P.S. GandhiP.S. GandhiMechanical Engineering Mechanical Engineering IIT BombayIIT Bombay

Concept of Stability, Poles, Concept of Stability, Poles, Zeros and Zeros and RouthRouth’’ss CriterionCriterion

Concept of stability

Very important characteristic of the Very important characteristic of the transient performance of the system.transient performance of the system.An LTI system is stable if the following two An LTI system is stable if the following two notions of system stability are satisfied:notions of system stability are satisfied:(i) When the system is excited by a (i) When the system is excited by a bounded input, the output is bounded.bounded input, the output is bounded.

(ii) In the absence of input, output tends (ii) In the absence of input, output tends towards zero (the equilibrium state of the towards zero (the equilibrium state of the system) irrespective of initial conditions( this system) irrespective of initial conditions( this is also called as asymptotic stability).is also called as asymptotic stability).

BIBO stability condition

For a single input, single output system with For a single input, single output system with transfer function:transfer function:

With With initial conditions assumed zeroinitial conditions assumed zero, the output of , the output of the system is given by:the system is given by:

Therefore,Therefore,

Where is the impulse response Where is the impulse response of the system. of the system.

11

11

.....( ) ( ) ;( ) .....

m mo m

n no n

b s b s bY s G s m nU s a s a s a

−

−

+ + += = <

+ + +

1( ) [ ( ). ( )]y t L G s U s−=

0

( ) ( ) ( )y t g u t dτ τ τ∞

= −∫1( ) ( ( ))g t L G s−=

BIBO stability condition (contd)

From this we get,From this we get,

If for the bounded input( ), and for the If for the bounded input( ), and for the bounded output( ), the condition we get bounded output( ), the condition we get from above is:from above is:

Which implies that stability is satisfied if the impulse Which implies that stability is satisfied if the impulse response is absolutely integrable, i.e., response is absolutely integrable, i.e., is finite. is finite.

0

| ( ) | | ( ) || ( ) |y t g u t dτ τ τ∞

≤ −∫1| ( ) |r t M≤ < ∞

2| ( ) |c t M≤ < ∞

1 20

| ( ) | | ( ) |y t M g d Mτ τ∞

≤ ≤∫

0

| ( ) |g dτ τ∞

∫( )g t

Poles and ZerosPoles and Zeros

Transfer function Transfer function G(sG(s))Poles: roots of denominator polynomialPoles: roots of denominator polynomialZeros: roots of numerator polynomialZeros: roots of numerator polynomialCharacteristic equation is obtained by making Characteristic equation is obtained by making denominator of the transfer function equal to denominator of the transfer function equal to zerozero

11

11

.....( ) ( ) ;( ) .....

m mo m

n no n

b s b s bY s G s m nU s a s a s a

−

−

+ + += = <

+ + +

Effect of Poles on StabilityEffect of Poles on Stability

The nature of impulse response The nature of impulse response g(tg(t) depends on the ) depends on the poles of the transfer function poles of the transfer function G(sG(s) which are the roots ) which are the roots of the characteristic equationof the characteristic equationThese roots may be both real and complex conjugate These roots may be both real and complex conjugate and may have multiplicity of various orders.and may have multiplicity of various orders.The nature of responses contributed by all possible The nature of responses contributed by all possible types of poles are shown in the following slides.types of poles are shown in the following slides.In each case find whether the system is stable or not In each case find whether the system is stable or not with the help of BIBO condition (with the help of BIBO condition (i.ei.e check whether check whether the area under the absolutethe area under the absolute--valued impulse curve is valued impulse curve is finite or notfinite or not). ).

System with single real System with single real pole in left half spole in left half s--plane.plane.

Impulse response is exponentially decaying (bounded Impulse response is exponentially decaying (bounded output).output).As BIBO condition satisfied, system is stable.As BIBO condition satisfied, system is stable.

jw

σS=-1

1( )G ss a

=+

System with single real System with single real pole in right half spole in right half s--plane.plane.

Impulse response is exponentially growing (unbounded Impulse response is exponentially growing (unbounded output).output).As BIBO condition not satisfied, system is unstable.As BIBO condition not satisfied, system is unstable.

jw

σS=1

1( )1

G ss

=−

System with a pair of complex System with a pair of complex poles in left half spoles in left half s--plane.plane.

Impulse response is exponentially decaying sinusoid Impulse response is exponentially decaying sinusoid (bounded output).(bounded output).As BIBO condition satisfied, system is stable.As BIBO condition satisfied, system is stable.

jw

σ

S=-0.5-j0.866

S=-0.5+j0.866

2 2( )2 n n

KG ss ζω ω

=+ +

System with a pair of complex System with a pair of complex poles in right half spoles in right half s--plane.plane.

Impulse response is exponentially growing sinusoid Impulse response is exponentially growing sinusoid (unbounded output).(unbounded output).As BIBO condition not satisfied, system is unstable.As BIBO condition not satisfied, system is unstable.

jw

σ

S=0.5-j0.866

S=0.5+j0.866

( ) ?G s =

System with a pair of System with a pair of complex poles on jw axis.complex poles on jw axis.

Impulse response is sinusoidal (bounded output).Impulse response is sinusoidal (bounded output).Although BIBO condition not satisfied, because of Although BIBO condition not satisfied, because of bounded output these systems are called marginally bounded output these systems are called marginally stable.stable.

σ

S=-j0.866

S=+j0.866

2 2

1( )0.866

G ss

=+

System with a double pair of System with a double pair of complex poles on jw axis.complex poles on jw axis.

Impulse response is linearly increasing sinusoid Impulse response is linearly increasing sinusoid (unbounded output).(unbounded output).As BIBO condition not satisfied, system is unstable.As BIBO condition not satisfied, system is unstable.

S=-j0.866 2poles

S=+j0.866 2poles

System with a pole at originSystem with a pole at origin

Impulse response is constant (bounded output).Impulse response is constant (bounded output).Although BIBO condition not satisfied, because of Although BIBO condition not satisfied, because of bounded output these systems are called marginally bounded output these systems are called marginally stable.stable.

jw

σS=0

1( )G ss

=

System with a double System with a double pole at originpole at origin

Impulse response is linearly increasing (unbounded Impulse response is linearly increasing (unbounded output).output).As BIBO condition not satisfied, system is unstable.As BIBO condition not satisfied, system is unstable.

jw

S=0

2

1( )G ss

=

ObservationsObservations

i.i. If all the roots of the characteristic If all the roots of the characteristic equation have negative real parts, the equation have negative real parts, the system is system is stablestable..

ii.ii. If any root of the characteristic equation If any root of the characteristic equation has a positive real part or if there is a has a positive real part or if there is a repeated root on the repeated root on the jwjw--axis, the system axis, the system is is unstableunstable..

iii.iii. If the condition (i) satisfied except for the If the condition (i) satisfied except for the presence of one or more non repeated presence of one or more non repeated roots on the jwroots on the jw--axis, the system is axis, the system is marginally stablemarginally stable. .

Effect of zeros on stabilityEffect of zeros on stability

We have seen how the closed loop poles We have seen how the closed loop poles effect the transient response of the system, effect the transient response of the system, let us see what is the effect on transient let us see what is the effect on transient response by the closed loop zeros.response by the closed loop zeros.We see this by adding a zero to the one of the We see this by adding a zero to the one of the systems that we have just seen.systems that we have just seen.

Effect of zeros on stabilityEffect of zeros on stability

Recall the impulse of the system with a pair Recall the impulse of the system with a pair of complex poles in left half sof complex poles in left half s--plane.plane.

jw

σ

S=-0.5-j0.866

S=-0.5+j0.866

Add a left half sAdd a left half s--plane plane zero to this systemzero to this system

We see no change in stability of the system except a We see no change in stability of the system except a change in the shape of the transient.change in the shape of the transient.

σ

S=-0.5-j0.866

S=-0.5+j0.866

S=-9

jw

Add a right half sAdd a right half s--plane plane zero to this systemzero to this system

We see no change in stability of the system except We see no change in stability of the system except a change in the shape of the transient either it is a change in the shape of the transient either it is right or left half sright or left half s--plane zero.plane zero.

σ

S=-0.5-j0.866

S=-0.5+j0.866

S=9

jw

Getting Response Getting Response using LTusing LT

It is difficult to find inverse It is difficult to find inverse LaplaceLaplace transform transform of commonly encountered functions in of commonly encountered functions in control engineering.control engineering.The alternative is to write this functions in The alternative is to write this functions in combination of easily recognizable forms combination of easily recognizable forms (which are similar to the laplace transforms (which are similar to the laplace transforms of known functions).of known functions).This can be done taking partial fractions of This can be done taking partial fractions of the given function.the given function.

PartialPartial--fraction expansion fraction expansion (F(s) having distinct poles)(F(s) having distinct poles)

Consider F(s) in factored form asConsider F(s) in factored form as

Then in partial fraction form it can be written Then in partial fraction form it can be written asas

Where Where aa’’ss are residues of the poles.are residues of the poles.

1 2

1 2

( )( ).....( )( )( ) ;( ) ( )( ).....( )

m

n

K s z s z s zB sF s m nA s s p s p s p

+ + += = <

+ + +

1 2

1 2

( )( ) .....( ) ( ) ( ) ( )

n

n

aa aB sF sA s s p s p s p

= = + + ++ + +

PartialPartial--fraction expansion fraction expansion (F(s) having distinct poles)(F(s) having distinct poles)

Now the residues are found asNow the residues are found as

Now applying inverse laplace transform Now applying inverse laplace transform

Gives Gives f(tf(t) as) as

( )( )( )

k

k ks p

B sa s pA s =−

⎡ ⎤= +⎢ ⎥⎣ ⎦

1 kp tkk

k

aL a es p

−− ⎡ ⎤=⎢ ⎥+⎣ ⎦

1 211 2( ) [ ( )] ..... : 0np tp t p t

nf t L F s a e a e a e t−− −−= = + + + ≥

PartialPartial--fraction expansion fraction expansion (F(s) having repeated poles)(F(s) having repeated poles)

Consider the function,Consider the function,

Partial fraction of this F(s) has three termsPartial fraction of this F(s) has three terms

Where Where bb’’ss are calculated as follows:are calculated as follows:

2

3

2 3( )( 1)

s sF ss+ +

=+

31 22 3

( )( )( ) 1 ( 1) ( 1)

bb bB sF sA s s s s

= = + ++ + +

33

1

( )( 1)( ) s

B sb sA s =−

⎡ ⎤= +⎢ ⎥⎣ ⎦

PartialPartial--fraction expansion fraction expansion (F(s) having repeated poles)(F(s) having repeated poles)

Solving we get Solving we get SoSo

HenceHence

1 2 31, 0, 2b b b= = =

32

1

23

1 21

( )( 1)( )

1 ( )( 1)2! ( )

s

s

d B sb sds A s

d B sb sds A s

=−

=−

⎡ ⎤= +⎢ ⎥

⎣ ⎦

⎧ ⎫⎡ ⎤⎪ ⎪= +⎨ ⎬⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭

1 1 12 3

1 0 2( )1 ( 1) ( 1)

f t L L Ls s s

− − −⎡ ⎤ ⎡ ⎤⎡ ⎤= + +⎢ ⎥ ⎢ ⎥⎢ ⎥+ + +⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2( ) (1 ) ; 0tf t t e t−= + ≥When there are multiple roots response has explicit time multiplication term

Solving LTI differential Solving LTI differential equationsequations

1.1. Convert the given D.E in to an Convert the given D.E in to an algebraic equation in algebraic equation in ‘‘ss’’ by applying by applying LaplaceLaplace transform to each term in transform to each term in the D.E.the D.E.

2.2. Obtain the expression for the laplace Obtain the expression for the laplace transform of the dependent variable.transform of the dependent variable.

3.3. By taking laplace transform using By taking laplace transform using partialpartial--fraction method we get time fraction method we get time response of the dependent variable.response of the dependent variable.

ExampleExample

What is the solution of the D.E What is the solution of the D.E

The The LaplaceLaplace transform of dependent transform of dependent variables with initial conditions are given byvariables with initial conditions are given by

.. .3 2 0,x x x+ + =

.(0) 1, (0) 2x x= =

.. .2( ) ( ) (0) (0)L x t s X s sx x⎡ ⎤ = − −⎢ ⎥⎣ ⎦

.( ) ( ) (0)L x t sX s x⎡ ⎤ = −⎢ ⎥⎣ ⎦

[ ( )] ( )L x t X s=

Example (Example (contdcontd))

Now using these relations and solving for Now using these relations and solving for X(sX(s) ) from the D.E we getfrom the D.E we get

By partialBy partial--fractions we getfractions we get

By taking inverse laplace transformBy taking inverse laplace transform

2

5( )3 2

sX ss s

+=

+ +

4 3( )1 2

X ss s

= −+ +

1 1 14 3( ) [ ( )]1 2

x t L X s L Ls s

− − −⎡ ⎤ ⎡ ⎤= = −⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

2( ) 4 3t tx t e e− −= −

Open-loop vs closed-loop Transfer Function

)(1)(sG

sGCLTF+

=

kcsmssG

++= 2

1)(

)(sGu y )(sG∑ yur +

-

)(sG∑ yu+

-

r )(sCcontroller plant

)()(1)()(sCsG

sCsGCLTF+

=

Unity feedback

All manipulationsalgebraic in s domain

Example

Open-loop vs closed-loop Transfer Function

)()(1)(

sGsHsGCLTF

+=

)(sGu y )(sG∑ yur +

-

Unity feedback

All manipulationsalgebraic in s domain

)(sHAnother configuration

PD control of massPD control of mass

The equation describing the simple mass system is,The equation describing the simple mass system is,where where ‘‘FF’’ is the input to the plantis the input to the plantand and ‘‘xx’’ is the output to the systemis the output to the system

‘‘rr’’ is the reference we need the system to followis the reference we need the system to followWith spring and damper, the system became PD controlledWith spring and damper, the system became PD controlledand now the system equation is,and now the system equation is,

(Refer block dia(Refer block diagram of gram of this systhis system from tem from previousprevious slides) slides)

../x F m=

...( )p dk r x k xx

m− −

=

x Fm r

2

( ) 1( )( )

X sG sF s ms

= =

Open loop transferfunction

let us apply laplace transforms on both sides,let us apply laplace transforms on both sides,

.. ..d p pm x k x k x k r⇒ + + =

2 .( ) ( ) ( ) p

d p

r kms X s k sX s k X s

s+ + =

2

.( )

( )p

d p

r kX s

s ms k s k⇒ =

+ +

PD control PD control

NoteStep input

Closed loop transferFunction???

To track the output, let us apply laplace transforms onTo track the output, let us apply laplace transforms onboth sides,both sides,

.. .. .d p p dm x k x k x k r k r⇒ + + = +

2 ( ) ( ) ( ) ( ). ( )d p p dms X s k sX s k X s R s k k sR s+ + = +

2

( )( ) ( )

p d

d p

k k sX sR s s ms k s k

+⇒ =

+ +

PD control PD control

Notedifference

Closed loop transferFunction: Poles and Zeros??

Now simply after substituting the values of the parameters,Now simply after substituting the values of the parameters,we can partially divide the above equation and apply inversewe can partially divide the above equation and apply inverseLaplace transforms to get the output time domain response,Laplace transforms to get the output time domain response,Which proves the simplicity of the method.Which proves the simplicity of the method.

Now let us go for PID controlling. The equation of theNow let us go for PID controlling. The equation of thesystem with PID controller is given bysystem with PID controller is given by

. ..( ) ( 0) ( )p d ik x r k x k x r dt m x− − − − − − =∫

PID controlPID control

Motor Dynamics

Armature Free Body Diagram

Electrical circuit Diagram

Motor with PD control

)(sGV θ)(sG∑ V+

-

r )(sCcontroller plant

)()(1)()(sCsG

sCsGCLTF+

=sRKKsBsJ

KsGatmm

t

)/()( 22 φ

φ++

=

Open loop motor

θ

skksC dp +=)(

Proportionalpart

Derivativepart

Motor position feedback

Performance Specs

Effect of PD gains on poles and zeros of the system.

We will discuss this first by taking only PD We will discuss this first by taking only PD control.control.Consider the plant whose open loop poles Consider the plant whose open loop poles are s=0 and s=are s=0 and s=--a (a (what is the physical sys?what is the physical sys?If a PD control is used with this plant it If a PD control is used with this plant it looks like:looks like:+

_

C(t)R(t) 1( )s s a+p dk k s+

Effect of PD gains on poles and zeros of the system (contd).

Now the closed loop transfer function of the system Now the closed loop transfer function of the system becomesbecomes

The poles of T are given byThe poles of T are given by

So depending on the gain values of the PD controller So depending on the gain values of the PD controller the poles can be positive real, negative real, complexthe poles can be positive real, negative real, complexWe already know how system with various closedWe already know how system with various closed--loop loop pole locations in spole locations in s--plane behaves.plane behaves.

2

1,2

( ) ( ) 42

d d pk a k a ks

− + ± + −=

2 ( )d p

d p

k s ks k a s k

+

+ + +

Consider a=2, i.e the open Consider a=2, i.e the open loop system is stable.loop system is stable.

Now choosing Now choosing kpkp=4,kd=1 gives impulse =4,kd=1 gives impulse response (because of resulting complex response (because of resulting complex closed loop poles).closed loop poles).

σ

S=-1.5-j1.323

S=-1.5+j1.323

S=-4

jw

Consider a=2, i.e the open Consider a=2, i.e the open loop system is stable.loop system is stable.

And choosing kp=4,kd=5 gives impulse And choosing kp=4,kd=5 gives impulse response (because of resulting response (because of resulting ––veve real real closed loop poles).closed loop poles).

jw

σS=-1

S=-6.37 S=-0.8

Consider a=Consider a=--2, i.e the open 2, i.e the open loop system is unstable.loop system is unstable.

Now choosing kp=4,kd=1 gives impulse Now choosing kp=4,kd=1 gives impulse response (because of resulting right half sresponse (because of resulting right half s--plane complex closed loop poles).plane complex closed loop poles).

jw

σ

S=1-j1.732

S=1+j1.732

S=-4

Consider a=Consider a=--2, i.e the open 2, i.e the open loop system is unstable.loop system is unstable.

And choosing kp=4,kd=5 gives impulse And choosing kp=4,kd=5 gives impulse response (because of resulting response (because of resulting ––veve real real closed loop poles).closed loop poles).

σ

S=-1.5-j1.323

S=-1.5+j1.323 jw

S=-0.8

ObservationsObservations

Thus we see that by properly choosing the Thus we see that by properly choosing the gains of PD controller a system that is gains of PD controller a system that is unstable in open loop can be converted to a unstable in open loop can be converted to a stable system in closed loop. stable system in closed loop. This is because the PD control gains effect the This is because the PD control gains effect the poles and zeros of the system.poles and zeros of the system.Similarly in case of PID controllers.Similarly in case of PID controllers.Now how to find for what set of values of Now how to find for what set of values of these gains is the closed loop system stable or these gains is the closed loop system stable or unstable????unstable????A graphical approach can easily determine A graphical approach can easily determine this.this.

Stability for higher order Stability for higher order systems?????....systems?????....

It It waswas difficult to find the poles and difficult to find the poles and zeros of a higher order systems when zeros of a higher order systems when classical control was developed. There classical control was developed. There was no was no MatlabMatlab or computersor computersAny other alternative for finding Any other alternative for finding system stability in such cases???..system stability in such cases???..

Routh stability criterionRouth stability criterion

It tells the number of closed loop poles It tells the number of closed loop poles lying in the right half slying in the right half s--plane with out plane with out actually solving for them.actually solving for them.This can be found from the coefficients This can be found from the coefficients of the characteristic equation.of the characteristic equation.The necessary condition for stability is The necessary condition for stability is that all coefficients be present and are that all coefficients be present and are positive. positive.

ProcedureProcedure

Write the characteristic equation of the Write the characteristic equation of the closed loop transfer functionclosed loop transfer function

as,as,

11

11

.....( ) ( ) ;( ) .....

m mo m

n no n

b s b s bC s G s m nR s a s a s a

−

−

+ + += = <

+ + +

ProcedureProcedure

Now evaluate bNow evaluate b’’s (the bs (the b’’s here are different s here are different from those in the transfer function) as from those in the transfer function) as follows:follows:

The evaluation of bThe evaluation of b’’s is continued until the s is continued until the remaining ones are all zeros.remaining ones are all zeros.We observe that bWe observe that b’’s are found by cross s are found by cross multiplying the coefficients in the previous multiplying the coefficients in the previous two rows.two rows.

1 2 0 31

1

a a a aba−

= 1 4 0 52

1

a a a aba−

= . . .

ProcedureProcedure

Now update the array with the values Now update the array with the values of bof b’’s founds foundAnd similarly find And similarly find cc’’ss, , dd’’ss , , ee’’ss …… by by cross multiplying the coefficients in the cross multiplying the coefficients in the previous two rows.previous two rows.Process is repeated until nth row has Process is repeated until nth row has been completed.been completed.

Routh stability criterion Routh stability criterion statesstates……

The number of poles of the closed loop The number of poles of the closed loop transfer function lying in the right half stransfer function lying in the right half s--plane = The number of sign changes in the plane = The number of sign changes in the first column of the array found.first column of the array found.Therefore the necessary and sufficient Therefore the necessary and sufficient condition for the closed loop system to be condition for the closed loop system to be stable is that all the coefficients of the stable is that all the coefficients of the characteristic equation be present and characteristic equation be present and positive and all the terms of the first column positive and all the terms of the first column of Routh array are of same sign.of Routh array are of same sign.

Application to control Application to control system analysissystem analysis

What is the range of values of What is the range of values of ‘‘KK’’ for which for which the closed loop system is stable?the closed loop system is stable?

Now the closed loop transfer function is,Now the closed loop transfer function is,

2( 2)( 1)K

s s s s+ + +

R(s) +

-

C(s)

2( 2)( 1)K

s s s s K+ + + +

2( 2)( 1)K

s s s s+ + +

SolutionSolution

The characteristic equation is The characteristic equation is

The array of coefficients become,The array of coefficients become,

For stability For stability ‘‘KK’’ must be +ve and all the terms must be +ve and all the terms in first column be +ve. Soin first column be +ve. So

4 3 23 3 2 0s s s s K+ + + + =

14 09

K> >

where where

now the case is similar to that of previous one, applying now the case is similar to that of previous one, applying Laplace transforms on both sides produces,Laplace transforms on both sides produces,

.. .

d p im x k x k x k xdt C⇒ + + + =∫p iC k r k rdt= + ∫

2( ) ( )id p

k Cms k s k X ss s

+ + + =

PID controlPID control

Next classNext class

Now, we can easily calculate the output time domain Now, we can easily calculate the output time domain response from above equation as described earlier.response from above equation as described earlier.

(Example for finding time domain response from equation (Example for finding time domain response from equation as above is already given in the partial fraction method as above is already given in the partial fraction method slides )slides )

3 2( ). ( )d p ims k s k s k X s C⇒ + + + =

3 2( )( )d p i

CX sms k s k s k

⇒ =+ + +

PID control (contdPID control (contd………….).)