RouthHurwitz Stability Criterion

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Routh-Hurwitz Stability Criterion
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Stability

• in order to know the location of the poles, we
  need to find the roots of the closed-loop
  characteristic equation.
• It turned out, however, that in order to judge a
  system's stability we don't need to know the
  actual location of the poles, just their sign.
  that is whether the poles are in the right-half
  or left-half plane.
• The Hurwitz criterion can be used to indicate
  that a characteristic polynomial with negative or
  missing coefficients is unstable.
• The Routh-Hurwitz Criterion is called a
  necessary and sufficient test of stability
  because a polynomial that satisfies the criterion
  is guaranteed to stable. The criterion can also
  tell us how many poles are in the right-half
  plane or on the imaginary axis.

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Stability

• need to construct a Routh array.

Consider the system shown in the Figure. The
closed-loop characteristic equation is

• The Routh array is simply a rectangular matrix
  with one row for each power of s in the
  closed-loop characteristic polynomial

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Stability
Table 1 Starting layout for Routh array
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Stability
Table 2 Completed Routh array
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Stability
The Routh-Hurwitz Criterion The number of roots
of the characteristic polynomial that are in the
right-half plane is equal to the number of sign
changes in the first column of the Routh Array.
If there are no sign changes, the system is
stable.
Example
Test the stability of the closed-loop system
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Stability
Solution Since all the coefficients of the
closed-loop characteristic equation s3 10s2
31s 1030 are present, the system passes the
Hurwitz test. So we must construct the Routh
array in order to test the stability further.
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Stability
For clarity, we can rewrite the array
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Stability
and now it is clear that column 1 of the Routh
array is

• and it has two sign changes (from 1 to -72 and
  from -72 to 103). Hence the system is unstable
  with two poles in the right-half plane.

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Stability
Special Case

• a zero may appear in the first column of the
  array
• a complete row can become zero

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Stability
Consider the control system with closed-loop
transfer function
Considering just the sign changes in column 1
Routh array will be

• If is chosen positive there are two sign
  changes. If is chosen negative there are also
  two sign changes. Hence the system has two poles
  in the right-half plane and it doesn't matter
  whether we chose to approach zero from the
  positive or the negative side.

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Stability
Consider the control system with closed-loop
transfer function
replace the zero row with a row formed from the
coefficients of the derivative
Routh array will be
There are no sign changes in the completed Routh
array, hence the system is stable.
Differentiate
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Example 1
Construct a Routh table and determine the number
of roots With positive real parts for the
equation
Solution

• Since there are two changes of sign in the first
  columm of
• Routh table, the equation above have two roots at
  right side (positive real parts).

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Example 2
The characteristic equation of a given system is
What restrictions must be placed upon the
parameter K in order to ensure that the system is
stable?
Solution
For the system to be stable, 60 6K lt 0, or k lt
10, and K gt 0. Thus 0 lt K lt 10