MECH303 2008-2009 Lecture Notes - Classical control theory PDF

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Classical Control Theory Laplace Transform 

£ [x (t )] exp( st ) x (t )dt



X (s )

0

The Laplace transform of a function converts this function from time domain to s-domain and quite often is easy to deal with mathematically than the original function itself. The function and its Laplace transform are equivalent. Classical control theory relies heavily on Laplace transforms. Some important results: £ [x (t )]  sX (s )  x (0) £ [x (t )] s 2 X (s )  sx (0) 

x (0)



sX (s )

if x (0) 0



s X (s ) if x (0) x (0) 0 2

£ [x (t )dt ]  X ( s) s

For a mass-spring-damper system of £ [ mx  cx  kx] ( ms 2  cs  k) X ( s)  ( ms  c) x(0)  mx(0)

important Laplace transform converts differentiation and integration with respect to t into polynomials or fractions of s. Thus a differential equation in t becomes an algebraic equation in s and hence is much easier to solve. Example: mx  cx  kx  f (t ) Laplace transform: ( ms  cs  k ) X ( s)  ( ms  c) x(0)  2

s-domain solution:

F ( s)  ( ms  c) x(0)  mx (0) 2 ms  cs  k 1 F (s )  (ms  c ) x (0)  mx (0) ] x(t ) £  [ ms 2  cs  k

X (s ) 

time-domain solution: Open-loop Control

mx (0) F ( s)

Reference input r(t)

Disturbing force d(t)

Control force f (t)

output x(t)

controller

plant

D(s)

R(s) C(s)

+

X(s) P(s)

G(s)

R(s)

X(s)

General rule with blocks Output = transfer function  input (for Laplace transforms) I ( s)

B(s) O ( s )  B( s ) I ( s )

Input-output relationship: X ( s) P( s)[C (s ) R (s )  D (s )] With disturbance X ( s) G( s ) R(s) X (s ) / R (s ) G (s ) Without disturbance G( s) C ( s ) P (s ) where

Closed-loop (Feedback) Control

O(s)

Error detector

Reference input r(t) +

Disturbing force d(t) output x(t) plant

Control force f (t) controller

-

e(t) b(t) sensor

D(s) E(s)

R(s)

+

C(s)

+

+

X(s) P(s)

-

H(s)

E(s)

R(s) +

X(s)

G(s)

-

H(s)

G (s )

Input-output relationship: X ( s) 1  G (s ) H ( s) R( s)

X ( s) G ( s)  R (s ) 1  G( s) H ( s)

G(s) : forward-path transfer function H(s) : feedback-path transfer function ( H ( s ) 1 is unity-feedback) G( s) 1  G (s )H (s ) : close-loop transfer function Some Performance Indicators max[ x (t )  (Maximum) overshoot: t

x ( )]

Peak time: the time taken for the response to reach its peak value

Delay time: the time taken for the response to reach 50% of its final value Rise time: the time taken for the response to rise from 5% to 95% (or 10% to 90%) of its final Settling time: the time taken for the response to decrease to and remain within a certain percentage (say 5%) of its final value Steady-State error: e() r ()  x () The final value theorem: e ( )  lim sE (s )  lim s 0

s 0

sR (s ) 1 G ( s)

e( )  lim sE( s)  lim s 0

s 0

s [1 GH  G ]R (s ) 1  G( s) H ( s)

general

(when H =1) very useful

Example: a mass-spring-damper system whose displacement is controlled by a controller producing force f (t) The equation of motion of the system is mx  cx  f (t )

c sensor

m x Controller

The force may be proportional to the error between the desired position and the actual position f (t ) ke(t ) k[ xi (t )  x (t )]

Laplace transforms: (ms 2  cs) X ( s)  F ( s )

F ( s) kE( s)

thus

G (s) 

1 ms  cs 2

E ( s )  X i ( s )  X (s )

Xi(s) +

-

1 ms  cs

k

X(s)

2

E(s)

F(s)

H(s) =1

The close-loop transfer function

X (s) k  2 X i ( s) ms  cs  k

Actually, the system equation is mx  cx  kx kxi (t )

A system’s performance can be assessed by its response to a unitstep input below

xi(t) u(t)

1

t k k X i ( s)  2 s (ms  cs  k ) ms  cs  k k 1 1 s  2 n    2 2 2 m s (s  2 ns   n ) s s  2 n s  n2

X (s ) 

2

Inverse Laplace transform of X(s) yields x( t) {1  exp(   n t)[cos d t 

n sin d t]}u( t ) d

It can be shown that Overshoot:

OS  x max(t )  1  exp(

Peak time:

tp 

π d

Rise time from 10% to 90%

tr 

2.16  0.60 n

5% settling time:

ts 

3.2 n

π 1-

2

)

(half period)

Steady-State error: 0 (why?) It can be seen that these performance indicators are interrelated so that specifying one of them may well specify others, which prevents good performance from being achieved for all performance indictors. This calls for more advanced control strategies than the proportional control. Steady-state Error For an open loop E ( s)  R( s) 

X ( s ) [1  G ( s )]R (s )

For a closed loop of unity feedback For a close loop of general feedback

E( s)  R( s)  X ( s) 

1 R( s ) 1  G (s )

1 G ( s )[ H ( s)  1] R ( s) E (s )  R ( s)  X ( s)  1  G ( s) H ( s)

If the input is a unity step, then 1 eopen ( ) lim sE( s) lim{ s[1  G( s)] } 1  G(0) s 0 s 0 s eclosed ( )  lim sE ( s ) lim{s s 0

s 0

1 1 1 } 1  G( s) s 1  G (0)

eclosed ( ) lim sE( s) lim{ s s 0

s 0

1  G (s )[H (s )  1] 1 1  G (0)[ H (0)  1] } 1  G (s )H (s ) s 1  G (0) H (0)

is usually called dc gain and is normally greater than 1. Therefore the steady-state error of an open loop can be significant while well-designed closed loop should have a small steady-state error. A modified controller with a right transfer function can also reduce the steady-state error. G (0 )

Example: An open loop system is shown below. Calculate its steady-state error to a unit-step input and suggest a simple way of reducing it without changing the plant and determine the new steady-state error. 8 s s 4

R(s)

X(s)

2

8

Answer: As G( s)  s 2  s  4 , e () 1  G(0) 1  2  1 (-100%), which is unacceptable. A simple way of reducing the steady-state error without changing the plant is to add a unity feedback to form a closed-loop system. 1 1 This makes e () 1  2 3 (33%), which is an improvement but still unsatisfactory. Two options are possible theoretically. (1) A new controller C(S) is added as shown below: open

closed

eopen () 1 C (0)G (0 ) 1  2C (0)

R(s)

8 s2  s  4

C (s )

X(s)

If we choose C (s ) 0.5  s , e ( )  0 . Note: Consideration of the steadystate error alone may not always be satisfactory (there are other performance indicators). The choice of C(s) is not unique. open

With this particular C(s),

X( s) C( s) G( s) R( s) (0.5  s)

1 8 s  s 4 s 2

which can be

converted to the following convenient form X (s ) 

4 8  s (s2  s  4) s 2  s  4

This means

1 0.25 , d  n 1    n  4 2 ,  2 n

2

2 1  0.25 2 1.94

in the table

of Laplace transforms. Inverse Laplace transform of X(s) yields x t( ) 1  1.02 exp(  0.5t ) sin(1.96t  ) 1.08 exp( 0.5t ) sin1.96t

hence x () 1 and e() 0 , where  is the phase angle. When x(t) is plotted, it may be seen that the overshoot is very big.

(2) (Non-unity) Closed-loop control: particular H(s), 8 1 8 X (s )  s  s  4  80.5 s s (s 2  s  8) . 1 2 s  s 4

H ( s ) 0.5 ,

eclosed ( ) 0 .

2

This means

 n  2 2 , 

With this

2 31 ,d  8 2

table of Laplace transforms. Inverse Laplace transform yields x (t ) 1  1.02 exp(  0.5t ) sin 2.78t When plotted, it may be seen that the over shoot is small.

in the...