Chapter 7

Steady State Error

Test waveforms for evaluating steady-state errors of position control systems

Steady-state error is the difference between the input and the output for a prescribed test input as t →∞

Test inputs for steady-state error analysis and design vary with target type

Steady-state error:

a. step input;b. ramp input

Closed-loop control system error:

a. general representation;b. representation for unity feedback systems

System with:

a. finite steady-state error for a step input;b. zero steady-state error for step input

For pure gain K as in (a), Css= K ess or ess =1/K Css there will always be errorBut if the pure gain is replaced by an integrator as in (b) the steady state error will be zero.

Feedback control system for Example 7.1

Problem Find the steady-state error for the system in(a) if T(s) = 5/(s2+7s+10) and the input is a unit step.Solution: Using E(s) = R(s)[1-T(s)]

We can apply the final value theorem To yield e(∞) =1/2

)107(

57)(

2

2

sss

sssE

)(lim)(0

ssEes

Steady-state error in terms of G(s)

E(s) =R(s) – C(s),

C(s) =E(s)G(s) so

E(s) = R(s)/1+G(s)

Using final value theorem)(1

)(lim)(lim)(

00 sG

sRsssEe

ss

Steady-state error in terms of G(s)

For step input, the steady state error will be zero if there is at least one pure integration in the forward path.

For ramp input, the steady state error will be zero if there is at least 2 pure integration in the forward path.

For parabolic input, the steady state error will be zero if there is at least 3 pure integration in the forward path.

)(lim1

1

)(1

)/1(lim)(lim)(

000 sGsG

ssssEe

sss

)(lim

1

)(1

)/1(lim)(lim)(

0

2

00 ssGsG

ssssEe

sss

)(lim

1

)(1

)/1(lim)(lim)(

2

0

3

00 sGssG

ssssEe

sss

Steady state error for systems with no integrations

Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function.

Solution First verify the closed loop system is stable.

For input 5u(t)

For input 5tu(t)

For input 5t2u(t)

21

5

201

5

)(lim1

5)(

0

sGe

s

0

5

)(lim

5)(

0ssG

es

0

10

)(lim

10)(

2

0sGs

es

Steady state error for systems with one integration

Problem Find the steady-state error for inputs 5u(t), 5tu(t), and 5t2u(t) to the system. u(t) is a step function.

Solution First verify the closed loop system is stable.

For input 5u(t)

For input 5tu(t)

For input 5t2u(t)

0

5 5( ) 0

1 lim ( )s

eG s

0

5 5 1( )

lim ( ) 100 20s

esG s

0

10

)(lim

10)(

2

0sGs

es

Static error constants

Position constant step input

Velocity constant ramp input

Acceleration constant parabolic input 2

0lim ( )as

K s G s

0lim ( )ps

K G s

0lim ( )vs

K sG s

Steady state error via static error constants

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs..

Solution First verify the closed loop system is stable.

Thus for step input,

For ramp input

For parabolic input

0

500 2 5lim ( ) 5.208

8 10 2ps

X XK G s

X X

0lim ( ) 0vs

K sG s

2

0lim ( ) 0as

K s G s

1( )

v

eK

1( )e

Ka

161.01

1)(

pKe

Steady state error via static error constants

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.

Solution First verify the closed loop system is stable.

Thus for step input,

For ramp input

For parabolic input

0lim ( )ps

K G s

0

500 2 5 6lim ( ) 31.25

8 10 12vs

X X XK sG s

X X

2

0lim ( ) 0as

K s G s

1 1( ) 0.032

31.25v

eK

1( )e

Ka

01

1)(

pKe

Steady state error via static error constants

Problem For the system of figure, evaluate the static error constants and find the expected error for standard step, ramp, and parabolic inputs.

Solution First verify the closed loop system is stable.

Thus for step input,

For ramp input

For parabolic input

0lim ( )ps

K G s

0lim ( )vs

K sG s

2

0

500X2X4X5X6X7lim ( ) 875

8X10X12as

K s G s

1( ) 0

v

eK

31 1( ) 1.14 10

975e X

Ka

01

1)(

pKe

Feedback control system for defining system type

System type is the value of n in the denominator, or the number of pure integrations in the forward path.

Relationships between input, system type, static error constants, and steady-state errors

Feedback control system for Example 7.6

Problem Find the value of K so that there is 10% error in the steady state..

Solution: Since the system is Type 1, the error must apply to a ramp input, thus

and

and K = 672 Applying Routh-Hurwitz we see that the system is stable at this gain.

1( ) 0.1

v

eK

0

X5lim ( ) 10

6X7X8vs

KK sG s

Feedback control system showing disturbance

1 2 2

2

1 2 1 2

( ) ( ) ( ) ( ) ( ) ( )

But ( ) ( ) - ( )

( )1So ( ) ( ) ( )

1 ( ) ( ) 1 ( ) ( )

C s E s G s G s D s G s

C s R s E s

G sE s R s D s

G s G s G s G s

Applying final value theorem we obtain

2

0 0 01 2 1 2

( )( ) lim ( ) lim ( ) lim ( )

1 ( ) ( ) 1 ( ) ( )

= e ( ) ( )

s s s

R D

sG sse sE s R s D s

G s G s G s G s

e

Previous system rearranged to show disturbance as input and error as output, with R(s) = 0

Assume step disturbance D(s) = 1/s

We get

This shows that the steady-state error produced by a step disturbance can be reduced by increasing the dc gain of G1(s) or decreasing the dc gain of G2(s)

10 0

2

1( )

1lim lim ( )

( )

D

s s

eG s

G s

Feedback control system for Example 7.7

Problem Find the steady state error due to a step disturbance for the shown system..

Solution: The system is stable, we find

10 0

2

1 1 1( )

1 0 1000 1000lim lim ( )( )

D

s s

eG s

G s

Forming an equivalent unity feedback system from a general non-unity feedback system

G(s) = G1(s)G2(s)

H(s) = H1(s)/G1(s)

Non-unity feedback control system for Example 7.8

Problem Find the system type, the appropriate error constant, and the steady state error for a unit step input..

Solution: The system is stable, we convert the system into an equivalent unity feedback system

The system is Type 0, and Kp= -5/4. The steady-state error is

3 2

( ) 100( 5)( )

1 ( ) ( ) ( ) 15 50 400e

G s sG s

G s H s G s s s s

1 1( ) 4

1 1 5 / 4p

eK

Steady-State error for systems in State Space

The Laplace transform of the error is E(s) = R(s) –Y(s), and since Y(s) = R(s)T(s).

Then E(s) = R(s)[1-T(s)]

Using T(s) =Y(s)/U(s) = C(sI-A)-1B + D

We have E(s) = R(s)[1-C(sI-A)-1B]

Applying final value theorem, we have

1

0 0lim ( ) lim ( )[1 ( ) ]s s

sE s sR s C sI A B

Steady-State error for systems in State Space

Example: Evaluate the ss error for the system described by

Solution: Substitute A,B, and C in

For unit step, R(s)=1/s and e(∞)=4/5, for unit ramp, R(s) = 1/s2 and e(∞)= ∞

5 1 0 0

0 2 1 ; 0 ; 1 1 0

20 10 1 1

A B C

3 20

3 2

3 20

4( ) lim ( ) 1

6 13 20

6 12 16 = lim ( )

6 13 20

s

s

se sR s

s s s

s s ssR s

s s s

1

0 0lim ( ) lim ( )[1 ( ) ]s s

sE s sR s C sI A B