The Routh Stability Criterion

7/26/2019 The Routh Stability Criterion

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The Routh Stability Criterion

Introduction

There are frequently questions about stability that you need to answer. Someof the questions might be.

Is this closed loop system stable?

For what range of gains is this system stable?

How stable is this closed loop system?

There are several techniques that provide information on stability. Two

examples are the root locus and the yquist Stability !riterion.

For the "oot #ocus$

o %laces where the locus enters the right half plane gives information on

stability.

o &ou can calculate gain for points where the locus enters the right half

plane ' giving information on range of gains.

o (amping ratio)s* of closed loop poles give information on relative

stability+ and distance into the left half plane can be ta,en as an

indication of relative instability.

For the yquist Stability !riterion

o The yquist Stability !riterion gives information on stability directly.

o It is possible to calculate the gain necessary for stability using

yquist plots or -ode plots.

o %hase margin and gain margin give information on relative stability.

However+ there is another technique available that can give information on

stability and range of gains for stability. That technique is the "outh Stability

!riterion. The "outh criterion can be used on small systems )Small here means


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systems of low order. &ou probably wouldnt use "outh on systems with /0 poles.*

There are many times that it can provide simple and clear answers to the first two

questions+ and we will examine the "outh !riterion in this lesson.

The "outh !riterion

To apply the "outh criterion+ you need to form an array from the polynomial

coefficients. Then+ after some computations+ you get information that will let you

determine how many roots of the polynomial are in the #H%. 1e can understand

this best with an example.

s32 s22 3s 2 40

#oo,ing at this polynomial+ it loo,s innocent enough. 5ll of the coefficients are

positive integers. &ou might not expect this to be the denominator of an unstablesystem. In fact+ this polynomial can be factored into this form.

)s 2 /*)s2' s 2 6*

The quadratic factor has two roots in the "H%.

1e will wor, with this polynomial+ and apply the "outh criterion to it as an

example. To begin+ we put the coefficients of the original polynomial into an array

)the "outh array*

s3 4 3

s2 4 40

s1

s0

The last two rows are filled in by doing some computation. !all the 747 coloredgreen the pivot element. The computations start by multiplying the two entries

colored green ' the lower 747 ' the pivot element and the 737. !all that result+ "4.

Then you multiply the two entries in blue. !all that result+ "/. Then you divide by

the green 747 ' the pivot element ' and place the result in the leftmost column.

)Since its a 747 here it wont ma,e any difference+ but it wont always be a 747.*

That computation is given by$


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"esult 8 9"4 ' "/:;9%ivot element:

The result is shown below ' with the result forming a new entry in the s1row.

Since there are no more elements in the first two rows+ that ends the

computations for the s

1

row.

s3 4 3

s2 4 40

s1 '<

s0

Finally+ we need to get the elements in the last row. 5s in the preceding row+ therewill really be only one element in this example+ but larger systems will require more

than that. The new pivot element will be the first element in the last row

modified. Thats the 7'


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ow+ what happens if the polynomial is a higher degree. #ets examine a

fourth order polynomial. Here is the polynomial.

s42 40s32 36s22 60s 2 4/>

1e form the array in the same way )using coefficients of every other power in

each row*+ but we now have three entries in the top row ' the coefficients of s4+ s2+

and s0.

s4 4 36 4/>

s3 40 60

s2

s1

s0

1e proceed the same way+ doing the cross multiplication. However+ in this case+ we

need to examine how we get 4/> in the second entry in the s2row.

s4 4 36 4/>

s3 40 60

s2 30 4/>

s1

s0

To get the 4/> in the second entry in the s2row we use the same pivot element.

It will always be the first element in the last row computed )or entered*. In this

case it will be 7407. 1hat we do is multiply the 40 and the 4/>. otice how we

multiply the first and the third column elements. )-efore+ we multiplied the first

and the second column elements.* 1e always use the same pivot element. ext+if there had been an element in the box below the 4/> entry+ we would have

multipled that by 4 ' the very first entry at the top left in the table. The

computations are$

)40x36 ' 4x60*;40 8 30


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)40x4/> ' 34x0*;40 8 )40x4/> ' 0*;40 8 4/>

ow+ we proceed as before+ and get the table below.

s

4

4 36 4/>s3 40 60

s2 30 4/>

s1 3.333

s0 4/>

The computations are$

)30x60 ' 40x4/>*;30 8 )4600 ' 4/>0*;30 8 /@0;30 8 3.333

)3.333x4/> ' 0*;3.333 8 4/>

Here+ there are no sign changes in the first column+ so all of the roots of the

polynomial are in the left half of the s'plane.

Thats basically the algorithm. There are some special cases+ and it is instructive

to examine at least one or tow higher order polynomial examples before we loo, at

those special cases.

Axample with %roblems B Cuestions

!onsider this polynomial.

s42 >s32 @s22 >s 2 30

Form the array as above )using coefficients of every other power in each row*+

with three entries in the top row ' the coefficients of s4+ s2+and s0.

s4 4 @ 30

s3 > >

s2 ??

s1


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s0

1hat is the the value of the term indicated by 7??7 ?

&ou should have gotten that one correct. If not+ go bac, to the point where we

discussed how to compute that coefficient. "emember the following points.

The 7>7 in the second row is the pivot element for the computation.

&our computation should end by saying )/> '>*;> 8 ??.

ow+ there is another question that needs to be answered.

Cuestion 4

Is there another element in the third row ' the s2row?

ow+ we still have not completed the table. &our table should loo, li,e this

one at this point+ and we need to consider what the first term in the penultimate

row is going to be. That is mar,ed as 7??7.

s4 4 @ 30

s3 > >

s2 6 30

s1 ??

s0

%roblem /

1hat is the the value of the term indicated by 7??7 ?

5nd+ we hope you got that one right also. If not+ remember the following.

The 767 in the third row is the pivot element for the computation.

&our computation should end by saying )/0 '4/0*;6 8 ??.


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Then+ answer the next question.

Cuestion /

Is there another element in the fourth row ' the s1row?

Finally+ you need to get the single element in the last row. &ou should have this

table at this point.

s4 4 @ 30

s3 > >

s2 6 30

s1 '/0

s0 ??

&ou should realiDe ' from the preceding examples ' that the last element is =ust

74307. Its the element in the s2row ' at the end.

5nd+ finally+ you get this table$

s4 4 @ 30

s3 > >

s2 6 30

s1 '/0

s0 30

5nd there are two sign changes+ so there are two roots of the polynomial in the"H%.

5 %oint to ote


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The polynomial in the example above can be related to a closed loop system.

!onsider the bloc, diagram below.

5ssume that the gain+ E+ is variable+ and that )s* is given by$

)s* 8 4;9)s 2 4*4:

Then+ the closed loop transfer function is given by$

CL)s* 8 E;9)s 2 4*42 E:

The polynomial in the example is the closed loop denominator for E 8 /G.

ow+ thin, about what happens when E is variable. Then the closed loop

denominator is given by$

s42 >s32 @s22 >s 2 4 2 E

1e can form the "outh array for this polynomial.

s4 4 @4 2

E

s3 > >

s2 64 2

E

s1 )/0'>'>E*;6

s0 4 2 E

The critical element is )4@ ' >E*;6. That element is negative whenever E is larger

than >+ and when that element is negative+ the closed loop system has poles in the

"H%+ and the closed loop system is unstable.


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The interesting and useful conclusion is that the "outh array can be used to

determine gain limits for stability ' at least for systems small enough that you can

do the computations for the "outh array.

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The Routh Stability Criterion