 Namaste! Welcome to the session Stability Analysis and Routh Criteria. At the end of this session, students should be able to apply Routh Criteria and analyze the stability of system. In this session, we are going to see Harwitz Criteria, Routh Criteria, stability analysis using Routh Harwitz Criteria. The transfer function of any linear closed loop system represented by where B and A are constants. Roots of denominator polynomial of a transfer function are called as poles and roots of numerator polynomials of a transfer function are called zeros. So, to find out closed loop poles, the denominator equation of transfer function which is also called as characteristic equation of a system is equated to 0. So, here denominator equation a 0 s raise to n plus a 1 s raise to n minus 1 so on plus n is equated to 0 which is also denoted as f of s. So, moving further, just recall Harwitz Criteria by having a pause here. So, I think you have recalled Harwitz Criteria. So, in Harwitz Criteria, it states that the necessary and sufficient condition to have all roots of the characteristic equation in the left half of the s plane is that the sub determinant d k where k 1 to n obtained from Harwitz determinant h must all be positive. So, we have seen in the last session how Harwitz determinant h is written from a characteristic equation as given here. So, it is written as d k Harwitz determinant h is written and then the Harwitz determinants are calculated as d 1, d 2, d 3, d 4 and so on, but there are disadvantages of Harwitz Criteria. For characteristic equation of higher order system to solve the determinants of higher order is very complicated and time consuming which we have seen in the last session. And for a system to be stable, number of roots located in right half of s plane cannot be judged by this method. For a system to be marginally stable, it is difficult to predict by Harwitz Criteria. So, to overcome the limitations of Harwitz method, a new method is suggested by mathematician Routh called Routh's Criteria also known as Routh-Harwitz Criteria. So, let us see Routh's Criteria. Edward John Routh was an English mathematician noted as the outstanding coach of students preparing for the mathematical triples examination of the University of Cambridge in the middle of 19th century. He also did much to systematize the mathematical theory of machine of mechanics and created several ideas critical to the development of modern control system theory. So, Edward Routh has contributed to Routh-Harwitz Criteria. So, Routh's Criteria also known as Routh-Harwitz Criteria or Routh's Routh suggested a method of arranging the coefficient of characteristics equation in a table. This arrangement is called as Routh's Array or Routh's Table. Let us write Routh's Array or Routh's Table. So, for a particular characteristics equation, the Routh's Table is written as for first row s raised to n, the coefficients are even coefficients a0, a2, a4, a6. Then for the second row s raised to n minus 1, the coefficients are a1, a3, a5, a7. Then for the row s raised to n minus 1, the b1, b2, b3, b4 are calculated by b1 is calculated by a1 into a2 minus a0 into a3 divided by a1. Then b2 is calculated by a1 into a4 minus a0 into a5 divided by a1. Then b3 is calculated by a1 into a6 minus a0 into a7 divided by a1. Then let us see how next row is calculated. So, here c1 is calculated by b1 into a3 minus a1 into b2 divided by b1. Then c2 is calculated by b1 into a5 minus a1 into b3 divided by b1. And c3 is calculated by b1 into a7 minus a1 into b4 divided by b1. So, for the next row, d1, d2, d3, d4 are calculated similarly. But what is stability condition by the Routh's Table or Routh's Array? So, the necessary and sufficient condition for a system to be stable is that all the terms in the first column of Routh's array must have same sign. There should not be any sign changes in the first column of Routh's array. System is unstable if any sign change occurs in the first column of Routh's table. The number of sign changes in the first column indicates number of roots in the right half of the s-plane. Now, let us see stability analysis using Routh's Herwitz criteria. So, let us see for a given equation is the system stable. So, let us find out the coefficients of this characteristic equation. So, by observing this equation, we have coefficients as a01, a18, a2, 12 and a34. So, let us write the Routh's array. So, here we have s cube. So, for the s cube row, the coefficients are a0, a2 as 1, 12 and there is no further element. So, we have written 0 here. So, for the second row for s square, the elements are 8 and 4 have as a1 and a3 and as there is no further element. So, here we have written 0. Then for the next row s raise to 1. So, here we need to calculate the element. So, which is nothing but 8 into 12 which is 96 minus 1 into 4 minus 4 divided by 8. So, it gives us 11.5. Then next element is calculated by 8 into 0 minus 1 into 0 divided by 8. So, obviously, it is 0 and further elements are also 0. So, for the next row s raise to 0, it is calculated by 11.5 into 4 minus 8 into 0 divided by 11.5 which gives us 4. So, further elements after calculation will be 0. So, this is what Routh's table for the given equation. So, you need to calculate the or write the Routh array up to s raise to 0. So, from the first column, we have got the values elements as 1, 8, 11.5 and 4. So, if you observe there is no sign change in the first column. So, what you can calculate? So, what we can conclude here is there is no sign change in the first column. So, the system is stable. So, let us say one more example where we are going to use Routh-Hurwitz criteria for stability analysis. For a given equation whether the system is stable or not. Here we have a equation 1000 divided by s plus 2 into s plus 3 into s plus 5 which is after calculation we will get 1000 divided by s raise to 3 plus 10 s square plus 31 s plus 1 0 3 0. So, for this equation we have the coefficients as a 0 1, then a 1 10, then a 2 31, then a 3 1 0 3 0. For the s cube row the elements will be 1 31 and 0. For the second row s square we have the elements 10 1 0 3 0 and 0. So, in the second row what we can do is we can divide it by 10. So, the element for the s square will be 1 and 1 0 3 and 0. So, for the third row which is s raise to 1. So, 1 into 31 31 minus 1 0 3 divided by 1. So, that will give us minus 72. Then next element is 10 into 0 which is 0 minus 0 divided by 10 that will give us 0. So, further elements will be 0. So, for the row s raise to 0 the values will be calculated by minus 72 into 1 0 3 minus 10 into 0 divided by minus 72. So, that will give us 1 0 3. So, we have calculated up to s raise to 0. So, if you observe the first column here the elements are 1 10 minus 72 and 1 0 3, but there is a sign change from the 10 to minus 72. So, it is changed from positive to negative and there is a further one more sign change from minus 72 to the positive 1 0 3. So, there are two sign changes. So, what we can conclude here for this particular question is there are two sign changes. So, the system is unstable with two poles in the right half of the s plane. So, in this way we can use Roth-Herwitz criteria for analyzing the stability of a system. These are references. Thank you.