lecture ruth herwitz stability criterion

8/16/2019 Lecture Ruth Herwitz Stability Criterion

1/22

Stability of Higher Order Systems

523

11035

1

1

+++

+=

sss

s

s R

sC )(

)(

)(

)23(

)1(10

)(

)(234

2

2

++++

+=

sssss

s

s R

sC

Un-Stable

Stable


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Routh-Hurwitz Stability Criterion This method yields stability information without the need to

solve for the closed-loop system poles. Using this method, we can tell how many closed-loop system

poles are in the left half-plane, in the right half-plane, and on

the jw-axis. (Notice that we say how many, not where.)

The method requires two steps:

1. Generate a data table called a Routh table.

2. interpret the Routh table to tell how many closed-loop system

poles are in the LHP, the RHP, and on the jw-axis.


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Routh’s Stability Condition

• If the closed-loop transfer function has all poles in the left half of the s-plane, the

system is stable. Thus, a system is stable if there are no sign changes in the first

column of the Routh table.

• The Routh-Hurwitz criterion declares that the number of roots of the polynomial

that are lies in the right half-plane is equal to the number of sign changes in the

first column. Hence the system is unstable if the poles lies on the right hand side

of the s-plane.


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Generating a basic Routh Table

• Only the first rows of the array are obtained from the characteristiceq. the remaining are calculated as follows!


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"#ample$%• &onsider the following characteristics equation'

• (e)elop Routh array and determine the stability of the system.


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"#ample$• &onsider the following system'

• (e)elop Routh array and determine the stability of the system.


8/22

*ind the stability of the continues system ha)ing the characteristic polynomial of a

third order system is gi)en below

• The Routh array is

• +ecause TO changes in sign appear in the first column, we find that two roots

of the characteristic equation lie in the right hand side of the s-plane. Hence the

system is unstable.

"#ample$


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• The Routh table of the gi)en system is computed and shown is the table below!

• *or system stability, it is necessary that the conditions 8 – k >0, and

1 + k > 0, must be satisfied. Hence the rang of )alues of a system

parameter must lie between -1 and / 0i.e., -1 < k < 81.

(etermine a rang of )alues of a system parameter 2 for which the system is stable.

"#ample$3


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"#ample$7


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(etermine the stability of the system ha)ing a characteristic equation gi)en below!

The Routh array is shown in the table!

here

There are TO sign changes due to the large negati)e number in the first column,

Therefore the system is unstable, and two roots of the equation lie in the right halfof the s-plane.

"#ample$7


13/22

(etermine the range of parameter K for which the system is unstable.

The Routh array of the abo)e characteristic equation is shown below!

here

• Therefore, for any )alue of K greater than zero, the system is unstable.

• 8lso, because the last term in the first column is equal to 2, a negati)e )alue

of 2 will result in an unstable system.

• &onsequently, the system is unstable for all )alues of gain 2.

"#ample$9


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Case-"! 4tability )ia Re)erse &oefficients 0:hillips, %;;%1.

• 8 polynomial that has the reciprocal roots of the original polynomial has its roots

distributed the same


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(etermine the stability of the of the closed-loop transfer function!

Table-1! The complete Routh table is

formed by using the denominator of

the characteristic equation T0s1.

• 8 zero appears only in the first column 0the s row1.

• =e#t replace the zero by a small number, >, and complete the table.

• 8ssume a sign, positi)e or negati)e, for the quantity >.

• hen quantity > is either positi)e or negati)e, in both cases the sign in the first

column of Routh table is changes twice.• Hence, the system is unstable and has two #oles in the right half-#lane.

Table-2! shows the first column of Table-1 along with the

resulting signs for choices of > positi)e and > negati)e.


16/22

$%am#le-&! (etermine the stability of the closed-loop transfer

function!

• *irst write a polynomial that has the reciprocal roots of the denominator of T(s).

• This polynomial is formed by writing the denominator of T(s) in re)erse order. Hence,

• The Routh table is

• Since there are T'O sign changes( the system is unstable and has T'O right-half-

#lane #oles.

• This is the same as the result obtained in the pre)ious "#ample.

• =otice that Table does not ha)e a zero in the first column.


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Case-""! "ntire Row is 5ero.

• 4ometimes while maing a Routh table, we find that an entire row consists of

zeros.

• This happen because there is an e)en polynomial that is a factor of the original

polynomial.

• This case must be handled differently from the case of a zero in only the first

column of a row.


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$%am#le-)! (etermine the stability of the system.

The characteristic equation q0s1 of the system is

here K is an ad?ustable loop gain.

The Routh array is then!

*or a stable system, the )alue of K must be!


20/22

$%am#le-)! (etermine the stability of the system.

• 8lso, when K * )( we obtain a row of zeros +Case-"",

• The au%iliary #olynomial( U(s)( is the equation of the row preceding the row of 5eros.

• The U(s) in this case, obtained from the s. row.

• The order of the au#iliary polynomial is always e/en and indicates the number of

symmetrical root #airs


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"#ample$;• &onsider the following characteristic equation.

(etermine the range of 2 for stability.

lecture ruth herwitz stability criterion

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