Lecture12-SteadyStateError

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Page : 1 EE406 Control Systems Lecture 12 : Steady State Error

UCSI University Faculty of Engineering

Kuala Lumpur, Malaysia Department of Mechatronics

Lecture 12

Steady State Error

Mohd Sulhi bin Azman Lecturer

Department of Mechatronics UCSI University

[email protected]

1 August 2011

Contents

• Steady state error

• System type

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Introduction

• A steady state error is the difference between

the input and the output of a prescribed test

input as t

→

∞.

• Steady state error constitute an extremely

important aspect of system performance, for it

would be meaningless to design for dynamic

accuracy if the steady output differed

substantially from the desired value.

Steady State Error : Step Input

• Consider the following system subjected to a step input. • We observe that:

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Steady State Error : Ramp Input

• The same case goes with a system subjected to a ramp input. We observe that Output 1 has a zero steady state error while Output 2 has a finite steady state error.

Evaluating Steady State Error

• Consider a unit feedback system shown below:

• We observe that:

( )

G s

( )

R s C s( )

( )

E s

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

C s G s

R s G s

C s E s G s

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Evaluating Steady State Error

• Therefore, the error signal is:

• Thus, the steady-state error may be found by use of the final value theorem:

• The above equation shows that the steady-state error depends upon the input R(s) and the forward transfer function G(s).

( ) ( )

1 ( )

R s E s G s = + 0 0 ( ) ( ) lim ( ) lim ( ) lim

1 ( ) ss

t s s

sR s

e s e t sE t

G s

→∞ → →

= = =

+

Steady-state Error : Unit Step Input

• Consider a unit-step input R(s)=1/s. • The steady state error is then:

• The term Kp is known as the position error constant,

where:

0 0

0

1 1 1

( ) lim ( ) lim

1 ( ) 1 lim ( ) 1 ss

s s

p s

s

e s sE t s

G s G s K

→ → →   = =  ⋅ = = + + +   0 lim ( ) p

s

K G s

→

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Steady-state Error : Ramp Input

• Given the ramp input, R(s)=1/s2_.

• The steady state error is then:

• The term Kvis known as the velocity error constant,

where:

[

]

2 0 0 0 0 0 0 1 ( ) lim ( ) lim

1 ( )

1 1 1 1 1

lim lim

1 ( ) ( ) 0 lim ( ) lim ( ) ss s s s s v s s s

e s sE t s

G s

s G s s sG s sG s sG s K

→ → → → → →   = =  ⋅  +   = = = = = + + + 0 lim ( ) v

s

K sG s

→

=

Steady-state Error : Parabolic Input

• Consider the parabolic input R(t)=1/s3_.

• The steady state error is then:

• The term Kais known as the acceleration error constant.

[

]

3

2

0 0 0

2 2 2 2

0

0 0

1 1

( ) lim ( ) lim . lim

1 ( ) 1 ( )

1 1 1 1

lim

( ) 0 lim ( ) lim ( ) ss

s s s

s

a

s s

s

e s sE t s

G s s G s

s s G s s G s s G s K

→ → → → → →   = =  ₊ = ₊   = = = = + + 2 0

lim ( ) a

s

K s G s

→

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System Type

• Consider the following system:

• System type is defined as the number of “n” integration in the open-loop transfer function G(s) as described above.

• Consequently, if:

– n=0 ~ type 0 – n=1 ~ type 1 – n=2 ~ type 2

Steady State Error for Various System Type

• Let us focus on finding the steady-state error

for system type 0.

• We note that:

– ess(position)=0 : apply step input

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Steady State Error for Various System Type

for system type 1.

• We note that:

– ess(velocity)=1/Kv : apply ramp input

– ess(acceleration)=∞ : apply parabolic input

Steady State Error for Various System Type

for system type 2.

• We note that:

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Summary of the Result

• Relationships between input, system type, static

error constant and steady-state errors.

Example 1

• Determine the steady-state error for the

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Solution to Example 1

• The transfer function (inside the box) is of type 0. Therefore, we can compute the steady-state error by applying the final value theorem.

• First, we apply the step input. That is, R(s)=5/s:

• In this case, the position error constant (K_p) is equal to 21. That is, Kp=21.

0

( ) ( ) lim ( )

1 lim ( )

5 5 5

0.238 1 20 21

120( 2) 1 lim

( 3)( 4) ss

s

sR s

e s sE s

G s s s s → → → = = + = = = = +  ₊  +  ₊ ₊   

Solution to Example 1

• Next, we apply the ramp input:

• Take note that the velocity error constant (Kv) is equal

to zero in this case. That is: Kv= 0.

• This is now matched with our table. 0

0 2

0

( ) ( ) lim ( )

lim ( )

5 5 5

0 0 120( 2)

lim

( 3)( 4) ss

s

sR s

e s sE s

G s

s s s

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Solution to Example 1

• Then, we apply the parabolic input:

• Take note that the acceleration error constant

(K

_a

) is equal to zero in this case. That is: K

_a

= 20.

0

3 2

0

( ) ( ) lim ( )

lim ( )

5 5 5

0 0

120( 2) lim

( 3)( 4) ss

s

sR s

e s sE s

sG s

s s s

s s s s → → → = = × = = = = ∞  ₊  ⋅  ₊ ₊   

Exercise

• A unity feedback system is described by the following forward transfer function (open loop):

• Determine the system type and the steady state error if the inputs are 15

u(t)

, 15

tu(t)

and 5

t

2

(t).

• Answer:

10( 20)( 30) ( )

( 25)( 35)

s s

G s

s s s

+ +

=

+ +

: ( ) 0

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Steady-state Error for Non-unity Feedback

• For a non-unity feedback systems, as shown

below, the difference between the input signal

R(s) and the feedback signal B(s) is the

actuating error signal E

_a

(s) :

( )

G s

( )

R s E sa( ) C s( )

( )

H s

( )

B s

Steady-state Error for Non-unity Feedback

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• Next, we perform the following (simplifying):

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Example 2

• For the system shown below, find system type

and the error constant subjected to step input.

• The error transfer function is:

• This is a type zero system. Then, finding Kp:

• And:

(

)

3 2

100 5 ( )

( )

1 ( ) ( ) ( ) 15 50 400 e

s G s

G s

G s H s G s s s s

+

= =

+ − − − −

(

)

3 2

0 0

100 5 5

lim ( ) lim ( )

15 50 400 4

p e e

s s

s

K G s G s

s s s

→ →

+

= = = −

− − −

(

)

1 1

( ) 4

1 _p 1 5 4

e

K

∞ = = = −

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Note of Error Constants

• The error constants K

_p

, K

_v

and K

_a

describes the

ability of a system to reduce or eliminate

steady-state errors.

• Systems with more than two integrations (n=2)

are not employed in practice because of:

– these are more difficult to stabilize.

– the dynamic errors for such systems tend to be larger than those for types 0, 1 and 2, although their steady-state performance is desirable.

Sensitivity

• Sensitivity is a ratio of the fractional change in the function to the fractional change in the parameter, as the fractional change in the parameter approaches zero. • Furthermore, sensitivity deals with the degree to which

changes in system parameters affect system transfer function and hence system performance.

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Sensitivity

• Consider the following function:

• If K=10 and a=100, then F equals:

• If a is tripled to 300 and K=10, then F equals:

• The percentage change of F is:

K F K a = + 10 0.091 10 100

F = =

+

10

0.032 10 300

F= =

+ 0.032 0.091 100 65% 0.091 − _× _{= −}

Sensitivity

• From the previous slide, we can conclude that changing (increasing) the values of “a” will result in the decrease of the value “F”. Thus, the function “F” is sensitive towards the change of the values of “a”.

• We can say that:

• Hence, the sensitivity of is:

0 0 0

Fractional change in

lim lim lim

Fractional change in

F F F F F F

S

F F

α _{∆ →}_α _α _{∆ →}_α _{α α} _{∆ →}_α _α α _α α

  _∆ _∆    _∂    =  = _∆ = _∆    = _∂              multiplication F F S F

α ∂_α α

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Example 3

• Consider the following system. Calculate the

sensitivity of the closed-loop transfer function

to changes in parameter a.

• The closed-loop transfer function (CLTF) is:

• The sensitivity is:

• Simplifying gives:

2

( )

K

T s

s

as

K

=

+ +

(

)

( )

2 2

2

( )

T s a

T s

a

Ks

a

S

K

a

T s

s

as

K

s

as

K







 







∂



−



=





=









∂







+ +







+ +





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Solution to Example 3

• Therefore, we conclude that an increase in K will

decrease (reduce) the sensitivity of the

closed-loop transfer function with respect to the

parameter a.

Example 4

• Consider the following system. Find the

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• This is a type zero system. The steady state

error for a type zero system is given as follows:

0 0

0

( )

lim

( )

lim

1

( )

1

lim

1

(

)(

)

ss s s s

R s

e

s

sE s

s

G s

s

ab

s

K

ab

K

s

a s b

ab

→ → →





=



⋅



+













=



⋅



=

+



+



+



+







• Sensitivity is defined as the change of one parameter with respect to another parameter. Now for this question, we are looking at the sensitivity of the changes of the steady-state error with respect to parameters Kand a.

• Now, let’s have a look at the change of steady state error with respect to parameter K :

(

)

2

error to K

K

e

K

ab

K

S

ab

error

K

ab

K

ab

K

ab

K

δ

− −

=





⋅

−

=

−

+





+





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• Now, let’s have a look at the change of steady

state error with respect to parameter

a

:

(

)

2 2

(

)

error to a

a

e

a

ab

K b ab

K

S

ab

error

a

ab

K

ab

K

ab

K

δ

− −

=





⋅

+

−

=

+





+





Differentiate e_ss(∞) with respect to the parameter a. Treat other parameters as constants. In this case, we use quotient rule in differentiating

the expression e_ss(∞).

Next Step

• Textbook reference : Chapter 7.

• Homework 11 has been posted on the course

website. Attempt them. You do not have to

submit Homework 11 as it will not be graded.