CONTROL SYSTEMS THEORY

Transient response

Chapter 4

Objectives

To find time response from transfer function

To describe quantitatively the transient response of a 1st and 2nd order system

To determine response of a control system using poles and zeros

Introduction In Chapter 1, we learned that the total

response of a system, c(t) is given by

In order to qualitatively examine and describe this output response, the poles and zeros method is used.

forced naturalc t c t c t

Poles & zeros

The poles of a TF are the values of the Laplace variable that cause the TF to become infinite (denominator)

The zeros of a TF are the values of the Laplace variable that cause the TF to become zero (numerator)

Poles & zeros

Example : Given the TF of G(s), find the poles and zeros

Solution : G(s) = zero/pole Pole at s=-5 Zero at s=-2

Poles & zeros

Zero (o), Pole (x) Transfer function = Numerator

Denominator = Zeros

Poles

Poles & zeros

Example : Given G(s), obtain the pole-zero plot of the system

Zero (o)Pole (x)

Poles & zeros

Exercise : Obtain and plot the poles and zeros for the system given

First order system

First order system with no zeros

First order system Performance specifications:

Time constant, t 1/a, time taken for response to rise to 63%

of its final value Rise time, Tr

time taken for response to go from 10% to 90% of its final value

Settling time, Ts time for response to reach and stay within

5% of final value

First order system System response

Second order system

Second order system

Second order system

Exercise : Is this system under/over/critically damped?

Second order system Performance specifications

damping ratio

% Overshoot = cmax – cfinal x 100

cfinal

Second order system Settling time, Ts

Peak time, Tp

nsT

4

a = 2ωn

21

n

pT

Second order system

2nd order underdamped response

Second order system

Second-order response as a function of damping ratio

Second order system

Second order system

Step responses of second-orderunder-damped systems as poles move:

a. with constant real partb. with constant imaginary partc. with constant damping ratio (constant on the diagonal)

Second order system

Exercise

Describe the damping of each system given the information below

Solution

Find value of zeta

2nd order general form

Exercise

Given these 2nd order systems, find the value of and . Describe the damping

Solution

Example

Given

Find settling time, peak time, %OS Hint :

Solution

Block diagram: Analysis

Finding transient responseFor the system shown below, find the peak time, percent overshoot and settling time.

Block diagram: Analysis

Answers:n=10

=0.25Tp=0.324

%OS=44.43Ts=1.6

Block diagram: Analysis and design

Gain design for transient responseDesign the value of gain, K, for the feedback control system of figure below so that the system will respond with a 10% overshoot

Block diagram: Analysis and design

Solution:Closed-loop transfer function is

Kss

KsT

5)(

2

K

Kn

n

2

5

Thus,

and

52

Block diagram: Analysis and design

Can be calculated using the %OS

= 0.591We substitute the value and calculate K, we getK=17.9

100/%ln

100/%ln22 OS

OS

Higher order systems

Systems with >2 poles and zeros can be approximated to 2nd order system with 2 dominant poles

Higher order systems

Placement of third pole. Which most closely resembles a 2nd order system?

Higher order systems Case I : Non-dominant pole is near

dominant second-order pair (=) Case II : Non-dominant pole is far from the

pair (>>) Case III : Non-dominant pole is at infinity

(=)

How far away is infinity? 5 times farther away to the LEFT from dominant poles

Exercises

Find , ωn, Ts, Tp and %OS

a)

b)

c)

T(s) = 0.04

s2 + 0.02s + 0.04

T(s) = 1.05 x 107

s2 + (1.6 x 103)s + (1.5 x 107)

T(s) = 16

s2 + 3s + 16

Solution part (a)

ωn = 4 ζ = 0.375 Ts =4s Tp = 0.8472 s %OS = 28.06 %

Solution part (b)

ωn = 0.2 ζ = 0.05 Ts =400s Tp = 15.73s %OS = 85.45 %

Solution part (c)

ωn = 3240 ζ = 0.247 Ts =0.005 s Tp = 0.001 s %OS = 44.92 %