Advanced Control Engineering - Chapter 5 pps

The Routh±Hurwitz stability criterion states:a For there to be no roots with positive real parts then there is a necessary, but not sufficient, condition that all coefficients in the cha

If the exponential terms decay as time increases, then the system is said to be stable Ifthe exponential terms increase with increasing time, the system is considered unstable.Examples of stable and unstable systems are shown in Figure 5.1 The motionsshown in Figure 5.1 are given graphically in Figure 5.2 (Note that (b) in Figure5.2 does not represent (b) in Figure 5.1.) The time responses shown in Figure 5.2 can

be expressed mathematically as:

For (a) (Stable)

x t o ( )

N (c) Stable

x t o ( )

Fig 5.1 Stable and unstable systems.

A A

A

t A

(a)

t

t t

5.1.1 Stability and roots of the characteristic equation

The characteristic equation was defined in section 3.6.2 for a second-order system as

The only difference between the roots given in equation (5.9) and those in equation(5.10) is the sign of the real part If the real part  is negative then the system is stable,but if it is positive, the system will be unstable This holds true for systems of anyorder, so in general it can be stated: `If any of the roots of the characteristic equationhave positive real parts, then the system will be unstable'

5.2 The Routh±Hurwitz stability criterionThe work of Routh (1905) and Hurwitz (1875) gives a method of indicating thepresence and number of unstable roots, but not their value Consider the character-istic equation

ansn‡ an 1sn 1‡


‡ a1s ‡ a0ˆ 0 (5:11)

The Routh±Hurwitz stability criterion states:

(a) For there to be no roots with positive real parts then there is a necessary, but not

sufficient, condition that all coefficients in the characteristic equation have thesame sign and that none are zero

If (a) above is satisfied, then the necessary and sufficient condition for stability is either

(b) all the Hurwitz determinants of the polynomial are positive, or alternatively

(c) all coefficients of the first column of Routh's array have the same sign The

number of sign changes indicate the number of unstable roots

The Hurwitz determinants are

D1ˆ a1 D2ˆ a1 a3

a0 a2

etc:

(5:12)

Routh's array can be written in the form shown in Figure 5.3

In Routh's array Figure 5.3

Routh's method is easy to apply and is usually used in preference to the Hurwitz

technique Note that the array can also be expressed in the reverse order,

commen-cing with row sn

··

q1 p1

Example 5.1 (See also Appendix 1, A1.5)Check the stability of the system which has the following characteristic equation

s4‡ 2s3‡ s2‡ 4s ‡ 2 ˆ 0 (5:15)Test 1: All coefficients are present and have the same sign Proceed to Test 2, i.e.Routh's array

5.2.1 Maximum value of the open-loop gain constant for the

stability of a closed-loop system

The closed-loop transfer function for a control system is given by equation (4.4)

Example 5.2 (See also Appendix 1, examp52.m)

Find the value of the proportional controller gain K1 to make the control system

shown in Figure 5.4 just unstable

Solution

The open-loop transfer function is

G(s)H(s) ˆs(s2‡ s ‡ 2)8K1 (5:25)The open-loop gain constant is

C

R(s) ˆ

K s(s 2 ‡s‡2)

Proportional Controller +

Multiplying numerator and denominator by s(s2‡ s ‡ 2)

s3‡ s2‡ 2s ‡ 2 ˆ 0factorizing gives

(s2‡ 2)(s ‡ 1) ˆ 0hence the roots of the characteristic equation are

s ˆ 1

s ˆ 0  jp2and the transient response is

c(t) ˆ Ae t‡ B sin(p2t‡ ) (5:37)From equation (5.37) it can be seen that when the proportional controller gain K1isset to 0.25, the system will oscillate continuously at a frequency ofp2rad/s

5.2.2 Special cases of the Routh array

Case 1: A zero in the first column

If there is a zero in the first column, then further calculation cannot normally proceed

since it will involve dividing by zero The problem is solved by replacing the zero with

a small number " which can be assumed to be either positive or negative When the

array is complete, the signs of the elements in the first column are evaluated by

allowing " to approach zero

Irrespective of whether " is a small positive or negative number in array (5.38), there

will be two sign changes in the first column

Case 2: All elements in a row are zero

If all the elements of a particular row are zero, then they are replaced by the

derivatives of an auxiliary polynomial, formed from the elements of the previous row

The elements of the s3row are zero in array (5.39) An auxiliary polynomial P(s) is

therefore formed from the elements of the previous row (s4)

i.e

P(s) ˆ 2s4‡ 12s2‡ 16dP(s)

The coefficients of equation (5.40) become the elements of the s3 row, allowing the

array to be completed

5.3 Root-locus analysis

5.3.1 System poles and zeros

The closed-loop transfer function for any feedback control system may be written inthe factored form given in equation (5.41)

The position of the closed-loop poles in the s-plane determine the nature of thetransient behaviour of the system as can be seen in Figure 5.5 Also, the open-looptransfer function may be expressed as

G(s)H(s) ˆK(s z01)(s z02) (s z0n)

(s p01)(s p02) (s p0n) (5:42)where z01, z02, , z0nare open-loop zeros and p01, p02, , p0nare open-loop poles

X

X

X X

Fig 5.5 Effect of closed-loop pole position in the s-plane on system transient response.

5.3.2 The root locus method

This is a control system design technique developed by W.R Evans (1948) that

determines the roots of the characteristic equation (closed-loop poles) when the

open-loop gain-constant K is increased from zero to infinity

The locus of the roots, or closed-loop poles are plotted in the s-plane This is acomplex plane, since s ˆ   j! It is important to remember that the real part  is

the index in the exponential term of the time response, and if positive will make the

system unstable Hence, any locus in the right-hand side of the plane represents an

unstable system The imaginary part ! is the frequency of transient oscillation

When a locus crosses the imaginary axis,  ˆ 0 This is the condition of marginalstability, i.e the control system is on the verge of instability, where transient oscilla-

tions neither increase, nor decay, but remain at a constant value

The design method requires the closed-loop poles to be plotted in the s-plane as K

is varied from zero to infinity, and then a value of K selected to provide the necessary

transient response as required by the performance specification The loci always

commence at open-loop poles (denoted by x) and terminate at open-loop zeros

(denoted by o) when they exist

Characteristic equation

1 ‡ G(s)H(s) ˆ 0Substituting equation (5.3) gives

1 ‡TsK ˆ 0

K Ts

+

–

Fig 5.6 First-order control system.

Roots of characteristic equation

When K is varied from zero to infinity the locus commences at the open-loop pole

s ˆ 0 and terminates at minus infinity on the real axis as shown in Figure 5.7.From Figure 5.7 it can be seen that the system becomes more responsive as K isincreased In practice, there is an upper limit for K as signals and control elementssaturate

Example 5.6Construct the root-locus diagram for the second-order control system shown inFigure 5.8

Open-loop transfer function

Open-loop poles

s ˆ 0, 4Open-loop zeros: none

–

Fig 5.8 Second-order control system.

Substituting equation (5.4) gives

s(s ‡ 4)ˆ 0

Table 5.1 shows how equation (5.7) can be used to calculate the roots of the

characteristic equation for different values of K Figure 5.9 shows the corresponding

root-locus diagram

In Figure 5.9, note that the loci commences at the open-loop poles (s ˆ 0, 4)when K ˆ 0 At K ˆ 4 they branch into the complex space This is called a break-

away point and corresponds to critical damping

Table 5.1 Roots of second-order characteristic equation for different values of K

K Characteristic equation Roots

5.3.3 General case for an underdamped second-order system

For the generalized second-order transfer function given in equation (3.43), equatingthe denominator to zero gives the characteristic equation

From Figure 5.10, Radius

If a system is(1) to be stable(2) to have acceptable transient response (  0:5)

then the closed-loop poles must lie in an area defined by

This is illustrated in Figure 5.11

5.3.4 Rules for root locus construction

Angle and magnitude criteria

The characteristic equation for a closed-loop system (5.24) may also be written as

The angle criterion

Equation (5.55) may be interpreted as `For a point s1 to lie on the locus, the sum of

all angles for vectors between open-loop poles (positive angles) and zeros (negative

angles) to point s1must equal 180.'

In general, this statement can be expressed as

 Pole Angles  Zero Angles ˆ 180 (5:57)

σ

60°

–60°

Acceptable Region (ζ≥ 0.5)

Unacceptable Region

jω

Fig 5.11 Region of acceptable transient response in the s-plane for   0.5.

Example 5.7Consider an open-loop transfer function

G(s)H(s) ˆ K(s ‡ a)

s(s ‡ b)(s ‡ c)Figure 5.12 shows vectors from open-loop poles and zeros to a trial point s1 FromFigure 5.12 and equation (5.57), for s1to lie on a locus, then

(1‡ 2‡ 3) (1) ˆ 180 (5:58)The magnitude criterion

If a point s1 lies on a locus, then the value of the open-loop gain constant K at thatpoint may be evaluated by using the magnitude criterion

Equation (5.56) can be expressed as

Equation (5.60) may be written as

K ˆProduct of pole vector magnitudesProduct of zero vector magnitudes (5:61)

For Example 5.7, if s1lies on a locus, then the pole and zero magnitudes are shown in

Figure 5.13 From Figure 5.13 and equation (5.61), the value of the open-loop gain

constant K at position s1 is

If there are no open-loop zeros in the transfer function, then the denominator of

equation (5.62) is unity

5.3.5 Root locus construction rules

1 Starting points (K ˆ 0): The root loci start at the open-loop poles

2 Termination points (K ˆ 1): The root loci terminate at the open-loop zeros when

they exist, otherwise at infinity

3 Numberof distinct root loci: This is equal to the order of the characteristic equation

4 Symmetry of root loci: The root loci are symmetrical about the real axis

5 Root locus asymptotes: For large values of k the root loci are asymptotic to

straight lines, with angles given by

 ˆ(1 ‡ 2k)(n m)where

k ˆ 0, 1, (n m 1)

n ˆ no of finite open-loop poles

m ˆ no of finite open-loop zeros

6 Asymptote intersection: The asymptotes intersect the real axis at a point given by

aˆ open-loop poles  open-loop zeros(n m)

7 Root locus locations on real axis: A point on the real axis is part of the loci if the sum

of the number of open-loop poles and zeros to the right of the point concerned is odd

8 Breakaway points: The points at which a locus breaks away from the real axis can

be calculated using one of two methods:

(a) Find the roots of the equation

dKds

sˆ b

ˆ 0where K has been made the subject of the characteristic equation i.e K ˆ (b) Solving the relationship

Xn 1

1(b‡ jpij)ˆ

Xm 1

1(b‡ jzij)where jpij and jzij are the absolute values of open-loop poles and zeros and

bis the breakaway point

9 Imaginary axis crossover: The location on the imaginary axis of the loci ginal stability) can be calculated using either:

(mar-(a) The Routh±Hurwitz stability criterion

(b) Replacing s by j! in the characteristic equation (since  ˆ 0 on the ary axis)

imagin-10 Angles of departure and arrival: Computed using the angle criterion, by ing a trial point at a complex open-loop pole (departure) or zero (arrival)

position-11 Determination of points on root loci: Exact points on root loci are found using theangle criterion

12 Determination of K on root loci: The value of K on root loci is found using themagnitude criterion

Example 5.8 (See also Appendix 1, examp58.m and examp58a.m)

A control system has the following open-loop transfer function

s(s ‡ 2)(s ‡ 5)(a) Sketch the root locus diagram by obtaining asymptotes, breakaway point andimaginary axis crossover point What is the value of K for marginal stability?(b) Locate a point on the locus that corresponds to a closed-loop damping ratio of0.5 What is the value of K for this condition? What are the roots of thecharacteristic equation (closed-loop poles) for this value of K?

Part (a)

Open loop poles: s ˆ 0, 2, 5 n ˆ 3

Open-loop zeros: none m ˆ 0

Asymptote angles (Rule 5)

Breakaway points (Rule 8)

Method (a): Re-arrange the characteristic equation (5.69) to make K the subject

6

b‡ 7b‡ 10 ‡ 2

b‡ 5b‡ 2

b‡ 2bˆ 032

Method (a) (Routh±Hurwitz)

s1 (70 K)/7

From Routh's array, marginal stability occurs at K ˆ 70

Method (b): Substitute s ˆ j! into characteristic equation From characteristicequation (5.69)

The root locus diagram is shown in Figure 5.14.

Part (b) From equation (5.52), line of constant damping ratio is

ˆ cos 1() ˆ cos 1(0:5) ˆ 60 (5:80)This line is plotted on Figure 5.14 and trial points along it tested using the angle

criterion, i.e

1‡ 2‡ 3ˆ 180

Hence point lies on the locus

Value of open-loop gain constant K: Applying the magnitude criterion to the above

s s ( +2)( +5) s

ζ= 0.5 ( = 60 )β °

Closed-loop poles (For K ˆ 11:35): Since the closed-loop system is third-order, thereare three closed-loop poles Two of them are given in equation (5.81) The third lies

on the real locus that extends from 5 to 1 Its value is calculated using themagnitude criterion as shown in Figure 5.15

of 0.25 and hence find(a) the value of K at this point(b) the value of K for marginal stabilitySolution

Asymptote angles (Rule 5)

Breakaway points: None, due to complex open-loop poles

Imaginary axis crossover (Rule 9)

Method (b)

(j!)3‡ 4(j!)2‡ 13j! ‡ K ˆ 0or

Plot line of constant damping ratio on Figure 5.16 and test trial points along it usingangle criterion.

At s ˆ 0:8 ‡ j2:9

104:5 ‡ 79:5 4 ˆ 180

Hence point lies on locus

Applying magnitude criterion

5.4 Design in the s-planeThe root locus method provides a very powerful tool for control system design Theobjective is to shape the loci so that closed-loop poles can be placed in the s-plane atpositions that produce a transient response that meets a given performance specifica-tion It should be noted that a root locus diagram does not provide informationrelating to steady-state response, so that steady-state errors may go undetected,unless checked by other means, i.e time response

5.4.1 Compensator design

A compensator, or controller, placed in the forward path of a control system will

modify the shape of the loci if it contains additional poles and zeros Characteristics

of conventional compensators are given in Table 5.2

In compensator design, hand calculation is cumbersome, and a suitable computerpackage, such as MATLAB is generally used

Case Study

Example 5.10 (See also Appendix 1, examp510.m)

A control system has the open-loop transfer function given in Example 5.8, i.e

G(s)H(s) ˆs(s ‡ 2)(s ‡ 5)1 , K ˆ 1

A PD compensator of the form

is to be introduced in the forward path to achieve a performance specification

Settling time (2%) less than 2 secondsDetermine the values of K1and a to meet the specification

Original controller

The original controller may be considered to be a proportional controllerof gain K and the

root locus diagram is shown in Figure 5.14 The selected value of K ˆ 11:35 is for a

damping ratio of 0.5 which has an overshoot of 16.3% in the time domain and is not

acceptable.Withadampingratioof0.7theovershootis4.6%whichiswithinspecification

This corresponds to a controller gain of 7.13 The resulting time response for the original

system (K=11.35) is shown in Figure 5.20 where the settling time can be seen to be 5.4

seconds, which is outside of the specification This also applies to the condition K=7.13

PD compensator design

With the PD compensator of the form given in equation (5.98), the control problem,

with reference, to Figure 5.14, is where to place the zero a on the real axis Potential

locations include:

(i)ii Between the poles s ˆ 0, 2, i.e at s ˆ 1

(ii)i At s ˆ 2 (pole/zero cancellation)

(iii) Between the poles s ˆ 2, 5, i.e at s ˆ 3

Table 5.2 Compensator characteristics CompensatorChar acter istics

PD One additional zero

PI One additional zero

One additional pole at origin PID Two additional zeros

One additional pole at origin

Example 5. 6Construct the root-locus diagram for the second-order control system shown inFigure 5. 8

Open-loop transfer function

Open-loop poles... class="page_container" data-page="21">

Closed-loop poles (For K ˆ 11: 35) : Since the closed-loop system is third-order, thereare three closed-loop poles Two of them are given in equation (5. 81) The third... data-page="6">

Example 5. 2 (See also Appendix 1, examp52.m)

Find the value of the proportional controller gain K1 to make the control system

shown in Figure 5. 4