chapter_6 (1)

Chapter 6: System Performance and Specifications

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Chapter 6: SYSTEM PERFORMANCE AND SPECIFICATIONS Before considering control system specifications and performance measures, lit would be helpful to first define and discuss the meanings of some of the basic control engineering terms we shall be using. The definitions will be with reference to the feedback control system shown in Fig. 6.1. For the

system of fig. 6.1, let )s(D)s(N

)s(GG

G= and )s(D)s(N)s(H

H

H= . One of the key things to note is the

fact that for a real system, the order of the numerator polynomial of the system's transfer function cannot exceed the order of the denominator polynomial. For the system of Fig. 6.1, the open loop transfer function and the closed loop transfer function are given by

+G(s)

R(s) C(s) -

H(s)

Fig. 6.1: FEEDBACK CONTROL SYSTEM

OLTF: )()()()(

)(sDsDsNsN

sGHHG

HG=

and

CLTF: )()()()(

)()()(1

)(sNsNsDsD

sDsNsGH

sG

HGHG

HG

+=

+

♦ System Order. The order of the system is given by the order of the denominator

polynomial of the closed loop transfer function (CLTF). It must be noted that the order of the denominator polynomial of the CLTF is the same as the order of the denominator polynomial of the open loop transfer function (OLTF).

♦ System Type Number. The system type number is defined only for unity feedback

systems. Consider a unity feedback system, i.e. a system for which H(s) = 1 and hence for which the open loop transfer function GH(s) equals G(s). Assume that the open loop transfer function (OLTF) can be put in the form

)ps(...)ps()ps(s)zs(...)zs()zs(k

)s(D)s(Nk

)s(G)s(HGr21

tm21

G

G

++++++

===

where n = t + r is the order of the denominator polynomial, m < n. The number t, of zero roots of the equation DG(s) = 0 is known as the system's type number. The type number has a lot of influence on the steady state error of the feedback system

♦ Characteristic polynomial and characteristic equation. The characteristic polynomial is the

numerator of 1 + GH(s). When the characteristic polynomial is set equal to zero, the resultant equation is known as the characteristic equation. The system response depends to a large extent on the roots of the characteristic equation.


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♦ System Poles And Zeros. Consider a system with transfer function F(s) = )s(D)s(N , where N(s)

and D(s) are polynomials in s. The zeros of the system are the roots of the equation N(s) = 0. Similarly, the poles of the system are similarly given by D(s) = 0.

For the system of Fig. 6.1, the open loop poles and zeros are obtained as indicated below.

♦ Open loop zeros are the roots of NG(s) NH(s) = 0

♦ Open loop poles are the roots of DG(s) DH(s) = 0 Similarly, the closed loop poles and zeros are given by ♦ Closed loop zeros are the roots of NG(s) DH(s) = 0

♦ Closed loop poles are the roots of the characteristic equation 1 + GH(s) = 0, i.e. the roots of

DG(s) DH(s) + NG(s) NH(s) = 0 Example 6.1: For the feedback control system with forward transfer function G(s) and feedback transfer function H(s) given by –

)2(

2)(4

)6(2.3)( 2 +=

+

+=

ssHand

ssssG

find the following – a) the open loop transfer function, the open loop poles and zeros of the system b) the characteristic polynomial and the characteristic equation of the system. c) the closed loop transfer function, the system order and the closed loop poles and zeros of the system. Solution: a) The open loop transfer function GH(s) is given by

GH(s) = G(s) H(s) = )2()4(

)6(4.6)2()4(

)6(4.6)2(

24

)6(2.322 ++

+=

++

+=

++

+sss

ssss

ssss

s .

The open loop zeros are the roots of the equation obtained by setting the numerator of GH(s) tto zero, i.e. the roots of 6.4(s + 6) = 0. Thus in this case, there is one zero at –6. The open loop poles are the roots obtained by setting the denominator of GH(s) to zero and finding the roots of the resultant equation. The poles are the roots of s(s + 4) (s + 2) = 0. The open loop poles are thus 0, -4 and –2. b) The characteristic polynomial is given by (the numerator) of 1 + GH(s) =

)86(4.384.6)86(

86)6(4.61

)2(2

4)6(2.31)(1 23

23

232 sssssss

ssss

sx

ssssHG

++

++++=

++

++=

++

++=+

i.e the characteristic polynomial is s3 + 6s2 + 14.4s + 38.4.


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The characteristic equation is obtained by setting the characteristic polynomial equal to zero, ie.

s3 + 6s2 + 14.4s + 38.4 = 0. c)

The closed loop transfer function is given by 4.384.146

)2()6(2.3)(1

)(23 +++

++=

+ sssss

sHGsG

The system order is three as the denominator of the closed loop transfer function is a third order polynomial. The closed loop poles are the roots of the equation obtained by setting the numerator of the closed loop transfer function to zero, i.e. the roots of 3.2 (s + 6) (2 + 2) = 0. Thus in this case, there are two zeros given by –6 and –2. The closed loop zeros are the roots obtained by setting the denominator of the closed loop transfer function to zero and finding the roots of the resultant equation. The poles are thus the roots of

s3 + 6s2 + 14.4s + 38.4 = 0. The three roots are –4.673, -0.667 + j2.787. The closed loop poles are thus –4.673, -0.667 + j2.787.

Response of First and Second Order Systems Consider a first order feedback system with closed loop transfer function given by

)1(

1)()(1

)(sp

kps

ksHG

sGτ+

=

+=

+

The single closed loop pole is at –p. The constant τ = p1 is the time constant of the system. The time

constant is thus the reciprocal of the magnitude of the system’s pole. If a unit step is applied to the

system we have r(t) = 1 and s1)s(R = . Thus the s-domain unit step response C(s) is given by

))(

11(1)(

)(pssp

ksps

ksC+

−=+

=

ess

Fig. 6.2: STEP RESPONSE OF FIRST ORDER SYSTEM

τ Time

Constant

0.632 css

1.0css

c(t)

r(t)

c(t), r(t)

Time, t


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The equivalent time domain response is of the form css [1 – e-tp ] = css [1 – e-t/τ ] where css = pk is the

steady state value of the response. The time constant is a measure of the decay rate of the transient response. The smaller the time constant the faster the system settles down. The step response is shown in fig. 6.2. Note that it takes τ seconds for the response r(t) to go from the initial value of zero to 63.2% of the final value css. Since most specifications are based on the response of second order systems, it is beneficial to have a detailed look at the response of second order systems too. Consider the second order feedback system with closed loop transfer function given by

CLTF = )s(HG1

)s(G+

The characteristic equation, given by 1 + GH(s) = 0, must be second order, i.e.

1 + GH(s) = b2s2 + b1s + b0 = 0 ……………………….. (6.1)

The roots of eqn. (6.1) are given by 2

20211

b2bb4bb −±−

. Three different responses, depending

on the roots, can be identified as indicated below. Fig. 6.3 also shows the three types of second order step response. ♦ Case 1: 4b2b0 > b1

2: The roots are complex and the system response is underdamped i.e. it has an oscillatory transient response.

♦ Case 2: 4b2b0 = b1

2: The roots are repeated and real. The system's transient response is said to be critically damped.

♦ Case 3: 4b2b0 < b1

2: There are two distinct real roots. The system's transient response is said to be overdamped.

Fig. 6.3: RESPONSE OF SECOND ORDER SYSTEMS

r(t) Critically Damped

ζ = 1

Overdamped ζ > 1

Underdamped ζ < 1

Time, t

r(t), c(t)

For second order systems, the characteristic equation is usually put in the standard form

s2 + 2ζωns + ωn2 = 0 ...................................... (6.2)

The constant ζ is known as the damping factor or damping ratio of the system and ωn is known as the undamped natural frequency of the system. From the above, it is clear that the value of the damping factor ζ determines the transient response type. The frequency of oscillation when the transient response is underdamped (i.e. ζ < 1) is given not by the undamped natural frequency ωn but by ωd, the damped natural frequency that is given by


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ωd = ωn√(1 - ζ2).

Note that if we divide eqn (6.1) by b2 we get

s2 + (b1/b2)s + (b0/b2) = 0 .................................. (6.3) By comparing the coefficients of eqns. 6.2 and 6.3, we can easily express ωn and ζ in terms of b2, b1 and b0 of eqn (6.1). Example 6.2: a) A feedback control system has characteristic polynomial given by 2s + 5. Find the time

constant of the system. b) A second order feedback control system has characteristic equation given by

1.5s2 + 2.6s + 4 = 0

Find the damping coefficient ζ, the undamped natural frequency ωn and the damped natural frequency ωd of the system. Solution: a) The characteristic polynomial may be expressed as 5(1 + 0.4s). Thus the time constant of the system is o.4 second. Alternatively, as the closed loop pole of the system is –2.5, the time

constant may be obtained as 5.2

1−

= 0.4 second.

b) The characteristic equation may be expressed as

1.5 (s2 + 1.733s + 2.667) = 0 or s2 + 1.733s + 2.667 = 0 Comparing this equation with the standard form s2 + 2ζωns + ωn

2 = 0 gives – ωn

2 = 2.667 i.e ωn = 667.2 = 1.633 rad/sec. and 2ζωn = 1.733 i.e 1.633 ( 2ζ ) = 1.733 i.e ζ = 0.531

The damped natural frequency ωd = 22 531.01633.11 −=ζ−nω = 1.384 rad/sec.

Control Objectives Control System Specifications. In order to be able to describe in an unambiguous way the standard of performance that is expected from a particular control system, it is necessary to have universally accepted measures of control system performance. These measures enable us to quantify the performance of a control system and thus enable us to specify the performance of the system. Both time domain and frequency domain specifications are used. Time domain specifications are defined in terms of the time response of the system while frequency domain specifications are defined in terms of the frequency response of the system. Regardless of whether time domain or frequency domain specifications are used, the following aspects of the control system's performance are the areas of major interest. ♦ Stability and relative stability. ♦ Transient performance i.e. the speed of response of the system (when the reference input is

varied). The speed of response must be appropriate for the application. ♦ Steady state performance i.e. the steady state error of the system.


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♦ Safety. Though safety is not explicitly considered in this course, it must be emphasised that as in

all engineering and industrial activities, safety is of paramount importance. All control systems must be designed to for safe operation and adequate safety and protective devises must be incorporated. All plants must be operated in a way that is consistent with good safety practices.

Other aspects of interest are ♦ Control effort, i.e. the energy expended in effecting the control strategy. ♦ Disturbance rejection i.e. how well the feedback system copes with noise and other disturbances.

Stability A control system is said to be unstable if its response grows in an unlimited way or if sustained oscillations result from the application of nonsinusoidal inputs. In practice, the magnitudes of the responses of unstable systems do not grow to infinity as theoretical results might suggest because of the limiting action of saturation that occurs in most systems. Instability is highly undesirable in control systems, as there is no control over the system behaviour when it is unstable. All effort must be made to ensure that the feedback system is stable at all times. The stability of a closed loop control system with forward transfer function, G(s), feedback transfer function H(s) and open loop transfer function GH(s), depends on the closed loop transfer function (CLTF). The CLTF is given by

CLTF = )s(GH1

)s(G)s(R)s(C

+=

The denominator polynomial of the CLTF, i.e. 1 + GH(s) is known as the characteristic polynomial of the system. Setting the characteristic polynomial to zero gives us the characteristic equation i.e.

1 + GH(s) = 0 The roots of the characteristic equation are known as the poles of the system. The stability of the closed loop system depends on these poles. The poles may be real or complex [note that complex roots occur in conjugate pairs]. However, system stability depends on the real parts of the poles. The values of the imaginary parts of the poles play no part in determining system stability. The condition for stability is as follows. For stability, all the roots of the characteristic equation, i.e. all the system poles, must have negative real parts. Note that if only one of the system poles has a non-negative real part, the system is unstable. Note also that the roots of the numerator polynomial of the CLTF, known as the zeros of the system play no part in determining the stability of the system. The effect of pole location on the transient component of the feedback system response is illustrated by Fig. 6.4. Poles with positive real parts produce transients that grow with time. The larger the magnitude of the real parts of these poles the faster the transients grow. Poles with zero real parts, i.e. imaginary poles, produce constant magnitude sinusoidal oscillations (which do not die down with time). However poles with negative real parts produce stable transient responses, i.e. transients which die down with time. The larger the magnitudes of the real parts of these stable poles, the faster the transients die down. From Fig. 6.4, it is also clear that while the imaginary parts of the poles do not affect the stability of the system, the have some effect on the transient response. The higher the magnitude of the imaginary part of a conjugate pole pair (note that complex roots occur in conjugate pairs), the higher the frequency of the oscillations in the associated transient component. As unstable systems must be avoided at all cost, we shall, later on in the course, devote more time to the issue of stability and also relative stability and how to ensure stability. A system that is only marginally stable could change into an unstable system due to system parameter variations due to for example ageing. Thus not only do we want our feedback system to be stable, but we also need to


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Imag.

XX

XX

X

X X XReal

XX

XX

X

Fig. 6.4: POLE LOCATIONS AND TRANSIENT RESPONSE

know how far or how close a stable system is from instability, hence the importance of the concept of relative stability. Tests For Stability. If we wish to investigate the absolute stability of system, that is we only wish to consider whether or not a system is stable, but do not necessarily wish to consider the system’s relative stability, we may use the any of the following stability techniques. 1: Test based on system poles. If we can explicitly solve the system’s characteristic equation for the system poles, it is easy to determine whether or not the system is stable. If all the poles have negative, non-zero real parts the system is stable. If one or more poles have zero or positive real parts, the system is unstable. While this method is straightforward, the problem is that for high order systems, it is not easy to manually determine the poles. 2:Test Based on Characteristic Polynomial Coefficients A necessary, but not sufficient condition for stability is that the characteristic polynomial must

• not have any zero coefficients and and

• have coefficients which are all of the same sign. If a characteristic polynomial does not satisfy the above then the system is definitely unstable. However, if it satisfies the above conditions, it does not then follow that the system is stable. Other tests must be carried out to check if the system is stable.


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3: Routh-Hurwitz Test. From the characteristic polynomial a0sn + a1sn-1 + … + an-2s2 + an-1s + an, form the Routh-Hurwitz array as follows. Note that for an nth order system, the Routh-Hurwitz array has at most (n+1) rows. The first two rows are constructed from the system’s characteristic polynomial coefficients. The other rows are derived from the preceding two rows. The elements of the Routh-Hurwitz array are formed from the coefficients of the characteristic polynomial as follows - ROW 1: a0 a2 a4 a6 .... ROW 2: a1 a3 a5 a7 ..... ROW 3: b1 b2 b3 ......... ROW 4: c1 c2 c3 .......... . . . ROW (n+1) : .. where

31

20

11 aa

aaa1b −=

51

40

12 aa

aaa1b −= etc

and

21

31

11 bb

aab1c −=

31

51

12 bb

aab1c −= etc.

The system is stable if and only if all the elements in the first column of the Routh-Hurwitz array have the same sign. The number of sign changes in the first column equals the number of poles with positive real parts, i.e. the number of unstable poles. The method outlined above need to be modified if either of the following conditions occur in the array. ♦ A zero entry in the first column of any derived row apart from the last one. This gives rise to

infinite entries in the following row. This situation could be handled in two ways. The first method involves replacing the zero entry with a small number ε, computing the subsequent rows and then taking the limit as ε approaches zero. In the second method, the original polynomial is multiplied by (s + a), a > 0. This adds one stable pole, and hence does not affect the number of unstable roots of the given system.

♦ A row of zero entries. This involves the use of auxiliary equations and is more difficult to

handle. The Lienard-Chipart method may be easier to use for such cases. 4: Lienard-Chipart Test. From the characteristic polynomial a0sn + a1sn-1 + ..... + an-2s2 + an-1s + an, form either the odd-numbered or even-numbered k x k Hurwitz determinants Hk for 1 < k < n from

a1 a3 a5 .................. a0 a2 a4 ................. 0 a1 a3 a5 ..............

Hk = 0 a0 a2 a4 .............. 0 0 a1 a3 a5 ........ 0 0 a0 a2 a4 ........ .....................................ak

The first two rows are constructed from the coefficients of the system’s characteristic polynomial. Note that the arrangement of the coefficients is different form the Routh-Hurwitz array. Other rows of the determinant, taken two at a time, are formed by right-shifting the preceding pair of rows. The


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rightmost elements are discarded. In forming the determinant, replace by 0 any coefficient which is absent in the given characteristic polynomial. If all the coefficients of the characteristic polynomial are positive, then a necessary and sufficient condition for stability (i.e. for all the roots of the characteristic equation to have negative real parts) is

• all the odd-numbered Hurwitz determinants must be greater than 0. or

• all the even-numbered Hurwitz determinants must be greater than 0. This means that we need to test only the odd-numbered determinants or only the even-numbered determinants but not both. Example 6.4: Investigate the stability of the closed loop system whose characteristic polynomial is given by – a: s5 + 3.4s4 + 6.73s3 – 12.33s2 + 22.4 using the characteristic polynomial coefficients tests. b: s4 + 2s3 + 5s2 + 8s + 10 using the Routh-Hurwitz method. c: s4 + 2s3 + 5s2 + 8s + 10 using the Lienard-Chipart test. Solution: a: the system is unstable as there is a one zero coefficient ( for s) . Also the coefficients are not all of the same sign. B Rows one and two of the Routh-Hurwitz table are obtained from the coefficients of the given characteristic polynomial, i.e. Row1: 1 5 10 Row 2: 2 8 0 The elements of row 3 are obtained as follows

12

)106(8251

21

1 =−−

=−=b and 102

)200(02

10121

2 =−−

=−=b

b3 is zero. Thus, row 3 is as follows – Row 3: 1 10 0 Similarly the elements of row 4 is obtained as follows –

121

)820(10182

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1 −=−

−=−=c . c2 is clearly zero and row 4 is as shown below –

Row 4: -12 0 0 The first element of row 5 is given by

1012

)1200(012

101121

1 −=−−

=−

−=d The other elements of row 5 are both

zero. Examining the first column of the table shows that there is one sign change from row 3 to row 4. Therefore the system is unstable. Furthermore we k now that the instability is caused by one unstable pole.


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c: Let us use the odd Hurwitz determinants i.e. H1 and H3, as in his case they are simpler to compute than the even ones. H1 = |2| = 2. This is positive, so we continue with the test and compute H3 which is given by –

8208

182

1052

8201051082

1 −===H = 2 (40 – 20) - 1 (64 – 0) = -44. As H1 is

negative the system is unstable.

Time Domain Specifications. For time domain specifications, the unit step is the most commonly used test input. The control system specifications are defined as follows in terms of the unit step response as shown in Fig.6.5. ♦ Overshoot (Peak Overshoot) Mp: The peak overshoot or simply overshoot is the maximum

difference between the transient response and the steady state response value, i.e. the difference between the first peak and the steady state response value. The overshoot provides some information about the stability of the system. The peak overshoot generally increases as the damping in the system is reduced, thus the higher the value of Mp, the more oscillatory the system response is. Note that critically and overdamped systems have no overshoot in their response.

For an underdamped second order systems whose transfer function can be put in the standard

form 22

2

2 nn

n

s ωξωω

++ (characteristic polynomial given by s2 + 2ζωns + ωn

2 ), the peak

overshoot can be shown to equal 21 ζ−

πζ−

e css. ♦ Peak Time tp: Peak time is the time it takes to reach the first peak, i.e. the maximum point. For

an underdamped second order systems whose transfer function can be put in the standard form

22

2

2 nn

n

s ωξωω

++ (characteristic polynomial given by s2 + 2ζωns + ωn

2 ),, the peak time is

given by dn

pt ωπ

ζω

π=

−=

21

♦ Settling Time ts: This the time taken for the response to reach and stay within a specified error band -usually 5%, but sometimes 2% or 1%, of the steady state value. Settling time is a measure of the system’s speed of response, that is, how quickly (or otherwise) the system’s response reaches its final value..

For a second order system with characteristic polynomial s2 + 2ζωns + ωn

2, the 5% settling time is approximately 3/ζωn. For 2% settling time the corresponding approximation is 4/ζωn.

♦ Delay Time td: The delay time is the time taken for the response to first reach 50% of its final

value. It is a measure of the control system's speed of response. ♦ Rise Time tr: This is the time taken by the response to rise from 10% to 90% of its final value.

It is also a measure of the system's speed of response. ♦ Steady State Error ess: Steady state error is a measure of the system's accuracy. The steady

state error is the error after the system's step input response has settled to a steady value, i.e. ess = [r(t) - c(t)].

∞→tlim


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♦ Subsidence Ratio: This is the ratio of the amplitudes of successive cycles of the transient part of the response. It is only relevant for underdamped systems. If the system’s step response has first and second peak values c1 and c2, then the subsidence ratio is (c1 – css) / (c2 – css) where css is the steady state value of the response.

A subsidence ratio value between 4:1 and 3:1 is desirable for most control systems.

♦ Time Constant τ: For a second or higher order systems, there may be two or more time

constants involved in the transient response. For such a system, time constant refers to the predominant time constant of the system. This is a measure of the speed of response of the overall system.

c(t), r(t)

r(t) Mp 1.0

ess 5% Band

0.9cs

c(t)

0.5cs

0.1cs

td tp ts Time t

tr Fig. 6.5: UNIT STEP RESPONSE

Integral performance indices are also used, but we shall not consider these in the course. Example 6.3: For the unity feedback system with forward transfer function given by

32

10)( 2 ++=

sssG

find the following – a) the peak time tp. and peak overshoot Mp b) the 5% settling time ts based on the exponential envelope of the system’s step response. Solution: a) The closed loop transfer function is given by –

CLTF = 132

10)(1

)(2 ++

=+ sssG

sG

Consider a unit step input, i.e. let r(t) = 1. Then R(s) = s1 . The system response C(s) is then given by


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C(s) = 132

1132 2

3212 ++

++=

++ ssksk

sk

sss10 .

Using partial fraction method, we obtain k1 = 1310 = 0.7692.

Thus C(s) = )132(

)132(7692.0132

7692.02

32

22

232

++

++++=

++

++

sssskskss

ssksk

s

Or C(s) = )132(

10)5385.1()7692.0(2

32

2

++

++++

ssssksk

Comparing coefficients of numerator polynomials gives – k2 = -0.7692 and k3 = -1.5385. The system response C(s) may then be expressed as –

C(s) = 22 464.3)1()464.3(222.0)1(7692.07693.

++

++−

ss

s0

The time domain equivalent c(t) is obtained by taking the inverse Laplace transform. c(t) = 0.7692 – 0.7692 e-t Cos (3.464 t) - 0.222 e-t Sin (3.464 t) or c(t) = 0.7692 - 0.8006 e-t Sin (3.464 t + 1.29)

At peak time, where a maximum occurs, we have dt

tcd )( = 0.

Now dt

tcd )( = 0.8006 e-t Sin (3.464 t + 1.29) - 2.7733 e-t Cos (3.464 t + 1.29)

When dt

tcd )( = 0, we have Tan (3.464 t + 1.29) = 3.464 i.e. (3.464 tp + 1.29) = Tan-1 (3.464)

or (3.464 tp + 1.29) = 1.29 + kπ ; k = 0, 1, 2, 3,.... k = 0 is unsuitable as it gives tp = 0. Trying k = 1 gives (3.464 tp + 1.29) = 1.29 + π, i.e. peak time tp = 0.907 second. The steady state value of the response css equals 0.7692. The peak overshoot Mp is then given by – Mp = c(tp) - css = -0.8006 e-0.907 Sin (3.464 x 0.907 + 1.29) = 0.3106. The exponential envelope of the transient response is given by -0.8008 e-t and 0.8008 e-t. At the 5% settling time, ts, we have 0.8008 e = 0.05 cst−

ss or = 0.048. Taking inverse logarithm gives ste−

-ts = In (0.048) = -3.036 i.e. settling time ts = 3.036 seconds. Steady State Errors. We shall now consider the steady state errors for unity feedback systems for the following standard inputs– • step input, i.e. r(t) = 1. • ramp or velocity input, i.e. r(t) = t.


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• parabolic or acceleration input, i.e. r(t) = ½t2.

r(t) r(t) r(t)

1

Time, t Time, t Time, tUnit Step - r(t) = 1. Unit Ramp - r(t) = t. Unit Parabolic - r(t) = ½t2.

Fig. 6.6: STANDARD INPUTS

The inputs to be considered are shown in Fig. 6.6. In particular we shall define the systems error constants and consider how these error constants enable us to determine the steady state errors of unity feedback systems without explicitly determining the system response first. For the unity feedback system with input r(t) and output c(t) we have

)s(R)s(G1

)s(G)s(C.e.i)s(G1

)s(G)s(R)s(C

+=

+=

For input R(s), the error E(s) is as you may recall given by

)()(1

1)( sRsG

sE+

=

From Laplace transform theory, the steady state time domain error, ess, is given by

)s(Eslim)t(elime0stss

→∞→==

Thus )s(R)s(G1

slime

0sss +=

→

We shall now consider the steady state error for step, ramp and parabolic inputs.

♦ Step Input r(t) = m, i.e. R(s) = sm

)0(1)(1lim

)(1lim

00 Gm

sGm

sm

sGse

ssss +=

+=

+=

→→

where m is the magnitude of the step input and G(0) is the value obtained after substituting s = 0 in G(s). Define the position error constant kp as

Kp = G(0) = G(s)|s=0. Then the steady state error ess for a step input of magnitude m is given by

pss K

me+

=1


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Note that for type 0 systems Kp is finite, hence ess must be nonzero, i.e. an error is unavoidable. For type numbers greater than 0, the position error constant ess is infinite and the steady state error is thus 0.

♦ Ramp r(t) = mt, i.e. R(s) = 2sm .

The steady state error for a ramp input of magnitude m is given by -

)(lim

)(lim

)(1lim

0020 sGsm

sGssm

sm

sGse

sssss→→→

=+

=+

=

Define the velocity error constant Kv as

Kv = )(lim0

sGss →

Then the steady state error ess for a ramp input of magnitude m is given by

vss K

me =

For type 0 systems Kv = 0 and hence the steady state error is infinite. for type 1 systems Kv is finite and hence there must be a nonzero steady state error. However for type 2 systems Kv is infinite, thus the steady state error in that case is zero.

♦ parabolic input r(t) = ½mt2, i.e. R(s) = 3sm

Consider a parabolic input of magnitude m. If we define the acceleration error constant Ka as

Ka = , )(lim 20

sGss →

then similarly we get the steady state error for a parabolic input of magnitude m as -

a

ss Kme =

The steady state error is infinite for types 0 and 1 systems, but finite for type 2 systems .for higher type systems, the steady state error is zero. TABLE 6.1: SUMMARY TABLE OF STEADY STATE ERROR ess

SYSTEM TYPE NO (l)

STEADY STATE ERROR ess FOR UNIT STEP INPUTS

STEADY STATE ERROR ess FOR UNIT RAMP INPUTS

STEADY STATE ERROR ess FOR UNIT

ARABOLIC INPUTS P 0

pK1

1+

∞ ∞

1

0

vK

1

∞

2

0

0

aK

1

Table 6.1 gives a summary of steady state error performance for types 0, 1 and 2 unity feedback systems. Even though the steady state error results above have been derived for unity feedback systems, the results could be applied to non-unity feedback system provided the error E(s) is taken as R(s) – H(s)C(s) rather than R(s) – C(s).


15

Example 6.5: For the unity feedback system with forward transfer function G(s) given by

G(s) = )25.1()75.0(

5.6++ sss

a: find the system order and type number. b: the position error constant kp, the velocity error constant kv and the acceleration error constant ka. c: the steady state error ess when (i) the input is a unit step (ii) a ramp of magnitude 2.5. Solution: a: The denominator of the open loop transfer function and hence that of the closed loop transfer function is a third order polynomial. Hence, the system is a third order system. Since there is one open loop pole at the origin, the type number is one. b: The position error constant kp is given by

kp = G(0) = ∞==++ 0

5.6)25.10()75.00(0

5.6

The velocity error constant kv ois given by

kv = 933.6)25.10()75.00(

5.6)25.1()75.0(

5.60

=++

=++→ sss

sslim

Lastly, the acceleration error constant ka is given by

ka = 09375.00

)25.1()75.0(5.6lim

)25.1()75.0(5.6

0

2

0==

++=

++ →→ sss

ssss

sslim

c: (i) For unit step input, the steady state error is given by ess = pk+1

1 . As kp is infinite,

the steady state error is zero. ii: For a ramp input of magnitude 2.5, the steady state error ess is given by

3606.0933.65.25.2

==vk

Other Error Indexes Although steady state error is of considerable usefulness, other error indices are also used. The commonly used ones are – 1: Integrated Absolute value of Errors (IAE) given by

∫=

T

0

dt|)t(e|IAE

2. Integral Time Absolute Error (ITAE), given by


16

∫=

T

0

dt|)t(e|tITAE

3. Integral Time Square Error (ITSE) given by

∫=

T

0

2 dt)t(etITSE

Frequency Domain Specifications Basically frequency domain specification are given on terms of either the closed loop or the open loop response of the control system when sinusoidal inputs are applied. The sinusoidal response can be obtained analytically from the system's transfer function or by carrying out a series of tests using the experimental setup of Fig. 6.7. In the tests of Fig. 6.7, a set of sinusoidal inputs of different frequencies is applied one at a time to the system. For each input, the output signal magnitude and phase angle (with respect to the input signal) are recorded. These form the frequency response of the system. In general, both the magnitude and phase angle of the response change as the frequency of the input signal is varied.

Fig. 6.7: Experimental Setup for Determination of Frequency Response.

Variable Frequency Sinusoidal Input |R| / 0

FEEDBACK SYSTEM UNDER TEST.

Steady State Sinusoidal Response. |C| / θ

Fig. 6.8 shows how the magnitude of the feedback systems output M(jω) changes with frequency (of the sinusoidal fixed magnitude input). The frequency domain specifications commonly used are as follows. ♦ Resonant Peak Mr: This is given by the maximum value of the closed loop magnitude response

(see Fig. 6.8). The resonant peak is a measure of the overshoot present in the system's transient response. Large values of Mr correspond to large overshoots.

♦ Resonant Frequency ωr: This is the frequency at which the resonant peak occurs. The resonant

frequency is a measure of the speed of response. The higher the value of the resonant frequency, the faster the transient response is.

♦ Bandwidth Bw: This is the range of frequencies over which the system responds satisfactorily.

The bandwidth is related to the system's speed of response. High bandwidths correspond to small rise times.

♦ Cut-Off Frequency ωχ: This is the frequency at which the closed loop response has a magnitude

that is 2

1 of its DC value i.e. value when frequency is zero.

♦ Cut-Off Rate: The cut-off rate is the slope of the closed loop magnitude response curve at the

cut-off frequency. Cut-off rate is a measure of the system’s ability to distinguish signal from noise.


17

In addition to the above there are two other specifications which are also commonly used. Both of them are measures of relative stability. These specifications are defined in terms of the open loop response. We shall consider them more fully later on in the course. ♦ Phase Margin. ♦ Gain Margin.

2M 0

| M(jω) |

Mr

Clo

sed

Loo

p M

agni

tude

M0

ωχ Bω Freq. ω ωr

Fig. 6.8: CLOSED LOOP FREQUENCY RESPONSE

chapter_6 (1)

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