in order to make e(inf) zero.

ProportionalError constant

Case 2: If r(k) is a unit ramp, R(z) = Tz/(z-1)2, & by the final value theorem of z-Transform

vz

z KzDzGzT

zDzGzTzze 1

)()(1lim)()(11

11lim)(

1

21

VelocityError constant

EECE 413, Fall, 200627

Final Value Theorem

• Steady state accuracy• Stability• Rise time• Overshoot• Settling time• Disturbance rejection• others

)(lim)(lim ssXxtxossst

If a system is stable and has a final, constant value

The final value theorem allows us to solve for that final value without extensive simulation. Useful for steady-state error estimation for control systems.

Time Domain Specifications

1

0.9

0.1

tr

Rise time

Mp (in %)overshoot

1% settling time

ts

Steady state error

+-1%

t

Transport delay – e-tds

td

EECE 413, Fall, 200628

Feedback Control System Block Diagram Properties

)()()(1)()(

)()( s

sGsDsGsD

sRsY

0)()(1 sGsD

)()()(1

1)()( sS

sGsDsRsE

Figure 2.8, p. 22r

CommandInput

++

Compensation

D(s)e

+

Disturbance, w

-+

+ Plant

G(s)y

Sensor

1+

Measurement noise, v

+

+

Transfer function of the closed-loop system, also called complementary sensitivity

The dynamic characteristics and stability are determined by the poles of the closed-loop transfer function of the characteristic equation.

The sensitivity equation is useful to determine the effect of disturbances on the system error.

Step Ramp Parabola

Type 0 1/(1+Kp) ∞ ∞

Type 1 0 1/Kv ∞

Type 2 0 0 1/Ka

EECE 413, Fall, 200629

Simplified Block Diagram in CT with PI Controller

sksk

skksD

asbsG ipi

p

)(,)(

ip

ip

bksbkasbksbk

sDsGsDsGsH

2)()(1)()()(

The CT block diagram of the previous control system may be simplified as,

R(s) Y(s)

The control problem is to choose an appropriate D(s) such that H(s) will have desired properties between R(s) & Y(s). We’ll focus on control system design of some 1st order systems G(s) with a proportional-integral (PI) controller, D(s).

The closed-loop system H(s) is a 2nd order system as its denominator is a polynomial s of degree 2.

)()(1)()()(sDsG

sDsGsH

EECE 413, Fall, 200630

Control System Stability

11)( 2

s

sH

ase at

1

tt eety 5.0)(

15.05.0

set

15.0

15.0

111

11)()()( 2

sssss

sRsHsY

15.05.0

se t

Example 1: Consider a closed-loop system with,

R(s) = 1 Y(s)

We have,

Using the Laplace transform table, we have,

0 1 2 3 4 5 6 7 8 9 100

2000

4000

6000

8000

10000

12000Step Response

Time (sec)

Ampl

itude

This system is said to be unstable because the output response y(t) goes to infinity or is unbounded as time t gets large. This happens because the denominator of H(s) has one positive root at s = 1.

Matlab code:step(tf(1,[1 0 -1]),10),grid on

EECE 413, Fall, 200631


231)( 2

ss

sH

tt eety 2)(

atat e

assae

as

1,1

1&11212,021

221

)1(221

21

11

211

231)()()( 2

ABBBBABABAss

BAsBAss

sBsAsB

sA

sssssssRsHsY

Example 2: Consider a closed-loop system with,

R(s) = 1 Y(s)

We have, Using the Laplace transform table, for an impulse, we have,

This system is said to be stable to both an impulse and a step because the output response y(t) is bounded as time t gets large. This happens because the denominator of H(s) has no positive roots.

Matlab code:figure(1),subplot(211),impulse(tf(1,[1 3 2]),10),grid on,subplot(212),step(tf(1,[1 3 2]),10),grid on

For a step we have,

t

t

at

eksk

sss

ess

ekass

kasss

215.05.0,2

21211

1111

1,21

111

tttt eeeety 22 5.05.015.01)(

0 1 2 3 4 5 6 7 8 9 100

0.1

0.2

0.3

0.4Impulse Response

Time (sec)

Ampl

itude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8Step Response

Time (sec)

Ampl

itude

EECE 413, Fall, 200632

L.H.P.

Stable Region

R.H.P.

Unstable Region


22

2

2)(

nn

n

sssH

DuCxyBuAxx

02 22 nnss

Consider a general 2nd order system,

R(s) = 1 Y(s)

The system is stable if the denominator of the system has no positive roots. It is unstable if it has positive roots.

Consider a general linear system characterized by a state space form.

1. It is stable if and only if all the eigenvalues of A are in the open left-half plane.2. It is marginally stable if and only if A has eigenvalues are in the closed left-half plane with some

(simple) eigenvalues on the imaginary axis.3. It is unstable if and only if A has at least one eigenvalue in the right-half plane.

EECE 413, Fall, 200633

Stability – Discrete SystemsA discrete-time system is said to be stable if its denominator of the system has no roots or poles outside the unit circle. It is unstable if it has poles or roots outside the unit circle.

x

x x

Stablex

x

x

Marginally Stable

Unstable

x

xxConsider a 1st order low-pass filter:

y = a*yp + (1-a)*upWhere y is the output, yp is previous output, a is the filter pole, & up is the previous input. If |a|<=1 y is bounded. If |a|>1 is not bounded.

EECE 413, Fall, 200634

2nd Order Control System Stability

22

2

2)(

nn

n

sssH

Consider a general 2nd order system,

R(s) = 1 Y(s)

The behavior of the system is as follows:

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Ampl

itude

The behavior of the system is fully characterized by ς, which is called the damping ratio, and ωn, which is called the natural frequency.

Matlab code:% Two_order.mw=1;z=[0:0.1:1];for i=1:11,tfbig(i)=tf(w,[1 2*z(i)*w w*w]);end,for i = 1:11, step(tfbig(i),12),grid on,hold onend

EECE 413, Fall, 200635

Control System Design with Time-Domain Specs

22

2

2)(

nn

n

sssH

nrt

8.1

Consider a general 2nd order system again,

R(s) = 1/s Y(s)

See page 21 of Franklin - Powell

r(t) = 1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

ς

1

0.9

0.1

tr

Rise time

Mp (in %)overshoot

1% settling time

ts

Steady state error

+-1%

t

nst

6.4

10,21 eM p

Time domain specs for simple systems:• Rise time•Settling time•Overshoot

For complex systems must use simulation.

EECE 413, Fall, 200636

PI Design Technique:

asbsG

)(

sksk

skksD ipi

p

)(

22

2

2)(

nn

n

sssH

Design D(s) with PI (proportional-integral) structure:

R(s)D(s)

E(s)G(s)

U(s) Y(s)+

-

With G(s) & D(s) as defined above, the system becomes a closed-loop system.

ip

ip

bksbkasbksbk

sDsGsDsG

sRsYsH

2)()(1)()(

)()()(

Compare this with the standard 2nd order system by equating the denominators:

ipnn bksbkasss 222 2

in

pn

bk

bka

2

2

bk

bak

ni

np

2

2

The key issue now is to choose parameters kp & ki such that the above resulting system has desired properties, such as prescribed settling time & overshoot.

EECE 413, Fall, 200637

Cruise-Control Example Design 1

mbs

msUsV

1

)()(

4.0

767.010*6.06.4

*6.46.4

s

nn

s tt

For the cruise-control system, we have:

Assume the mass of the car m = 3000 kg and the friction coefficient b = 1. Design a PI controller for it such that the speed of the car will reach the desired speed 90 km/h in 10 seconds (i.e., the settling time is 10 s) and the maximum overshoot is less than 25%.

6.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

x

To achieve an overshoot < 25%, we obtain from the Mp figure that:

To be safe, we choose

To achieve a settling time of 10 s, we use

EECE 413, Fall, 200638

Cruise-Control Example Design 1 – cont.

000333.030001

30001

300011

)()(

bas

mbs

msUsV

bk

bak

ni

np

2

2

767.010*6.06.4

*6.46.4

s

nn

s tt

The transfer function of the cruise-control system is below. Assume the mass of the car m = 3000 kg and the friction coefficient b = 1:

Again, using the formulae derived,

The final cruise-control system:

1765

30001767.0

2760

30001

30001767.0*6.0*22

22

bk

bak

ni

np

70 MPHu Speed v

30001

30001

s

car

sssD

sksk

skksD ipi

p

17652760)(

)(

ss 17652760

EECE 413, Fall, 200639

Cruise-Control Example Design 1 – Results

10-1

100

101

-30

-20

-10

0

10

mag

nitu

de in

dB

frequency in rad/sec

0 2 4 6 8 10 120

0.5

1

1.5Step Response

Time (sec)

Am

plitu

de

70 MPHu Speed v

30001

30001

s

car

ss 17652760

ts spec = 10 sts result < 7 s

ωn spec = 0.767ωn result ~ 1.8 rad/s

% cruise_validate.ma = 1/3000;b = a;g1 = tf(b,[1 a]); d1 = tf([2760 1765],[1 0]);sys1 = d1*g1;sys1 = feedback(sys1,1); figure(1)subplot(211)step(sys1,12);grid on subplot(212)[mag,phase,w] = bode(sys1);semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid onylabel('magnitude in dB')xlabel('frequency in rad/sec')

Risky design. High control bandwidth gives poor disturbance rejection.

EECE 413, Fall, 200640

Cruise-Control Example Design 2

mbs

msUsV

1

)()(

46.0

82.08*7.06.4

*6.46.4

s

nn

s tt

For the cruise-control system, we have:

Assume the mass of the car m = 1000 kg and the friction coefficient b = 100. Design a PI controller for it such that the speed of the car will reach the desired speed 100 km/h from 0 in 8 seconds (i.e., the settling time is 8 s) and the maximum overshoot is less than 20%.

7.0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

x

To achieve an overshoot < 20%, we obtain from the Mp figure that:

To be safe, we choose

To achieve a settling time of 8 s, we use

EECE 413, Fall, 200641

Cruise-Control Example Design 2 – cont.

1.01000100,001.0

1000100

100011

)()(

bas

mbs

msUsV

bk

bak

ni

np

2

2

82.08*7.06.4

*6.46.4

s

nn

s tt

The transfer function of the cruise-control system is below. Assume the mass of the car m = 1000 kg and the friction coefficient b = 100:

Again, using the formulae derived,

The final cruise-control system:

670

10182.0

1050

101

1000182.0*7.0*22

22

bk

bak

ni

np

100 km/Hu Speed v

101

10001

s

car

sssD

sksk

skksD ipi

p

6701050)(

)(

ss 6701050

EECE 413, Fall, 200642

0 2 4 6 8 10 120

0.5

1

1.5Step Response

Time (sec)

Am

plitu

de

10-1

100

101

-20

-15

-10

-5

0

5

mag

nitu

de in

dB


Cruise-Control Example Design 2 – Results100 km/H

u Speed v

101

10001

s

car

ss 6701050

ts spec = 8 sts result < 7 s


% cruise_validate.m%a = 1/10;b = 1/1000;g2 = tf(b,[1 a]); d2 = tf([1050 670],[1 0]);sys2 = d2*g2;sys2 = feedback(sys2,1);figure(2)subplot(211)step(sys2,12);grid on subplot(212)[mag,phase,w] = bode(sys2);semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid onylabel('magnitude in dB')xlabel('frequency in rad/sec')

Design risk. High control bandwidth gives poor disturbance rejection.

EECE 413, Fall, 200643

Problem 2.1 in Text2

10)(s

sG

bsks

kb

kbssss

nn

nn

10,2

1022

222

Design feedback with lead compensation for the open-loop system:

kbssk

kbssk

bsskbssk

sGsDsGsD

bssk

sbssksGsD

1010

1010

101

10

)()(1)()(

1010)()(

2

2

R(t)

2

10s

The rise time should be <= 1 sec & the overshoot should be <= 10%.

D(s) G(s)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

6667.71*6.06.4

*6.46.4

s

nn

s tt

x

58.0

8778.510

,2.96667.7*6.0*222

n

n

k

b

2.98778.5)(

s

sbs

kssD

EECE 413, Fall, 200644

10-1

100

101

102

103

-100

-50

0

50

mag

nitu

de in

dB


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5Step Response

Time (sec)

Am

plitu

de

Problem 2.1 in Text – Results

2

10s2.9

8778.5s

s

ts spec = 1 sts result < 0.9 s

ωn spec = 7.6667 rad/sωn result ~ 9.0 rad/s

% Prob2_1.mk = 5.8778;b = 9.2;g1 = tf(10,[1 0 0]); d1 = tf([k 0],[1 b]);sys1 = d1*g1;sys1 = feedback(sys1,1); figure(1)subplot(211)step(sys1,2);grid on subplot(212)[mag,phase,w] = bode(sys1);semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid onylabel('magnitude in dB')xlabel('frequency in rad/sec')

Slightly high control bandwidth may give poor disturbance rejection.

EECE 413, Fall, 200645

Problem 2.2 (mod) in Text2

5)(s

sG

bsks

kb

kbssss

nn

nn

5,2

522

222

Design feedback with lead compensation for the open-loop system:

kbssk

kbssk

bsskbss

k

sGsDsGsD

bssk

sbssksGsD

55

55

51

5

)()(1)()(

55)()(

2

2

R(t)

2

5s

The rise time should be <= 1 sec & the overshoot should be <= 10%.

D(s) G(s)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

6667.71*6.06.4

*6.46.4

s

nn

s tt

x

58.0

7556.115

,2.96667.7*6.0*222

n

n

k

b

2.97556.11)(

s

sbs

kssD

EECE 413, Fall, 200646

10-1

100

101

102

103

-100

-50

0

50

mag

nitu

de in

dB


0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.5

1

1.5Step Response

Time (sec)

Ampl

itude

Problem 2.2 (mod) in Text – Results

2

10s2.9

7556.11s

s

ts spec = 1 sts result < 0.9 s


% Prob2_1.mk = 5.8778;b = 9.2;g1 = tf(10,[1 0 0]); d1 = tf([k 0],[1 b]);sys1 = d1*g1;sys1 = feedback(sys1,1); figure(1)subplot(211)step(sys1,2);grid on subplot(212)[mag,phase,w] = bode(sys1);semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid onylabel('magnitude in dB')xlabel('frequency in rad/sec')

Slightly high control bandwidth may give poor disturbance rejection.

EECE 413, Fall, 200647

PID Control

KsDtKetu )()()(

sTKsDde

TKtu

i

t

i

)()()(0

sT

sTK

sesusD d

i

11)()()(

sKTsDteKTtu dd )()()(

PlantPIDe(t)r(t) y(t)

Proportional control term:

Integral control term:

Derivative control term:

Where Ti is called the integral (or reset) time, Td the derivative time, and K the position feedback gain. The combined transfer function is:

EECE 413, Fall, 200648

PID Control• Proportional feedback control can lead to reduced errors to disturbances but still

maintains errors bounded.• It can reduce the rise time, but at the cost of a larger overshoot.• Integral control can eliminate steady state error, but can contribute to oscillation as

contributes an additional pole.• Derivative control adds damping at the expense of disturbance noise sensitivity.

Usually this term or a component is low-pass filtered.

Root Locus• Root locus shows how changes in the system’s open-loop characteristics influence

the closed-loop dynamics characteristics. • Requires a variable open-loop parameter which produces a root-locus of

characteristic equation roots. • Allows study of the closed-loop roots while only knowing the factors (poles & zeros) of

the open-loop system.• The term D(s)G(s) (open-loop transfer function) must be formulated as Kb(s)/a(s). • See drawing rules, p. 26. It is useful to help visualize the system.

EECE 413, Fall, 200649

Matlab Root Locus

210ss

103)(,

210)(

ssKsD

sssG

Example 2.2 (p. 29)1. Plot the root locus using matlab for the open-loop

system shown.2. Find the gain K associated with the point of

maximum damping and plot the step response with that value of K.

e(t)r(t) y(t)

103

ssK

-10 -5 0-4

-3

-2

-1

0

1

2

3

40.20.40.580.720.830.91

0.96

0.99

0.20.40.580.720.830.91

0.96

0.99

246810

Root Locus

Real Axis

Imag

inar

y A

xis

0 0.5 1 1.5 20

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Ampl

itude

% example2_2.m%numD = [1 3];denD = [1 10];numG = 10;denG = [1 2 0];sys = tf(numD,denD)*tf(numG,denG);k = 0:0.1:4;figure(1)subplot(121)rlocus(sys,k),grid on% [K,p] = rlocfind(sys);subplot(122)K = 3.7;sysCL = feedback(K*sys,1);step(sysCL,2),grid on

EECE 413, Fall, 200650

Stability Margins

1,180

180,1

ggggg

ppppp

jGjKjGjKanglePM

jGjKanglejGjK

GM

Gain margin (GM), phase margin (PM), & V margin (VM)

All root locus points have the property that |KG(s)| = 1 & (KG(s)) = 180 degrees. At the point of neutral stability these root-locus conditions hold for s = jw, so |KG(s)| = 1 & (KG(s)) = 180 degrees.

The GM is the factor by which the gain < the neutral stability value when the phase = 180 degrees. The PM is the amount by which the phase of G(s) exceeds – 180 degrees when |KG(s)| = 1. VM is a combination of both.

PM is generally related to the damping of a system. For a 2nd order system, is commonly used. Thus, the speed of response of the system is associated with the GM while the stability is associated with the PM.

100PM

GM is the maximum additional gain you can apply to the closed-loop system such that it will still remain stable. Similarly, PM is the maximum phase you can add to the closed-loop system such that it will still remain stable.

EECE 413, Fall, 200651

Example 2.3 211)(

ss

KsGFor the plant: determine the PM and GM for the system with unity feedback and a) K=1, b) determine if the system is stable if K = 5, & c) find what value of K is required to achieve a PM of 45 & 70.

-150

-100

-50

0

50

100

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = 6.02 dB (at 1 rad/sec) , Pm = 21.4 deg (at 0.682 rad/sec)

Frequency (rad/sec)

margin(tf(1,[1 2 1 0]))GM

PM

Unstable line < -180

Read system bandwidth frequency following ‘crossover’ @ -3 dB

a) K=1

EECE 413, Fall, 200652

-150

-100

-50

0

50

100

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = -7.96 dB (at 1 rad/sec) , Pm = -23.2 deg (at 1.52 rad/sec)

Frequency (rad/sec)

Example 2.3 – cont. 211)(

ss


margin(tf(1,[1 2 1 0]))GM

PM


20*log10(1/5) ~ -14 dB -> phase < -180 or system is unstable with K = 5. For stability both PM & GM must be stable.

b) K=5 stable?

Note that change in gain only moves magnitude plot up or down, but doesn’t affect phase plot.

K=5

K=1

EECE 413, Fall, 200653

-150

-100

-50

0

50

100

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-270

-225

-180

-135

-90

Pha

se (

deg)

Bode DiagramGm = 20 dB (at 1 rad/sec) , Pm = 68.2 deg (at 0.193 rad/sec)

Frequency (rad/sec)

Example 2.3 – cont. 211)(

ss


hold on,margin(tf(5,[1 2 1 0])),margin(tf(1,[1 2 1 0])),margin(tf(.5,[1 2 1 0])),margin(tf(.2,[1 2 1 0])),hold off


c) K=? for PM = 45 & 70?

K=5 crossover

K=1 XoverK=.5 XoverK=.2 Xover

K=1 PM=21K=.5 PM=45K=.2 PM=68 K=5 PM=-22

Note that change in gain only moves magnitude plot up or down, but doesn’t affect phase plot.

EECE 413, Fall, 200654

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

Example: Root Locus Design Method for DC Motor Position Control

230,5.0 n

5.0

From the main problem, the dynamic equations in transfer function form are the following:

and the system schematic looks like:

With a 1 rad/sec step reference, the design criteria are: Settling time less than 0.04 seconds Overshoot less than 16% No steady-state error to a reference No steady-state error due to a disturbance Now let's design a controller using the root locus method

Matlab Code:% motor.mJ=3.2284E-6;b=3.5077E-6;K=0.0274;R=4;L=2.75E-6;num=K;den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2) 0];motorTF=tf(num,den); gain = 0:0.1:20;rlocus(motorTF,gain)

First, try simple P (proportional controller).Note motor parameters given.

x

23004.0*5.06.4

*6.46.4

s

nn

s tt

EECE 413, Fall, 200655

Example: Root Locus Design Method for DC Motor Position Control – cont.

-15 -10 -5 0

x 105

-250

-200

-150

-100

-50

0

50

100

150

200

250Root Locus

Real Axis

Imag

inar

y A

xis

Very large pole

-250 -200 -150 -100 -50 0 50

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0.2

0.25Root Locus

Real Axis

Imag

inar

y A

xis

Close up view, note bandwidth spec = 230 rad/s

With P only controller, motor will never be fast enough to satisfy bandwidth of 230 rad/s.Also, recall that we need an integrator in the controller (not just in the system) to remove steady-state error due to a disturbance.

EECE 413, Fall, 200656

-300 -250 -200 -150 -100 -50 0

-80

-60

-40

-20

0

20

40

60

80 0.5

0.5

Root Locus

Real Axis

Imag

inar

y A

xis

-300 -250 -200 -150 -100 -50 0

-200

-150

-100

-50

0

50

100

150

200

Root Locus

Real Axis

Imag

inar

y A

xis


Close up view, note bandwidth spec = 230 rad/s

With PI only controller, motor will never be fast enough to satisfy bandwidth of 230 rad/s.In order to pull the root locus further to the left, to make it faster, we need to place a second open-loop zero, resulting in a PID controller. After some experimentation, we can place the two PID zeros at s=-60 and s=-70.

PI controller

gain = [0:.1:20];contr=tf([1 20],[1 0]);rlocus(contr*motorTF,gain)

PID controllergain = [0:.01:20];numc=conv([1 60],[1 70]);denc=[1 0];contr=tf(numc,denc);rlocus(contr*motorTF,gain)

EECE 413, Fall, 200657

motor step response, k = 0.1565, dominant poles are -230 -101.756 -59.6074

Time (sec)

Am

plitu

de

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.2

0.4

0.6

0.8

1

1.2

1.4

System: sys_clTime (sec): 0.0402Amplitude: 1.01



Overshoot criteria met @ 12% @ 0.013 s and no steady state error.

If you recall, we need the settling time and the overshoot to be as small as possible, particularly because of the effect of the extra zero. Large damping corresponds to points on the root locus near the real axis. A fast response corresponds to points on the root locus far to the left of the imaginary axis. To find the gain corresponding to a point on the root locus, we can use the rlocfind command. We can find the gain and plot the step response using this gain all at once.

PID controller

gain = [0:.01:20];% contr=tf([1 20],[1 0]);% rlocus(contr*motorTF,gain)numc=conv([1 60],[1 70]);denc=[1 0];contr=tf(numc,denc);rlocus(contr*motorTF,gain)sgrid(0.5,0)sigrid(230)[k,poles] = rlocfind(contr*motorTF,-230)sys_cl=feedback(k*contr*motorTF,1);t=0:0.001:.1; step(sys_cl,t),grid ontitle(['motor step response, k = ',num2str(k),... ', dominant poles are ',num2str(poles(2:4)')])

Settling time criteria met @ 1% @ 0.04 s.

sss

sssss

4200130420013070601565.

2

EECE 413, Fall, 200658

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

System: dist_clTime (sec): 0.04Amplitude: 0.0206

motor step response

Time (sec)

Am

plitu

de


The disturbance response may be found by computing the closed-loop disturbance transfer function and plotting its step response. As you recall the controller must minimize disturbance.

The response to a step disturbance reaches a steady-state value of zero, and in fact, stays within 0.02 (or 2%) after 0.04 seconds. Therefore, all the design requirements have been met.

dist_cl=feedback(motorTF,k*contr);step(dist_cl,t),grid ontitle('motor step response')

EECE 413, Fall, 200659

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

10

20

30

40

50

60

70

80

90

100

Mp,

%

Example: Bode Plot Design Method for DC Motor Position Control

505.0*100100100

230,5.0

PM

PMn

5.0

From the main problem, the dynamic equations in transfer function form are the following:

and the system schematic looks like:

With a 1 rad/sec step reference, the design criteria are: Settling time less than 0.04 seconds Overshoot less than 16% No steady-state error to a reference No steady-state error due to a disturbance Now let's design a controller using the bode plot method

Matlab Code:% motor_bode.mclear allJ=3.2284E-6;b=3.5077E-6;K=0.0274;R=4;L=2.75E-6;num=K;den=[(J*L) ((J*R)+(L*b)) ((b*R)+K^2) 0];motorTF=tf(num,den);contr=1;margin(contr*motorTF),

First, try simple P (proportional controller).Note motor parameters given. Try to get PM = 50 & GM = 3 dB @ 230 rad/s

x

23004.0*5.06.4

*6.46.4

s

nn

s tt

EECE 413, Fall, 200660

-300

-200

-100

0

100

Mag

nitu

de (d

B)

100

102

104

106

108

-270

-225

-180

-135

-90

Phas

e (d

eg)

Bode DiagramGm = 92.2 dB (at 9.28e+003 rad/sec) , Pm = 61.9 deg (at 31.6 rad/sec)

Frequency (rad/sec)

Example: Bode Plot Design Method for DC Motor Position Control – cont.

Again, while from the root locus analysis we removed steady state error, the PM potential is too low. Can’t satisfy the bandwidth = 230 rad/sec while satisfying GM & PM.

Increasing gain K raises magnitude curve without changing phase curve. However, mag crossover moves to the right reducing the PM in this case. Can’t use P controller since the PM will go negative long before GM is satisfied, therefore, try PI.

Increasing gain

Decreasing gain

Increasing gainDecreasing gain

P controller

-300

-200

-100

0

100

Mag

nitu

de (d

B)

100

102

104

106

108

-270

-225

-180

-135

Phas

e (d

eg)

Bode DiagramGm = 88.6 dB (at 7.55e+003 rad/sec) , Pm = 29.7 deg (at 35.3 rad/sec)

Frequency (rad/sec)

PI controller

motorTF=tf(num,den);contr=1;margin(contr*motorTF),

contr=tf([1 20],[1 0]);margin(contr*motorTF),

EECE 413, Fall, 200661

motor step response, k = 0.09

Time (sec)

Ampl

itude

motor step response

Time (sec)

Ampl

itude

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1-0.02

0

0.02

0.04

0.06System: dist_clTime (sec): 0.0241Amplitude: 0.0202

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

0.5

1

1.5



-150

-100

-50

0

50

100

150

Mag

nitu

de (d

B)

100

102

104

106

108

-180

-135

-90

Phas

e (d

eg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 50.2 deg (at 234 rad/sec)

Frequency (rad/sec)

Example: Bode Plot Design Method for DC Motor Position Control – cont.

sss

sssss 42001301565.42001301565.70601565.

2

This is quick & dirty solution starting with root locus solution. Bandwidth is close to 230 rad/sec and PM is good. However, overshoot doesn’t meet specs<16%.

PID controller

sss

sssss 1400024009.1400024009.14010009.

2

numc1=conv([1 60],[1 70]);numc2=conv([1 100],[1 140]);denc=[1 0];contr1=tf(numc1,denc);contr2=tf(numc2,denc);margin(0.1565*contr1*motorTF),hold onmargin(0.09*contr2*motorTF),hold off

sys_cl=feedback(k2*contr2*motorTF,1);t=0:0.001:.1;step(sys_cl,t),grid ontitle(['motor step response, k = ',num2str(k2)])dist_cl=feedback(motorTF,k2*contr2);step(dist_cl,t),grid ontitle('motor step response')

EECE 413, Fall, 200662

Lead Compensation

maxmax

max 1,sin1sin1

and

Lead compensation has a transfer function of: For small α, it is an approximation of PD control. The frequency response of a typical lead compensator is shown in the figure below:

Here parameters K, φmax and ωmax are to be determined from design specifications.

1,11)(

ssKsD

EECE 413, Fall, 200663

Uncompensated and lead-compensated systems

The Key Idea

The key idea using the lead compensation is to enlarge the gain crossover frequency and add additional phase. Thus, if the desired gain crossover frequency is larger than that of KG(s), the lead compensation should be used.

EECE 413, Fall, 200664

Lead Compensation Design Procedure

maxmax

max 1,sin1sin1

Step 1: Determine open-loop gain K to satisfy requirements on steady state error.Step 2: Find new open-loop crossover frequency from desired ωn=ωmax, the point the phase lead is added.Step 3: Evaluate the PM of KG(ω) at the desired cross over frequency ωn.Step 4: Allow for some extra PM (5 to 12 degrees), and determine the needed phase lead φmax. Step 5: Compute

Step 6: Verify the design using matlab. Redo if needed.

EECE 413, Fall, 200665

Lead Compensator Design Example

9.05.05.05.0)(

zz

zzG

29.1105.1

2020

9.020205.0

2020

5.020205.0

|)()(/2/2

www

ww

ww

ww

zGwGwTwTz

An electric motor can be modeled with a sampling period T = 0.1 s as follows:

Design a digital control system with a lead compensator such that the resulting system output tracks a step reference with a settling time less than 1s with an overshoot < 20% and steady state error < 12.5%.

Converting this to w-plane we have:

EECE 413, Fall, 200666

Lead Compensator Design Example

50100 PM

sn t

6.4

overshootx

20%

35.079.015.01

5.05.0)1()1(

,7)()(1125.0

11)(

KKxGDK

eeK

Ke

p

pp

Settling time Open-loop gain crossover frequency ωg ~ ωn = 9

1 second 2.915.06.4

xn ωg ~ ωn = 9 Step 2

Steady state error – Step 1: Determination of the static gain K for this type 0 system.

EECE 413, Fall, 200667

Bode DiagramGm = 9.95 dB (at 12.7 rad/sec) , Pm = 43.8 deg (at 5.72 rad/sec)

Frequency (rad/sec)

-60

-40

-20

0

20

40

Mag

nitu

de (d

B)

System: sysFrequency (rad/sec): 8.93Magnitude (dB): -5.42

10-2

10-1

100

101

102

103

90

180

270

360

Phas

e (d

eg) System: sys

Frequency (rad/sec): 9.05Phase (deg): 199

Step 3:

K = 1K = 0.35

Step 4:Φmax = Desired PM – Original PM + Allowance = 50 + (180 – 199) + 4 = 35

% lead_example.m%numG = 20*[-1 20];denG = conv([1.5 10],[1.9 2]);sys = tf(numG,denG);figure(1)margin(sys),hold onnumG1 = 0.35*20*[-1 20];sys1 = tf(numG1,denG);margin(sys1),hold off

EECE 413, Fall, 200668

Step 5

1213.027.0

1213.035.011)(

ww

wwKwD

213.01,27.0

180/35sin1180/35sin1

sin1sin1

maxmax

max

Finally, we obtain the lead compensator in w-domain:

Which can be converted back to z-domain using the inverse bilinear transformation. T = 1/30/9.2 = 0.0036 s. Use T = 0.01.

16.016.2148.1848.1

111

01.02058.0

111

01.02213.0

35.0|)()(112

zz

zzzz

wDzDzz

Tw

The above is the digital lead compensator. However, nothing is certain without verification. We need to first verify our design in frequency domain. More importantly, it should also meet the design specifications in time domain.

EECE 413, Fall, 200669

Verification in w-domain (pseudo s-domain

-60

-40

-20

0

20

Mag

nitu

de (d

B)

10-2

10-1

100

101

102

103

90

180

270

360

Phas

e (d

eg)


Frequency (rad/sec)

Resulting gain crossover frequency is 9.21 rad/s & PM = 52.7Desired gain crossover frequency is 9.2 rad/s & PM = 54

% lead_example.m%numG = 20*[-1 20];denG = conv([1.5 10],[1.9 2]);sys = tf(numG,denG);figure(1)margin(sys),hold onnumG1 = 0.35*20*[-1 20];sys1 = tf(numG1,denG);margin(sys1),hold offnumD = 0.35*[0.214 1];denD = [0.058 1];sysC = tf(numD,denD);figure(2)margin(sysC*sys)

EECE 413, Fall, 200670

Verification in z-domain% lead_example.m%numG = 20*[-1 20];denG = conv([1.5 10],[1.9 2]);sys = tf(numG,denG);figure(1)margin(sys),hold onnumG1 = 0.35*20*[-1 20];sys1 = tf(numG1,denG);margin(sys1),hold offnumD = 0.35*[0.214 1];denD = [0.058 1];sysC = tf(numD,denD);figure(2)margin(sysC*sys)% verification in z domainfigure(3)margin(c2d(sysC*sys,0.04)),hold onmargin(c2d(sysC*sys,0.02)),hold onmargin(c2d(sysC*sys,0.01)),hold off

-30

-20

-10

0

10

20

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

103

-90

0

90

180

270

360

Pha

se (d

eg)


Frequency (rad/sec)

Sampling rate of 0.01 matches specs well, however reducing the sampling rate doesn’t change the 0 dB magnitude crossover, but reduces the phase margin noticeably. Note that the 4 degree continuous PM allowance forced the discrete PM to be very close to 50.

EECE 413, Fall, 200671

Verification in Time Domain

Steady state error = 12.5%

Step Response

Time (sec)

Ampl

itude

0 0.5 1 1.5-0.2

0

0.2

0.4

0.6

0.8

1

1.2

System: untitled2Time (sec): 1.24Amplitude: 0.875




Settling time = 0.63s

Overshoot = 23.4%

EECE 413, Fall, 200672

Lag Compensation1,

11)(

ssKsDLag compensation has a transfer function of:

For large α, it is an approximation of PI control. The frequency response of a typical lag compensator is shown in the figure below:

Key idea: Lag compensator reduces the gain crossover frequency to where the phase margin is required to be.

EECE 413, Fall, 200673

PID CompensationNote that the lead compensation is the approximation of PD control and the lag compensation approximates PI control. PID compensation is nothing more than the combination of lead compensation (PD) and lag compensation (PI), which has the following frequency responses:

Design procedures for the lag compensation and PID compensation can be found in almost any textbook on introduction to classical control.

EECE 413, Fall, 200674

function cf = optim_hmwk1(xin);%sys1 = tf(50,[1 0 0]);gc1 = tf(xin(1)*[1 xin(2)],1); [Gm,Pm,Wcg,Wcp] = margin(gc1*sys1);[mag,phase,w] = bode(gc1*sys1);i = find(mag >= 0.7);[y,i] = min(mag(i)); [y,t] = step(feedback(gc1*sys1,1),20);if max(y)>1.22, cf = abs(Pm - 45)/100 + abs(0.92 - w(i)) + 15*(max(y)-1.22);else cf = abs(Pm - 45)/100 + abs(0.92 - w(i));end[xin, cf]save hmwk1_param xin

Optimization of Controllers2

50)(s

sG Assume a plant: with time domain specifications: ts = 10 sec & Mp <= 22%.

How can we avoid the tedious manual iterations needed to solve? Optimization using matlab’s fminsearch is one approach.

load hmwk1_param% xin = [0.0130 0.7588]; xout = fminsearch('optim_hmwk1',xin);

dpdp KsKsKKsD )(Assume a controller:

The 2nd form allows specifying zero.

General steps:•Determine specifications•Generate plant representation•Determine and generate controller•Calculate numerically specifications•Generate optimization cost function•Code cost function in matlab function script•Code calling function in matlab script

Phase margin/100 ~ damping coefficient

For bandwidth check

Do step for overshoot check

EECE 413, Fall, 200675

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



10-2

10-1

100

101

-30

-20

-10

0

10

20

30

40

50

60

70

mag

nitu

de in

dB


-50

0

50

100

150

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

-180

-90

0

90

Pha

se (

deg)

Bode DiagramGm = Inf , Pm = 63.7 deg (at 0.699 rad/sec)

Frequency (rad/sec)

gcgpgc*gp

Optimizer Example 1 Results load hmwk1_param sys1 = tf(50,[1 0 0]); figure(1) margin(sys1),grid on figure(2) gc1 = tf(xin(1)*[1 xin(2)],1); margin(gc1),grid on,hold on margin(sys1),grid on,hold on margin(gc1*sys1),grid on,hold off legend('gc','gp','gc*gp') figure(3) [mag,phase,w] = bode(gc1*sys1); semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid on ylabel('magnitude in dB') xlabel('frequency in rad/sec') figure(4) step(feedback(gc1*sys1,1));grid on

BW ~ 0.92

GM = 63.7Mp = 22%

EECE 413, Fall, 200676

Optimizer Example 2sksk

sD ip )(

Generate a PI controller for the following plants:Tune the system to the time domain specifications: ts = 10 sec & Mp <= 22%. Iterate to match them as closely as possible. Solve using pole placement. Verify using step responses. Calculate an appropriate sampling rate. What are the steady state errors of the systems?

532)(,

52)(,

52)(

sssG

ssG

ssG

n

T*301

EECE 413, Fall, 200677

function out = optim_hmwk1b(xin);%sys1 = tf(2,[1 5]);gc1 = tf(xin(1)*[1 xin(2)],[1 0]); [Gm,Pm,Wcg,Wcp] = margin(gc1*sys1); [y,t] = step(feedback(gc1*sys1,1),10);wn = 0.657;z = 0.6;[y1,t] = step(tf(wn*wn,[1 2*z*wn wn*wn]),10);out = abs(Pm - 50) + sum(abs(y - y1)); [xin, out]save hmwk1_param xin

Optimization of Controllers, Example 252)(

s

sGAssume a plant: with time domain specifications: ts = 10 sec & Mp <= 22%.

load hmwk1_paramxout = fminsearch('optim_hmwk1',xin);

Assume a PI controller:

The 2nd form allows specifying zero.

Phase margin/100 ~ damping coefficient

Do step comparison for bandwidth & overshoot check

sksk

sksk

sD ipip

)(

Cost function

EECE 413, Fall, 200678

Step Response

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 80

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


10-1

100

101

102

103

104

105

-160

-140

-120

-100

-80

-60

-40

-20

0

20

mag

nitu

de in

dB


-200

-150

-100

-50

0

50

100

Mag

nitu

de (

dB)

10-1

100

101

102

103

104

105

-180

-135

-90

-45

0

Pha

se (

deg)

Bode DiagramGm = Inf , Pm = 82.9 deg (at 0.627 rad/sec)

Frequency (rad/sec)

gcgpgc*gp

Optimizer Example 2 Results load hmwk1_param sys1 = tf(2,[1 5]); figure(1) margin(sys1),grid on figure(2) gc1 = tf(xin(1)*[1 xin(2)],[1 0]); margin(gc1),grid on,hold on margin(sys1),grid on,hold on margin(gc1*sys1),grid on,hold off legend('gc','gp','gc*gp') figure(3) [mag,phase,w] = bode(gc1*sys1); semilogx(w,20*log10([mag(:),ones(length(w),1)*0.7])),grid on ylabel('magnitude in dB') xlabel('frequency in rad/sec') figure(4) step(feedback(gc1*sys1,1));grid on

BW ~ 0.92

GM = 82.9

ts <= 10 s

EECE 413, Fall, 200679

State-Space Design• Estimator/observer design• Control law design• Combine/integrate observer &

control law• Tune the control system to

achieve desired performance• Note that the order is different

from the text, because the observer dynamics must be incorporated into the control law.

GuFxx Hx

x̂

plant

+- rEstimator-K

uy

Control law

State space formulation

xHyLGuxFx ˆˆˆ is the estimator formulation where L is a proportional gain in the form of a vector or matrix depending on the dimensions of y.

xLHFx ˆˆ is the error equation from which the dynamics may be obtained.

The characteristic equation of the error is: det[sI – (F – LH)] = 0.

Estimator/Observer Design

EECE 413, Fall, 200680

State-Space Block Diagram

GuFxx s1

Hx

plant

+- rEstimator-K

u y

Control law

State space formulation

xHyLGuxFx ˆˆˆ

is the estimator formulation where L is a proportional gain in the form of a vector or matrix depending on the dimensions of y.

uG xdot x H y

F

J

L

s1 x̂x̂-K

F-LH-GK

Plant

Controller/Estimator

Control engineer’s job is to: 1) determine plant description [ F,G,H,J] matrices, 2) Calculate and justify L matrix for estimator, and 3) calculate and justify K matrix for controller to satisfy requirements.

- r+

EECE 413, Fall, 200681

Estimator/Observer Design• The 2 general approaches to observer design are:

– Pole placement– Optimal design (Kalman filter)

• Both rely on selection of the observer eigenvalues or observer poles.

• The observer poles are selected to be 2-10 faster than the controller poles. This allows the control derived poles to dominate. It allows a faster decay of estimator errors compared with the desired dynamics.

• However, if sensor noise is large the estimator poles may need to be less than 2 times the speed of the controller poles. This yields a system with lower bandwidth and more noise smoothing.

• The trade-off is that if the estimator poles are too fast the system becomes more noise sensitive.

Matlab code for pole placement:L = acker(F’,H’,p)’L = place(F’,H’,p)’Where p is the vector of desired poles which is the same dimension as the state vector.

Matlab code for optimal estimation theory:L = kalman(sys,Rw,Rv)Where sys is the system formulation, Rw is the process noise intensity, and Rv is the sensor noise intensity. A possible approach is to let Rv = 1 & vary Rw. Thus, Rw can be the only tuning variable in this approach.

EECE 413, Fall, 200682

State Space Control Law• Two major approaches

– Pole placement– Linear quadratic regulator (LQR)

• Both rely on pole or eigenvalue selection and state estimation (although output feedback is occasionally used).

• The control law is a linear combination of all the state variables. For an nth order system there are n feedback gains & and n poles of the system. By choosing proper values of Ki we should be able to place the poles at the desired locations.

• It is assumed that the system is controllable, ie., it is affected by control inputs.

n

n

x

xx

KKKKxu

2

1

21

GKxFxxKxu

GuFxx

,,System equation

Control law

substitutionThe characteristic equation of the closed-loop system is det[sI - (F – GK)] = 0.

Matlab code for pole placement:K = acker(F,G,p)K = place(F,G,p)Where p is the vector of desired poles which is the same dimension as the state vector. Note the difference between control law & observer design.

Matlab code for lqr:K = lqr(F,G,Q,R))Where Q is the nxn state weighting matrix, R is the mxm control weighting matrix. Since we are concerned now with SISO systems R may be set to 1. Q becomes a tuning matrix to produce the desired response.

EECE 413, Fall, 200683

LQR• When selecting the desired root locations, remember that

the control effort required is related to how far the open-loop poles are moved by the feedback.

• When a zero is near a pole, the system may be nearly uncontrollable (because of near cancellation) & moving such a pole may require a large control effort.

• Pole placement may only fix undesirable aspects of the open-loop response & move poles that are near zeros will allow smaller gains and smaller control actuators.

• LQR achieves a balance between good system response and control effort required. It consists of calculating the gain K by minimizing a cost function J:

dtRuuQxxJ TT

0

EECE 413, Fall, 200684

LQR/Observer Closed-Loop

JuHxyGuxFx

,ˆ̂

LyxGKLHFx

LyGuxLHFx

xHyLGuxFx

ˆˆ

ˆˆ

ˆˆˆ

xLHFx ˆˆ

The observer becomes part of the system & contributes dynamics to the system response. Consequently, the observer poles must be considered in the control law. The control matrix becomes F – LH – GK.

Observer matrix

ExampleLinear systemdescription

0,01

,10

,0010

JH

GF8.0,3

,20 22

zwss

n

nn is the desired characteristic equation

clear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*4;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = acker(F',H',po)';K = acker(F,G,p);p = eig(F-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,

Results for pole placement:w = 3.0000 3.0000ze = 0.8000 0.8000L = 19.2000 144.0000K = 9.0000 4.8000

%lqr codeQ = [1 0;0 0];Q = [100 0;0 5];R = 1;K = lqr(F-L*H,G,Q,R);p = eig(F-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,

Results for lqr:w = 3.1623 3.1623ze = 0. 7906 0. 7906L = 19.2000 144.0000K = 10 5

EECE 413, Fall, 200685

SS Pole Placement Results

Note that increasing magnitude of observer poles reduces overshoot.

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4LQR step response, desired/observer poles = 3 24

Time (sec)

Am

plitu

de

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4LQR step response, desired/observer poles = 3 12

Time (sec)

Ampl

itude

EECE 413, Fall, 200686

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Ampl

itude

SS LQR Resultsclear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*8;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = place(F',H',po)';Q = [1 0;0 0];Q = [10 0;0 5];Q = [100 0;0 5];R = 1;K = lqr(F,G,Q,R);p = eig(F-L*H-G*K);p = eig(F-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,hold off

Q = [1 0;0 0]; Q = [10 0;0 5]; Q = [100 0;0 5];

Note the effect of the state weights. Heavier weights give a faster response. The observer poles are fixed at 8x the previous desired response value of 3.

EECE 413, Fall, 200687

Closed-Loop Feedback

JuHxyGuxFx

,ˆˆ

LrxGKLHFx

xHrLxGKxFx

xHyxHyxKu

yrLGuxFx

ˆˆ

ˆˆˆˆ

ˆˆˆˆ,ˆ

ˆˆˆ

The observer now estimates the error = r-y, where r is the setpoint & y is the feedback measurement. The system control matrix is still F – LH – GK.

Same Example with pole placement

Linear systemdescription 0,01

,10

,0010

JH

GF8.0,3

,20 22

zwss

n

nn is the desired characteristic equation

% lqr_example_series.m%clear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*6;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = place(F',H',po)';K = acker(F-L*H,G,p);K = acker(F,G,p);p = eig(F-L*H-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);% sysCL = -feedback(sys*rsys,1,+1);sysCL = -sys*rsys/(1 - sys*rsys);


L = place(F',H',po)';Q = [1 0;0 0];Q = [10 0;0 5];Q = [100 0;0 5];R = 1;K = lqr(F,G,Q,R);p = eig(F-L*H-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);

Results for lqr:w = 21.8632 21.8632ze = 0.7730 0.7730L = 28.8000 324.0000K = 10.0000 5.0000

System equations

EECE 413, Fall, 200688

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

SS Pole Placement Feedback Results% lqr_example_series.m%clear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*6;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = place(F',H',po)';K = acker(F-L*H,G,p);K = acker(F,G,p);p = eig(F-L*H-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,


EECE 413, Fall, 200689

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Ampl

itude

SS LQR Feedback Results% lqr_example_series.m%clear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*6;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = place(F',H',po)';Q = [1 0;0 0];Q = [10 0;0 1];Q = [100 0;0 5];R = 1;K = lqr(F,G,Q,R);p = eig(F-L*H-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,

Q = [100 0;0 5];

This is a tracking controller. Note how the heavier weight on the 2nd state reduces the overshoot.

Q = [1 0;0 0]; Q = [10 0;0 1];

Q = [100 0;0 50];

EECE 413, Fall, 200690

SS Pole Placement Feedback Prob. 2.9

0,01,10

,0410

,4

JHGF

TT dc

Characteristic equation from denominator.

Given equations2 states, since 2nd order system

State space matrix values.

4

1)()(

)()()()(4

),()()(4)(4

22

2

ssTsTssTsTss

sTsTsssTT

dcdc

dcdc

Taking Laplace transforms:

L = 20 96K = 4 4

% Prob2_9.m%clear *F = [0 1;-4 0];G = [0;1];H = [1 0];J = 0;po = [-10;-10];p = [-2+2*i;-2-2*i];L = acker(F',H',po)';K = acker(F,G,p);

EECE 413, Fall, 200691

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

SS Pole Placement Feedback Prob. 2.9% lqr_example_series.m%clear *F = [0 1;0 0];G = [0;1];H = [1 0];J = 0;wn = 3;wno = wn*6;z = 0.8;po = roots([1 2*wno*z wno*wno]);p = roots([1 2*wn*z wn*wn]);L = place(F',H',po)';K = acker(F,G,p);p = eig(F-L*H-G*K);[w,ze] = damp(p)sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(1)step(sysCL,10),grid on,


This is a stabilizing controller. This is not shown in the text.

Observer poles

EECE 413, Fall, 200692

SS with Integral Control

t

ii edtxeHxrx

xx

Kxx

KoKiu

ruGx

xGH

xx

ii

ii ,010

00

Integral control is accomplished by augmenting (adding) the state vector with the desired integral xi. It obeys the differential equation:

Augmentation is shown below:

A pole placement example will illustrate the procedure. The advantage of this procedure is to add in the integral controller to an automatic design method.

EECE 413, Fall, 200693

SS Pole Placement with Integrator Results% lqr_example_integrator.m%clear *F = [0 1;-4 0];G = [0;1];H = [1 0];J = 0;Fbar = [0 -H;[0;0],F];Gbar = [0;G];Hbar = [0 H];Jbar = [0 0];po = [-10;-10];p = [-2+2*i;-2-2*i;-4];L = acker(F',H',po)';K = acker(Fbar,Gbar,p);Ki = K(1);Ko = K(2:3);A = [Fbar-Gbar*[Ki 0 0],-Gbar*Ko;-G*Ki,L*H,F-L*H-G*Ko];B = [1;0;0;0;0];C = [Hbar 0 0];D = 0;sysCL = ss(A,B,C,D);figure(1)subplot(211)step(sysCL,10),grid on,ylabel('Setpoint Input')subplot(212)B2 = [0;G;0;0];sysCL1 = ss(A,B2,C,D);step(sysCL1,10),grid on,ylabel('Disturbance Input')figure(2)B3 = [1;G;0;0];sysCL2 = ss(A,B3,C,D);step(sysCL,sysCL2,10),grid on,ylabel('SP/Disturbance Input')legend('setpoint','combination')

0 1 2 3 4 5 6 7 8 9 100

0.5

1

1.5Step Response

Time (sec)

Set

poin

t Inp

ut

0 1 2 3 4 5 6 7 8 9 10-0.05

0

0.05

0.1

0.15Step Response

Time (sec)

Dis

turb

ance

Inpu

t

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

Step Response

Time (sec)

SP

/Dis

turb

ance

Inpu

t

setpointcombination

EECE 413, Fall, 200694

Prob 2.20 modified

501015000)( 2

sssssGDetermine if G(s) is stable. If stable, how to stabilize to PM = 45 using bode

plots? Using pole placement & estimator place poles at [-4+4i,-4-4i,-5,-5].

% Prob_2_20_mod.m%clear *sys = tf(5000*[1 1],[1 60 500 0 0]);figure(1)margin(sys),grid

-150

-100

-50

0

50

100

150

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

103

104

-270

-225

-180

-135

-90

Phas

e (d

eg)

Bode DiagramGm = 14.5 dB (at 21 rad/sec) , Pm = 35.7 deg (at 7.84 rad/sec)

Frequency (rad/sec)

% Prob_2_20_mod.m%clear *sys = tf(5000*[1 1],[1 60 500 0 0]);figure(2)margin(0.607*sys),grid

-200

-100

0

100

200

Mag

nitu

de (d

B)

10-2

10-1

100

101

102

103

104

-270

-225

-180

-135

-90

Phas

e (d

eg)

Bode DiagramGm = 18.8 dB (at 21 rad/sec) , Pm = 45 deg (at 5.4 rad/sec)

Frequency (rad/sec)

EECE 413, Fall, 200695

% Prob_2_20_mod.m%clear *sys = tf(5000*[1 1],[1 60 500 0 0]);figure(1)margin(sys),gridfigure(2)margin(0.607*sys),grid[F,G,H,J] = ssdata(sys);p = [-2+1i,-2-1i,-10,-15];po = 6*p;L = acker(F',H',po)';K = acker(F-L*H,G,p);K = acker(F,G,p);sys = ss(F,G,H,J);rsys = ss(F-L*H-G*K,L,-K,J);sysCL = -feedback(sys*rsys,1,+1);figure(3)subplot(211)step(sysCL,5),grid on,subplot(212)pzmap(sysCL)

Prob 2.20 modified – cont.

Step Response

Time (sec)

Ampl

itude

Pole-Zero Map

Real Axis

Imag

inar

y Ax

is

-102

-101

-100

-20

-10

0

10

20

System: sysCLPole : -2 - 1i

Damping: 0.894Overshoot (%): 0.187

Frequency (rad/sec): 2.24

System: sysCLPole : -15

Damping: 1Overshoot (%): 0

Frequency (rad/sec): 15

System: sysCLZero : -10

Damping: 1Overshoot (%): 0

Frequency (rad/sec): 10

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

0.5

1

1.5

System: sysCLTime (sec): 1.06Amplitude: 1.28

System: sysCLTime (sec): 2.79Amplitude: 1.01

EECE 413, Fall, 200696

Prob 3.12

1)(s

sG

TsTsG/2/2)(

bsasksD

)(

a) Design a continuous lead compensation D(s) for the plant G(s) so that the complex roots are at approximately s = -4.4 +- j4.4 rad/s

b) Implement a delay of T/2 & determine the approximate effect.

c) Determine the revised root locations for sample rates of ws = 5, 10, & 20 Hz, where T = 1/ws.

% Prob_3_1.m%clear *sysG = tf(1,[1 0 0]);sysD = tf([1 2.5],[1 15]);% part ak = 0:400;sys = sysG*sysD;pdes = -4.4 + j*4.4;figure(1)rlocus(sys,k), hold on,gridaxis([-16 0 0 8])[k,poles] = rlocfind(sys,pdes);p = rlocus(sys,k);i = find(imag(p)>0);plot(real(p(i)),imag(p(i)),'+'),hold onxlabel(['real axis, k = ',num2str(k)])% part bT = 1/20;sysGh = tf(2/T,[1 2/T]);sysh = sysGh*sys;p = rlocus(sysh,k);i = find(imag(p)>0);plot(real(p(i)),imag(p(i)),'+'),hold onT = 1/10;sysGh = tf(2/T,[1 2/T]);sysh = sysGh*sys;p = rlocus(sysh,k);i = find(imag(p)>0);plot(real(p(i)),imag(p(i)),'+'),hold onT = 1/5;sysGh = tf(2/T,[1 2/T]);sysh = sysGh*sys;p = rlocus(sysh,k);i = find(imag(p)>0);plot(real(p(i)),imag(p(i)),'+'),hold on

-16 -14 -12 -10 -8 -6 -4 -2 00

1

2

3

4

5

6

7

80.350.580.760.86

0.92

0.96

0.984

0.996

246810121416

Root Locus

real axis, k = 92.7216

Imag

inar

y Ax

is20 Hz

10 Hz

5 Hz

RL pole placement using rlocfind

Decreasing sample rate degrades performance.

Zero a @ 2.5 & pole b @ 15 found by iteration. At this time can use rlocfind to get k.

EECE 413, Fall, 200697

Prob. 3.2

aeeKbuusEasKsUbsbsasK

sEsUsD ooo

)()()()()(

Tkxkxkx )1()()(

)1(11)(

1)(

11)1(

ke

bTaTKke

bTKku

bTku o

o

Repeat Example 3.1, but use the approximation that

Which is the backward rectangular version of Euler’s method. Compare the resulting difference equations with the forward rectangular Euler method. Also compute the numerical value of the coefficients for both cases vs. sample rate for ωs = 1 to 100 Hz. Assume the continuous values from eq. 3.8. Note that the coefficients of interest are given in eq. 3.7 for the forward rectangular case as (1 – bT) and (aT – 1).

Starting with

Plug in the approximation above for each derivative

)()1()()()1()( kaeTkekeKkbu

Tkuku

o

Or change k->k+1 & k-1->k to solve for u(k+1)

)1()(1)(1)1( keKkeaTKkubTku ooCompare with forward rectangular from eq. 3.7

0 10 20 30 40 50 60 70 80 90 100-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

disc

rete

pol

e

sampling rate - Hz

a = 2;b = 10;Ko = 70;Ws = 1:100;T = 1./Ws;cl = 1./(1+b*T);plot(Ws,cl,'-'),hold oncl = 1 - b*T;plot(Ws,cl,'--'), hold onplot(Ws,-ones(length(Ws),1),'r--'), hold offaxis([0 100 -3 1])ylabel('discrete pole')xlabel('sampling rate - Hz')

EECE 413, Fall, 200698

Prob. 3.315125)(

sssD

c1 c2 c3

1 - bT Ko(aT -1 ) Ko

1/(1 + bT) -Ko /(1 + bT) Ko(1+aT) /(1 + bT)

For the compensation:

Use Euler’s forward rectangular method to determine the difference equations for a digital implementation with a sample rate of 80 Hz. Repeat the calculations using the backward rectangular method & compare the difference equation coefficients.

For u(k+1) = c1 u(k) + c2 e(k) + c3 e(k+1)

Forward rectangularBackward rectangular

For a=1, b=15, Ko=25

Forward rectangularBackward rectangular

c1 c2 c3

0.8125 -24.6875 25

0.8421 -21.0526 21.3158

EECE 413, Fall, 200699

-150

-100

-50

0

50

100

Mag

nitu

de (d

B)

101

102

103

104

-180

-170

-160

-150

Phas

e (d

eg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 29.4 deg (at 432 rad/sec)

Frequency (rad/sec)

Prob. 3.8

bsasKsD

)(

2

500)(s

sG The read arm on a computer disk drive has the transfer function

Design a continuous lead compensation so that the closed-loop system has a bandwidth of 100 Hz and a phase margin of 50. Modify the matlab file fig32.m so that you can evaluate the digital version of your lead compensation using Euler’s forward rectangular method. Try different sample rates, and find the slowest one where the overshoot does not exceed 30%.

Simplest is to start with a single lead with 10:1 lead ratio (b/a = 10) for ~ 50 PM, where

% Prob_3_8.m%clear *sys = tf(500,[1 0 0]);figure(1)margin(sys),hold onxin = [1000 125 1250];k = xin(1);a = xin(2);b = xin(3);sysD = tf([1 a],[1 b]);margin(sysD*sys),hold onsysD = tf(k*[1 a],[1 b]);margin(sysD*sys),hold offfigure(2)sysCL = feedback(sysD*sys,1);margin(sysCL)

For bandwidth of 100 Hz = 630 rad/s, pick crossover frequency of 400 rad/s, so a = 400/sqrt(10) ~ 125 & b = 400*sqrt(10) ~ 1250, then iterate on k until find max phase @ k ~ 1000. Then plug into optimizer until convergence to PM & crossover frequency.

EECE 413, Fall, 2006100

Prob. 3.8 – optimizer code

function out = Optim_Prob_3_8(xin);%sys = tf(500,[1 0 0]);k = xin(1);a = xin(2);b = xin(3);sysD = tf(k*[1 a],[1 b]);sysCL = feedback(sysD*sys,1);[gm,pm,wcg,wcp] = margin(sysCL);out = abs(pm - 54) + abs(wcp - 2*pi*100);[xin,out]

% main_optim_prob_3_8.m%xin = [1000 125 1250];xin = fminsearch('Optim_Prob_3_8',xin);

Once the starting point is established of [1000 125 1250], it is a simple thing to optimize to find a closed-loop solution to the system. The final solution is [612.1357 231.2867 678.5847]. The cost function must cause convergence to PM = 50 + Allowance = 54 & crossover frequency = 100 Hz. The absolute values of the deviations from these values are used as shown below. The command fminsearch accesses a Nelder-Mead simplex optimization package.

The main script initializes & starts the optimization. The function Optim_Prob_3_8 calculates the cost function for each trial. Each trial set of three values is passed in as a vector. The cost function (out) is a scalar value that can be compared to other trials to pick the best. Even with the optimization there are an infinite combination of solutions, also, the optimizer may get stuck on a local minimum. If so, modify the initialization & restart. The global minimum is very close to 0 in this case.

-60

-40

-20

0

20

Mag

nitu

de (

dB)

101

102

103

104

-180

-135

-90

-45

0

Phas

e (d

eg)

Bode DiagramGm = Inf dB (at Inf rad/sec) , Pm = 54 deg (at 628 rad/sec)

Frequency (rad/sec)

EECE 413, Fall, 2006101

Prob 3.8 – Different Sampling Rates

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

[1000 125 1250][1.2459 0.1036 1.1]*1e3

0 1 2 3 4 5 6 7 8

x 10-3

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2Step Response

Time (sec)

Am

plitu

dexin = [1.2459 0.1036 1.1000]*1e3;kk = xin(1);a = xin(2);b = xin(3);tf = 0.01;Ws = 500;T = 1/Ws;[numGd,denGd] = c2dm(500,[1 0 0],T,'zoh');numDd = kk*[1 -(1-a*T)];denDd = [1 -(1-b*T)];numd = conv(numGd,numDd);dend = conv(denGd,denDd);[numcld,dencld] = feedback(numd,dend,1,1);td = 0:T:(tf*Ws-1)*T;dstep(numcld,dencld,td),grid on,hold onWs = 600;T = 1/Ws;numDd = kk*[1 -(1-a*T)];denDd = [1 -(1-b*T)];numd = conv(numGd,numDd);dend = conv(denGd,denDd);[numcld,dencld] = feedback(numd,dend,1,1);td = 0:T:(tf*Ws-1)*T;dstep(numcld,dencld,td),grid on,hold onWs = 800;T = 1/Ws;numDd = kk*[1 -(1-a*T)];denDd = [1 -(1-b*T)];numd = conv(numGd,numDd);dend = conv(denGd,denDd);[numcld,dencld] = feedback(numd,dend,1,1);td = 0:T:(tf*Ws-1)*T;dstep(numcld,dencld,td),grid on,hold off

800 Hz 600 Hz 500 Hz

Note: the optimized gains improve the settling time. Also, 500 Hz is unstable -> -(1-b*T) > 1.

continuous

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