Control chap6

CONTROL SYSTEMS
THEORY
Forced Response Errors
CHAPTER 6
STB 35103

Objectives
 To

find the steady-state error for a unity
feedback system
 To specify a system’s steady-state error
performance
 To design system parameters to meet
steady-state error performance
specifications

Introduction


In chapter 1, we learnt about 3
requirements needed when designing a
control system






Transient response
Stability
Steady-state errors (SSE)

Up until now we only covered until
transient response and stability

Review on transient response


We learned in Chapter 4, there are 4 types
of transient response for a second-order
system.





Overdamped
Underdamped
Undamped
Critically damped

Review on transient response


An example of elevator response



The transient response for elevator can be
considered as overdamped. The system is
stable but has steady-state error

Introduction


What is steady-state error?


Steady-state error is the difference between
the input and output for a certain test input as

t →∞



Test input used for steady-state error analysis
and design are
Step
 Ramp
 Parabola


Introduction


Test waveforms

Introduction


Example of systems tested using the test
signal.


Targeting system:


Targeting a static target. (e.g. a stopping car). We
test the system using step input because the position
of the car is in constant position.



Targeting a car moving with constant velocity. We
test the system using ramp input because the car is
moving in constant velocity.



Targeting an accelerating car. We test the system
using parabola input because the car is accelerating.

Introduction


We are only concerned with the difference
between the input and the output of a
feedback control system after the steady
state has been reached, our discussion is
limited to stable systems where the
natural response approaches zero when
(time) t approaches infinity.

SSE for unity feedback system


Unity feedback system can be represented
as



Steady state error can be calculated from
a system’s closed-loop transfer function,
T(s), or the open-loop transfer function,
G(s), for unity feedback systems.



Closed loop transfer function, T(s) is calculated
by solving the unity feedback system using the
block diagram reduction method for feedback
system.

1

T ( s) =

G ( s)
1 + G ( s ) × (1)



Open-loop transfer function for a unity
feedback system is the value of G(s) multiply
1.

1

1



Steady state error in terms of T(s).


To find E(s), the error between the input, R(s) and
output, C(s), we write

E ( s) = R(s) − C ( s)
= R ( s ) − R ( s )T ( s )
= R ( s ) 1 − T ( s ) 




)
We can find final value of the error, e(∞ in terms of
T(s) using

e ( ∞ ) = lim sR( s ) 1 − T ( s ) 


s →0



We can only use this equation if T(s) is stable, E(s) has no
poles in the right-half plane or poles on the imaginary axis
other than the origin



Example 7.1


Find the steady state-error for a unity feedback
system that has T(s) = 5/(s2+7s+10) and the
input is a unit step.



Solution:
R(s) =unit step = 1/s
T(s) = 5/(s2+7s+10), we must check the
stability of T(s) using Routh table or poles.



Example 7.1 (cont.)


We know from the unity feedback system

E ( s) = R(s) − C (s )


C ( s) = R( s )T ( s )

So, E(s) can be calculated using both equation

E ( s) = R(s) − C (s)
= R ( s ) − R ( s )T ( s )
= R ( s ) 1 − T ( s ) 





Example 7.1 (cont.)


E(s) in example 7.1 is

1
5

E ( s ) = 1 − 2
s  s + 7 s + 10 


1  s 2 + 7 s + 10
5
=  2
− 2

s  s + 7 s + 10 s + 7 s + 10 
2
 s2 + 7s + 5 
1
s + 7s + 5
=  2
=
s  s + 7 s + 10  s s 2 + 7 s + 10

(

)



Example 7.1 (cont.)


Before calculating the final value of the error
we must check the position of E(s) poles

s2 + 7s + 5
s 2 + 7s + 5
E ( s) =
=
2
s ( s + 2 ) ( s + 5)
s s + 7 s + 10

(



)

The poles for E(s) are at (0,0), (-2,0) and
(-5,0). Since all the poles are not on the right
half plane or the imaginary axis we can use the
equation to calculate final error value in terms
of T(s).



Example 7.1 (cont.)

e ( ∞ ) = lim sR ( s ) 1 − T ( s ) 


s →0

5
1

= lim s  ÷1 − 2

s →0
s   s + 7 s + 10 

5 5 1

= 1 −  =
=
 10  10 2



Steady state error in terms of G(s)
 We can find final value of the error, e(∞
)
in
terms of G(s) using

sR ( s )
e ( ∞ ) = lim
s →0 1 + G ( s )


We are going to use three types of input R(s);
step, ramp and parabola. So the final value of
the error for this types of input can be
described as



Step input e(∞
)

e ( ∞ ) = estep ( ∞ ) =


1
sR(s)

1 + lim G ( s )
s →0

)
Ramp input e(∞

1
sR(s)
e ( ∞ ) = eramp ( ∞ ) =
lim sG ( s )
s →0



)
Parabola input e(∞

1
sR(s)
e ( ∞ ) = e parabola ( ∞ ) =
lim s 2G ( s )
s →0

Steady state error with no integration
 Example 7.2




Find the steady-state errors for inputs of 5u(t),
5tu(t), and 5t2u(t) to the system below.

No integration


Solution hint
5u(t) = unit step = 5(1/s)
 5tu(t) = ramp = 5(1/s2)
 5t2u(t) = parabola = 5(2/s3) = 10(1/s3)




Example 7.2 (cont)

5
5
5
e ( ∞ ) = estep ( ∞ ) =
=
=
1 + lim G ( s ) 1 + 20 21
s →0

5
5
e ( ∞ ) = eramp ( ∞ ) =
= =∞
lim sG ( s ) 0
s →0

10
5
e ( ∞ ) = e parabola ( ∞ ) =
= =∞
2
lim s G ( s ) 0
s →0



Try to solve steady state errors for
systems with one integration in Example
7.3.



From the previous slides, the final error
value for three kinds of input; step, ramp
and parabola, are as follows

1
e ( ∞ ) = estep ( ∞ ) =
1 + lim G ( s )

position constant, K p

1
e ( ∞ ) = eramp ( ∞ ) =
lim sG ( s )

velocity constant, K v

s →0

s →0

1
e ( ∞ ) = eramp ( ∞ ) =
lim s 2G ( s )
s →0

acceleration constant, K a



Steady state error via static error
constants


Example 7.4 (Figure 7.7 (a) )



Solution


First step is to calculate the static error constants.

500( s + 2)( s + 5)( s + 6) 500(0 + 2)(0 + 5)(0 + 6)
K p = lim G ( s ) = lim
=
= 5.208
s →0
s →0 ( s + 8)( s + 10)( s + 12)
s (0 + 8)(0 + 10)(0 + 12)
s (500)( s + 2)( s + 5)( s + 6)
K v = lim sG ( s ) = lim
=0
s →0
s →0
( s + 8)( s + 10)( s + 12)
s 2 (500)( s + 2)( s + 5)( s + 6)
K a = lim s G ( s ) = lim
=0
s →0
s →0
( s + 8)( s + 10)( s + 12)
2



Next step is to calculate the final error value.

1
Step input, e(∞) =
= 0.161
1+ K p
1
Ramp input, e(∞) =
=∞
Kv
1
Parabola input,e(∞) =
=∞
Ka


Try to solve the remaining problems in Figure
7.7 (a) and (c).



System Type


We are still focusing on unity negative
feedback system.



Since steady-state errors are dependent upon
the number of integrations in the forward path,
we give a name to this system attribute.



Below is a feedback control system for
defining system type.







We define the system type to be the value of n
in the denominator.
Type 0 when n = 0
Type 1 when n = 1
Type 2 when n = 2



Relationship between input, system type,
static error constant, and steady-state
errors can be summarized as



Steady-state error specifications.


We can use the static error constants to
represent the steady-state error characteristic
of our system.



Conclusion that we can made based on static
error constants.



Problem: What information is contained in the
specification Kv = 1000.



Kv = 1000

Solution:
1. The system is stable.
2. The system is of Type 1, since only Type 1
have Kv that are finite constant


3.

A ramp input is the test signal. Refer to table.

4.

The steady-state error between the input
ramp and the output ramp is 1/Kv per unit of
slope.

Control chap6

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Control chap6