Control Systems Lecture (4) Constantinos Siettos

(1)

Control Systems

Lecture (4)

Constantinos Siettos

(2)

Feedback Control Systems

Control of a System’s Response

Transient Steady State

(3)

Why Feedback?

Steady-State Error



E(s)  R(s) 1  K



E(s)  R(s) 1  K s



e(t)  1 1  K



e(t)  e

^{K t}

Y(s) Y(s)

For

R(s)=1/s For R(s)=1/s

  ^lim

_t

  ^lim

_s ₀

 

f f t sF s

 

  

The difference between input and output for a predetermined input for



t  

(4)

Steady-State Error- Proportional Control

Control Gc(s)

system G

_p

(s)

feedback H(s)=1 reference input, R(s)

error, E(s) output, Y(s)

+ -

error, u(s)

   

     

   

1

p c

c p

G s G s T s Y s

R s G s G s

 

    

   

1

_c _p

E s R s

G s G s

 

 

₀

 

lim lim

ss t s

e e t sE s

 

 

(5)

Steady-State Error with Proportional Controller - Step Input

   

1

_c _p

E s R s

G s G s

         

       

0 0

0

1 1 1

lim lim lim

lim lim

ss t s s

c p

s c p s c p

A

e e t sE s s s

G s G s

A A

G s G s G s G s

  



   



  

Constant of steady state error

Α

(6)

Steady-State Error with Proportional Controller-Ramp Input

   

1

_c _p

E s R s

G s G s

         

   

2

0 0

0

lim lim lim

lim

ss t s s

c p

s c p

A

e e t sE s s s

s sG s G s A

sG s G s

  



   



Constant of velocity error

Α

(7)

Steady-State Error with Proportional Controller

   

 

1

p c

c p

G s k T s Y s

R s k G s

 



E.g. for a 1st order process

 

( )

p

G s K

 s a



     

c

R s s a E s s a k K

 

     

 

^c c

Y s Kk

T s  R s  s a k K

 

Control Gc(s)=kc

system G

_p

(s)

feedback H(s)=1 reference input, R(s)

error, E(s) output, Y(s)

+ -

error, u(s)

(8)

Steady-State Error, Use of Matlab

clear all Kc=1;

K=1;

a=0.5;

sys1=tf(1,1);

sys2=tf(K,[1 a]);

sysh=tf(1,1);

sys3=series(sys1,sys2);

sys4=feedback(sys3,sysh) step(sys4)

T=0:0.01:4

U=ones(length(T),1);

LSIM(sys4,U,T) U=0.35*T;

LSIM(sys4,U,T)

(9)

Steady-State Error-

Simple Design Example

   

 

1

_p

E s R s

 KG s



Find Κ so that the steady-state error is 10%

1. For R(s) step ?

   

       

     

0 0

1 1 0

5 5

1 1

6 7 8 6 7 8

lim lim

s s

e t s s

s s

K K

s s s s s s s s

 

    

 

     

(10)

   

 

1

_p

E s R s

 KG s



1. For R(s) ramp ?

   

       

     

       

2

0 0

1

1 5 5

1 6 7 8 6 7 8

1 336

5 5 0 1 6 7 8

lim lim

.

s s

e t s s

s s

K s K

s s s s s s s

K K

 

    

 

     

 

So Κ=672

Steady-State Error-

Simple Design Example

Find Κ so that the steady-state error is 10%

(11)

Control System of the Direction of Automotive Robot

  1

p

G s K

 s

 

 

1 ²

c

G s K K

  s

   

1

_c _p

E s R s

G s G s

  1. For R(s) step and Κ

²

=0:

 

₀ ₀

1 1 1

1 1

lim lim

s s

A

A A

e t s s

K K K K

K K

s s

 

 

    

  

 

2. For R(s) step and Κ

²

>0:

 

₀ ₀

1 2

2 1

0 1 1

1 1

lim lim

s s

A s A

e t s

K s K K

K K

K s  s s  s

 

    

   

       

y

(12)

  1

p

G s K

 s

 

 

1 ²

c

G s K K

  s

   

1

_c _p

E s R s

G s G s

 

3. For R(s) ramp and Κ

²

>0:

 

2

0 0

2 2

1 2

1

1 1

lim lim

s s

A

A A

e t s s

K K s K s K K K K

K s  s  s

 

    

   

       

y

Control System of the Direction of

Automotive Robot

(13)

  1

p

G s K

 s

 

 

1 ²

c

G s K K

  s

   

1

_c _p

E s R s

G s G s

 

3. For R(s) ramp and Κ

²

>0:

 

2

0 0

2 2

1 2

1

1 1

lim lim

s s

A

A A

e t s s

K K s K s K K K K

K s  s  s

 

    

   

       

y

Control System of the Direction of

Automotive Robot

(14)

  1

p

G s K

 s

 

 

1 ²

c

G s K K

  s

 

2

e t A

   K K

clear all K1=1;

K2=0.5;

Tau=0.1

sys1=tf([Κ1 Κ2],[1 0]);

sys2=tf(K,[tau 1]);

sysh=tf(1,1);

sys4=feedback(sys3,sysh) T=0:0.01:10

U=sin(T);

LSIM(sys4,U,T)

0 1 2 3 4 5 6 7 8 9 10

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8

1 Linear Simulation Results

Time (sec)

Amplitude

Control System of the Direction of

Automotive Robot-Use of Matlab

(15)

0 05

4 3

3 :

^{n s}^T

.

s

n s

s

n

T e

T T



 





 

  



1 

2





 

n

T

p



P.O. 100e



1 ²



M

_pt

1 e



1 ²

Percent Overshoot is defined as:

P.O. = [(M_pt – fv) / fv] * 100%

M_pt = The peak value of the time response fv = Final value of the response

) 1 sin(

1 )

(   









 e

^

t

y

ⁿ^t _n

1 

2

     cos

^¹



Performance of Feedback System in Time

Field

(16)

Performance of Feedback System in Time Field

) 1 sin(

1 )

(   









 e

^

t

y

ⁿ^t _n

1 

2

     cos

^¹



Oscillation Period:

2

n

1 T 

 

 

Time that the output takes the maximum value :



²



²



²



2

1 1 1 0

1 , 0, 1, 2, ...

1 ( )

ⁿ ⁿ^t

sin

_n _n ⁿ^t

cos

_n

n

dy t e t e t

dt

t n n

 

        





 

 

       



  



The first time that the edge appears for n=1:

1 

2





 

n

T

p

 

₂ ² ₂

 

2

n

n n

Y s R s

s s



 

  

(17)

clear all

t=[0:0.1:12];

Num=1;

For zeta=[0.1 0.2 0.5 0.9 1 2];

Den1=[1 2*zeta 1];

Sys=tf(num,den);

[y,tout]=step(sys,t);

plot(tout,y);

Pause(1) hold on;

end

0 2 4 6 8 10 12

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

ζ

Performance of Feedback System in Time

Field-MATLAB

(18)

An example: Reading the Hard-Disc

H(s)=1 R(s)

Κ Y(s),

Rotation Velocity + -

-

5 +

D(s)

 ¹ ²⁰ 

s s 

Design: the maximum overshoot <5%, Τs<250, Overshoot in step disturbance <5x10-3

(19)

   ⁵   

₂

⁵  

₂ ² ₂

 

20 5 20 5 2

n

n n

K K

Y s R s R s R s

s s K s s K s s



 

  

     

1 2

1 1 005 .

M

pt

e







 





3 250

s

n

T ^  ^

1 2

100 e 5









2

_n

20 10

n

 

   

maple( 'solve(100*exp(-10*pi/

(sqrt(5*x))/sqrt(1-100/(5*x)))<5),x' )

An example: Reading the Hard-Disc

H(s)=1

R(s) Κ Y(s),

Rotation Velocity + -

-

5 +

D(s)

 ¹²⁰

s s

(20)

clear all

t=[0:0.01:1];

K=25;

Num=[5*K];

Den1=[1 20 5*K];

Sys=tf(num,den);

plot(tout,y),grid;

Pause(1) hold on;

end

 

₂

⁵  

20 5

Y s K R s

s s K

  

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

0 0.2 0.4 0.6 0.8 1 1.2 1.4

time

y(t)

H(s)=1

R(s) Κ Y(s),

Rotation Velocity + -

-

5 +

D(s)

 ¹²⁰

s s

An example: Reading the Hard-Disc

(21)

Control of the Insouline injection

 

   

     

2 1 2

1 1

K s

s s K s

Y s R s R s

s s s K s

K s s



 

 

   

 

What is the value of Κ so that the maximum value of response in a unit step disturbance do not exceed 5x10-3,

Time period of settling 250ms

Pump

Κ Body

H(s)=1

R(s) E(s) Y(s)

Sugar concentration

+ -

insouline

   

 

2

p

1 G s s

s s

 



(22)

   

     

2

1 2

Y s K s R s

s s K s

 

  

What is the value of Κ so that the maximum value of response in unit step disturbance does not exceed 2%, settling time 4ms

clear all

t=[0:0.1:12];

Num=[K 2K];

Den1=[1 (1+K) 2*K];

Sys=tf(num,den);

plot(tout,y);

   

 

2

2 2 2

n

n n

a s a

Y s R s

s s



 

 

 

Control of the Insouline injection

Pump

Κ Body

H(s)=1

R(s) E(s) Y(s)

Sugar concentration + -

insouline

   

 

2

p 1 G s s

s s

 



(23)

Integral Indices of Performance – Optimal Design

1. Integral of the absolute value of the error (IAE)

 

0

IAE e t dt



 

e(t) denotes the difference between set point and output

• Criteria that optimize the closed loop response of set-point changes or disturbances.

• The optimal values of the control parameters minimize the integral criteria

• The 3 most used criteria are the following:

(24)

2. Integral of the squared error (ISE)

 

²

0

ISE e t dt



 

   

3. Integral of the time-weighted absolute error (ITAE)

 

0

ITAE t e t dt



 

Integral Indices of Performance – Optimal Design

• Criteria that optimize the closed loop response of set-point changes or disturbances.

• The optimal values of the control parameters minimize the integral criteria

• The 3 most used criteria are the following:

(25)

H(s)=1 R(s)

Κ Y(s), bank angle

+ - _

^s ¹⁰^¹⁰

_ ^{s s}  ¹² ^ ² 

Activation of the Wind Slope (BANK)

Plane Dynamics

Gyroscope Gain

Integral Indices of Performance –

Optimal Design-Matlab

(26)

Aim: Optimization of the closed loop response in a step change with IAE criterion

clear all kiter=0;

for K=0.1:0.001:2;

sys1=tf(K,1);

sys2=tf(10,[1 10]);

sys3=tf(12, [1 2 0]);

sysopen=series(sys12,sys3);

sysclosed=feedback(sysopen,1);

%H=get(sysclosed);

y=step(sysclosed);

kiter=kiter+1;

Ksave(kiter)=K;

e(kiter)=sum(abs(y-1));

[K e(kiter)]

end

plot(Ksave,e)

Integral Indices of Performance – Optimal Design-Matlab

H(s)=1

R(s) Κ Y(s), bank angle

+ -

^ ^

12 2 s s

^s¹⁰^¹⁰

Activation of the Wind Slope

Plane Dynamics

Gyroscope Gain

(27)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 0

100 200 300 400 500 600

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

10 11 12 13 14 15 16 17 18 19 20

0 1 2 3 4 5 6 7

0 0.2 0.4 0.6 0.8 1 1.2

1.4 Step Response

Time (sec)

Amplitude

0 10 20 30 40 50 60 70 80 90 100

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

2 Step Response

Time (sec)

Amplitude

0 50 100 150 200 250

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8

2 Step Response

Time (sec)

Amplitude

Integral Indices of Performance – Optimal Design-Matlab

Aim: Optimization of the closed loop response in a step change with IAE criterion

(28)

Y _s(s)

C(s) U(s)

Y _sp (s)

G _c(s) ^Y(s)

D(s)

G _a(s) G _p(s)

G _s(s)

G _d (s)

+ -

E(s)

+ +

Feedback Control loop

1 )

( )

(

) ( )

( )

( ) (

 

s G

s Y

s c

a p

c a

p sp

1 )

( )

(

) ( )

( ) (

 

s G

s D

s Y

s c

a p

d

(29)

Characteristic Equation

1 0 ( ) ( ) ( ) ( )

p a c s

G s G s G s G s  

The transient response of the system is determined from the place of poles which are the roots of the characteristic equation

A state of a system is stable if bounded inputs result to bounded outputs

The complex s-plane seperates in two regions: one stable whchcorresponds to the left semi-plane and the unstable region which corresponds to the right semi-plane

s-plane

stable unstable

x x

x

x x

x

j



x x

(30)

Poles and Stability

Re Im

Time

y

Re Im

Time

y

Re Im

Time

y

(31)

Linear Control System– PID Controllers

0

0 1 ( )

( ) ( ) ( )

where ( ) ( )

t

c D

I

s s

c t c K e t e t dt de t

dt

e t R y t

 

 

     

 

 



 

 



  



 s

K s s

E

s s C

G _D

I c

c 

 1 1

) (

) ) (

(

(32)

Linear Control Systems– PID Controllers

e(t)- the error from setpoint [e(t)=R

_s

-y

_s

] K

_c

- the controller gain is a tuning parameter



I

- the reset time is a tuning parameter and determines the amount of integral action.