control chap6

CONTROL SYSTEMS THEORY

Forced Response Errors

CHAPTER 6STB 35103

Objectives To find the steady-state error for a unity

feedback system To specify a system’s steady-state error

performance To design system parameters to meet

steady-state error performance specifications

Introduction In chapter 1, we learnt about 3

requirements needed when designing a control system

Transient response Stability Steady-state errors (SSE)

Up until now we only covered until transient response and stability

Review on transient response We learned in Chapter 4, there are 4 types

of transient response for a second-order system.

Overdamped Underdamped Undamped Critically damped

Review on transient response An example of elevator response

The transient response for elevator can be considered as overdamped. The system is stable but has steady-state error

Introduction What is steady-state error?

Steady-state error is the difference between the input and output for a certain test input as

Test input used for steady-state error analysis and design are

Step Ramp Parabola

t

Introduction Test waveforms

Introduction Example of systems tested using the test

signal. Targeting system:

Targeting a static target. (e.g. a stopping car). We test the system using step input because the position of the car is in constant position.

Targeting a car moving with constant velocity. We test the system using ramp input because the car is moving in constant velocity.

Targeting an accelerating car. We test the system using parabola input because the car is accelerating.

Introduction We are only concerned with the difference

between the input and the output of a feedback control system after the steady state has been reached, our discussion is limited to stable systems where the natural response approaches zero when (time) t approaches infinity.

SSE for unity feedback system Unity feedback system can be represented

as

Steady state error can be calculated from a system’s closed-loop transfer function, T(s), or the open-loop transfer function, G(s), for unity feedback systems.

SSE for unity feedback system

Closed loop transfer function, T(s) is calculated by solving the unity feedback system using the block diagram reduction method for feedback system.

1 ( ) (1)

G sT s

G s

1

SSE for unity feedback system

Open-loop transfer function for a unity feedback system is the value of G(s) multiply 1.

1

1

SSE for unity feedback system Steady state error in terms of T(s).

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

We can find final value of the error, in terms of T(s) using

We can only use this equation if T(s) is stable, E(s) has no poles in the right-half plane or poles on the imaginary axis other than the origin

( )e

0

lim ( ) 1s

e sR s T s

( ) ( ) ( )

( ) ( ) ( )

( ) 1

E s R s C s

R s R s T s

R s T s

SSE for unity feedback system Example 7.1

Find the steady state-error for a unity feedback system that has T(s) = 5/(s2+7s+10) and the input is a unit step.

Solution:R(s) =unit step = 1/s

T(s) = 5/(s2+7s+10), we must check the stability of T(s) using Routh table or poles.

SSE for unity feedback system Example 7.1 (cont.)

We know from the unity feedback system

So, E(s) can be calculated using both equation

( ) ( ) ( )E s R s C s ( ) ( )C s R s T s

( ) ( ) ( )

( ) ( ) ( )

( ) 1

E s R s C s

R s R s T s

R s T s

SSE for unity feedback system Example 7.1 (cont.)

E(s) in example 7.1 is

2

2

2 2

2 2

2 2

1 5( ) 1

7 10

1 7 10 5

7 10 7 10

1 7 5 7 5

7 10 7 10

E ss s s

s s

s s s s s

s s s s

s s s s s s

SSE for unity feedback system Example 7.1 (cont.)

Before calculating the final value of the error we must check the position of E(s) poles

The poles for E(s) are at (0,0), (-2,0) and (-5,0). Since all the poles are not on the right half plane or the imaginary axis we can use the equation to calculate final error value in terms of T(s).

2 2

2

7 5 7 5

2 57 10

s s s sE s

s s ss s s

SSE for unity feedback system Example 7.1 (cont.)

0

20

lim 1

1 5lim 1

7 10

5 5 11

10 10 2

s

s

e sR s T s

ss s s

SSE for unity feedback system Steady state error in terms of G(s)

We can find final value of the error, in terms of G(s) using

We are going to use three types of input R(s); step, ramp and parabola. So the final value of the error for this types of input can be described as

( )e

0

( )lim

1s

sR se

G s

SSE for unity feedback system

Step input

Ramp input

Parabola input

( )e

0

1

1 limstep

s

e eG s

( )e

0

1

limramp

s

e esG s

( )e

2

0

1

limparabola

s

e es G s

sR(s)

sR(s)

sR(s)

SSE for unity feedback system Steady state error with no integration Example 7.2

Find the steady-state errors for inputs of 5u(t), 5tu(t), and 5t2u(t) to the system below.

Solution hint 5u(t) = unit step = 5(1/s) 5tu(t) = ramp = 5(1/s2)

5t2u(t) = parabola = 5(2/s3) = 10(1/s3)

No integration

SSE for unity feedback system

SSE for unity feedback system Example 7.2 (cont)

0

0

2

0

5 5 5

1 lim ( ) 1 20 21

5 5

lim ( ) 0

10 5

lim ( ) 0

step

s

ramp

s

parabola

s

e eG s

e esG s

e es G s

SSE for unity feedback system Try to solve steady state errors for

systems with one integration in Example 7.3.

SSE for unity feedback system From the previous slides, the final error

value for three kinds of input; step, ramp and parabola, are as follows

0

0

2

0

1

1 lim

1

lim

1

lim

step

s

ramp

s

ramp

s

e eG s

e esG s

e es G s

position constant, pK

velocityconstant, vK

acceleration constant, aK

SSE for unity feedback system Steady state error via static error

constants Example 7.4 (Figure 7.7 (a) )

SSE for unity feedback system

Solution First step is to calculate the static error constants.

0 0

0 0

22

0 0

500( 2)( 5)( 6) 500(0 2)(0 5)(0 6)lim lim 5.208

( 8)( 10)( 12) (0 8)(0 10)(0 12)

(500)( 2)( 5)( 6)lim lim 0

( 8)( 10)( 12)

(500)( 2)( 5)( 6)lim lim

(

ps s

vs s

as s

s s sK G s

s s s s

s s s sK sG s

s s s

s s s sK s G s

s

0

8)( 10)( 12)s s

SSE for unity feedback system

Next step is to calculate the final error value.

Try to solve the remaining problems in Figure 7.7 (a) and (c).

1Step input, ( ) 0.161

1

1Ramp input, ( )

1Parabola input, ( )

p

v

a

eK

eK

eK

SSE for unity feedback system System Type

We are still focusing on unity negative feedback system.

Since steady-state errors are dependent upon the number of integrations in the forward path, we give a name to this system attribute.

SSE for unity feedback system Below is a feedback control system for defining

system type.

We define the system type to be the value of n in the denominator.

Type 0 when n = 0 Type 1 when n = 1 Type 2 when n = 2

SSE for unity feedback system Relationship between input, system type,

static error constant, and steady-state errors can be summarized as

SSE for unity feedback system Steady-state error specifications.

We can use the static error constants to represent the steady-state error characteristic of our system.

Conclusion that we can made based on static error constants.

Problem: What information is contained in the specification Kv = 1000.

SSE for unity feedback system

Kv = 1000

Solution:1. The system is stable.2. The system is of Type 1, since only Type 1

have Kv that are finite constant

SSE for unity feedback system

3. A ramp input is the test signal. Refer to table.

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