Lecture- 8

2
 Routh-Hurwitz Stability Criterion

• It is a method for determining continuous system

stability.

• The Routh-Hurwitz criterion states that “the number of

roots of the characteristic equation with positive real
parts is equal to the number of changes in sign of the
first column of the Routh array”.
 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.
 Example: Generating a basic Routh Table.

• Only the first 2 rows of the array are obtained from the characteristic eq. the remaining
are calculated as follows;
 Four Special Cases or Configurations in the First
Column Array of the Routh’s Table:

1. Case-I: No element in the first column is zero.

2. Case-II: A zero in the first column but some other elements of the row

containing the zero in the first column are nonzero.

3. Case-III: Entire Row is zero

Case-I: No element in the first column is zero.
Example-1: Find the stability of the continues system having the characteristic
equation of

The Routh table of the given system is computed as;

• Since there are no sign changes in the first column of the Routh table, it means
that all the roots of the characteristic equation have negative real parts and hence
this system is stable.
Example-2: Find the stability of the continues system having the characteristic
polynomial of a third order system is given below

• The Routh array is

• Because TWO 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.
 Example-3: Determine a rang of values of a system parameter K for which the
system is stable.

• The Routh table of the given system is computed and shown is the table below;

• For system stability, it is necessary that the conditions 8 – k >0, and 1 + k > 0,
must be satisfied. Hence the rang of values of a system parameter k must be lies
between -1 and 8 (i.e., -1 < k < 8).
 Example-4: Find the stability of the system shown below using Routh criterion.

The close loop transfer function is shown in the figure

The Routh table of the system is shown in the table

Because TWO 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.
 Example-5: Find the stability of the system shown below using Routh criterion.

• The Routh table of the system is

• System is unstable because there are two sign changes in the first column of the
Routh’s table. Hence the equation has two roots on the right half of the s-plane.
Case-II: A Zero Only in the First Column

1. Stability via Epsilon Method.

2. Stability via Reverse Coefficients (Phillips, 1991).
 Case-II: Stability via Epsilon Method

• If the first element of a row is zero, division by zero would be required to

form the next row.
• To avoid this phenomenon, an epsilon, ε, (a small positive number) is
assigned to replace the zero in the first column.
• The value ε is then allowed to approach zero from either the positive or
the negative side, after which the signs of the entries in the first column
can be determined.
 Case-II: Stability via Epsilon Method
Example-6: Determine the stability of the system having a characteristic equation given below;

The Routh array is shown in the table;

Where

There are TWO sign changes due to the large negative number in the first column,
Therefore the system is unstable, and two roots of the equation lie in the right half of the s-plane.
 Example-7: Determine the range of parameter K for which the system is unstable.

The Routh array of the above characteristic equation is shown below;

Where

• Therefore, for any value of K greater than zero, the system is unstable.
• Also, because the last term in the first column is equal to K, a negative value
of K will result in an unstable system.
• Consequently, the system is unstable for all values of gain K.
 Example-8: Determine the stability of the of the closed-loop transfer function;

Table-1: The complete Routh table is Table-2: shows the first column of Table-1 along with the
formed by using the denominator of resulting signs for choices of ε positive and ε negative.
the characteristic equation T(s).

• A zero appears only in the first column (the s3 row).

• Next replace the zero by a small number, ε, and complete the table.
• Assume a sign, positive or negative, for the quantity ε.
• When quantity ε is either positive or negative, in both cases the sign in the first
column of Routh table is changes twice.
• Hence, the system is unstable and has two poles in the right half-plane.
 Case-II: Stability via Reverse Coefficients (Phillips, 1991).

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

distributed the same—right half-plane, left half plane, or imaginary axis—because

taking the reciprocal of the root value does not move it to another region.

• If we can find the polynomial that has the reciprocal roots of the original, it is possible

that the Routh table for the new polynomial will not have a zero in the first column.

• The polynomial with reciprocal roots is a polynomial with the coefficients written in

reverse order.

• This method is usually computationally easier than the epsilon method.

 Example-9: Repeated example-8: Determine the stability of
the closed-loop transfer function;

• First 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 reverse order. Hence,

• The Routh table is

• Since there are TWO sign changes, the system is unstable and has TWO right-half-
plane poles.
• This is the same as the result obtained in the previous Example.
• Notice that Table does not have a zero in the first column.
 Case-III: Entire Row is Zero.
• Sometimes while making a Routh table, we find that an entire row consists of

zeros.

• This happen because there is an even 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.
 Example-10
• Determine the number of right-half-plane poles in the closed-loop
transfer function.

• First we return to the row immediately above the row of zeros and form
an auxiliary polynomial, using the entries in that row as coefficients.

• Next we differentiate the polynomial with respect to s and obtain

• Finally, we use the coefficients of above equation to replace the row of

zeros. Again, for convenience, the third row is multiplied by 1/4 after
replacing the zeros.
 Example-10
• The remainder of the table is formed in a straightforward manner
by following the standard form .

• All the entries in the first column are positive. Hence, there are no
right–half-plane poles.