cntrl systems

8/12/2019 Cntrl Systems

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VTUEdusat Programme5th

Semester Mechanical - ME 55 Control Engineering 1

Dr. D.V. GIRISHProfessor & HeadDepartment of Mechanical Engineering

Malnad College of EngineeringHassan 573201

[email protected] : 08172 -245683Phone (O) 08172-245319 (R) 08172-266102

Mobile : 9448639079


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Transfer Functions:

It is defined as the ratio of the Laplace transform of output (response) to theLaplace transform of input (excitation) assuming all the initial conditions tobe zero.

Fig (a) System in time domain Fig (b) System in Laplace domain

Fig: Transfer Functions of a system

If G(S) be the transfer function of the system, we can write mathematically

This Transfer function is a property of the system itself, independent of the

input or driving function. The T.F includes the units necessary to relate the

input to the output, However, it does not provide any information concerning

the physical structure of the system. i.e., the T.F of many physically different

systems can be identical

STEPS TO OBTAIN TRANSFER FUNCTION

a) Write the appropriate equation which defines the behaviour of the

element.

b) Transform this equation assuming all initial conditions to be zero.

c) Form the ratio of output C(S) to input R(S)

g (t)

r(t) C(t)

G (S)

R(S) C(S)

L.T. of output

L.T. of InputG(S) =

all initial conditions are Zero

C(S)

R(S)=

C(S)

R(S)= G(S)


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Consider a spring-mass-damper (k-m-c) system on which the force F actsand displacement x of the mass is the output.

Draw the free body diagram as shown Equation of Motion

= F cx kx = mx

Taking laplace transform of each term of this equation (assuming Zero initialcondition), we can write,

F(s) = ms2 X(s) + cs X(s) + kX(s)

Now, taking the ratio of X(s) to F(S) we can write the transfer function of the

system

Here the highest power of the complex variable S, in the denominator of thetransfer function determines the order of the system. Thus the k-m-csystem under consideration is a second order system.

Similarly we can write for k-m systemi.e., c = 0

and if m=0, we can write

x

F

M

0

Kx Cx

x

F

M

0

mx + cx + kx = F.

=..

X (S)

F (S)= = G(S) =

1

ms2+CS+K

X (S)

F (S)=

1

ms2+K

X (S)

F (S)=

1

CS+K

K C

.

. ..


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Block diagram Reduction:

We know that,Input-Output behavior of a Linear System or Element of a Linear System isgiven by Transfer Function,

G(s)= C(s)/R(s)Where, R(s) = Laplace transformation of the input Variable

C(s) = Laplace transformation of the output Variable

A Convenient graphical representation of this behaviour, i.e., short handpictorial representation of the cause and effect relationship between theinput and out put of a physical system is known as BLOCK DIAGRAM

This is shown in Fig (i)

(Input) (Output)Fig (i)

Here, the signal into the black represents the input R(S) and signal out of the

black represents the output C(S), while the block itself stands for transferfunction G(S).

The flow of information (Signal) is unidirectional from input to the output,

with the out put being equal to the input multiplied by the transfer functionof the block.

A complex system comprising of several elements is represented by theinterconnection of the blocks for Individual elements.

The blocks are connected by lines with arrows indicating the unidirectional,flow of information from the out put of one block to the input of the other. Inaddition to this summing or differencing of signals is indicated by thesymbols shown in the fig (ii) (a) (b) & (c), while take off point of a signal isrepresented by fig(iii)

i.e., in summing point two or more signals can be added or subtracted.

G (S)

R(S) C(S)

Fig (ii) (a)

A A + B

++

B

Fig (ii) (b)

A A - B

+-

B

AB+C

C

Fig (ii) (c)

A ++

B-

Fig (iii)


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The points at which the output signal of any block can be applied to two ormore points is known as Take off Point.

(This output is analogous to voltage but not to current)

Forward Path:

The direction of flow of signal from Input to output is Known as ForwardPath.

Feedback Path:

The direction of flow of signal from output to Input is Known as FeedbackPath.

The following Equations refer to a Tension Regulating Apparatus such asused in the Paper Industry.(a) Main Input x =Fr A.,(b) Lever Measurement. e = (x-y)/2

(c) The change in torque provided by motor tm = [Km/(1+p) e(d) Roll tension Fc = tm/R(e) Tension of Control Spring y = 2 Fc/K

Draw individual Block Diagram and Determine overall Transfer function

Output

Input =

C(S)

R(S) G1(S) G2(S) G3(S)=

R(s)G (S)

H (S)

C(s)

Input Output

G1(S)

R (S) R(S) G1(S)

G2(S)

R (S) G1(S) G2(S)

G3(S)C (S)

= R (S) G1(S) G2(S) G3(

Output

Input =C(S)

R(S)

= ?

Feed Forward

Feed back


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Solution: Individual Block Diagrams:

(a) Main Input x =Fr A.,

(b) Lever Measurement. e = (x-y)/2

(c) The change in torque provided by motor tm= (Km/(1+p))e

(d) Roll Tension, Fc = tm/R,

(e) Tension of Control Spring, y = 2 Fc/K,

Complete Block diagram and overall transfer function:

Overall Transfer Function

CANNONICAL FORM OF CLOSED LOOP SYSTEM:

If a Block diagram which consists of a forward path having one block, a feed

back path having one block, a take off point and a summing point, itrepresents a Cannonical form of a closed loop system

Km

(1+

)

tm1/2A

Fr e1/R

2/K

Fcx-y

A

Fr x

eX +

1/2

Y -

X - Y

Km/(1+p)

2/KFc

y

1/Rtm Fc

Fc

Fr= = ?

e

t


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R (s) Laplace Transform of Reference Input (t)

C (s) Laplace Transform of Controlled Output c(t)

E (s) Laplace Transform of Error signal e(t)

B (s) Laplace Transform of Feed back signal b(t)

G (S) Equivalent forward path Transform function = (c(s)/E(s))

H (S) Equivalent feedback path Transform function = (c(s)/B(s))

BLOCK DIAGRAM ALGEBRA

Block diagrams of some of the control systems turn out to be very complexsuch that the evaluation of their performance requires simplification (orreduction) of block diagrams which is carried out by block diagram

rearrangements, using the rules ofBLOCK DIAGRAM ALGEBRA

Some of the important Block diagram rearrangements are discussed here

Block diagram Reduction:

(i) Block diagram of a closed loop system

In the above figure, we can write

E (S) = R (S) B (S) &

B (S) = C (S) . H (S)

R(s)G (S)

C(s)E(s)

H (S)

+

+B(S)

G (S)R (S) + C (S)

B(S)H(s)

E S


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C (S) = G (S) . E (S)

C (S) = G (S) . [R (S) B (S) ]

C (S) = G (S) . R (S) - G (S) H (S) . C (S)C (S) + G (S) . H (S) . C (S) = G (S) . R (S)

= C (S) [ 1 + G (S) H (S)] = G (S) . R (S)

C (S) / R (S)

= G (S) / 1 + G (S) H (S)

Similarly it can be shown for a positive feed back

C (S) / R (S) = G (S) / 1 G(S) H (S)

In General we can write

C (S) / R (S) = G (S) / 1 G (S) H (S)

In Block diagram representation we can write

C (S)

R (S)

= Transfer Function =G(S)

1+G(S) H(S)

G (S)R (S) + C (S)

B(S) H(s)

E(S)


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RULES OF BLOCK DIAGRAM ALGEBRA (Fourteen)

Original Block Diagram Equivalent Block Diagram Transformation

1)Rearranging Summing

Points

2) Splitting the SummingPoints

3)

4)

5)

G (S)

1 G (S) H (S)

C (S)R (S)

A + A -B AB+C

B C

-

+

+

A + A+C AB+C

BC

-

+

-

A-B+C

-

A +

B

A + A - B AB+C

B

-

+

-

Interchanging theBlocks

G1 G2

A G1G2A

G1G2

A G1G2A

G1 G2

A G1G2A

Combining theBlocks in Cascade

G1 G2

A G1G2A

A G1

G1

G1

A A(G1 +G2)

A G2

Combining the

Blocks in Parallel

OR Eliminating aForward Loop

G1+G2

A G1+G2A


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6)

7)

8)

9)

10)

11)

G1A A G1 - B

B

A G1

-

+ Moving a

Summing Point

ahead of a Block

G1+

A A G1 - BA-B/ G1

1/ G1

BB / G1

-

G1

A - B A G1B G1

+

B

-

AMoving aSumming Point

Beyond a Bloc

-

B G1

G1+

G1

A G1A G1B G1

BB

G1

A G1A

A G1

G1

G1

A

A

A G1

A G1

Moving a take off

point ahead of a

Block

G1

A G1

A

A G1

1/G1

A G1

A G1

A

AMoving a take offpoint beyond a

Block

A -B

A +

-

B

A -B

A+

A -BB

A -B

B

-

+Moving a take offpoint before (ahead

of) a summing Block

Moving a take off

point after (beyond)a Summing block

A +- (AB)

A

(AB)

-

++ A

BA -B +

A -B


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12)

13)

(No. 10, 11, 12 & 13 to be used very carefully or better to be avoided)

14)

(No. 6, 7, 8, 9 & 14 are very important / to be used frequently)

Procedure for reduction ofBlock Diagram

Step. 1 : Reduce the Cascade BlocksStep. 2 : Reduce the parallel BlocksStep. 3 : Reduce the internal Feed back loopsStep. 4 : It is advisable to shift take off points towards right

and summing points towards left. It is always betterto avoid rules 10 and 11[i.e. shafting take off point before a summing block andshifting of take off point after summing block]

Step. 5 : Repeat steps 1 to step 4 until the simple form isobtained

A G1 + A G2

G1

G2

A G1

A

A

+Removing a Block

from a Forward Path

B = (A-B G2) G1B = (AG1 - B G1G2)

G1

G2

A

-

+ BRemoving a Block

from a Feed BackLoop

B

G21/G2 G1

-

+

B = (A / G2B) G1 G2B = (AG1 - B G1G2)

B = (A-B H1) G1B = (AG1 - B G1H1)

G1

H1

A

-

+ B

+

A Eliminating a

Feedback Loop

A G1A G2

G2 1/G2 G1

A G1 + A G2

A G2+

AAA

G11+ G1H1

B


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Step. 6 : Find transfer function of the over all system using theformula C(S) / R (S)

Any complicated system can be brought into simple form by reduction ofblock diagram using block diagram Algebra, discussed earlier.

EXERCISES

Ex - 01

Ex - 02

G 1

H 1

G 2

R

H 1

G1+G2

+

R +

R G 1+ G 2

1-(G 1+ G 2) H 1

C=

G1+G2

1-(G 1+ G 2) H 1R

C

G 1

G 2

H 1

R C


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G 2

G 1

G 1

H 1

R C

R C

H 1G 1

G1+ G 2

R C

H 1 G 1

G1 + G2

1 (G 1+G 2)

G 1 + G 2

1-(G 1 + G 2) H 1 G 1 1

G1 + G2

CR .


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Ex - 03

Example 04:(Jan/Feb-2003)

G2

R G 1

1- G 1 H 1

C

G 1 + G 2

1-G 1 H 1

R C G 1 + G 2

1-G 1 H 1

C

R

=

G 2

G 1

H1

RC

CG 1

1- G 1 H 1G 1

R G 1

1- G 1 H 1G 2

C

R=

H 2

H 1

G 1 S2 G 2 G 3 G 4S1

C(S)R(S)


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Example: 05July/Aug-2003

H 1

G 1 S2 G 2 G 4

C s

H 2

S1

1/G4

G 4R s

G 2 G 3 G 4

1- G 2 G 3 G 4 H 1S1

H 2/ G 4

G1

C sR s

G 1G 2 G 3 G 4

1-G 2 G 3 G 4 H 1+ G 1 G 2 G 3 H 2

C(s)R(s)

H3

G2G1 G3

H 1

H2

C(s)R(s)

R(s)

G 1G 2 G 3 G 4

1-G 2 G 3 G 4 H 1

G 1G 2 G 3 G 4 H 2

1-G 2 G 3 G 4 H 1 G 4

C(s)

1 +


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Simplify further . . . . . .

Example: 06

R (s) G 2G 1

H 1

H3/G3

G3

1+G3 H2C (s)

G 1

H 1

G 2 G 3 G 2 G 3 H 3

1+G 3 H 2 1+ G 3 H 2 G 3

R(s) C(s)1 +

G1 G2 G4

G5

G6

H1

H2

-

- -

C(s)

G3

R(s)

1/G3H3

G2G1

C(s)

H1

H2

R(s)G3


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G1 G21+ G1 G2 H1

G3G4 + G5 G6

H2

R sC(s)

G1 G2 (G3G4 + G5)

1 + G1 G2 H1

G1 G2 (G3G4 + G5)

1+ G1 G2 H1

G6

H2

1 +

C (S)R (S)

G1 G2 (G3G4 + G5)

1+G1 G2 H1+ G1 G2 (G3G4 + G5) H2x G6

C (S)R (S)


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SYSTEM STABILITY CRITERION

A control system, if it is to be of any practical value, must be STABLE.

This means that, in response to some input, the system will not oscillate

violently or drive itself to some limiting value of the controlled variable butrather will attain some useful response.

More specifically a stable system is one in which the transients decay, that

is, the transient response disappears with increasing values of time.

This latter statement constitutes the basic concept of stability. There are so

many techniques for determining the stability, all are based on the previous

definition.

For example, consider the following equations:

(i) My + Ky = 0 (Spring mass system)

= Mp2 + K = 0

=jn

(ii) cy + Ky = 0

= cp + K = 0 )(rootc

kP

y = C1ept = c1 e

(iii) My + Cy +ky = 0

= Mp2 + Cp + K = 0

m

mkc

m

cP

2

4

2

2

12

= -a jd Solution = y = e-at {C1 Sind + C2 cos dt}

21

m

k

jP

m

kn

tc

k

..

.

..

.


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Equation 2, and 3, are stable (but not eq.1), since transients will die as

t , because of exponentially decreasing terms.

Now consider

Here characteristic equation is p-1 = 0

p=1

and transient solution is, Yt = c,e1.t

Here it can be seen that the response does not vanish, but instead grows

rapidly with increasing values of time (i.e, yt as t )

The response of the actual system would not of course approach infinity butrather would proceed to some extreme limiting value determined by thephysical nature of the system. Such a response is of limited practical value

Again consider the equation

Ch Eq. P2 - 2 P + 5 = 0

= 1j2

and the transient solution is Yt= e1t (C1 Sin 2t+ C2 cos 2t)

Here again an oscillatory response is indicated with an amplitude whichincreases with time rather than decreases. The physical system would tend

to oscillate violently, perhaps destroying itself

These above examples illustrates the response of systems when the roots ofthe characteristic equation are -ve or +ve real numbers or complex numberswith +ve real parts, indicating STABLE and UNSTABLE systemsrespectively.

Here, it should be noted that the system input has no effect on stability.Therefore a linear system which is STABLE to one input is STABLE to allinputs.

So far we have discussed the possibility of having a complex root with +ve or-ve real parts.

xydt

dy

0522

2

ydt

dy

dt

yd

2

5442 p


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But what about the possibility of having complex roots in which the realpart is a zero?

Now again consider My + Ky = 0Ch Eq., =

Mp2 + K = 0=jn

Solution, : C1 sin nt + C2 cos nt

The response turned out to be a persistent oscillation as shown.

The amplitude of oscillation neither decaying nor growing with time. Asystem with this response is said to possess LIMITED STABILITY

Hence border line case between absolute stability and instability is called as

LIMITED STABILITY

Consider a 100th degree characteristic equation, suppose that 99 of itsroots are either -ve real number or pairs of complex numbers with -ve realparts, but that remaining single root is a positive number is the system isSTABLE ?

No the system is UNSTABLE

Why?

Because there will be one exponential term in the transient solution whichwill grow with time and so the transient solution will approach infinity as

time approaches infinity.This above example will emphasise the basic concept of stability and statesthat

For absolute stability, all roots must be negative real numbers or

complex numbers with -ve real parts.

Since solving a higher degree characteristic equation for its roots is timeconsuming, the Routh / Hurwitz criterion can be used to determine whetheror not there are +ve roots with out actually solving it.

We Know,Every system has to pass through a Transient Stage for a small period beforereaching Steady State.

So, Naturally the question comes in to playWill the system reach itssteady state after passing through transients

21

m

kjP

..


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To find an answer to the above question study of STABILITY is of utmostimportant

STABILITY?

Even after excitation by a bounded input, output must be bounded. In the absence of Input, Output must be zero irrespective of Initial

Conditions. If its impulse response approaches zero as time approaches infinity.

A Control System, if it is to be of any practical value must be stable Thismeans

In response to some input the system will not oscillate violently ordrive itself to some limiting value of the controlled value, but ratherwill attain some useful response

i.e. A System is Stable if its impulse response approaches zero as timeapproaches infinity. The Stability of a System is determined by its responseto inputs or disturbances.

In general,

A Stable System is one that will remain at rest unless excited by an external

source and will return to rest if all excitations are removed.

Such condition requires that the co-efficient of t in the exponential terms ofthe Transient Solution be negative real numbers or Complex numbers with

negative real parts.

Ex: e-5t decays while e+5t grows as time advances.

These co-efficient are of course the roots of the characteristic equation.

Therefore,

From the Physical view point, a stable system is one in which the transientdie out and the system Settles down to some useful response.

Mathematically, a system is said to be stable if the roots of the characteristicequation are negative real number or Complex numbers with negative realparts.

In a vast majority of practical systems, the following statements on stabilityare quite useful.

(i) If all the roots of the characteristic equation have ve real parts thesystem is STABLE.


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(ii) If any root of the characteristic equation has a +ve real part or if there

is a repeated root on the j-axis, the system is unstable

(iii) If condition (i) is satisfied except for the presence of one or more non

repeated roots on the j-axis the system is limitedlySTABLE

In further subdivision of the concept of stability, a linear system ischaracterized as:

(i) Absolutely stable with respect to a parameter of the system if it isstable for all values of this parameter.

(ii) Conditionally stable with respect to a parameter, if the system isstable for only certain bounded ranges of values of this parameter.

It follows from the above discussion that stability can be established by

determining the roots of characteristic equation. Unfortunately, no generalformula in algebraic form is available to determine the roots of characteristicequation of higher than second order. Though the various numericalmethods exist for root determination of characteristic equation, these arequite cumber some even for third and fourth order systems.

However, simple graphical and algebraic criteria have been developed whichpermit the study of stability of a system with in the need of actuallydetermining the roots of its characteristic equation. These criteria answerthe question, whether a system be stable or not, in YES or NO form

The roots of characteristic equations of several systems are given below.Determine in each case if the set of roots represents stable, marginally stable

or unstable systems.

(a) -1, -4, -6, -8, (b) -3, +3, -2, -6,

(c) -5, -4, 0, -6, (d) -2+j, -2-j, -3+j, -4+j,(e) -2+j, -2-j, 2j, -2j (f) 4, -3, -2, 6,

(g) -4, -6, 8, 5, (h) -3+2j, -3-2j, -2, -4,(i) j, j, -2, 2, (j) -2, -2+j, -2-j, -4,

Comment on the STABILITY of the following:

(a) (d) (h) & (j) represents stable system since all the rootshave negative real parts.

(c.) (e.) represents marginally stable systems sinceall the roots have non +ve real parts, that iszero or ve.

(b) (f) (g) and (i) represents unstable systems since each hasroot with a +ve real part


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ii. A system has poles at -2, -4, 6, and zeros 2, 4, -3, is the system is stable?

The system is STABLE since the poles which are the roots of the systemcharacteristic equation have negative real parts. The fact that the systemhas a zero with a +ve real part does not affect its stability.

iii. (S+2).(S+3)2.(S-4) = 0, is the system is stable The characteristic

equation has the roots -2, -3, -3 and +4 and therefore represents anUNSTABLE system since there is a +ve real root.

(a) An integrator (it may be written as dy/dt=x)(b) A step input(c) Abounded input that produces an unbounded output.(d) x(t)= Cos wt(e) x(t) = e-t sin4t

iv. (a) The characteristic equation of an integrator is S=0. Since the rootdoes not have a ve real part an integrator is NOT STABLE. Sincethere is no root with a +ve real part an integrator is MARGINALLYSTABLE.

(b) Since a step function gives the system is MARGINALLY STABLE.

(c) The system is UNSTABLE

(d) The system is UNSTABLE since there is no decay

(e) The system is STABLE because there is decay exponentially

Rouths Creterion

E.J. Routh (1877) developed a method for determining whether or not anequation has roots with + ve real parts with out actually solving for theroots.

A necessary condition for the system to be STABLE is that the real parts ofthe roots of the characteristic equation have -ve real parts. This insures thatthe impulse response will decay exponentially with time.

If the system has some roots with real parts equal to zero, but none with +vereal parts the system is said to be MARGINALLY STABLE.

It determines the poles of a characteristic equation with respect to the leftand the right half of the S-plane with out solving the equation.

The roots of this characteristic equation represent the closed loop poles. Thestability of the system depends on these poles. The necessary, but not

S

SF )(


24/38



sufficient conditions for the system having no roots in the right half S-Planeare listed below.

(i) All the co-efficients of the polynomial must have the same sign.(ii) All powers of S, must present in descending order.

(iii) The above conditions are not sufficient

In a vast majority of practical systems. The following statements on stabilityare quite useful.

(i) If all the roots of the characteristic equation have ve real parts thesystem is STABLE.

(ii) If any root of the characteristic equation has a +ve real part or if there

is a repeated root on the j-axis, the system is unstable

(iii) (iii) If condition (i) is satisfied except for the presence of one or morenon repeated roots on the j-axis the system is limitedlySTABLE

In this instance the impulse response does not decay to zero although it isbounded. Additionally certain inputs will produce outputs. Thereforemarginally stable systems are UNSTABLEThe Routh Stability criterion is a method for determining system stabilitythat can be applied to an nth order characteristic equation of the forman sn + an-1 sn-1 + an-2 sn-2 + an-3 sn-3 +. a1 s1 + a0 = ZERO

The criterion is applied through the use of a Routh Array (Routh table)

Defined as follows:

Sn an an-2 an-4 .

Sn-1 an-1 an-3 an-5 .

Sn-2 b1 b2 b3

Sn-3 c1 c2 c3

Sn-4 d1 d2

.

.

S2 e1 a0

S1 f1

S0 a0


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Where an an-1,. a0 are the co-efficients of characteristic equation andb1 b2C1 C2 evaluated as follows

1

3211

n

nnnn

a

aaaab

1

5412

n

nnnn

a

aaaab

This Process is continued till we get a ZERO as the last co-efficient in thethird row.

In a similar way the co-efficient of 4th, 5th, 6th ..nth and (n+1)th rows areevaluated.

1

2131

1b

baabc nn

1

31512

b

baabc nn

. . . . . . .et.al

d1 =

d2 =

This table is continued horizontally and vertically until only zeros areobtained. Any row can be multiplied by a constant before the next row iscomputed with out disturbing the properties of the table.

This process is continued until s0 is obtained, which is equal to a0.

The ROUTH STABILITY CRITERION is stated as follows,

All the terms in the first column of Rouths Array should have same sign,

and there should not be any change of sign.

This is a necessary and sufficient condition for the system to be stable

On the other hand any change of sign in the first column of Rouths Arrayindicates,

(i) The System is Unstable, and

(i) The Number of changes of sign gives the number of roots lying in theright half of S-Plane


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Examples:

Ex. 01:S3+6S2+12S+8=0 using Rouths method

S3 1 12 0

S2 6 8 0

S1

6

64

6

18126

0 0

S0

8

6

64

086

64

0 0

There are no sign changes in the first column of the array and so there areno roots with +ve real parts and hence, the system in question is stable.Ex. 02:

S5+3S4+7S3+20S2+6S+15=0

S5 1 7 6

S4 3 20 15

S3 1/3 1 0S2 11 15 0

S1 6/11 0 0

S0 15

System is stable

Ex. 03:S4+2S3+3S2+8S+2=0

S4 1 3 2

S3 2 8 0

S3 1 4 0

S2 -1 2 0

S1 6 0 0

S0 2


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There are two sign changes (plus to minus & minus to plus) in the firstcolumn, showing that there are two roots with +ve real parts. Therefore the

system is unstable.

Ex. 04:3S4+10S3+5S2+5S+2=0

S4 3 5 2

S3 10 5 0

S3 (2) (1) 0

S2 7/2 2 0

S1 -1/7 0

S0 2

Here two roots are +ve (2 changes of sign) and hence the system is unstable.

Ex. 05:Examine the stability of

S5+2S4+4S3+8S2+3S+1=0

S5 1 4 3

S4 2 8 1

S3 0 2.5 0

S1 - - - - - -

How to proceed? What to do?

Ex. 06:S5+2S4+2S3+4S2+4S+8=0

S5 1 2 4

S4 2 4 8

(S4) (1) (2) (4)

S3 0 0 0

How to proceed? What to do?

Rouths array failed

Rouths array failed


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Special cases:Occasionally, in applying the Routh stability criterion certain difficultiesarise causing the break down of the Rouths test.

The difficulties encountered are generally of the following types.

Difficulty 01:

When the first term in any row of the Routh array is zero while rest of therow has at least one non zero term.

[Because of this zero term, the terms in the next row become infinite andRouths test breaks down]

Ex 05, refers to above difficulty. Now how to solve?

Difficulty 02:

When all the elements in any one row of the Routh array are zero.

Because of a zero row in the array the Rouths test breaks down.

Ex. 06 refers to this difficulty.

Now, How to solve?

Difficulty 01:

Let me repeat the problem No.5

i.e., S5+2S4+4S3+8S2+3S+1=0

We have already written the Rouths array and it is given as

S5 1 4 3

S4 2 8 1

S3 0 2.5 0

S1 - - - - - -

Here it indicates that first term in horizontal 3rd row of Routh's array is zerowhile rest of the row has atleast one non zero term

To remove this difficulty, the following methods can be used,

METHOD 1

(i) Replace zero by a small +ve number and complete the Rouths array(ii) Examine the signs of the first column of Rouths array by letting 0

Rouths array failed


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Following the above method, continue the Rouths array

S5 1 4 3

S4 2 8 1

S3

2.5 0S2

581

01.

0

S1

58

5.258

0 0

S0 1

Now, lt 0

58= t18

8

5

Therefore, the sign is ve.

Again,

It

58

5.258

It

58

5.258 2

0 0

= It 5.25

5.12

0


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Therefore, the array is given by

S5 1 4 3

S4 2 8 1

S3 2.5 0

S2 - 1 0

S1 2.5 0 0

S0 1 0 0

There are two changes of sign in the first column, and hence the system is

UNSTABLE (i.e. having two poles in the right half of S-plane)

METHOD 2

(i) Modify the original characteristic equations by replacing S by 1/Z

(ii) Apply the Rouths test on the modified equation in terms of Z

(iii) The number of Z-roots with +ve real parts are the same as the number ofS-roots with +ve real parts

(iv) Use same Routh criterion to determine status of the system with givenpolynomial for the same above

For the problem discussed above, (i.e, example 05) let us try this method 2

ExampleS5+2S4+4S3+8S2+3S+1=Zero

Replace S, by 1/Z, ( i.e. put S=1/z) The polynomial becomes

0138421

2345

ZZZZZOR

012483 2345 ZZZZZ


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Rouths array will be

Z5 1 8 2

Z4

3 4 1

Z3

6.67 1.67 0

Z2

3.25 1 0

Z1

-0.382 0 0

Z0

1 0 0

There are two changes of sign in the first column of the Rouths array which

tells us that there are two Z-roots in the right half Z - plane. Therefore the

number of S-roots in the right half S - plane is also two

Hence the system is UNSTABLE

Note: This method works in most but not all cases.

Difficulty 02:

Let me repeat the problem No.6i.e., S5+2S4+2S3+4S2+4S+8=0

We know that Rouths array will be

S5 1 2 4

S4

2 4 8

S3

0 0 0

Here it indicates that all the elements in the 3rd row of the routh array is

zero

This condition indicates that there are symmetrically located roots in the S-

plane (pair of real roots with opposite signs and/or pair (pairs) of conjugate

roots on the imaginary axis and / or complex conjugate roots forming

quadrates in the S-plane)

Rouths array failed


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To eliminate the above difficulty

(i) Form an equation by using the co-efficients of the array which is justabove the row of zeros. This equation is known as auxiliary equation.

[This polynomial gives the number and location of root pairs of

characteristic equation which are symmetrically located in theS-plane. The order of the auxiliary polynomial is always even]

(ii) Replace the row of zeros in the Rouths array by a row of co-efficientsof the polynomial generated by taking the first derivative of theauxiliary polynomial.

(iii) Continue the Rouths array and status of the polynomial may bedetermined by using the same Rouths criterion discussed earlier.

Now, the auxiliary equation is formed from the coefficient of the S4

row, which is given by

A(S) = 2S4+4S2+8

The derivative of the above polynomial with respect to S, is

The zeros in the S3 row are now replaced by the co-efficients 8 and 8 and

continue the Rouths array

S5

1 2 4

or

S5

1 2 4

S4 2 4 8 S4 1 2 4

S3 8 8 0 S3 4 4 0

S2

2 8 0 S2

1 4 0

S1

- 24 0 S1

-12 0

S0

8 S0

4

Here, there are two changes of sign and hence, the system is UNSTABLE

Solve: (Home Work)

(i) S6+S5-2S4-3S3-7S2-4S-4=0

(ii) S6+2S5+8S4+12S3+20S2+16S+16=0

SSdS

SdA88

)( 3


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Ex: . . . . . ?

The characterstic equation of a given system is S4+6S3+11S2+6S+K=0.

What restrictions must be placed upon the parameter K, in order to insure

that the system is stable. The routh table for the system is

S4 1 11 K

S3 6 6 0

S2 10 K 0

S1

10

660 k0 0

S0 K

For the system to be stable, the following the following restrictions must beplaced upon the parameter K:

i.e., 60-6K >0, or K0

Thus K must be greater than zero and less than 10

K>0K


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S1

5.5

433 k0

S0 K

The system will stable if no change of sign occurs in the first column.

K > 0 and5.5

433 k> 0 = 33-4K > 0 or 4K < 33 K< 33/4

or K < 8.25 Range of K, is 0 < K < 8.25

Example:(i) S2+KS+(2K-1) = 0

S

2

1 (2K-1)

S1

K 0

S0

2K-1

2K-1 > 0, K> 0

2K > 1

K >

If K > system is stable

(ii) S4+6S3 + 11S2 + 6S = K = 0

S4 1 11 K

S3 6 6 0

S2 10 K 0

S1

10

660 k0 0

S0 K

For system to be stable

60 60 K > 0 or K < 10 and K > 0

Range of K, V 0 < K < 10


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HURWITZ STABILITY CRITERION

The Hurwitz Stability Criterion is another method for determining whether ornot all the roots of a characteristics equation have ve real parts. Thiscriterion is applied through the use of determinants formed from the co-

efficients of characteristic equation.

It is assumed that the first co-efficient, an is positive. The determinants ifor i = 1, 2, 3, .n-1 are formed as the Principal minor determinants of thefollowing arrangement (Called Hurtwitz determinant)

W.K.T. Characteristic equation

=an sn + an-1 sn-1 + an-2 sn-2 +. a1 s1 + a0 =0

Then n, can be written as,

The determinates are thus formed as below

1 = an-1

2 =an-1 a0

an an-2

For Ex:If n=3, then Hurwitz determinant can be written as

WKT Characteristic Eq: a3S3+a2S2+a1S1+a0 = 0

H =

a2 a0 0

a3 a1 00 a2 a0

then 1 = a2,

n =H =

an-1 a n-3 o . o

a n a n-2 o . o

o a n-1 a n-3 ----------- o

o a n a n-2 ----------- o

-----------------------------------o ----------------------------------an

ao if n, odda1 if n, even

a1 if n, odda0 if n, even

H =n


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2 =a2 a0

a3 a1

3 =a2 a0 0

a3 a1 00 a2 a0

Thus all the roots of characteristic equation have ve real part if

1 = a2 > 02 = a2 a1a3 a0 > 0

3 = a1 a2 a0a02 a3 > 0

Then the system is STABLE

Again for Ex: if n = 4, The Characteristic Equation:

= a4S4 + a3S3 + a2S2 + a1S1+a0 = 0

H =

a3 a1 0 0

a4 a2 a0 00 a3 a1 0

0 a4 a2 a0

The disadvantages of Hurwitz Criterion are as follows

(i) It is very complicated and time consuming for solving higher ordersystem.

(ii) This method is unable to find the exact number of poles located inthe right half of S-plane

(iii) It is very tough to predict marginal stability

EX: 01s4 + 3 s3 + 6 s2 + 9 s+ 12 =0

The Hurwitz arrangement is given below

3 9 0 0

1 6 12 0

0 3 9 0

0 1 6 12

for stability 1,2,3& 4 all must be greaterthan zero


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1 = 3>0

2 = 18 - 9 = 9 > 0

3 = 3(54-36) 9(9-0) + 0( ) = 54 - 81 = -27 0

System is UNSTABLE

Ex- 02

S2 + KS + (2K-1) = 0

H =

K 0

1 (2K-1)

in order for these determinants be +ve, it is necessary that

K > 0 and 2K 1> 0

i.e., K >

If K > 1/2 , then the system is stable

EXERCISES CONDITION?

4 s3 + 3s2 + 2s + 5 = 0(March 2001)

S6 + 2s5 + 8s4+ 12s3 +20S2 + 16S +16=0(Aug- 2003) (Feb- 2004)

s4+ 5s3 +5S2 + 4S +K = 0(Feb- 2005)

s5 + 4s4+ 12s3 +20S2 + 30S +100 = ZERO(Aug- 2005)

1 = K, 2 = K (2K-1)-0= 2K

2- K


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CHARACTERISTIC EQUATION

The denominator polynomial in terms of S of a transfer function is knownas characteristic polynomial. If this polynomial is equated to Zero,characteristic equation will be obtained.

The characteristic polynomial of the transfer function G(S) of the equation

, is ansn + an + sn-1 + . . . + a0

and characteristic equation is given by

anSn + an + Sn-1 + . . . . + a0 = ZERO

Solving the characteristic equation of a transfer function we get poles of the

transfer function.

)(

)()(

SR

SCSG

cntrl systems

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