 I am Professor Vivek Sartain of Department of Mechanical Engineering, Valshan Institute of Technology, Sulema. Already, we have finished two sessions or two videos in which we have learnt the necessary condition and sufficient condition for the correct characteristic equation to be stable. And we have categorically mentioned that for any mathematical point of view, for any mathematical point of view, you must have both the things satisfied, necessary as well as sufficient condition. Now today, we are going to learn the Routh stable and analyze the given characteristic equation using that particular set table. So at the end of the session, you will be able to prepare Routh stable. I am sure that you can do it. And from that, you can draw the conclusion also. And after that, we are going to discuss about the various errors that are involved in the Routh stability criteria. Now you can see, today's task is preparation of Routh stability table or Routh stable. We have seen that necessary condition and sufficient conditions are same for any polynomial. The only thing is that with quadratic, up to quadratic, not with quadratic, up to quadratic, necessary condition is also sufficient condition. Necessary condition is also sufficient condition because we have taken example of a quadratic. Because in a quadratic, advantage is that whenever we write a quadratic equation, if the signs of all are positive or negative, roots are always going to be negative. That is, they are on the left side, guaranteed. You take any quadratic. And if the roots are complex, they will be on the left half of the S plane. So necessary and sufficient condition is satisfied that its quadratic is complete and roots are on the left side. Whether they are real or complex, it does not matter, but system is going to be stable. Then we have learned a cubic equation in the previous video. And for the cubic, initially we were having some trouble because we saw that a cubic equation was given and it was a complete equation. This contained all the terms of q, square, s, s to 0 and sorry s square and s and s to 0 term. But the problem is, problem was that was not a stable system. And how we could identify? We identified it by finding out its factors. And after finding the factors, we obtained the roots. The possibility for the roots for a cubic are various. All the three may be real. All the three may be real. One may be real. Two may be imaginary. Or one is real. Two imaginary roots may be same. Then there may be two real roots same. One is another third real root. Out of this, whatever the commission is there, whatever the values of roots we get, we are comfortable to call it stable when all the roots are on the left half of the S plane. But for that particular problem that we had discussed, the polynomial was complete. But in the sense, it contained cubic term, square term, ss to 1 term and ss to 0 term. And another problem, necessary condition was that all the coefficients must have same sign. So it also satisfies that condition. So when my equation or a polynomial, which is greater than degree 2, see what is important word, greater than degree 2 is not guaranteed stable if it satisfies necessary condition. So this is what we are going to learn today. In the previous video, we have tested it that yes, we are getting roots on the right hand side. Then today we are learning the basis for it. And if that is the case, how to do it? Now just imagine it was a cubic equation, we could factor it, suppose it is a six dollar equation. How to factor it? It is a big task. If it is a 15 dollar equation, how to factor it? So it is difficult. So for a 15 dollar equation, if I ask you that check whether the system is stable or not, then what you will do? You will factor it. You will get all the 15 roots, then we will plot them on the paper and see that all the roots are coming on the left hand side and therefore system is stable. So that is a very big task. So even if you do it computationally with the computer, it takes a lot of computational effort. And always remember, some things which can be done elegantly should be done elegantly. That is called as a way of solving a problem. Otherwise by say enumeration we can solve the problem, means just go on finding out the factors and take the values or suppose the range is given to you, put on the values of that particular expression and check whether we get the particular say value positive or negative, but that is not the way to solve the problem. So Routh developed, Routh developed a table and he made it very necessary condition that your equation must be, must satisfy the necessary condition and always remember, we prepare Routh's table, even the system is stable, not stable or marginally stable because from the Routh's table we get the idea whether my system is marginally stable or not. By finding out the values of roots on the left hand side, whether they are there or not, I don't get an idea whether my system is going to be marginally stable or not or what is the critical frequency of vibration for the system that is allowed. But if I prepare Routh's table, I can get a better idea. So how the Routh's table is prepared? So first thing is he says that write the equation that is given to you a polynomial equation in decreasing order, in decreasing order means what? So if it is a, for your understanding I have taken 8th order polynomial. You write here 8th order polynomial, 8th order here then 7th, 6th, 5th, 4th, 3rd, 2nd, 1 and 0. Now remember I may give you a numerical or a problem in which some part is missing, some part is missing. So as far as your analysis is concerned, you may say that sir as there is one term is missing, the system is going to be unstable, okay no problem, system is unstable, I agree. But you do the table, complete it and then we can see that yes if it is unstable whether it is a marginally stable or unstable. So that idea we can get only with the help of Routh's criteria. So for that Routh says that you write the particular say equation in the descending order. Then what to do if any one term is missing, you write 0 for that, you write 0 as a coefficient for that. Suppose S dash to 6 term is missing, you write a0, S dash to 5, a1, S dash to 7, 0, S dash to 6 and further. But assume that this is given to you. Then what is the stable structure, then you write here the powers of S in the decreasing order. So S dash to 8, S dash to 7 and so on we move in this way, 8, 7, 6, 5, 4, 3, 2, 1 and 0. Then what we do, what we do as this is a polynomial, it is a complete polynomial whether it is not complete, the given problem is not complete, we are making it complete by making the coefficient 0, that is mandatory. So as far as this equation is stable is concerned, this is my complete polynomial. Some terms may be 0, say coefficient, suppose this constant term is not there, so we have to write that 0, suppose S dash to 4 coefficient is not present, S dash to 4 term is absent, you write 0 into S dash to 4. So for me all the terms are present. So if it is 8th order equation, 1, 2, 3, 4, 5, 6, 7, 8 and 9 terms are there, that is simple. So for a quadratic we have got 3 terms, for a cubic we have got 4 terms, for a quartic we have got 5 terms, similarly nth order polynomial we will have n plus 1 terms. So 8th order polynomial we will have 9 terms, is it not? So he says that after writing this in the vertical direction, S dash to 8 to S dash to 0, what we have to do, we have to first write the coefficients of odd number terms. This is 1st term, this is 3rd term, this is 5th term, this is 7th term and this is 9th term. See I have underlined with blue color the terms, see I have not taken the looking at the power, power may be even, power may be even, but this is odd term, why odd term? Because if I write here, this is my 1st term, this is my 2nd term, this is my 3rd term, this is my 4th term, this is my 5th term, 6th, 7th, 8th and 9th term. So out of this 9 terms, how many odd terms I have got? 1, 3, 5, 7 and 9, 1, 3, 5, 7 and 9. So what I will do as a first case, I will write here in the first line, the all odd term number coefficients a naught, then a 2, then a 4, then a 6, 1, 2, 3, 4 and a 8. So this is my 1st term, 3rd term coefficient, 5th, 7th and 9th. Then what I tell you, then you write down the even number terms, then you write the even number terms, so these are the even number terms, I will make them circle even number terms. So they are a 1, a 3, a 5 and a 7, when it is not there, you can just put a dash or you can take it at 0 for your calculation. Then I will say that if you prepare this, you make the next row by multiplying this, subtracting this and divide it by this pure. So the first element in this row, first element in this row will be a 1, a 2, minus a naught, a 3 divided by a 1. For this a 1, a 4, minus a naught, a 5, a 1, a 4, minus a naught, a 5 divided by still a 1. So likewise you can prepare that a 1. Then we will call these numbers as b 1, b 2 and so on. Then for our next table, this is our first row and this is our second row. For next, this is first row and this is second row and you can proceed in this particular way. So you can very easily prepare the row stable and do the calculations. So this is just say structure I have given, very simple. First write odd number terms, coefficients, then even number coefficients. Then for this third row, a 1, a 2, minus a naught, a 3 divided by pure element. For second term, a 1, a 4, minus a naught, a 5 divided by a 1. Third, a 1, a 6, minus a naught, a 7 divided by a 1 and likewise. What will happen? The last answer, s 0 will be always equal to a 8. Remember. So this is your check that you can make. If you get this a 8 here, then your table preparation is correct. Otherwise, there is some mistake in the calculation. So this part is important in our further discussions because we are going to discuss how to do the analysis of this particular table and what are the comments that we can make on Routh's table. Now you have got a mastery over this particular analysis. If you want to go for further reading, you can go for Ravin, Buxian, Barpatet. Thank you for patience listening. In the next two videos, we will discuss type 1 and type 2 errors that we come across in the stability analysis. Thank you.