 I am Mr. Sarvajne Gandhi, working as assistant professor in Department of Mechanical Engineering from Valshan Institute of Technology, Sallapur. So in this particular session, we are going to continue with rules to construct root locus. At the end of the session, students will be able to understand the rules to construct root locus and analyze the nature of root locus. Rule number 6, breakaway point. Breakaway point is a point on root locus where multiple roots of characteristic equation occur for particular value of k. So we have three general predictions about existence of breakaway point. First, if there are adjacently placed poles on the real axis and the real axis between them is a part of root locus, then minimum one breakaway point exists between the adjacently placed poles. For example, the open loop transfer function given to us is as k divided by s into s plus 1. So from this open loop transfer function, we have two poles which are at 0 and minus 1 respectively and the root locus is going to exist between 0 and minus 1. So from each pole, one branch is going to start, correct? So we have two poles, so two branches are going to start and the note is that the branch approaches towards the point from the poles and that point is called as breakaway point. Second, if there are adjacently placed zeros on the real axis and real axis between them is a part of root locus, then minimum one breakaway point exists between adjacently placed zeros. So for this, we have open loop transfer function as k into s plus 2 into s plus 4 divided by s square into s plus 6. So if we try to plot the roots on the s plane, so we are going to have two roots at origin, one root at minus 2, one root at minus 4 and minus 6 respectively. So if we try to see that there are two adjacently placed zeros at minus 2 and minus 4 and in between that we have root locus. So the minimum one breakaway point is going to exist between them and we have to take a note that branch always moves from the point towards zeros. Third point, if there is a zero on the real axis and to the left of that zero, there is no pole or zero existing on the real axis and complete real axis to the left of this zero is a part of root locus. Then there exists minimum one breakaway point to the left of that zero. For that we have one open loop transfer function given to us as k into s plus 4 divided by s into s plus 2. So if we try to plot it on s plane, so we see that the root locus is going to exist between zero and minus 2 and from minus 4 to infinity. So if we apply this third condition then minimum one breakaway point is going to exist to the left of minus 4. Rule number 7, it tells us the intersection of root locus with imaginary axis. So for that we have four steps. In step number 1, we are going to construct the characteristic equation that is 1 plus g of s into h of s equal to 0 from the open loop transfer function that is given to us. In step number 2, we are going to construct the routes array in terms of k. In step number 3, we are going to determine k marginal value. So in this we have two conditions. If the k marginal value is positive, the root locus is going to have intersection with imaginary axis and we need to find the auxiliary equation for that from the row which is immediately above the term involving k. And second case, if the k marginal is negative, the root locus is not going to have intersection with imaginary axis that implies all the routes are going to lie on the left hand side of the s plane and the system is stable. In step number 4, if the k marginal value is positive and we have found out the auxiliary equation of it, so we are going to equate it to 0 and we are going to find the routes out of it. The routes which are found out of auxiliary equation represents the intersection with imaginary axis. Rule number 8, angle of arrival and angle of departure. Angle of arrival is applicable when the routes are complex conjugate 0s. Angle of departure is applicable when the routes are complex conjugate poles. The open loop transfer function given to us is k into s plus 2 divided by s into s plus 4 into s square plus 2 s plus 2. So if we see the number of 0s present r1 and it is going to lie at minus 2, number of poles are 4 and those are going to lie at 0 minus 4 minus 1 minus j minus 1 plus j. So the 0s are represented by a circle which is at minus 2 and the poles are represented by a cross which is at 0 minus 4 and plus or minus minus 1 plus or minus j. So these are the complex conjugate poles that we have drawn. So angle of departure is applicable when we have complex conjugate poles. We are going to consider this complex conjugate pole as a reference and we are going to join all the routes that are present in this s plane to this particular complex conjugate pole. So we are going to start with minus 4 and we are going to join to this complex conjugate pole with dotted line and similarly continue for the others. Fine. It is pole so it is labeled as p. The count is 1, 2 and 3. This is 0 that is represented by z1. So here in this case we need to measure the angle. In this case we need to measure the angle from the real axis in anticlockwise direction. So for this pole p1 we are going to measure the angle from this real axis up to this point. So that represent phi of p1. Similarly for this point 2 we are going to have a line which is parallel to the real axis and from that we are going to measure angle phi p2. Similarly here we have phi z1 and here we have phi p3. So we are going to use basic trigonometry and from that we can find the value of phi p1 as this is 90, this is 90 plus this theta. So we are going to have 90 plus tan inverse of 1 by 1. So it will be 90 plus 45. So it will be 135 degrees. So for phi p2 directly we can see that angle is 90 so it is 90 degrees. And for phi p3 the angle is tan inverse of tan inverse of 1 by 3. So the angle comes out to be 18.43. So from this we are going to do the summation of phi p that is 135 degrees plus 90 degrees plus 18.43 degrees. So the phi p is going to sum up to 243.43 degrees. Similarly phi of z1 is equal to tan inverse of 1 by 1 so that is 45 degrees. So summation of phi of z is equal to 45 degrees. So phi angle is equal to summation of phi p minus summation of phi z. So we have 243.43 degrees minus 45 degrees. So it is going to come 198.43 degrees. And angle of departure is represented as phi of d that is equal to 180 degrees minus this phi. So it is 180 minus 198.43. So it is coming as minus 18.43 degrees. And that is for a complex conjugate pole minus 1 plus j. So that is represented over here with line and since the angle is negative we are going to measure in clockwise direction so the angle is 18. So somewhere it is going to come over here and we are going to have a line over here representing the angle as minus 18.43 degrees. And for this complex conjugate pole the angle will be same but the sign will change. So here the angle will be plus 18.43 so that is angle of departure. If we have complex conjugate zeros then we are going to have angle of arrival that is phi of a is equal to 180 plus phi. Rest all process is same. These are my references. Thank you.