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Control Engineering

Steady State Error

GUC Faculty of Engineering and Material ScienceDepartment of Mechatronics

Control Engineering ENME 503Dr. Ayman Ali El-Badawy

Tutorial #7

So to calculate the steady state error, you need to know the following:

1. Input type (step, ramp, …)2. Is the system unity feedback or not

Y RE

)()()( tytrte )()()( sY sRsE

)(lim0

ssEes

ss

The definition of the error signal is or

then by using the final value theorem, the steady state error is

Where R is the reference input and Y is the system output

Steady State Error

Y(t)







1. Input type



2. Is the system unity feedback or not

1. Unity feedback case

G

sRsR

G

GsR

G

GsRsY sRsE

1

)()(]

11[)(

1)()()()(

G

sRsE

1

)()(

G

ssRssE e

ssss

1

)(lim)(lim

00

Applying the final value theorem, we get





And, for various inputs, the result of this limit is summarized in the following table

But how to determine the system type???

For unity feedbacksystems, only !!



System type (for a unity feedback)Y+

_G(s)

R

E

system type is obtained from the denominator of the openloop transfer function, G(s)And it equals the degree of the free S in the denominator

So, the system type can be defined to be the degree of the polynomialfor which steady state error is a non-zero finite constant.





2. General Case (Non-unity feedback)

)(1

1)( sR

GH

GGH sE

)(lim0

ssE es

ss

Applying the final value theorem, we get

)(1

1)(]

11[)(

1)()()()( sR

GH

GGH sR

GH

GsR

GH

GsRsY sRsE

Important Note:

For the case of non-unity feedback, you can not use the summarized table

given in slide 5



Problem 1

• Determine the system type and error constant (Kp, Kv or Ka).

450

)()(

410)()(

2

ssH b

ssH a



Solution

As you can see, the system is unity feedback, hence we can use the table in

slide 5. But first we need to get the system type:

Then the open loop transfer is

,410)()( ssH a3

5)(

ssG

3

)4.0(50

3

)410(5

s

s

s

sGH

As you see, there is no free s in the denominator, hence the system is type 0.

And, by looking in the table of slide 5, you see that the system to have a

steady state error, the input has to be step input, in this case you can get theerror constant Kp. For other inputs, the system has infinite steady state error

And the error constant is

3

20

3

)4.0(50limlim

00

s

sGK

ssp





3

20,0

3

201

1

4103

51

11.

4103

51

1)(:

)(1

limlim

lim

00

.

0

.

p

ssss

sss

K and Typeof systemtheTherefore

ss

ss

s

se

ssRinputstepfor

sRGH

se

Another Solution



Then the open loop transfer is

3

5)(

s

sG

)3()5.12(20

3

)450

(5

3

)450(5

2

22

2

2

ss

sss

s

ssGH

As you see, the degree of the free s in the denominator is 2, hence the

system is type 2.And by looking in the table of slide 5, you see that the system to have a

steady state error, the input have to be acceleration input, in this case youcan get the error constant Ka. For other inputs, the system has zero steady

state error

And the error constant is

3.83

3

250

)3(

)5.12(20lim

)3(

)5.12(20limlim

2

02

22

0

2

0

s

s

ss

ssGH sK

sssa

,450

)()(2

ssH b





3

250,2

3

250

1

3

2500

1

4503

5

11.

450

3

51

1)(:.

1

00

1

450

3

5

11.

450

3

51

1)(:

0

01

1

450

3

51

11.

450

3

51

1)(:

)(1

2203

2

0

.

3

02

2

0

.

2

2

0

2

0

.

0

.

limlim

limlim

limlim

lim

a

ssss

ssss

ssss

sss

K Typeof systemtheTherefore

ss

ss

ss

se

ssRinputaccfor

Typeof notissystemtheTherefore

sss

ss

ss

se

ssRinputrampfor

Typeof notissystemtheTherefore

ss

s

ss

se

ssRinputstepfor

sRGH

se

Another Solution





How to calculate the transfer function for afeedback system with disturbance?

Problem: We have 2 inputs!

1. Reference input

2. Disturbance

)(sW

)(sR

cG pG

)(sY



What to do??? Assume disturbance input = 0

Get T.F. for the reference input.

Then assume reference input = 0

Get T.F. for the disturbance.

The general transfer function is sum of both transfer functions!!

)(sW

)(sR

cG pG

)(sY





)(sW

)(sR

cG pG

)(sY

For a feed back system with disturbance

)(1

)(1

)( sW GG

GsR

GG

GGsY

pc

p

pc

pc

Here, the closed loop T.F. is



Problem 2







Problem 3

W

Y

+

_

+

+

D(s)1

s2

k

+

_

Consider the system shown in the Figure below, which represents control

of the angle of a pendulum which has no damping.

R

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