CHAPTER 6

STEADY-STATE ERROR

Analysis Objectives

of Control Systems?

1. Producing the desired transient response:

2. Achieving stability: A system that can produce a

consistent/steady output is a stable system..

3. Reducing steady state error: Steady state response only

exists for stable systems.

How to Measure SSE?

The system is stimulated with some standard input, typically,

a step function of time or a ramp

The system comes to a steady state

ERROR = Difference between the Input and Output

DEFINITION

Steady state error

The difference between the input and the output of a

system after the natural response has decayed to zero - i.e.

the steady state has been reached.

or

The difference between the input and the output for a

prescribed test input as t → ∞

Test waveforms for steady-state errors analysis and design for

position control systems

Test inputs for steady-state error analysis

and design vary with target type

In order to explain how these test signals are used, let

us assume a position control system where the output

position follows the input commanded position.

Step Input represent constant position and thus are useful

in determining the ability of the control system to position

itself with respect to a stationary target such as a satellite in

geostationary orbit.

Ramp Input represent constant velocity inputs to a

position control system by their linearly increasing input to

track a constant velocity target. For example, a position

control system that tracks a satellite that moves across the

sky at a constant angular velocity would be tested with

ramp input

Parabolas represent constant acceleration inputs to

position control systems and can be used to represent

acceleration target such as the missile to determine the

steady state error performance.

Tracking

system

EVALUATING STEADY STATE

ERRORS.

Let us examine the concept of steady state errors.

In Figure (a) a step input shows two possible

outputs. Output 1 has zero steady state error and

output 2 has a finite (limited) steady state error,

e2()

In Figure (b) a ramp input is compared with

output 1 , which has zero steady state error and

output 2 has a finite steady state error, e2() as

measured vertically between the input and output

2 after the transients have died down. If the

output slope is different from that of the input,

output 3 will result. Here the steady state error is

infinite (immeasurable or endless) after the

transient have died down and t approaches

infinity.

(a) general representation

(b) representation for unity feedback systems

The most general block diagram with unity feedback

Error is the difference between the input and the output of a system

The error, E(s) is formed by taking the difference between the input and the output

as shown in Figure (a).

Here we are interested in the steady state, or final value of e(t). For unity

feedback systems (where the feedback, H(s) = 1), E(s) appears as shown in

Figure (b).

Steady State Error in Terms of T(s)-CLOSED LOOP TRANSFER FUNCTION-

Steady State Error in Terms of T(s)

A system’s SSE can be calculated from a system’s closed-loop TF, T(s)

Closed-loop control system error

representation for unity feedback

systems

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

E(s) = R(s) - C(s) (6.1)

But C(s)= R(s)T(s) (6.2)

Finally substituting Eq (7.2) into Eq. (7.1) and solving for E(s) yields

)](1)[()( sTsRsE (6.3)

)](1)[(lim)(0

sTssRes

(6.5)

Tutorial Exercise!Find the steady state error for CLTF with unit step input

)107(

5)(

2

sssT

)](1)[(lim)(0

sTssRes

2

1)(, eSo

Steady State Error in Terms of G(s)

-OPEN LOOP TRANSFER FUNCTION-

Steady State Error in Terms of G(s)Many unity feedback system with a forward transfer function is configured as a G(s).

Closed-loop control system error

representation for unity feedback

systems

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

E(s) = R(s) - C(s) (7.1)

But C(s)= E(s)G(s) (7.2)

Finally substituting Eq (7.2) into Eq. (7.1) and solving for E(s) yields

)(1

)()(

sG

sRsE

(7.3)

0

)(lim)(lim)(

st

ssEtee (7.4)

Apply final value theorem and letting t approach infinity

By substituting Eq. (7.3) into Eq. (7.4) we obtain,

)(1

)(lim)(

0 sG

ssRe

s

(7.5)

Steady State Error

for Step, Ramp and Parabolic Input

Now we can calculate the steady state error, e(∞), given the input, R(s), and the system, G(s).

Step input: with R(s) = 1/s, we find

Ramp input: with R(s) = 1/s2, we find

Parabolic input: R(s) = 1/s2, we find

Time for Some Exercise!Find the steady state errors for inputs 5u(t), 5tu(t) and 5t2u(t).

Revision:

Laplace Transform Table

16

Time for More Exercise!Find the steady state errors for inputs 5u(t), 5tu(t) and 5t2u(t).

Static Error Constants

and System Type

STATIC ERROR CONSTANT AND SYSTEM TYPE

Steady state error performance specifications are called static error

constants.

In the previous section we derived the following relationships:-

. For a step input, u(t),

For a ramp input, tu(t),

,2

1 input, parabolic aFor 2u(t)t

Each term in the denominator are taken to the limit. We call these limits static

error constant. Individually, their names are:

STATIC ERROR CONSTANT AND SYSTEM TYPE

Depending the form of G(s), the value of steady state error (SSE) could be zero, finite or infinity.

Since the static error constant appears in the denominator of the SSE, the value of the SSE

decreases as the static error constant increases.

Previously we have evaluated SSE by using final value theorem.

Alternatively we can make use of the static error constant. Please refer example 7.4

To determine SSE using Static Error Constants

Thus, for a step input:- e(∞)= pK1

1

Thus, for a ramp input:- e(∞)= vK

1

Thus, for a parabolic input:- e(∞)= aK

1

What is SYSTEM TYPE?

Feedback control system for defining system type

A system with n = 0 is a Type 0 system. If n = 1 or

n=2 , the corresponding system is a Type 1 or Type

2 system respectively.

Relationships between Input, System Type, Static Error

Constants, and Steady-state Errors

Steady-state Error Specifications

For example, if a control system has the specification KV = 1000, we can draw

several conclusions:

1. The system is stable

2. The system is of Type 1, since only Type 1 systems have Kv’s that are finite constants.

Recall that Kv = 0 forType 0 system, whereas Kv= ∞, forType 2 systems.

3. A ramp input is the test signal. Since Kv is specified as a finite constant, and the steady

state error for a ramp input in inversely proportional to Kv, we know the test input is a

ramp.

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

input slope.

STEADY STATE ERROR SPECIFICATIONSStatic error constant can be used to specify the study state error characteristics of control

systems.

Tutorial Questions

Tutorial Exercise Q2Compute the steady state error for unit step and ramp input

)77(

4)(

2

2

ss

ssT

)](1)[(lim)(0

sTssRes

)(,eSo

sunitstepswhereRsTsRsE

1)()],(1)[()(

Tutorial Exercise!For the transfer function below, solve for test input 5u(t)

(Use final value theorem method)

)22(

50/1)(

2

sssG

Laplace transform of 5u(t) is 5/s, so

Tutorial Exercise Q4A control system has Kp=1000.

List FOUR (4) conclusions about this system

1. The system is stable

2. The system is of Type 0, since only Type 0 systems have

finite Kp Recall that Kp = ∞ for Type 1 system, whereas

Kv= ∞, forType 2 systems.

3. A step input is the test signal.

4. The steady state error per unit step is

1/1+Kp=1/(1+1000)=1/1001

Tutorial Exercise Q5For the transfer function below, determine system type. Solve for test input 5u(t), 5tu(t)

Reduce the block diagram

System is Type 1

Next, calculate error for step and ramp input

But for Type 1 system, e step(∞)=0, e ramp(∞)=?

Use static error constant method.

Tutorial Exercise Q6 (a)Identify system type and find value of K so there is 10%$ error in the steady state

)18)(14(

)12()(

ss

sKsG

Tutorial Exercise Q6 (b)Identify system type and find value of K so there is 10%$ error in the steady state

)2610)(1(

)4()(

2

sss

sKsG