CH4 Trans Resp

KJM 597: Transient Response Analysis

SYSTEM PERFORMANCE The time response of a control system consists of two parts: the transient and steady state response. The transient response is the response that disappears with time. The steady state response is that which exists a long time following any input signal initiation. The ability to adjust the transient and steady state response performance is a distinct advantage of feedback control systems. To analyse and design a control system, we must measure it performance. Then based on the desired performance, the system parameters may be adjusted to provide the desired response. The most important characteristic of the dynamic behaviour of a control system is the absolute stability i.e. whether the system is stable or unstable. A control system is stable if the output eventually comes to its equilibrium state when the system is subjected to an initial condition. If the system at steady state does not exactly agree with the input, the system is said to have steady state error. This error is indicative of the accuracy of the system. First order system Physically, this system may represent the RC circuit, thermal system, etc. The block diagram is as shown below. The input-output relationship is given by

11

)()(

+=

TssRsC

The system responses are analysed using inputs such as unit step, unit-ramp and impulse and assuming the initial conditions are zero.

Ts1+ -

R(s) C(s)

1


Unit-step response of first order system

s1

ssR 1)( =Laplace transform of unit step function is , and substituting , then

sTssC 1

11)(+

=

expanding

TssTsT

ssC

111

11)(

+−=

+−=

inverse Laplace

Ttetc /1)( −−= , for 0≥t

2


Unit-ramp input

2

1s 2

1)(s

sR =Laplace transform of unit ramp function is , then , and substituting

2

11

1)(sTs

sC+

=

expanding

11)(

2

2 ++−=

TsT

sT

ssC

inverse Laplace

TtTeTttc /)( −−−= , for 0≥t

The error signal is then )(te

)1(

)()()(T

teT

tctrte−−=

−=

3


Tt

e−

as t approaches infinity, approaches zero, thus the error signal approaches T or )(teTe =∞)( .

Unit impulse Laplace transform of unit impulse function is 1 , and substituting , then 1)( =sR

11)(+

=Ts

sC

inverse Laplace

TteT

tc /1)( −= , for 0≥t

The response curve is as shown

4


Second Order System An example of a second order system is the servo system. The system responses are analysed using inputs such as unit step, unit-ramp and impulse and assuming the initial conditions are zero. Servo system: The simplified equation is:

TcBcJ ++ &&& Laplace transform and assuming initial conditions are zeros

)()()(2 sTsBsCsCJs =+ or

)(1

)()(

BJsssTsC

+=

or

)/()/(/

)()(

22 JKsJBsJK

KBsJsK

sRsC

++=

++=

5


(a) servo system (b) block diagram (c) simplified block diagram

The equation can be rewritten as:

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛−+

⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛++

=

JK

JB

JBs

JK

JB

JBs

JK

sRsC

22

2222

)()(

In transient response it more convenient to write:

2nJ

K ω= σζω 22 == nJB,

where σ is called the attenuation, , the undamped natural frequency and ζnω , the damping ratio. The damping ratio is the ratio of the actual damping B to the critical damping JKBc 2= or

JKB

BB

c 2==ζ

6


)2(

2

nssn

ζωω++ -

R(s) C(s)

Let us consider a second order system and determine its response to a unit step input. A

closed loop feedback system is shown below and the out put is

)s(Rs2s

)s(R)s(G1

)s(G)s(C 2nn

The dynamic b haviou of

2

2n

ω+ζω+ω

=+

=

e r the second order system can be described in terms of two arameters i.e. ς and . p nω

If 10 << ς , the closed loop poles are complex conjugates and lie in the left-half s plane.

r the res o a u ree dif ases: nderdamped (

The system is then called underdamped and the transient is oscillatory. We will conside ponse of the system t nit-step input for th ferent cu 10 << ζ ), critically damped ( 1=ζ ) and overdamped )1( >ζ cases.

(a) Underdamped case 10 << ζ

( ) (( ))dndn jsjssRsC n

ωζωωζωω

−+++=

2

)()(

21 ζwhere ωω =d −n

r step input fo

7


( )

2222

22

22

2

221

221

2)(

nn

n

nn

n

nn

n

nn

n

sssss

s

sss

s

ssssC

ωζωζω

ωζωζω

ωζωζωωζω

ω

++−

+++

−=

+++

−=

++=

inverse Laplace transform

⎟⎟⎠

⎞⎜⎜⎝

⎛

−+−= − ttetc dd

tn ωζ

ζωζω sin1

cos1)(2

the error signal therefore

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+=

−=

− tte

tctrte

ddtn ω

ζ

ζωζω sin1

cos

)()()(

2

for 0≥t

(b) Critically damped case 1=ζ

When the two poles are equal, then the system is said to be critically damped. for step input

( ) sssC

n

n2

2

)(ωω+

=

Inversen Laplace transform

( )tetc ntn ωω +−= − 11)( for 0≥t

(c) Overdamped case 1>ζ

( )( )

( )( )ssssC

sssRsC

nnnn

nnnn

n

n

11)(

11)()(

22

2

22

2

−−+−++=

−−+−++=

ζωζωζωζω

ω

ζωζωζωζω

ω

Inverse Laplace

8


( )

( )⎟⎟⎠

⎞⎜⎜⎝

⎛−

−+=

−−−−

−+−+=

−−

⎟⎠⎞⎜

⎝⎛ −−−

⎟⎠⎞⎜

⎝⎛ −−−

212

2

1

22

1

22

21

2

2

121

1121

11211)(

se

se

e

etc

tsts

t

t

n

n

ζ

ω

ζζζ

ζζζωζζ

ωζζ

( ) ns ωζζ 12

1 −+= ( ) ns ωζζ 122 −−=where and

Figure: Unit-step response curves of the system Definitions of transient-response specifications. The desired performance characteristics of control systems are specified in term of time domain quantities. The quantities are as follows: 1. Delay time td. The delay time is the time required for the response to reach half the

final value the very first time. 2. Rise time, tr. The rise time is the time required for the response to rise from 10% to

90%, 5% to 95%, or 0% to 100% of its final value. For undamped second-order systems, the 0% to 100% rise time is normally used

9


3. Peak time, tp. The peak time is the time required for the response to reach the first peak of the overshoot.

4. Maximum overshoot, Mp. The maximum overshoot is the maximum peak value of the

response curve measured from unity. If the final steady state value of the response differs from unity, then it common to use the maximum percent overshoot. It is define by

100)(

)()(x

cctc p

∞

∞−Maximum percent overshoot %

The amount of the maximum (percent) overshoot directly indicates the relative stability of the system

5. Settling time, ts. The settling time is the time required for the response curve to reach

and stay within a range about the final value specified by the absolute percentage of the final value (usually 2% or 5%).

Second order systems and transient-response specifications The rise time, peak time, maximum overshoot, and settling time can be obtained in terms of and . ζ nω

1)( =tc(a) rise time, t is obtained by r

10


⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+−== −

rdrdt

r ttetc rn ωζ

ζωζω sin1

cos11)(2

since 0≠− rnte ζω

0sin1

cos2

=−

+ rdrd tt ωζ

ζω

or

σω

ζζ

ω drd t −=

−−=

21tan

thus

d

d

drt ω

βπσ

ωω

−=⎟

⎠⎞

⎜⎝⎛−

= −1tan1

(b) peak time,

pt

The peak time can be obtained by differentiating with respect to time )(tc

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−−+

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+= −− ttette

dtdc

dddt

ddt

nnn ω

ζ

ζωωωζ

ζωζω ζωζω cos1

sinsin1

cos22

Evaluated at ptt =

( ) 01

sin2

=−

= −

=

pn

p

tnpd

tt

etdtdc ζω

ζ

ωω

11


0sin =pd tω ,.....2,,0 ππω =pd t or

πω =pd t , hence for the first peak

dpt

ωπ

=

(c) Maximum overshoot, Mp

The maximum overshoot occurs at the peak time

dptt ω

π==

πζ

ζ

ωπζω

πζ

ζπ

⎟⎟⎠

⎞⎜⎜⎝

⎛

−−

⎟⎠⎞⎜

⎝⎛−

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

−+−=

−=

21

2

sin1

cos

1)(

e

e

tcM

dn

pp

(d) Settling time, ts

Figure: Pair of envelope curves for the unit step response curve of the system.

12


ns Tt

ζωσ444 === (2% criterion)

ns Tt

ζωσ333 === (5% criterion)

13


Example

6.0=ζ1. Consider the system shown in figure where and rad/s 5=nω .Obtain the rise time, t, peak time tp, maximum overshoot M and settling time tp s when the system is subjected to a unit step input.

2. For the system shown in Figure (a), determine the values of gain K and velocity

feedback constant Kh so that the maximum overshoot in the unit-step response is 0.2 and the peak time is 1 sec. With these values of K and Kh, obtain the rise time and settling time. Assume that. J= 1 kg-m2 and B = 1 N-m/rad/sec.

14


Impulse response of second-order systems For unit impulse input r(t), the corresponding Laplace transform is unity, . The unit response is

1)( =sR)(sC

22

2

2)(

nn

n

sssC

ωζωω

++=

The inverse Laplace of the equation as follows For 10 <≤ ζ ,

tetc ntn n 2

21sin

1)( ζω

ζ

ω ζω −−

= − , for 0≥t

1=ζFor

t

nntetc ωω −= 2)( for 0≥t

For 1>ζ

tntn nn eetcωζζωζζ

ζ

ω

ζ

ω ⎟⎠⎞⎜

⎝⎛ −+−⎟

⎠⎞⎜

⎝⎛ −−−

−−

−=

22 1

2

1

2 1212)( for 0≥t

Figure : Unit impulse response curves

15


Tutorial 1. Determine the values, of K and k of the closed-loop system shown in Figure Q1 so that

the maximum overshoot in unit-step response is 25 % and the peak time is 2 sec. Assume that J = 1 kg-m2.

Figure Q1

2. Consider the system shown in Figure . The damping ratio of this system is 0.158 and

the undamped natural frequency is 3.16 rad/sec. To improve the relative stability, we employ tachometer feedback. Figure shows such a tachometer-feedback system.

Determine the value of Kh, so that the damping ratio of the system is 0.5. Then obtain the rise time , peak time , maximum overshoot , . and settling time in the unit-step response

strt pt pM

Figure Q2

16


3. Figure 3 is a block diagram of a space-vehicle attitude-control system. Assuming the time constant T of the controller to be 3 sec and the ratio to be JK / 9

2 rad2/sec2,

Figure Q3

4. Consider the system shown in Figure Q4. Determine the value of k such that the damping ratio ζ is O.5. Then obtain the rise time , peak time , maximum overshoot , . and settling time in the unit-step response.

ptrt

stpM

Figure Q4

17


Higher Order Systems The sum of response of higher order systems in general can be considered as the summation of the responses first order and second order systems. Consider the general system shown

G(s)

H(s)

C(s)

+-

)()(1)(

)()(

sHsGsG

sRsC

+=

in general, G(s) and H(s) are given as;

)()()(

sqspsG =

)()()(

sdsnsH = and

or

)()()()()()(

)()(

snspsdsqsdsp

sRsC

+=

nnnn

mmmm

asasasasbsbsbsb

++++++++

=−

−−

−

11

10

11

10

............

In general, for step input

∑∑== ++

−+++

++=

r

k kkk

kkkkkkq

j

j

sscsb

pjsa

sasC

122

2

1 21)(

)(ωωζ

ζωωζ)2( nrq =+ where

Assuming all closed loop poles is distinct. The higher order terms is composed of a number of terms involving simple functions found in the responses of the first and second order terms.

18


Inverse Laplace

∑∑∑=

−

=

−

=

− −+−++=r

kkk

tk

r

kkk

tk

q

j

tpj tectebeaatc kkkkj

1

2

1

2

1

1sin1cos)( ζωζω ωζωζ for t>0.

Dominant closed-loop poles The closed-loop poles that have dominant effects on the transient response behaviour are called the dominant closed loop poles.

> 0.4 and tFigure 5-17 Region in the complex plane satisfying the conditions < 4/ζ σ . s

19


Stability Analysis in the complex plane If any of the pole lie in the right half of the s-plane, with increasing time they give rise to the dominant mode, the transient response increases with increasing amplitude. This represents an unstable system. Stability Analysis The stability of a linear closed-loop system can be determined from the location of the closed-loop poles in the s-plane. If any of these poles lie in the right-half s-plane, then with the increasing time they give rise to the dominant mode, and the transient response increases monotonically or oscillate with increasing amplitude. Routh’s stability criterion Routh’s stability criterion tells us whether or not there are unstable roots in the polynomial equation without actually solving them.

)()(

)()(

11

10

11

10

sAsB

asasasabsbsbsb

sRsC

nnnn

mmmm

=++⋅⋅⋅++++⋅⋅⋅++

=−

−−

−

Where the a’s and b’s are constants and nm ≤ . The locations of the roots of the characteristic equation (the denominator) determine the stability of the close loop. Procedure of the Routh’s stability criterion 1. Write the polynomial in in s in the following form:

01

110 =++⋅⋅⋅++ −

−nn

nn asasasa

0≠na , any zero root has been removed. where coefficient are real quantities.

2. If any of the coefficient are zero or negative in the presence of at least one positive coefficient there is a root or roots that are imaginary or that have positive real parts.

3. Arrange the coefficients of the rows and columns in the following pattern

20


10

11

212

43213

43212

75311

6420

gsfs

ees

ccccsbbbbsaaaasaaaas

n

n

n

n

⋅⋅⋅⋅

⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅

−

−

−

4. The coefficients are evaluated as follows

,...., 3,21 bbb

1

30211 a

aaaab

−=

1

50412 a

aaaab

−=

1

50613 a

aaaab

−=

….. coefficients ,...., 3,21 ccc

Special cases

a. If the first term in any row is zero, but remaining terms are not zero or there is no remaining term, then the zero is replaced with a very small positive number ε and the rest of the array is evaluated.

b. If the coefficients in any derived row are zero, it indicates that there are roots of equal magnitude laying radially opposite in the s plane, that is , two real roots with equal magnitude and opposite signs and/or two conjugate imaginary roots. The evaluation of the rest of the array can be continued by forming an auxiliary polynomial with the last row and by using the coefficients of the derivative of this polynomial in the next row.

Application of Routh’s Stability Criterion. The limitation of this criterion is that it does not suggest how to improve the relative stability or how to improve the relative stability. However, it is possible to determine the effects of changing one or two parameters of a system by examining the values that cause that instability.

21


Tutorial 1. Determine the range of K for stability of a unity feedback control system whose open-

loop transfer function is

( )( )21)(

++=

sssKsG

2. Consider the unity-feedback control system with following open-loop transfer

function:

( )( )32110)(

+−=

ssssG

Is this, system stable?

3. Consider the following characteristic equation:

( ) 025942 234 =+++++ ssKss

Using Routh stability criterion, determine the range of K for stability. 4. Consider the closed-loop system shown in Figure Determine the range of K for

stability. Assume that K > 0.

22


Effects of Integral and Derivative Control Actions on System Performance In this section, we shall investigate the effects of integral and derivative control actions on the system performance. Here we shall consider only simple systems so that the effects of integral and derivative control actions on system performance can be clearly seen. Integral Control Action. In the proportional control of a plant whose transfer function does not possess an integrator 1/s, there is a steady-state error, or offset, in the response to a step input. Such an offset can be eliminated if the integral control action is included in the controller. In the integral control of a plant, the control signal, the output signal from the controller, at any instant is the area under the actuating error signal curve up to that instant. The control signal u(t) can have a nonzero value when the actuating error signal e(t) is zero, as shown in Figure (a). This is impossible in the case of the proportional controller since a nonzero control signal requires a nonzero actuating error signal. (A nonzero actuating error signal at steady state means that there is an offset.) Figure (b) shows the curve e(t) versus t and the corresponding curve u(t) versus t when the controller is of the proportional type. Note that integral control action, while removing offset or steady-state error. may lead to oscillatory response of slowly decreasing amplitude or even increasing amplitude, both of which are usually undesirable

Proportional Control of Systems. We shall show that the proportional control of a system without an integrator will result in a steady-state error with a step input. We shall then show that such an error can be eliminated if integral control action is included in the controller.

23


Figure Consider the system shown in Figure . Let us obtain the steady-state error in the unit-step response of the system. Define

1)(

+=

TsKsG

Since

)(11

)()(1

)()()(

)()(

sGsRsC

sRsCsR

sRsE

+=−=

−=

the eiror E(s) is given by

)(

11

1)()(1

1)( sR

TsK

sRsG

sE

++

=+

=

For the unit-step input , we have ssR /1)( =

sKTsTssE 1

11)(+++

=

The steady-state error is

11

11lim)(lim)(lim

+=

+++

===∞→∞→∞→ KKTs

TsssEteesstss

Such a system without an integrator in the feed forward path always has a steady-state error in the step response. Such a steady-state error is called an offset. Figure shows the unit-step response and the offset.

24


Integral Control of Systems. The controller is an integral controller. The closed-loop transfer function of the system is

KTssK

sRsC

++=

)1()()(

Hence

KTssTss

sRsCsR

sRsE

+++

=−

=)1(

)1()(

)()()()(

Since the system is stable, the steady-state error for the unit-step response can be obtained by applying the final-value theorem, as follows:

01)1()(lim 2

2

0=

+++

==→ sKsTs

TssssEesss

Integral control of the system thus eliminates the steady-state error in the response to the step input. This is an important improvement over the proportional control alone, which gives offset. Response to Torque Disturbances (Proportional Control).

25


The proportional controller delivers torque T to position the load element, which consists of moment of inertia and viscous friction. Torque disturbance is denoted by D. Assuming that the reference input is zero or R(s) = 0, the transfer function between C(s) and D(s) is given by

pKbsJssDsC

++= 2

1)()(

pKbsJssDsC

sDsE

++−=−= 2

1)()(

)()(

The steady-state error due to a step disturbance torque of magnitude T is given by d

p

dd

pssss K

Ts

TKbsJs

sssEe −=++

−==

→→ 200lim)(lim

At steady state, the proportional controller provides the torque -Td, which is equal in magnitude but opposite in sign to the disturbance torque Td. The steady-state output due to the step disturbance torque is

p

dssss K

Tec =−=

The steady-state error can be reduced by increasing the value of the gain Kp. Increasing this value; however, will cause the system response to be more oscillatory. Response to Torque Disturbances (Proportional-Plus-Integral Control). To eliminate offset due to torque disturbance, the proportional controller may be replaced by a proportional-plus-integral controller.

26


If integral control action is added to the controller, then, as long as there is an error signal, a torque is developed by the controller to reduce this error, provided the control system is a stable one. Figure shows the proportional-plus-integral control of the load element, consisting of moment of inertia and viscous friction. The closed-loop transfer function between C(s) and D(s) is

i

pp T

KsKbsJs

ssRsC

+++=

23)()(

In the absence of the reference input, or r(t) = 0, the error signal is obtained from

)()(23

sD

TK

sKbsJs

ssE

i

pp +++

−=

If this control system is stable, that is, if the roots of the characteristic equation

27


023 =+++i

pp T

KsKbsJs

Ti has negative real parts, and then the steady-state error in the response to a unit-step disturbance torque can be obtained by applying the final-value theorem as follows:

01lim)(lim23

00=

+++

−===

→→ sTK

sKbsJs

sssEe

i

pp

ssss

Thus steady-state error to the step disturbance torque can be eliminated if the controller is of the proportional-plus-integral type. Note that the integral control action added to the proportional controller has converted the originally second-order system to a third-order one. Hence the control system may become unstable for a large value of Kp since the roots of the characteristic equation may have positive real parts. (The second-order system is always stable if the coefficients in the system differential equation are all positive.) It is important to point out that if the controller were an integral controller, as in Figure, then the system always becomes unstable because the characteristic equation

023 =++ KbsJs will have roots with positive real parts. Such an unstable system cannot be used in practice. Note that in the system of Figure the proportional control action tends to stabilize the system, while the integral control action tends to eliminate or reduce steady state error in response to various inputs. Derivative Control Action. Derivative control action, when added to a proportional controller, provides a means of obtaining a controller with high sensitivity. An advantage of using derivative control action is that it responds to the rate of change of the actuating error and can produce a significant correction before the magnitude of the actuating error becomes too large. Derivative control thus anticipates the actuating error, initiates an early corrective action, and tends to increase the stability of the system.

28


Although derivative control does not affect the steady-state error directly, it adds damping to the system and thus permits the use of a larger value of the gain K, which will result in an improvement in the steady-state accuracy. Because derivative control operates on the rate of change of the actuating error and not the actuating error itself, this mode is never used alone. It is always used in combination with proportional or proportional-plus-integral control action. Proportional Control of Systems with Inertia Load. Before we discuss the effect of derivative control action on system performance, we shall consider the proportional control of an inertia load. Consider the system shown in Figure. The closed-loop transfer function is obtained as

p

p

KJsK

sRsC

+= 2)(

)(

Since the roots of the characteristic equation

02 =+ pKJs are imaginary, the response to a unit-step input continues to oscillate indefinitely, as shown in Figure (b).

29


Control systems exhibiting such response characteristics are not desirable. We shall see that the addition of derivative control will stabilize the system. Proportional-Plus-Derivative Control of a System with Inertia Load. Let us modify the proportional controller to a proportional-plus-derivative controller whose transfer function is ( )sTK dp +1 .The torque developed by the controller is proportional to . Derivative control is essentially anticipatory, measures the instantaneous error velocity, and predicts the large overshoot ahead of time and produces an appropriate counteraction before too large an overshoot occurs.

( eTeT dp &+ )

Consider the system shown in Figure (a). The closed-loop transfer function is given by

( )pdp

dp

KsTKJssTK

sRsC

++

+= 2

1)()(

The characteristic equation is

02 =++ pdp KsTKJs now has two roots with negative real parts for positive values of .J, K , and Tp d. Thus derivative control introduces a damping effect. A typical response curve c(t) to a unit step input is shown in Figure (b). Clearly, the response curve shows a marked improvement over the original response curve shown in Figure 5(b). Proportional-Plus-Derivative Control of Second-Order Systems. A compromise between acceptable transient-response behavior and acceptable steady-state behavior may be achieved by use of proportional-plus-derivative control action. Consider the system shown in Figure. The closed-loop transfer function is

pd

dp

KsKBJssKK

sRsC

++++

=)()(

)(2

30


The steady-state error for a unit-ramp input is

pss K

Be =

The characteristic equation is is,

0)(2 =+++ pd KsKBJs The effective damping coefficient of this system is thus B + Kd rather than B. Since the damping ratio of this system is ζ

JKKB

p

d

2+

=ζ

it is possible to make both the steady state error ess for a ramp input and the maximum overshoot for a step input small by making B small Kp large and K large enough so that d

is between 0.4 and 0.7. ζ

31


STEADY STATE ERRORS IN UNITY FEEDBACK CONTROL SYSTEMS Classification of control systems Control systems may be classified by the ability to follow step inputs, ramp inputs, parabolic input and so on.

text

+-

Consider the unity feedback with the following open loop transfer function

( )( ) ( )( )( ) ( )111

111lim)(

210 +⋅⋅⋅++

+⋅⋅⋅++=

→ sTsTsTssTsTsTK

sGp

Nmba

s

A system is called type 0, type 1, type 2, ……, if N = 0, N = 1, N = 2,…., respectively. This classification is different from that of the order of the system. As the type number is increased, accuracy is improved by aggravates the stability problem. Steady state errors

)(1)(

)()(

sGsG

sRsC

+=

Error signal transfer function

)(11

)()(1

)()()(

)()(

sGsRsC

sRsCsR

sRsE

+=−=

−=

Using final value theorem

32


)()(1

1)( sRsG

sE+

=

the steady state error is

)()(lim)(lim)(lim

00 sGTsssRssEtee

sstss +===

→→∞→

Static Position Error Constant K . p Steady state error of the system for unit step input

ssGsssEe

ssss1

)(1lim)(lim

00 +==

→→

)0(11G

ess +=

The static position error constant K is defined p

)0()(lim0

GsGKsp ==→

Thus, steady state error is given as

11+

=p

ss Ke

For a type 0 system,

( )( )( )( ) K

sTsTsTsTK

K ba

sp =⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

210

For a type 1 or higher system,

( )( )( )( ) ∞=

⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

210 sTsTs

sTsTKK N

ba

sp , for 1≥N

For unit step input, the steady state error are as follows sse

Kess +

=1

1 for type 0 systems

33


0=sse for type 1 or higher systems

Static Velocity Error Constant Kv

The steady state error of the system with a ramp input is given as

200

1)(1

lim)(limssG

sssEessss +

==→→

)(1lim

0 ssGe

sss →=

The velocity error constant is defined vK

)(lim0

ssGKsv →

=

Thus, the velocity error constant in terms of the static velocity error is given vK

vss K

e 1=

The term velocity error is used to express the steady state error for ramp input. For a type 0 system,

( )( )( )( ) 0

1111

lim21

0=

⋅⋅⋅++⋅⋅⋅++

=→ sTsT

sTsTsKK ba

sv

For type 1,

( )( )( )( ) K

sTsTssTsTsK

K ba

sv =⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

210

For type 2 or higher,

( )( )( )( ) ∞=

⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

210 sTsTs

sTsTsKK N

ba

sv for 2≥N

The steady state error for unit ramp input are as follows

34


∞==v

ss Ke 1 for type 0 systems

KKe

vss

11== for type 1 systems

01==

vss K

e for type 2 or higher

Static Acceleration Error Constant Ka. The steady state error of the system with a unit parabolic input (acceleration input), which is defined by

2)(

2ttr = , for 0≥t

for 0<t 0= is given by

)(lim11

)(1lim 2

0

30 sGsssGse

ssss

→→

=+

=

The static acceleration error constant Ka is defined by the equation

)(lim 2

0sGsK

sa →=

The steady state error is then

vss K

e 1=

The values of Ka are obtained as follows For a type 0 system

( )( )( )( ) 0

1111

lim21

2

0=

⋅⋅⋅++⋅⋅⋅++

=→ sTsT

sTsTKsK ba

sa

35


For a type 1 system,

( )( )( )( ) 0

1111

lim21

2

0=

⋅⋅⋅++⋅⋅⋅++

=→ sTsTs

sTsTKsK ba

sa

For type 2 system,

( )( )( )( ) K

sTsTssTsTKs

K ba

sa =⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

212

2

0

For type 3 or higher system,

( )( )( )( ) ∞=

⋅⋅⋅++⋅⋅⋅++

=→ 11

11lim

21

2

0 sTsTssTsTKs

K Nba

sa for 3≥N

Thus, the steady state error for the unit parabolic input is

∞=sse for type 0 and type 1 systems

Kess

1= , for type 2 systems

0=sse for type 3 or higher systems

Summary

Steady state error Type of input Type-0 system Type-1 system Type 2 system∞

Unit-step 0 0

pK+11

Unit-ramp 0

vK1 ∞

Unit-parabolic

aK1 ∞ ∞

)(lim0

sGKsp →

= (lim0

ssGKsv →

= ) )(lim 2

0sGsK

sa →=

36


Example : Derivative Error Compensation

)1( +ssKτ

The figure shown above, the type of compensation has been introduced by using a PD controller. The plant transfer function is

( )1)(

+=

ssKsGτ

The controller transfer function is

( ) sKKsKKKsTKsD DcDccDc +=+=+= 1)( The control signal is

dttdeKteKtu Dc)()()( +=

The open-loop transfer function of the overall system

( )( )1

)()(++

=ss

sKKKsGsD Dc

τ

For the system under consideration, velocity error constant

csv KKssGK ==→

)(lim0

Therefore, the steady error to velocity input

cvss KKK

e 11==

37


The characteristic equation of the system is

( ) 012 =+++ cD KKsKKsτ or

012 =+⎟⎠⎞

⎜⎝⎛ +

+ττ

cD KKsKKs

where

⎥⎥⎦

⎤

⎢⎢⎣

⎡ +=

τζ

c

D

KKKK1

21

τω c

nKK

= and

example 2. Reconsider the design problem. The position control system of has open-loop transfer

( )2.3614500)()(+

=ss

KsGsD A

Let us consider an application wherein static accuracy requirement is very high: steady state error to unit-ramp is required to be less than 0.025 deg (0.000436 rad) Solution

6.2293000436.0

1==VKRequired

The design is satisfied if

1.1844500

2.3616.2293=

×=AK

For this value of , the characteristic equation of the system becomes AK

08284502.3612 =++ ss therefore

2.910828450 ==nω

38


198.02.9102

2.361=

×=ζ

%53100

21/ =×= −− ζπζeM p Relative stability is obviously very poor. To improve damping and peak overshoot while maintaining at 2293.6, we propose the replacement of amplifier with gain by a PD controller

vK AK

sKKsD Dc +=)(

With the PD controller, the open loop transfer function becomes

)2.361()(4500

)()(+

+=

sssKK

sGsD Dc

The closed-loop transfer function

( ) cD

Dc

R

L

KsKssKK

ss

450045002.361)(4500

)()(

2 ++++

=θθ

Velocity error constant

6.22932.361

4500== c

vK

K 1.184=cK when

For this value of , the characteristic equation of the system becomes cK

( ) 082845045002.3612 =+++ sKs D therefore

2.910828450 ==nω

DD KK 472.2198.0

2.910245002.361

+=×+

=ζ

this clearly shows the positive effect of on damping. For critical damping DK

324.0472.2

198.01=

−=DK

39


This system no longer represents a standard second order system, the transient response is also effected by the zeros of the transfer functions at Dc KKs /−= . In general, if is large, zero will be close to the origin in the s-plane, the overshoot will be increased substantially and damping ratio

DKζ no longer gives an accurate estimate

on the peak overshoot of the out put. a=tf(828450,[1 361 828450]) b=tf([1458 828450],[1 1819.2 828450])

40


Derivative Output Compensation The reason behind using derivative of the actuating error signal is to improve the damping of the system can be extended to the output signal.

Eliminating minor feedback loop,

( )2

1

1)(

KKssKKsG++

=τ

The velocity error constant

1

1

0 1)(lim

KKKKssGK

sv +==

→

The closed loop transfer function is given by

( ) 122

1

1)()(

KKsKKsKK

sRsC

+++=τ

and the characteristic equation is

( )01 1212 =+

++

ττKKsKKs

The natural frequency and damping ratio of the compensated system are given as

τω 1KK

n = ⎥⎦⎤

⎢⎣⎡ +

=τ

ζ 2121 KK and

41


For specified and vK , we can write ζ

⎥⎥⎦

⎤

⎢⎢⎣

⎡=

τζ t

vKK

K21

Example 3

( )2.36145000+ss 10

1=n

Rθ Lθ

When ‘S’ is open

( )2.3614500

)(+

=ss

KsG A

Assume that the steady state error to unit ramp input is required to be less that 0.025 deg (0.000436 rad). Solution

6.2293000436.0

1==vKRequired

This requirement is satisfied if

1.1844500

2.3616.2293=

×=VK

For this value of , the characteristic equation of the system becomes AK

08284502.3612 =++ ss This gives

42


198.0 ,2.910 == ζωn Relative stability is obviously very poor. Now, closing the switch ‘S’.

( )A

A

KssKsG450002.361

4500)(++

=

The closed-loop transfer function

( ) At

A

R

L

KsKsK

ss

450045002.3614500

)()(

2 +++=

θθ

Velocity error constant

t

AV K

KK450002.361

4500+

=

The characteristic equation of the system is

( ) 04500450002.3612 =+++ At KsKs where

A

t

KK

45002450002.361 +

=ζAn K4500=ω and

The tachogenerator feedback increases the damping of the system; it however reduces the system . vK

0324.0=tKWith , we get critical damping when 1.184=AK . We get

4.455=vK Therefore

deg 0.126 rad 0022.0

==rampunitsse and

43


Integral Error Compensation

⎟⎟⎠

⎞⎜⎜⎝

⎛+

sTK

ic

11

The plant transfer function is

( )1)(

+=

ssKsGτ

The controller transfer function is

sK

KsT

KsD ip

ip +=⎟⎟

⎠

⎞⎜⎜⎝

⎛+=

11)(

The open-loop transfer function of the overall system is

( )1)(

)()( 2 +

+=

ssKsKK

sGsD ip

τ

The characteristic equation of the closed loop is

023 =+++ ip KKsKKssτ Applying Routh criteria this equation yields the result that the system is stable for

τ/0 pi KK << Example

( )1)(

+=

ssKsGτ

44


a=tf(828450,[1 361 828450]) c=tf([66276 662760],[1 361.2 66276 662760])

45

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