Lecture 9 examples of steady-state error

1

Examples of steady-state error

Example 1

The forward transfer functions of the negative unity feedback system are as

follows:

2( )

( 1)(0.5 1)G s

s S s

Determine the steady-state error for the following inputs, respectively.

Check the stability of the system before applying the final-value theorem.

2 21 11. ( ) 1( ) 2. ( ) 3 3. ( ) 4. ( ) 1( ) 32 2

r t t r t t r t t r t t t t

Answer:

1. Check the stability of the system

3 2 3 2

( ) 2 4( )

1 ( ) 0.5 1.5 2 3 2 4

G ss

G s s s s s s s

The characteristic equation 3 23 2 4 0s s s

3

2

1

0

1 2

3 4

2

3

4

s

s

s

s

The system is stable.

2. Manner 1.

2 2

3 2

( 3 2)( ) ( ) ( ) ( ) ( ) ( ) ( )

3 2 4r

s s sE s R s C s R s s R s R s

s s s

12

1 3 20 0

( 3 2) 1lim ( ) lim 0

3 2 4ssr r

s s

s s se s E s s

s s s s

22

2 3 2 20

( 3 2) 3lim 1.5

3 2 4ssr

s

s s se s

s s s s

32

3 3 2 30

( 3 2) 1lim

3 2 4ssr

s

s s se s

s s s s

4Superposition 4 1 2 3 0 1.5ssr ssr ssr ssre e e e

Lecture 9 examples of steady-state error

2

Manner 2

Because the system is a unity feedback system

The error definition between ( ) ( ) ( )e t r t c t and ( ) ( ) ( )e t r t b t are same.

Open loop transfer function 2

( )( 1)(0.5 1)

G ss s s

1 typesystem. 2K

Find error constants pK vK aK

0 0

2lim ( ) limps s

K G ss

0lim ( ) 2vs

K sG s

2

0lim ( ) 0as

K s G s

1

10

1 1ssr

p

Re

K

2

31.5

2ssr

v

Ve

K

3

1

0ssr

a

Ae

K

Superposition 4 1 2 3 0 1.5ssr ssr ssr ssre e e e

Example 2

The control system is showed in Fig. example 2.

Find the steady-state error of the system.

5.0

s

1

)12.0(

10

ss

1.0n

tr 1 c

_

e

Fig. example 2

Lecture 9 examples of steady-state error

3

Answer:

1. Check the stability of the system

Open loop transfer function 2

1 10 10(0.5 1)( ) ( 0.5)

(0.2 1) (0.2 1)

sG s

s s s s s

Close loop transfer function 3 2

( ) 5 10( )

1 ( ) 0.2 5 10

G s ss

G s s s s

The characteristic equation 3 20.2 5 10 0s s s

3

2

1

0

0.2 5

1 10

3

10

s

s

s

s

The system is stable.

2. Find the steady-state error of the system.

Manner 1.

3 2

3 2

( ) 1 1 0.2

1 10( ) 1 ( ) 0.2 5 101 ( 0.5)

(0.2 1)

RE s s s

R s G s s s s

s s s

3 2

3 2 20 0

0.2 1lim ( ) lim 0

0.2 5 10ssr R

s s

s s se sE s s

s s s s

Manner 2.

2 type system

( ) 1r t t 0ssre

When ( ) 0r t 3 2

10

( ) 10(0.2 1)

1 10( ) 0.2 5 101 ( 0.5)(0.2 1)

nE s ss s

N s s s s

s s s

3 20 0

10 0.1lim ( ) lim 0

0.2 5 10ssn n

s s

se sE s s

s s s s

Superposition 0 0 0ss ssr ssne e e