UNIT II - manojmechkul.files.wordpress.com · • Time response: The variation of output with respect to time is called time responsetime response. • Time response analysis : A

UNIT IIUNIT II

Time Domain Analysis and DesignTime Domain Analysis and Design

Topics to be coveredTopics to be covered

• Standard test inputs, p ,• Time response of first order and second order systems,• Steady state analysis: steady state error and error constants,

transient response specifications• Stability analysis – Routh-Hurwitz criterion

R t L t h i D i f t i R t• Root Locus technique, Design of compensators using Root Locus

• Time response: The variation of output with respect to time is called time responsetime response.

• Time response analysis : A input signal is applied to a system and the performance of the system is evaluated by studying the system response in time domainresponse in time domain.

• To analyze and design a control system it is necessary to define and measure its performance.Then,based on the desired performance of control system the system parameters may be adjusted to provide thecontrol system the system parameters may be adjusted to provide the desired performance.

• In order to analyze time response, first step is always obtain mathematical model of systemmathematical model of system.

• Before proceeding with the time response analysis of control system, it is necessary to test the stability of the system.

• System stability can be tested through indirect tests without actually• System stability can be tested through indirect tests without actually obtaining transient response. If the system is unstable no need to proceed with transient response analysis

Introduction• Control systems are generally called upon to perform both under transient(dynamic) and steady conditions.

• Analysis of response means to see the variation f i h i Thiof output with respect to time. This output 

behavior with respect to time should be within specified limits to have satisfactory performancespecified limits to have satisfactory performance of the system.

• The system stability, system accuracy andThe system stability, system  accuracy and complete evaluation is always based on the time response analysis and corresponding results.

Time ResponseTime Response

Definition: The response given by the systemDefinition: The response given by the system which is function of the time , to the applied excitation is called time response of a controlexcitation is called time response of a control system.

IntroductionThe time response of a control system consists The time response of a control system consists of two parts:of two parts:

1.Transient response ‐ from initial state to the final state – purpose of p p

control systems is to provide a desired response.

d2. Steady‐state response‐ the manner in which the system output 

behaves as t approaches infinity – the error after

the transient response has decayed, leaving onlyp y , g y

the continuous response. 

• Control systems have transient phenomenon before steady state is reached.

• Because of mass ,inertia , inductance etc the response of control system cannot follow sudden changes in the input instantaneously and hence transients are present. The control y pof transient response is necessary.

• Transient response gives how closely or quickly the system follows changes in the inputfollows changes in the input.

• Steady state response of control system gives the final f haccuracy of the system.

• If the steady state response of the output is not equal to the desired output , the system is said to have steady state error. p , y y

• Therefore,the study of CS in time domain means study of transient and steady state response of the system.

Performance of Control SystemsSpecifications (time domain) 

St d d i t i l d i d iStandard input signals used in design 

‐ actual signals unknown 

‐ standard test signals: 

step, ramp, parabola, impulse, etc. sinusoid        

(study freq. response later)

Transient responseTransient response 

• Steady‐state response 

Relate to locations of poles and eros• Relate to locations of poles and zeros 

Test InputTest Input

• Control systems are used to control some physical variable That variable may be a temperature somewhere thevariable. That variable may be a temperature somewhere, the attitude of an aircraft or a frequency in a communication system. Whatever the variable, it is important to control the

i bl lvariable accurately. • If you are designing a control system, how accurately the

system performs is important. If it is desired to have thesystem performs is important. If it is desired to have the variable under control take on a particular value, you will want the variable to get as close to the desired value as possible Certainly you will want to measure how accuratelypossible. Certainly, you will want to measure how accurately you can control the variable. Beyond that you will want to be able to predict how accurately you can control the variable.

• To be able to measure and predict accuracy in acontrol system, a standard measure ofy ,performance is widely used. That measure ofperformance is steady state error - SSE - andsteady state error is a concept that assumes thesteady state error is a concept that assumes thefollowing: The system under test is stimulatedwith some standard input. Typically, the test inputis a step f nction of time b t it can also be ais a step function of time, but it can also be aramp or other polynomial kinds of inputs.

• The system comes to a steady state, and theThe system comes to a steady state, and thedifference between the input and the output ismeasured.Th diff b h i h d i d• The difference between the input - the desiredresponse - and the output - the actual response isreferred to as the error.

First Order SystemFirst Order System

as1R(s) C(s)E(s)

as

Test signal is step function, R(s)=1/s

First – order systemA first-order system without zeros can be represented by the following transfer function

11

)()(

+=

ssRsC

τ

p y g

1)( +ssR τ

Given a step input, i.e., R(s) = 1/s , then the system output (called step response in this system output (called step response in this case) is

ττ 111

)1(1)(

11)(

++=

+=

+=

sssssR

ssC

τ

First – order systemFirst  order systemTaking inverse Laplace transform, we have the step response

τt

etc−

−=1)(Time Constant: If = t, So the step response is c(t) =

(1− 0.37) = 0.63τ

( )

a is referred to as the time constant of the response. In other words, the time constant is the time it takes for other words, the time constant is the time it takes for the step response to rise to 63% of its final value. Because of this, the time constant is used to measure how fast a system can respond The time constant has how fast a system can respond. The time constant has a unit of seconds.

First – order system

Plot c(t) versus time:

First – order systemUnit Ramp Response

First – order system

First – order system

Time Constant , 1

=τRise Time Tr:

h ( b l ) d d h

a

The rise-time (symbol Tr units s) is depend as the time taken for the step response to go from 10% to 90% of the final value. Tr= 2.2

a

Settling Time Ts:

Defined the settling-time (symbol Ts units s) to be the time taken for the step response to come to within p p2% of the final value of the step response. Ts= 4

a

Problem

A t h t f f ti G( ) 50A system has a transfer function G(s)=    50s+50

Find the time constant, Tc, Settling time, Ts and Rise Time, Tr

Find Tc, Tr and Ts

Second – Order System

Second-order systems exhibit a wide range of responses which t b l d d d ib dmust be analysed and described. Whereas for a first-order system, varying a single parameter changes the speed of response, changes in the parameters of a g p p , g psecond order system can change the form of the response.

F l d d di lFor example: a second-order system can displaycharacteristics much like a first-order system or, depending on component values, display damped or pure oscillations for itscomponent values, display damped or pure oscillations for its transient response.

Second – Order System

- A general second-order system is characterized by the following transfer function:g

- We can re-write the above transfer function in the following form (closed loop transfer function):

Second – Order System

- referred to as undamped the natural frequency of the second order system which frequency of the second order system, which is the frequency of oscillation of the system without damping.

- referred to as the damping ratio of the second order system, which is a measure of the degree of resistance to change in the the degree of resistance to change in the system output.

Poles;Poles are complex if ζ< 1!

Second – Order System

- According the value of ζ, a second-order system can be set into one of the four categories:can be set into one of the four categories:

1. Overdamped - when the system has two real di i l (ζ 1)distinct poles (ζ >1).2. Underdamped - when the system has two complex conjugate poles (0

Second – Order System- Given a step input, i.e., R(s) =1/s, then the system output (or step response) is;( p p ) ;

- Taking inverse Laplace transform, we have the step response;

Where;

Second – Order System

Second – Order SystemThe typical step response of an underdamped second-order system;

Second – Order System

• rise time (time to rise past command), Tr• peak time (time to first peak), Tp• percent overshoot, Mp (Maximum)• settling time (time to settle into bounds), Ts

1) Rise time, Tr. The time required for the waveform to go from 0.1 of the final value to 0.9 of the final value.

i l i l l i hi b i i dA precise analytical relationship between rise time and damping ratio cannot be found.

Second – Order System2) Peak time, Tp - The peak time is the time required for the response to reach the first peak, which is given by;

3) Percent overshoot, %OS - The percent overshoot is defined as the amount that the waveform at the peak time overshoots the steady-state value, which is expressed as a percentage of the steady-state value.

For given %OS the damping ratio can - For given %OS, the damping ratio can be solved from the above equation;

Second – Order System

100*max% CfinalCOS − 100*%Cfinal

fOS =

4) Setting time, Ts - The settling time is the time required for the amplitude of the sinusoid to decay required for the amplitude of the sinusoid to decay to 2% of the steady-state value.

nsT ζω

4=

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T βπ −=

Rise time (0 to 100%):

drT ω

h

21 ζωω −= nd

Where

, ζnd

⎟⎟⎠

⎞⎜⎜⎝

⎛= − d

ζω

β 1tan ⎟⎠

⎜⎝ nζω

β

Second – Order SystemExample: Find the natural frequency and damping ratio for the system with transfer functiony

36)( =sG

Solution:

362.4)( 2 ++=

sssG

Compare with general TF

ωn= 6

ξ =0.35

Second – Order SystemExampleExample:: Given the transfer function

Second  Order System

ps TOSTfind ,%,

Solution:

750;10 ξω

sTOSsT 4750%8382%5330

75.0;,10 == ξωn

sTOSsT ps 475.0%,838.2%,533.0 ===

Poles and ZerosPoles and Zeros

Overdamped Response

1>ξ 1>ξ

)146.1)(854.7(9

)99(9)( 2 ++

=++

=ssssss

sC

s= 0; s = s= 0; s = --7.854; s = 7.854; s = --1.146 1.146 ( two real poles)( two real poles)

tt KKK 14618547)( tt eKeKKtc 146.13854.7

21)(−− ++=

OVERDAMPEDOVERDAMPED RESPONSE !!!RESPONSE !!!OVERDAMPEDOVERDAMPED RESPONSE !!!RESPONSE !!!

Underdamped Response

10

UNDERDAMPEDUNDERDAMPED RESPONSE !!!RESPONSE !!!UNDERDAMPEDUNDERDAMPED RESPONSE !!!RESPONSE !!!

Undamped Response

0=ξ

tKKtc 3cos)( 21 +=s = 0; s = s = 0; s = ±± j3 j3 ( two imaginary poles)( two imaginary poles)

UNDAMPEDUNDAMPED RESPONSE !!!RESPONSE !!!

Critically Damped System

1=ξ

tt teKeKKtc 333

21)(−− ++=

S = 0; s = S = 0; s = --3,3,--3 3 ( two real an equal poles)( two real an equal poles)

CRITICALLY DAMPEDCRITICALLY DAMPED RESPONSE !!!RESPONSE !!!

Second – Order System

Second – Order System

Second – Order SystemExample: Describe the nature of the second-order system response via the value of the damping ratio for th t ith t f f tithe systems with transfer function

12)(1 =sG128

)(.1 2 ++=

sssG

16168

16)(.2 2 ++=

sssG

20820)(.3 2 ++

=ss

sG

Stability of a Linear feedback SystemStability of a Linear feedback System

• Stability implies, small changes in the system input(in initialconditions or in system parameters) do not result in largechanges in the system output.

• After the transient behavior, if the system reaches to steadystate value ,then the system is called as a stable system.

• Almost every working system is designed to be stable.• Unstable closed loop system is generally of no practicalUnstable closed loop system is generally of no practical

value.• If the system is stable, then for what range of system

parameters it is stable?parameters it is stable?• How it can be made stable by varying the system

parameters ,if it is unstable?• All these things are important while designing a system• All these things are important while designing a system

• Stabilty of a feedback system is related with the roots of characteristic equation of system T.F.of characteristic equation of system T.F.

• Routh- Hurwitz method is useful for assessing system stability.

• It is useful to determine how the roots of characteristic equation move around S-plane as we change the system parameter The locus of the roots in the s plane can beparameter.The locus of the roots in the s-plane can be determined from the graphical method.

• The graph of roots by varying one system parameter is g p y y g y pknown as root locus plot.It’s a powerful tool for designing and analysing feedback control systems.

• This method provides graphical information and• This method provides graphical information and approximate sketch can be used to obtain qualitative information about stability and performance of the system.

Stability DefinitionsStability DefinitionsStability DefinitionsStability Definitions

• Bounded Input Bounded Output Stability:p p yA system is BIBO stable if, for every bounded input, the output remains bounded with increasing time ( ll t l t li i th l ft h lf f th(all system poles must lie in the left half of the s‐plane).

• Marginal Stability:Marginal Stability:A system is marginally stable if some of the poles lie on the imaginary axis, while all others are in the 

f h l l hLHS of the s‐plane.  Some inputs may result in the output becoming unbounded with time. 

Stability AnalysisStability AnalysisStability AnalysisStability Analysis

• To test the stability of a LTI system we need only y y yexamine the poles of the system, i.e. the roots of the characteristic equation.

• Methods are available for testing for roots with positive real parts, which do not require the actual solution of the characteristic equationsolution of the characteristic equation.

• Also, methods are available for testing the stability of a closed-loop system based only on the loopof a closed loop system based only on the loop transfer function characteristics.

R h H i S bili C i iRouth‐Hurwitz Stability Criterion• A quick method for checking BIBO stability.• Assume the characteristic polynomial is

1)( asasasasQ nn ++++= −

where a0 ≠ 0 . • A necessary (but not sufficient) condition for all roots to

011)( asasasasQ nn ++++= − L

A necessary (but not sufficient) condition for all roots to have non‐positive real parts is that all coefficients have the same sign.

• For the necessary and sufficient conditions first form theFor the necessary and sufficient conditions, first form the Routh array.

Routh‐Hurwitz Stability CriterionTh R th AThe Routh Array

1)( asasasasQ nn ++++= − L

sn a a a a

011)( asasasasQ nn ++++ − LHurwitz Determinants are formed as follows

s an an‐2 an‐4 an‐6 …sn‐1 an‐1 an‐3 an‐5 an‐7 …sn‐2 b b b b 321 − nnnn aaaab

where

s b1 b2 b3 b4 …sn‐3 c1 c2 c3 c4 …: : : : 541 − aaaa

1

3211

−

−−−=n

nnnn

aaaaa

b

: : : :s2 k1 k2s1 l1

1

5412

−

−−−=n

nnnn

aaaaa

b

etc.s l1s0 m1

Routh‐Hurwitz Stability CriterionTh R th AThe Routh Array

• In a similar manner, elements in the 4th row, c1, c2 , … are calculated based on the two previous rows.

21311 b

baabc nn −−

−=

31512

1

bbaab

c

b

nn −− −=

• The elements in all subsequent rows are calculated in the same manner

1b

in the same manner.

hh b lb lRouthRouth‐‐Hurwitz Stability CriterionHurwitz Stability CriterionNecessary and sufficient conditions:Necessary and sufficient conditions:• If all elements in the first column of the Routharray have the same sign, then all roots of the y g ,characteristic equation have negative real parts.

• If there are sign changes in these elements, then the number of roots with non‐negative real parts is equal to the number of sign changes.El i h fi l hi h d fi• Elements in the first column which are zero define a special case.

Routh‐Hurwitz Stability CriterionE l 1Example 1:

10532)( 234 ++++= sssssQ 10532)( ++++ sssssQ

s4 2 3 10 0 1010001010

77101033

2211==

−−==−−==

−−== bbbbs

s3 1 5 0 0s2 b1 b2 0 4343..6677

10103535

1111

11

2211

==−−

−−−−==ccs

s1 c1 0s0 d1

10104343..66

00))4343..66((1010

77

11==

−−==

−−

dds 4343..66

The characteristic equation has two roots with positive real parts since the elements The characteristic equation has two roots with positive real parts since the elements of the first column have two sign changes. (2,1,of the first column have two sign changes. (2,1,‐‐7,6.43,10)7,6.43,10)

Routh‐Hurwitz Stability CriterionSpecial Case 1:Special Case 1:

• A zero in the first column:• Remedy: substitute ε for the zero element, finish the Routh array, and then let ε → 0 .finish the Routh array, and then let ε → 0 .

23)( 3 +−= sssQ 2231

−→

−−=b

ε (negative)

s3 1 ‐3 0 02

0(ε) 2 0 221

1

1

=⋅

=

→

bc

bεε

s 0(ε) 2 0s1 b1 00

11 b

There are two roots with positive s0 c1 real parts (1, ε, ‐2/ε , 2)

Routh‐Hurwitz Stability CriterionS i l C 2Special Case 2:

• An all zero row in the Routh array corresponds to i f i h i ipairs of roots with opposite signs.

• Remedy:form an auxiliary polynomial from the coefficients in the– form an auxiliary polynomial from the coefficients in the row above.

– Replace the zero coefficients from the coefficients of the diff ti t d ili l i ldifferentiated auxiliary polynomial.

– If there is not a sign change, the roots of the auxiliary equation define the roots of the system on the imaginary axis.

Routh‐Hurwitz Stability CriterionSpecial Case 2 (example):Special Case 2 (example):

1)( 34 −−+= ssssQ 1)( −−+= ssssQ

12Auxiliary polynomial

s4 1 0 ‐1 0

)1( 2 −sd

12 −sthen

s 1 0 1 0s3 1 ‐1 0 02 1 1 0 2 s

)1( =dssd

d1 = –1

s2 1 ‐1 0s1 0 00 d

2d1   1

The system has one root with a positive real part ( 1, 1, 1, 2, ‐1).

s0 d1

y p p ( , , , , )The root is found from the auxiliary eq.  s2 – 1 = 0 , s = ± 1

Routh‐Hurwitz Stability CriterionP t R T tParameter Range Test

• The Routh‐Hurwitz stability criterion may beThe Routh Hurwitz stability criterion may be used to find the range of a parameter for which the closed‐loop systems is stablewhich the closed loop systems is stable.

• Leave the parameter as an unknown coefficient in the characteristic polynomialcoefficient in the characteristic polynomial, form the Routh array, check the range of the parameter such that the first column does notparameter such that the first column does not change sign. 

Routh‐Hurwitz Stability CriterionP t R E lParameter Range Example

)( 234 KsssssQ ++++= 6116)( 234

660 Ks4 1 11 K 0s3 6 6 0 0

10660

11=−= KdKc

Th f bilis2 10 K 0s1 c1 0

0>KThen for stability,

s 1s0 d1

100