Time Response

CONTROL PRINCIPLES

BENE 2323

CHAPTER 4: TIME RESPONSE

Chapter 4 : TIME RESPONSE

Learning Outcomes

Students should be able to:

Find the time response from the transferfunction

Use poles and zeros to determine theresponse of a control system

Describe quantitatively the transientresponse of first-order and second ordersystems

Topics

Introduction

Poles, Zeros and System Response

Poles of a transfer function

Zeros of a Transfer function

Poles and Zeros of a First-Order system

First-Order Systems

Time constant

Rise time

Settling time

Topics Second-Order Systems Overdamped Response Underdamped Response Undamped Response Critically Damped Response

General Second Order Systems Natural Frequency Damping Ratio

Underdamped Second-order Systems Peak Time Overshoot Settling Time Rise Time

Response Characteristics And System Configurations

In Chapter 2, we learned how transfer functions canrepresent linear, time-invariant systems.

In chapter 3 systems were represented directly intime domain via the state and output equations.

After domain mathematical representation ofsubsystem, subsystem is analyzed for transient andsteady state response to see if these characteristicsyield desired behavior.

Chapter 4 is devoted to the analysis of systemtransient response.

Introduction


- Poles of a Transfer Function

- Zeros of a Transfer Function

- Poles and Zeros of a First-order

System

PART 1

The output response of a system is the sum of tworesponses: the forced response and the naturalresponse.

Forced response is also called the steady state erroror particular solution.

Natural response is called the homogenous solution.


1. Value of Laplace transform variable, s that causethe transfer function to be come infinite; or

2. Any roots of the denominator of transferfunction that are common to roots ofnumerator.

Poles of a Transfer Function

1. Value of Laplace Transform variable, s that causeto become zero; or

2. Any roots of numerator of the transfer functionthat common to roots of denominator.

Zeros of a Transfer Function

Example:

Given a transfer function as shown in Figure 4.1 (a).

Figure 4.1 (a)

A pole exists at s = -5 and a zero exists at 2.

Poles and Zeros of a First-Order System

Poles and Zeros of a First-Order System

Figure 4.1

Given a transfer function as shown in Figure 4.1 (a).

)5()2()(

++

=ss

ssC

5)5()2()(

++=

++

=s

Bs

Ass

ssC

Properties Of Poles And Zeros

Figure 4.1

55

35

2

5)(

++=

++=

sss

Bs

AsC

( )( ) 5

2

52

0

=++

=s

s

sA

( )53

25=

+=

ss

sB

tetc 553

52)( +=



Figure 4.1

From the development summarized in Figure4.1(c), we draw the following conclusions:

1. A pole of the input function generates the formof the forced response (i.e., the pole at the origingenerated a step function at the output).

2. A pole of the transfer function generates theform of the natural response (i.e., the pole at 5generated e-5t).


From the development summarized in Figure 4.1(c),we draw the following conclusions:

3. A pole on the real axis generates a exponentialresponse of the form , where is the polelocation on the real axis. Thus, the farther tothe left a pole is on negative real axis, the fasterexponential transient response will decay tozero (i.e., again the pole at 5, see Figure 4.2 forthe general case).

4. The zeros and poles generate the amplitudesfor both the forced and natural responses.


Technique of using poles to obtain the form of thesystem response.

Each pole of the system transfer function that is on thereal axis generates an exponential response that is acomponent of the natural response.

The input pole generates the forced response.


Figure 4.2

Given the system of Figure 4.3, write the output, c(t), ingeneral terms. Specify the forced and natural parts of thesolution.

Example 1

Figure 4.3

By inspection, each system poles generates anexponential as part of the natural response. The inputspole generates the forced response.Thus,

++

++

++=

Response Natural ResponseForced

4321

)5()4()2( )( sK

s

Ks

Ks

KsC

Solution : Example 1

Taking inverse transform, we get

+++=


54

43

221

)( ttt eKeKeKKtc

Solution : Example 1 (continue)

A system has a transfer function,

Write, by inspection, the output, c(t), in general terms if the input is a unit step.

)10)(8)(7)(1()6)(4(10)(++++

++=

ssss

sssG

Example 2

++

++

++

++=


)10()8()7()1( )( sE

s

Ds

Cs

Bs

AsC

++++=


1087

)( tttt EeDeCeBeAtc


First Order Systems- Time Constant

- Rise Time

- Settling Time

PART 2

First Order Systems

A first order system without zeros can be described by the transfer function shown in Figure 4.4(a).

Figure 4.4 (a) First-order systems (b) Pole plot

If the input is a unit step , where

The Laplace transform of the step response is C(s), where

Taking the inverse transform, the step response is given by,

ssR 1)( =

)()()()( assa

sRsGsC+

==

atnf etctctc

=+= 1)()()(

First Order Systems

Equ. 4.1

where the input pole at the origin generated the forceresponse, cf (t) = 1 , and the system pole at a, asshown in Fig.4.4(b), generated the natural response, cn(t) = -e at .

First Order Systems

Equation 4.1 is then plotted as shown in Figure 4.5.

First Order Systems

Lets examine the significance of parameter a, the only parameter needed to describe the transient response. When

or

Now we define three transient response performance specifications.

at 1=

37.01/1

== =

eeat

at

63.037.011)( 1/1 === = etc at

First Order Systems

Equ. 4.2

Equ. 4.3

1/a is the time constant of the response.

From Eq.(4.2), the time constant can bedescribed as the time for e-at to decay to 37% ofits initial value.

Alternately, from Eq.(4.3), the time constant isthe time it takes for the step response to rise to63% of its final value (Fig.4.5).

The reciprocal of the time constant has the units(1/seconds), or frequency.

Time Constant

We can call the parameter a the exponentialfrequency.

Since, the derivative of e-at is a when t = 0, a is theinitial rate of change of the exponential at t = 0.

Thus, the time constant can be considered atransient response specification for a first-ordersystem, since it is related to the speed at which thesystem responds to a step input.

Time Constant

The time constant can also be evaluated fromthe pole plot (Fig.4.4(b)).

Since, the pole of the transfer function is at a,we can say the pole is located at the reciprocal ofthe time constant, and the farther the pole fromthe imaginary axis, the faster the transientresponse.

Formula for time constant is,

Time Constant

aconstant time

=1

Rise time, Tr is defined as the time for thewaveform to go from 0.1 to 0.9 of its finalvalue.

Rise time is found by solving Eq.(4.2) for thedifference in time at

c1(t) = 0.9 and c2(t) = 0.1. Hence,

aaaTr

2.21.03.2==

Rise Time, Tr

where,

and9.01)(1 == atetc 1.01)(2 == atetc

at

at

e

e

at

at

3.23.2)9.01ln(

9.019.01

=

==

=

=

at

at

e

e

at

at

1.01.0)1.01ln(

1.011.01

=

==

=

=

Rise Time, Tr

Settling time is defined as the time for the response to reach, and stay within 2% of its final value.

Letting c(t) = 0.98 in Eq.(4.1) and solving for time, t , we find the settling time to be:

Formula for settling time is,

aaaTs

491.3)98.01ln(=

=

Settling Time, Ts

aTs

4=

Since the transfer function is a representation ofthe system from input to output, the systemsstep response can lead to a representation eventhough the inner construction is not known.

With a step input, we can measure the timeconstant and steady-state value, from which thetransfer function can be calculated.

First-Order Transfer Functions via Testing

Consider a simple first-order system,

whose step response is,

If we can identify K and a from laboratory testing, we can obtain the transfer function of the system.

)/()( asKsG +=

)(//

)()( asaK

s

aKass

KsC

+=

+=


For example:

Assume the unit step response given in Figure 4.6.

It has the first-order characteristics no overshoot and nonzero initial slope.


Figure 4.6

From the response we measure the time constant, that is, the time for the amplitude to reach 63% of its final value.

Since the final value is about 0.72, the time constant is evaluated where the curve reach 0.63 0.72 = 0.45, or about 0.13 second.

Hence, time constant = 0.13 =

therefore,


a

1

7.713.01

==a

To find K, we realize that the forced response reaches a steady-state value of

Substituting the value of a = 7.7, we find

Thus, the transfer function for the system is

*Note :Actually, the response of Figure 4.6 was generated using the transfer function of

72.0/ =aK

First Order Systems

0.72 7.7 0.72 5.54K a= = =

)7.7(54.5)(+

=ss

sG

)7(5)(+

=ss

sG

A system has a transfer function,

Find the time constant, Tc ; settling time, Ts ; and rise time, Tr.

)50(50)(+

=ss

sG

Exercise 1

Second Order Systems- Introduction

- Overdamped Response

- Underdamped Response

- Undamped Response

- Critically Damped Response

PART 3

A second-order system exhibits a wide range ofresponses that must be analyzed and described.

Varying a first-order systems parameter simplychanges the speed of the response; changes inthe parameters of a second-order system canchange the form of the response.

Introduction

Example, a second-order system can displaycharacteristics much like a first-order system or,depending on component values, display dampedor pure oscillations for its transient response.

Numerical examples of the second-order systemresponses are shown in Figure 4.7. All examplesare derived from Fig. 4.7(a), the general case,which has two finite poles and no zeros.

Introduction

Figure 4.7 second order systems, pole plots and step responses

The term in the numerator is simply a scale or input-multiplying factor that can take on any value withoutaffecting the form of derived results.

By assigning appropriate values to parameters a and b, wecan show all possible second-order transient responses.

The unit step response then can be found using C(s) =R(s) G(s) ,

where, followed by a partial-fraction expansionand the inverse Laplace transform.

ssR 1)( =

Introduction

For this response, Figure 4.7(b)

)146.1)(854.7(9

)99(9)( 2 ++=++= sssssssC

Overdamped Response

This function has a pole at the origin that comesfrom the unit step input and two real poles thatcome from the system.

The input pole at the origin generates theconstant forced response; each of the twosystem poles on the real axis generates anexponential natural response whose exponentialfrequency is equal to the pole location.

Overdamped Response

The output initially could have been written as ;

This response is called overdamped (Figure 4.7(b).

It seems that the poles could tell the form of the response without the tedious calculation on the inverse L-transform.

tteKeKKtc 146.13

854.721)( ++=

Overdamped Response

For this response, Figure 4.7(c)

This function has a pole at the origin that comes from the unit step input and two complex poles that come from the system.

Compare the response of the second-order system to the poles that generated it.

)92(9)( 2 ++= ssssC

Underdamped Response

First compare the pole location to the time function, andthe compare the pole location to the plot. FromFig.4.7(c), the poles that generate the natural response

are s = -1 j .

Comparing these values to c(t) in the same figure, thereal part of the pole matches the exponential decayfrequency of the sinusoids amplitude, while the imaginarypart of the pole matches the frequency of the sinusoidaloscillation.


Lets compare the pole location to the plot.


Figure 4.8 shows a general, underdamped response for a second-order system.


The transient response consists of anexponentially decaying amplitude generated bythe real part of the system pole times asinusoidal waveform generated by the imaginarypart of the system pole.

The time constant of the exponential decay isequal to the reciprocal of the real part of thesystem pole.

The value of the imaginary part is the actualfrequency of the sinusoid, depicted in (Fig.4.8).


This sinusoidal frequency is called dampedfrequency of oscillation,d.

Finally, the steady-state response (unit step) wasgenerated by the input pole located at the origin.

The type of response shown in Figure 4.8 is calledan underdamped response, one whichapproaches a steady-state value via a transientresponse that is a damped oscillation.


To demonstrate how a knowledge of therelationship between the pole location and thetransient response can lead rapidly to theresponse form without calculating the inverseLaplace transform.

Example 3

By inspection, write the form of the step response of the system in Figure 4.9.

Example 3 - Form of underdamped response using poles

Figure 4.9

First, we determine that the form of the forcedresponse is a step.

Next, find the form of the natural response.

Factoring the denominator of the transferfunction in Figure 4.9, we find the poles to be s= -5 j13.23.


The real part, -5, is the exponential frequency for thedamping. It is also the reciprocal of the time constant ofthe decay of the oscillations.

The imaginary part, 13.23, is the radian frequency forthe sinusoidal oscillations.

Using previous discussion and Figure 4.7(c) as a guide,we obtain:


c(t) = K1 + e-5t ( K2 cos 13.23t + K3 sin 13.23t )

= K1 + K4 e-5t (cos 13.23t - )

where,

= tan-1 K3 / K2 , K4 = and c(t) is a constant plus an exponentially damped

sinusoid.

23

22 KK +


For this response, Figure 4.7(d)

This function has a pole at the origin that comes from the unit step input and two imaginary poles that come from the system.

)9(9)( 2 += sssC

Undamped Response

The input pole at the origin generates theconstant forced response, and the two system poleson the imaginary axis at j3 generate a sinusoidalnatural response whose frequency is equal to thelocation of the imaginary poles.

Undamped Response

Hence, the output can be estimated as c(t) = K1 + K4 cos( 3t - ).

The response type shown in Figure 4.7(d) is calledundamped.

Note that the absence of a real part in the pole paircorresponds to an exponential that does not decay.

Mathematically the exponential is e-0t =1 .

Undamped Response

For this response, Figure 4.7(e)

)96(9)( 2 ++= ssssC

Critically Damped Response

This function has a pole at the origin that comes fromthe unit step input and two multiple real poles thatcome from the system.

The input pole at the origin generates the constantforced response, and the two poles on the real axis at -3generate a natural response consisting of an exponentialand an exponential multiplied by time, where theexponential frequency is equal to the location of the realpoles.


Hence, the output can be estimated as c(t) = K1 + K2 e-3t

+ K3 t e-3t .

This type of response, shown in Figure 4.7(e), is calledcritically damped.

Critical damped responses are the fastest possiblewithout the overshoot that is characteristic of theunderdamped response.


Overdamped responses :

Poles : Two real at -1 , -2 Natural response :

cn(t) = K1 e-1t + K2 e

-2t

Two exponentials with time constants equal to the reciprocal of the pole locations, or

cn(t) = K1 e-1t + K2 e

-2t

Summary of Second-Order Systems

Underdamped responses :

Poles :Two complex at -d jd Natural response : Damped sinusoid with anexponential envelope whose time constant is equalto the reciprocal of the poles real part. The radianfrequency of the sinusoid, the damped frequency ofoscillation, is equal to the imaginary part of thepoles, or

cn(t) = A e-dt cos d t -


Undamped responses :

Poles :Two imaginary at j1 cn(t) = A cos (1 t - ) Natural response : Undamped sinusoid with radianfrequency equal to the imaginary part of the poles,or

cn(t) = A cos (1 t - )


Critically damped responses :

Poles :Two real at -1 cn(t) = K1 e

-1t + K2 te-1t

Natural response :One term is an exponential whosetime constant is equal to the reciprocal of the polelocation.Another term is the product of time, t , andan exponential with time constant equal to thereciprocal of the pole location, or

cn(t) = K1 e-1t + K2 te

-1t


The step responses for the four cases of dampingdiscussed in this section are superimposed in Figure 4.10.

Notice that the critically damped case is the divisionbetween the overdamped cases and the underdampedcases and is the fastest response without overshoot.


Figure 4.10

General Second Order System

- Natural Frequency

- Damping Ratio

PART 4

In this section we define two physically meaningfulspecifications for second-order system.

Natural Frequency,wn the frequency of oscillation of the system without

damping.

The General Second Order System

Damping Ratio,

we define the damping ratio, , to be:

constant timelexponentiaperiod natural

21

d)(rad/seconfrequency naturalfrequencydecay lexponentia

pi

=

=


Damping Ratio,

Consider the general system

Without damping, the poles would be on the jw axis, andthe response would be an undamped sinusoid. For thepoles to be purely imaginary, a=0. Hence, by definition,the natural frequency,wn, is the frequency of oscillation ofthis system. Since the poles of this system are on the jwaxis at ,

bassb

sG++

= 2)(

bj bwn =


Damping Ratio,

Hence,

Assuming an underdamped system, the complex poleshave a real part, equal to a/2. The magnitude of thisvalue is then the exponential decay frequency describedin section 4.4. Hence,

2nwb =

nn w

a

wradfrequencyNaturalfrequencydecaylExponentia 2

sec)/(

===

nwa 2=


Damping Ratio,

Our general second-order transfer function finally looks like this:

Solving the poles of the transfer function, G(s) yields:

The various cases of second-order response are a function of and are summarized in Figure 4.11.

122,1 = nn wws


2

2 2( ) 2n

n n

G ss

=+ +

Given the transfer function as shown below, find

and wn.

362.436)( 2 ++= sssG

Example 4

Compare with the general second-order transfer function,

362.436)( 2 ++= sssG

22

2

2)(

nn

n

wws

wsG

++=


Therefore

362 =nw

6= nw

2.42 =nw

35.062

2.4=

=


For each of the systems shown in Figure 4.12,finds the value of and give the kind of responseexpected.

Exercise 2

Underdamped Second Order

System

- Peak Time

- Overshoot

- Settling Time

- Rise Time

PART 5

A common model for physical problem.

A plot of this response appears in Figure 4.13 for variousvalues of , plotted along a time axis normalized to thenatural frequency.

UNDERDAMPED SECOND ORDER SYSTEMS

We have defined two parameters associated with second-ordersystems, and wn. Others parameters associated with the undampedresponse are percentage overshoot (%OS), peak time (Tp), settlingtime (Ts) and rise time (Tr). These specifications are defined as follow(refer Figure 4.14):

Underdamped Second Order Systems

1. Peak time, Tp : The time required to reach the first,or maximum peak.

2. Percent overshoot, %OS : The amount that thewaveform overshoots the steady state, or final valueat the peak time.

3. Settling time, Ts : The time required for thetransients damped oscillations to reach and staywithin +2% of the steady state value.

4. Rise time, Tr : The time required for the waveform togo from 0.1 of the final value to 0.9 of the final value.


Evaluation of Tp

Evaluation of %OS

where

21 pi

=

n

pw

T

100100% )1(max2

=

= pieC

CCOS

final

final

)1(max

2

1 pi += eC


and for the unit step used;

The inverse of this equation allows to solve for by:

1=finalC

)100/(%ln)100/ln(%

22 OSOS

+

=

pi


Evaluation of Ts For criteria between +2%

For criteria between +5%

n

sw

T

4=

n

sw

T

3=


Underdamped Second Order SystemsEvaluation of Tr A precise analytical relationship between rise time and damping ratio,

cannot be found. However, we can find the rise time using computer.

Figure 4.15 illustrates the resulting curve obtained relating thevariation of wntr and the damping ratio, .

*Note: Fig. 4.16 can be approximated by the following polynomials

Underdamped Second Order Systems If a simple expression is desired forapproximations, we can obtain the values of tr overthe desirable range of damping factors, from 0.4 to0.7, and determine an average value over thisrange.

The value of wntr for a damping of 0.4 is 1.463; thevalue of wntr for a damping of 0.7 is 2.126.Therefore, an average value rise time over thisdamping factor range is given by, tr = 1.80/wn.


Figure 4.17 Pole plot for an underdamped second-order system

wd is the imaginary part of the pole and called the damped frequency ofoscillation, and

is the magnitude of the real part of the pole and called as exponentialdamping frequency.

d

Given the transfer function

Find , wn, Tp, %OS, Ts and Tr.

10015100)( 2 ++= sssG

Example 5 - Finding Tp, %OS and Ts from transfer function

1002 =nw10= nw

152 =nw

21015

=

75.0=


Natural frequency, Damping ratio,

21 pi

=

n

pw

T

275.0110 =

pipT

475.0= pT


100% )1(2

= pieOS

100% )75.0175.0(2

= pieOS

838.2% = OS

Peak Time, Overshoot,

nsw

T

4=

1075.04

=sT

5333.0= sT


#From Fig. 4.15, we get Tr=2.3,Tr = 2.3/wnTr = 2.3/10Tr = 0.23 s

Settling time,

Exercise 3

Find , wn, Tp, %OS, Ts and Tr.

a)

b)

16316)( 2 ++= sssT

04.002.004.0)( 2 ++= sssT

Given the pole plot shown in Figure 4.17, find , wn, Tp, %OS and Ts.

Example 6 - Finding Tp, %OS and Ts from pole location

cos=)

37

cos(tan 1=)166.1cos(=

394.0=


22 37 +=nw616.7= nw

21 pi

=

n

pw

T


7pipi

==d

pw

T

sec449.0= pT

Percent Overshoot,

100% )1(2

= pieOS

100% )394.01394.0(2

= pieOS%26% =OS


Settling time

n

sw

T

4=


344

==d

sT

sec333.1=sT

Exercise 4

Given %OS = 12% and Ts = 0.2s. Find :

a) Location of poles

b) Transfer function of the system

Exercise 5

Given the system shown below, find J and D to yield 20% overshoot and a settling time of 2s for a step input of Torque(t).

Thank you for your attention.

Time Response

Documents

system response poles

system response poles

transfer function poles

transfer function poles

transfer function zeros

transfer function zeros

natural response

output response