Routh Hurwitz CriterionBefore discussing the RouthHurwitz Criterion, firstly we will study the stable, unstable and marginally stable system.
Statement of RouthHurwitz CriterionRouth Hurwitz criterion states that any system can be stable if and only if all the roots of the first column have the same sign and if it does not has the same sign or there is a sign change then the number of sign changes in the first column is equal to the number of roots of the characteristic equation in the right half of the splane i.e. equals to the number of roots with positive real parts. Necessary but not sufficient conditions for StabilityWe have to follow some conditions to make any system stable, or we can say that there are some necessary conditions to make the system stable. Consider a system with characteristic equation:
If all the coefficients have the same sign and there are no missing terms, we have no guarantee that the system will be stable. For this, we use Routh Hurwitz Criterion to check the stability of the system. If the abovegiven conditions are not satisfied, then the system is said to be unstable. This criterion is given by A. Hurwitz and E.J. Routh. Advantages of Routh Hurwitz Criterion
Limitations of Routh Hurwitz Criterion
The Routh Hurwitz CriterionConsider the following characteristic Polynomial When the coefficients a0, a1, ......................an are all of the same sign, and none is zero. Step 1: Arrange all the coefficients of the above equation in two rows: Step 2: From these two rows we will form the third row: Step 3: Now, we shall form fourth row by using second and third row: Step 4: We shall continue this procedure of forming a new rows: ExampleCheck the stability of the system whose characteristic equation is given by s^{4} + 2s^{3}+6s^{2}+4s+1 = 0 SolutionObtain the arrow of coefficients as follows Since all the coefficients in the first column are of the same sign, i.e., positive, the given equation has no roots with positive real parts; therefore, the system is said to be stable.
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