 I am Mr. Sarvajne Gandhi, working as assistant professor in Department of Mechanical Engineering from Balchan Institute of Technology, Solapur. In this session, we are going to study regarding steps to construct root locus and plot the root locus using MATLAB. At the end of the session, students will be able to apply the rules to construct root locus and solve the root locus problems and analyze the nature of root locus. So, we have step number 5 where we are going to find breakaway point. So, for finding breakaway point step one is to have the characteristic equation that is 1 plus g of s into h of s. So, in this case, we have 1 plus k divided by s into s plus 5 into s plus 10. So, if we try to simplify, we have 1 plus k into s square plus 5 s into s plus 10. So, if we further try to simplify, we will get 1 plus k divided by s cube plus 15 s square plus 50 s. So, we have characteristic equation as s cube plus 15 s square plus 50 s plus k equal to 0. So, then we have step number 2 where we are going to take the terms with k on one side. So, we are going to have k as minus s cube minus 15 s square minus 50 s. So, in step number 3, we are going to have d k by d s. So, it will be minus 3 s square minus 30 s minus 50 and you are going to equate it to 0. So, you have 3 s square plus 30 s plus 50 equal to 0. So, if you try to find the roots of this, you are going to have s 1 as minus 2.113 and s 2 as minus 7.88. Out of this, which one is to be selected? Minus 2.11 or minus 7.88. So, you need to see these points on the root locus that you have drawn in step number 2. If you see that minus 2.11 is going to lie somewhere over here. So, it is a part of root locus. So, from that we can say that minus 2.11 is a valid breakaway point and 7.8 lies somewhere over here. So, it is not a part of root locus. So, that point is invalid. Other method to check whether this point is valid or invalid is to put the value of s as minus 2.11 in this step number 2, where we have k. So, if we are going to get the value of k as positive, it implies that that is valid breakaway point. If we are getting the k value as negative, that implies that it is invalid breakaway point. So, if we put the value of s as minus 2.11 in this equation, you are going to get positive value of k and it will be 48.112 and that implies that it is a valid breakaway point. This is invalid. Fine. Then we have step number 6, where you are going to have intersection with imaginary axis. So, for that we require the characteristic equation that we have already found that is s cube plus 15s square plus 50s plus k equal to 0. So, from that you are supposed to construct Routh's array. The highest power is 3, then s raised to 2, then we have 1, 15, 50 and k. So, for s raised to 1, you are going to have 750 minus k divided by 15 and for s raised to 0, you are going to have k and you are going to select the element which has the k involved in it. So, that is 750 minus k divided by 15 and you are equating it to 0. So, you are going to get k marginal value as 750. Since k marginal value is positive, so that implies that it is going to have intersection. Correct? It is going to have intersection and for that you need to consider the auxiliary equation which is a row which is present immediately above that element. So, it will be 15s square plus k equal to 0. So, we have auxiliary equation as 15s square plus k equal to 0. We are going to keep the value of k as k marginal and it is equated to 0. So, we have 15s square plus 750 equal to 0. If you try to solve, you are going to have two roots plus or minus 7.07j which is intersection with imaginary axis. Then we have step number 7. So, in this case, we do not have complex conjugate either poles or zeros. So, this step is not applicable for us. Then we have step number 8. So, in this step number 8, we are going to have the plot on graph paper. In step number 8, we are going to draw the final plot on s plane. From the open loop transfer function, we know that the poles are at 0. It is at minus 5 and it is at minus 10 and the root locus is existing between 0 and minus 5 and from minus 10 up to minus infinity. We know the centroid is located at minus 5 and from here, the asymptotes is going to pass with angle of 60, 180 and 300. So, this represents theta 1 which is at 60 degree. This represents theta 2 which is at 180 and this represents theta 3 which is at 300 degrees. Then we came to know that the breakaway point which is a valid breakaway point is at minus 2.11. So, that is breakaway point and we came to know that the K marginal value is 750 and it is going to have the intersection with imaginary axis at plus or minus 7.07 j. So, we are going to draw a root locus which is going to pass through this breakaway point and it is going to have intersection with imaginary axis at plus or minus 7.07 j. So, from here, the root locus is going to start and it is going to cut this imaginary axis at 7.07 j. Similarly, it is going to pass through this breakaway point and have intersection with imaginary axis at minus 7.07 j. So, if you see p minus z, so from that we know three branches are approaching infinity. So, this is the first branch which is approaching infinity, this is the second branch which is approaching infinity and this is the third branch which is approaching infinity. Step number 9, we are going to comment on stability. The K marginal value is 750 that represent the system is marginally stable when K value is less than 750 and greater than zero. So, we say that the system is stable. K value is greater than 750 then we say that the system is unstable. These are my references. Thank you.