 Namaste, welcome to the session stability analysis and Harwitz criteria. At the end of this session, students will be able to apply Harwitz criteria and analyze the stability of system. In this session, we are going to see Harwitz criteria and how to analyze stability of a system using Harwitz criteria. The transfer function of any linear closed loop system represented by y of s divided by x of s equal to b 0 s raise to m plus b 1 s raise to m minus 1 plus so on b m minus 1 s plus b m divided by a 0 s raise to n plus a 1 s raise to n minus 1 plus a n minus 1 s plus a n where b and a are constants here. Roots of denominator polynomial of a transfer function are called pulls whereas, roots of numerator polynomials of a transfer function are called zeros. To find out closed loop pulls, the denominator equation of transfer function which is also called as characteristics equation of a system should be equated to 0. So, here the denominator from the above transfer function is equated to 0. So, roots of the characteristics equation are the closed loop pulls of the system which decide the stability of the system. Now take a pause here and recall how pulls of the system on s plane decide the stability of the system. So, I think you have recalled from the last session when pulls are on the left side of the s plane then it is a stable system, when pulls are on the right hand side of the s plane then system is unstable, when pulls are there on the imaginary axis without repeating the pulls then system is marginally stable or critically stable system and there are other conditions there are two necessary, but insufficient conditions for the roots of the characteristics equation to lie in left hand side of s plane for the system to be stable. The conditions are first condition is all the coefficients of polynomial should have the same sign and second condition is all coefficients of the polynomial should present in the equation that is none of the coefficients vanish, but these two conditions are not sufficient to find out a system is stable or not. According to these two conditions find out whether given system is stable or not by observing this equation. So, after observing this equation and remembering the previous condition you can say that system is not stable because here we have negative value all the coefficients are not positive. So, one of the coefficient is negative. So, system does not satisfy the necessary condition for stability. Now, let us say one more example, so for given equation is the system stable you can observe that all the powers of s are there, but s raise to 2 is not there. So, s square is absent so system does not satisfy the necessary condition for stability. So, a system is unstable if all signs of coefficient of the denominator of the closed loop transfer function are not same and if powers of s are missing from the denominator of the closed loop transfer function then the system is either unstable or sometimes it may be marginally stable, but what if all coefficients are positive and no power of s is missing then how to find out whether system is stable or not. So, for that we have Hurwitz criteria. So, what is Hurwitz criteria? Adolf Hurwitz was a German mathematician who worked on algebra, analysis, geometry and number theory. In the field of control system and dynamical system theory Hurwitz derived the stability criteria for determining whether a system is stable in 1895. So, let us see Hurwitz criteria. Hurwitz criteria states that the necessary and sufficient condition to have all the roots of the characteristics equation in left half of s plane is that the sub determinants d k where k may be 1, 2, 3 up to n obtained from Hurwitz determinant h must be positive. So, all the determinants in Hurwitz criteria must be positive. So, for this characteristic equation how to write Hurwitz table? So, Hurwitz table is given as d k is also nothing, but Hurwitz table h is equal to now here first row is written from the coefficients of the equation. So, first row here is having all the odd coefficients a 1, a 3, a 5, a 7 and so on. Then second row of the Hurwitz table is written from all the even coefficients a 0, a 2, a 4, a 6 and so on. The third row of Hurwitz table is written by including 0 first and then copying first row as it is. Then next row is written by entering 0 first and copying the second row as it is and third row the next row is written as including 0 and copying the third row and so on. So, this is how we can write Hurwitz table from the characteristics equation of a system. Now, how to find out different determinants of Hurwitz criteria? So, here d 1 is 1 by 1 matrix of Hurwitz table. So, here you can see d 1 is equal to a 1. Then how to write d 2 determinant 2? It is nothing but 2 by 2 matrix a 1, a 3, a 0, a 2. Then third determinant of Hurwitz is written as d 3 is equal to 3 by 3 matrix. Then fourth determinant of Hurwitz criteria is written as 4 by 4 matrix from the Hurwitz table. Now, let us see stability analysis using Hurwitz criteria. So, for example, find out the stability of the system for given equation. The equation human is s cube plus 8 s square plus 14 s plus 24. So, let us find out the coefficients of equation. So, here you can find out a 0 is 1, a 1 is 8, a 2 is 14. a 3 is 24. Now, we know that Hurwitz criteria and Hurwitz table where first row is including all the odd coefficients from the characteristics equation. Second row including all the even coefficients of characteristics equation. Then third row is written by including 0 first and copying the first row as it is. Then next row is written as including 0 first and copying the second row and so on. So, let us write Hurwitz table for the given equation. So, you can see here Hurwitz table for given equation. So, first row should be all the odd coefficient of the equation. So, a 1 here is 8, then a 3 is 24, there is no a 5, so it is 0. Then second row includes includes all the even coefficients where a 0 is 1, a 2 is 14 as there is no a 4. So, here it is 0. Then to write the third row from this equation, we have to include 0 here, then first row as it is we have to copy. So, in the first row we have 8, 24 and 0, but here we are including 8 and 24 because there are no further coefficient to make the Hurwitz table 4 by 4 matrix. So, from this equation we have Hurwitz table which is of 3 by 3. So, from this Hurwitz table for the given equation, d 1 is 1 by 1 matrix where d 1 is equal to 8, then d 2 is 2 by 2 matrix where the coefficients are 8, 24, 1, 14 and d 3 is the whole table as it is which is 3 by 3 matrix. So, after calculating d 1, d 2, d 3 we have here 8 which is positive, d 2 88 which is also positive, d 3 2112 which is also positive. So, what we can conclude is as all the determinants of Hurwitz are positive, so this system is stable. So, if one of the determinant is negative then system could have been unstable. So, in this way we can analyze characteristics equation of a system and find out whether it is stable or unstable from Hurwitz criteria. These are references. Thank you.