 Professor Vivek Satay of WIT, Solapur. Already we have discussed four videos on the stability criteria for control systems. Now today is the fifth video in which we are going to discuss type two error that we encounter in preparing a roast table. And that is very important because after finishing this, now we will be comfortable to solve all the problems in this chapter as well as in the root locus analysis. As usual, there is a learning outcome, means when your session is over, you will be able to identify type two difficulty and analyze the same. What is the meaning of analyze? Analyze means you have to make a comment, system is stable, unstable, marginally stable, and so on, okay? Now, first of all, I will say see that type one error that we have already discussed, we have already discussed and you have learned it, I will just revise it. In type one error, we have seen that necessary condition is satisfied by the equation, that is it has, it is a complete polynomial, and there are signs of the, all the terms are also positive or negative, that is same sign. But as it is more than two degree equation, there is no necessity that it is a sufficient condition. So we have to prepare a roast table because we cannot factorize the equation of a higher degree. So there is a method for it, that is called the roast table. So if preparing the roast table, the procedure is known to you, odd number terms we write as the first row, then even number term is the second row, and there is a procedure that is taking the element in the first column, second element as the first pure element, we complete the roast table. But in the preparation of roast table, if we come across a zero, then it becomes very difficult to complete the next rows because for the next row that zero is a pure element, which is to be divided. That is, that comes in the denominator, and zero divided error is a big issue in mathematics. Zero divided error is a big issue. So how to resolve it? Though there are two ways we have seen, we can replace this zero by an epsilon, which is a very small positive number, and work with epsilon. Then in the first column, we'll get some numbers and some epsilon terms, and we take the limit of that as it tends to say zero, and if that terms comes as negative, then we count the number of sign changes in the first column. If the number of sign changes are there, it must be because the system is unstable. Always remember, there must be some sign change. As you have encountered with type one difficulty, now type one difficulty means we have encountered a zero in the first column, as well as at least one element in that row contains one non-zero element. That is very important. That is very important because today we are going to study, learn the type two error in which we get first element zero in the first column. That is some element in the column one as zero. But additionally, all the row elements are zero. But that is not the case with type one. In the type one, only first column element is zero, and all other elements are not simultaneously zero. At least one is non-zero. So this is very important. You must fit in your mind. At least one element must be non-zero. But for type two difficulty, all the elements must be zero. So that was our criteria. And when type one difficulty arises without any problem, we say that system is unstable. Why it is unstable? Because the roots are going to be on the right-hand side. And how many roots? If it is a complex pair, two roots. If it is a complex quadruple, four roots. If it is a real number, real root, two or one, depending upon whether I get a my sign change in the, say, very last column, first column or not. So there will be two changes, two roots will be there. And most probably we'll get that roots as a complex conjugates. And they are on the right-hand side of the S plane. So system is guaranteeing unstable. Now, we are taking today a different problem. Now, this problem is a fifth-order polynomial, which is given to you. Now, if I ask you, check whether the problem satisfies both the necessary conditions. You'll say that, yes, sir, no problem. It is a fifth-order polynomial. All the terms from 5, 4, 3, 2, 1, 0 are present. And all the coefficients have the same sign. So it is an equation which is eligible for necessary condition. And sufficient condition, we cannot take, say, immediately. We have to prepare it able. We have to see that whether one difficulty arises or not. Then we start preparing. Once again, I do the revision for you. All number terms are written over here. All number terms, this is the first term, third term, then one, first, second, third, fourth. This is the fifth term. And this is the second term. This is the fourth term. And this is the sixth term. So all number terms are 187. And even number terms are 487. So these two, that we get from the data. This we get from the data. That is a given equation. Now with the help of these two, we form the third row. How to find the third row? Once again, I have a small revision again. This is my POD element. This is my POD element. So this A4 into 8 is 32. Minus 1 into 8 is 8 divided by 4. So 32 minus 8 is 24 divided by 4 is going to be 6. So I get 6 here. Similarly for this, 4 into 7 is 28. Minus 4 is 24 divided by 4. Again 6. So I get 6 and 6. Then next third row I have to obtain. Now for this row, fourth row, this is the first row. This is second row. So this is the POD element. So 6 into 8 minus 6 into this divided by 6 is 1. And this is also 1 that we get. And next row, this is the first row. This is second row. And if I do this, I will get 6 minus 6, 0. And for this, there is no term, so it is 0. So all the terms in this row are 0. Now see here, what is important for type 2 difficulty? In type 2 difficulty, we must come, must get a situation in which entire row is 0. So my this entire row is 0. And in type 1 difficulty, only this is 0 and at least one is non-zero. Okay. Now if this entire row is 0, this is called as type 2 difficulty. But there is a different, say, estimation or there is different, say, command that we have to put when the system has got type 2 difficulty. As we said, the type 1 difficulty ensures that system is guaranteed unstable. Listen, type 1 difficulty is guaranteed unstable. Type 2 difficulty is also unstable, no doubt. But there is a possibility that system may become marginally stable. So marginal stability, marginal stability is an important aspect of a control system which is unstable, no doubt, but it is still stable. For example, something, for example, in our IC engines, cam is, say, moving the valves. It is oscillating, is it not? But that oscillations are fixed. Means it is oscillating. So there is continuous variation in the output, but output variation has got certain fixed order. So that is called as a stable system. But in a control architecture, if we require a system in which oscillations are required there, then it is a stable system. So I require oscillation of fixed frequency and I know that that is my stability. So whenever we encounter type 2 difficulty, we have got, we have got marginally stable system. Now what is the meaning of marginally stable system? Means there are roots lying on the imaginary axis. So whenever I draw the pole zero plot, say real axis and imaginary axis I show over here. So my roots will lie here. There may be pair of roots here. Or the situation may come which is very funny that I will get four roots like this. That is minus a plus or minus b like this. So these roots are called as quadruples. Means it is minus a plus bi, minus a plus bi, minus a minus bi, plus a plus bi, plus a minus bi. So this is a mirror image of this. This is a mirror image of this and this is a mirror image of this. So this is called as a quadruple. So these are the possibilities that I can get when I have type 2 difficulty. Out of this whenever my roots lie on the imaginary axis I know that system is going to be marginally stable. Now the problem is how to solve it. So that is very simple method. You identify the row which is having all the zeros. Then look at this row, a row above this row. So this is this row. So what are the coefficients one and one? What is the power here two? So I will get in equation s square for this and plus one for this because here powers change by two. If it is a fifth power, this is a third power term, one power term. If it is a fourth power term, this is second power term, this is zero power term. So if it is a second power term, this is zero power term. So I will get s square plus one is equal to zero. This is called as auxiliary equation. This is called auxiliary equation. So you differentiate this auxiliary equation with respect to s. I will get two s is equal to zero. Now you replace the coefficient of this s which is two by this and there is no constant term. So zero will be there as it is. Zero will be there as it is. Now this is two here and zero here. Now you complete your table. Now you can complete because there is no problem in it. After completing your table, you can say see that all the roots are not lying on, no root is lying on the right hand side. Means roots are lying on the imaginary axis and our roots are on the left hand side. So because of this, if the roots are there on the imaginary axis, system is marginally stable. So we make an important comment that it is marginally stable. But when it is marginally stable, means if it moves to the right, it is unstable. If it comes to the left, it is stable and on this it is on the word of instability. Then what is the frequency for that? If I want to find out the frequency for that, so on this axis I got omega. So frequency is obtained by substituting this equation equal to zero. So I will get s square is equal to minus one, s is equal to plus or minus i. So plus or minus i means one. So one is my radian per second is my frequency. Now see what we have seen. Type two difficulty arises when we get entire row zero in the row stable preparation. Then that is resolved by takes looking at the row just above it. And that row is always in even power row. It never comes to the non-power row that is the mathematics there. Then we form an auxiliary equation which is s square plus one. We take its first derivative that becomes two s plus zero and we replace the coefficients of the row where the zeros were obtained by these coefficients that is two and zero. And then we further complete our set out stable and do the analysis. But always remember with type two difficulty system is marginally stable. And for marginal stability we can find out the frequency and that is obtained by equating the auxiliary equation equal to zero. And the root of that particular equation that is s square is equal to minus one that is plus or minus i. I get the value of one. So one radian per second is the frequency for that is the critical frequency or natural frequency of this particular system. And this is required for us to draw the root locus in our root locus chapter. So I think with this five videos we have got a clear cut idea about discussing the transient behavior then prepared necessary condition, sufficient condition for a quadratic or up to quadratic. Then if it is a polynomial of more degree we have to satisfy the sufficient condition. And for sufficient condition it is not possible to always calculate the roots or factorize the particular equation. So we have used the root stable. Root stable may encounter two difficulties, type one difficulty. In type one difficulty first element is zero in the first column somewhere. And all other elements are not necessarily zero at least one is non-zero. Then we replace zero by epsilon or we replace s by one upon z and do the regular thing. Type one difficulty systems are always unstable. Then sometimes we come across type two difficulty where entire row becomes zero. Then we see the row above it which is always a even powered row. Then we form a polynomial which is known as auxiliary polynomial. We take its first derivative and its coefficients are replaced by the row of zero zeros and we prepare a table. So when a system is having a type two difficulty chances are there that system is marginally stable and its frequency is obtained by equating the called auxiliary equation equal to zero and solving for s. So this is all about our say transient system. I think you have got a clear cut idea about this. Now say those who are interested you can say refer to the references that are given in say this particular say slide, Ravan, Bakshi and Barapatte. Thank you for your patience listening. Good day. Thank you.