 I am Professor Vivek Sartain of Balchan Institute of Technology, Solapur. In the previous three videos, we have discussed about the stability criteria. In the very first video, we have seen that necessary condition is given as in a two form, to say two characteristics. The first characteristic was it must be a complete polynomial, that is the first necessary thing and second necessary thing was that all the coefficients must have same or similar sign, it may be plus or it may be minus, so that is necessary condition. And also we have seen that this necessary condition is also sufficient condition up to quadratic, so that already you have learnt. So, any quadratic we can take, if it satisfies this condition, then it is also a sufficient condition for it, but the problem is not only quadratic, now we go for the cubic equation. So, we encounter a cubic equation, which was a complete equation and at the same time, it had say all the coefficients of the terms positive, but the roots were not on the left half of S plane, so necessary condition and sufficient condition are to be satisfied and then only we can make a comment. Now what happens, when I have got a polynomial of 4 degree, 5 degree, 7 degree or higher degree polynomial, it is very difficult for me to factorize it and identify all the factors and then check it, whether they are on the left half side of or lying on the imaginary axis and then make a comment on the stability, so instead of that, there is a very elegant method that is known as the Routh's table. So, the preparation of Routh's table we have seen in the last lecture, that we have to write the equation in the descending order, then we have to prepare a vertical column of the variables power or order from highest order to the lowest order that is as to 0, that is a constant term, then we have to write the odd number terms and then even number terms and then we have to discuss it. But the problem is whenever we prepare Routh's table, if my pivot element becomes 0 by which I divide, in that my case was the, I1 was the pivot element, if that becomes 0 in the calculation, it becomes a 0 divide error and you know that in my 0 divide error, there is a NIN, so computer will hang or we cannot do any calculations, so that is called as a type I error and that error we are going to study in detail. So, in after learning today's, going through today's session, you will be able to identify type I error and then comment on the stability, okay. Now, as you are well versed with the preparation of Routh's table, I have given here a equation which is a sixth order polynomial, I can call this as it is a QS and today we are going to discuss type I error, type I error, right, okay. Now, first of all I will see that whether my equation satisfies necessary condition or not. So, necessary condition is it is a sixth order polynomial, so power of 6, power of 5, power of 4, power of 3, power of 2, power of 1 and power of 0, this is 5, okay, is present and all the coefficients are having same sign, positive, so necessary condition is satisfied, so I will tick mark that yes, necessary condition is satisfied. Now, what about sufficient condition, that is a big question, if I ask you to say factorize this expression, it will take half an hour to factorize this because to factor a quadratic is simple, even cubic is also difficult because we have to use Chinese remainder theorem or synthetic division method to find, but after fourth and fifth order and sixth order or eighth order, it is very complicated or cumbersome process and basically when we discuss about stability, we are not interested in root, we are interested only in making a comment that whether the system is stable or not, okay. So, now I will prepare a table so that you will have again the revision exactly what we have studied in our previous video, so this is my polynomial, I will identify, so S6 is the maximum power I have, so I will write S6, 5, 4, 3, 2, 1, 0, so this is the table preparation that I have written, then my first term, third term, fifth term and seventh term because it is a sixth order polynomial, sixth order polynomial, so there will be seven terms, as there are seven terms, this is the first term, this is the third term, fifth and seventh term, so these are odd number terms, these are odd number terms and these are blue colored terms or even number terms, so this is second, this is fourth and this is sixth, okay, so what we will do now? So the odd number terms coefficients we will write because every term has got a power of S, this 5 is actually 5 into S raise to 0, 5 into S raise to 0 because S raise to 0 is 1, now what I will do? I will write the coefficients of this odd number terms, odd number terms not odd power terms, always students make mistake in preparing our table, they see that what is the odd power, the odd power is 5 but odd term is 1 that is S raise to 6, so we are interested in odd number terms, not odd power terms, suppose it is a fifth order polynomial, this itself is odd term and odd power, odd term as well as odd power, if it is a sixth order this is even power but term is odd, so what is one of the coefficients? 1, 1 and 3, so we will write here, this is 5, I have to write it, 1, 1, 3 and 5, okay, so we will write 1, 1, 3 and 5, second 2, 2 and 4, 2, 2 and 4, this is nothing means dash we can put it, now what we will do? We will use our Rouse criteria, I will explain here how we have done it, so Rouse criteria says that you have to multiply this term, that is the bold line I have made with a blue color, you have to multiply this, then you have to subtract say this red colored dotted line product, so take a product of these two terms, subtract from it the product of these two terms, it is a, sorry, this is first and this is second, okay, so this is the first that we take and this is the second, so you multiply this 2 into 1 and you subtract this and divide it by 2, this is my pure, so what I will get, 2 into 1 is 2, minus 1 into 2 is 2 divided by 2, answer is 0, okay, I will get 0, second term I will calculate for your clarity because you will come to know about the procedure, so this is 2 into 3, what is 2 into 3? 6, then 1 into 4 is 4 divided by 2, so 6 minus 4 is 2 divided by 2 is equal to 1, so I will get 1 here, 0 here, I will take the last term, 2 into 5 minus 1 into 0, there is nothing number here, so 2 into 5 is 10 minus 0 divided by 2, so this is 5, I will get 5 here, okay, now you, so with this simple example you have understood the preparation of Routh's table for this given problem, now what I told you, if this is my first row and second row with the help of these 2 rows, I have evaluated the third row, which is equal to 0, 1 and 5, but now suppose I want to go for the fourth row, I want to go for fourth, this is first, second and third row, this is third row, this is the fourth row, this is my first row and this is my second row, but if I take this as the first and this as second, the pure element is not 0, pure element is 0, so whatever the calculations we do, 2 into 1, sorry, 0 into 2 minus 2 into 1 upon 0, it will be infinity, so everywhere it will be infinity, everywhere it will be infinity, then we cannot complete the Routh's table, okay, so whenever in the preparation of Routh's table, we come across a situation that in the first column, this is my first column of the calculation, this is the first column, second, third, fourth, depending upon how many say, what is the degree of the equation, it will be, it will depend on that, okay, in the first column, if we encounter 0 or in the calculation where 0 comes, then what happens? Routh's table fails, means we cannot go further, we cannot go further, but you know that mathematics is not that way, say, a very simple subject to accept the, say, the repeat immediately, it says that, okay, no problem in limit, in calculus, we handle such situations by using the method of small positive delta or small positive number epsilon, so if it fails, we cannot complete the table, one thing is clear, whenever type 1 difficulty occurs, type 1 difficulty occurs, system is unstable, means this is necessary condition is satisfied, both the condition is satisfied, and in the Routh's table, first column, we are encountering an element 0 here, and these elements are not 0, means at least one is non-zero, so what is the condition, this first column element must be 0, and out of these remaining elements, at least one, see my word is important, at least, at least one element, at least one element, element is not equal to 0, means if suppose this is 0, and this is 5, no problem, if this is 1, and this is 0, no problem, if this is 0, and this is minus 5, no problem, because we are not worried about even minus number also, because our condition is that, all the elements in the row should not be 0, means my type 1 difficulty will be there, only when I get first element 0, and not all elements are 0, at least one element is non-zero, this is called type 1 difficulty, so how to resolve this, mathematics says that, you just delete this, and write in epsilon, write a small number epsilon, so what is value of epsilon, epsilon is a very small number greater than 0, epsilon is not 0, it is not negative number also, it is a positive number, and you work with that epsilon, now what will happen, I will get my some expression with epsilon into 2, in minus 2 upon epsilon, do not worry, so you will get some number here, in terms of epsilon, last number is going to be our constant term that is 5 year, last term, so what will happen, I will get some numbers here, 1, 2, and some term with epsilon, then epsilon square will be there, whatever it may be there, I will get a table, first column with numbers and epsilon, now the question is, what to do with this, then row says that, no problem, you go on substituting epsilon as a very small number, which is a positive number, and check whether you get number here positive or negative, so we work with this, and we suppose, after putting epsilon or taking limit as epsilon tends to 0, if I get this as a negative number, this as a positive number, and this as a positive number, then he says that, just go on changing, say looking at the sign changes, so this is positive, negative, positive, positive, so positive to positive, no sign change, positive to negative, one sign change, then again negative to positive, second sign change, then again positive, so there is no sign change, so when there are two sign changes, now how many sign changes are there, two sign changes, from plus to minus, and minus to plus, so this plus to minus is one change, and this minus to plus is a second change, so there are two sign changes, so row's stability criteria is so elegant it says that the number of sign changes in your column one number of sign changes in column one gives you that so many roots are on the right hand of the s-plane means out of the six roots two roots are there on the right hand right half of the s-plane and four are there on the left half of the s-plane so see without factorizing without factorizing we got an idea that how many roots are there on the right side we are not worried about the roots but we know that these are the roots that are present on the right side so we get a clear picture about the stability of the system remember whenever type 1 difficulty arises the system is going to be unstable so type 1 system is always unstable and it gives you idea that how many roots are there on the right side and how many roots are there on the left side so how many roots are there on the right side the number of sign changes that we have in column number one is the number of roots that are present in the right side so this problem will give the idea that there are six roots out of this four are on the left hand and two are on the right side their system is marginally system is unstable this is one way there is another procedure he says that if you are working with this say epsilon and if you are getting lot of trouble if what you do you put s is equal to 1 upon z you put s is equal to 1 upon z in your equation then you will get a polynomial in z you will get a polynomial in z and then do it regularly as you are done for your regular table because 1 upon z will not take care about the infinity because infinity division problem was there 0 now it will be converted into 0 problem so substituting for s is equal to 1 upon z we can definitely solve the problem in a regular fashion because there will be no change in the nature of roots what is important nature of roots we are not finding out the values nature of roots of the polynomials will not alter it will remain the same so now with this we have got the idea about type one error and type one error gives you the idea that system is always going to be unstable and when the system is going to be unstable we complete our table in such a way that we will get an idea that how many roots are there on the right side and how many roots are there on the left side so even if we know whenever we have encountered a 0 in the first column that system is going to be unstable we complete the roots table by two methods one is called an epsilon method but that is substituting that 0 by replacing that 0 by epsilon and working as usual that is we prepare a table and then we take the limit of the terms where the epsilon is present as epsilon tends to 0 and if that value comes the terms have to be negative then we count the number of sign changes plus plus minus plus plus so plus two minus is one sign change minus two plus is another sign change and number of sign changes will give you that how many roots are there on the right hand side and we can make a comment that system is unstable because type one difficulty is there but out of six roots four roots are there on the left hand side two roots are there on the right hand side this is useful because in the root locus we are going to use this very effectively and we can convert our system into a stable system by either adding in pole or zero in the particular second whole equation that we are going to study in detail in our root locus as well as in bode plot analysis so those who are interested they can refer to this standard books Raven, Bakshi and Barapate and say you can get number of solve problems there so thank you for patient listening in the next video we will discuss about the type two error and that will finish our say transient response analysis for the control systems.