 I am Mr. Sarvajne Gandhi, working as assistant professor in Department of Mechanical Engineering from Valtran 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, general steps to solve the root locus problems. Step number one, general information about the number of poles, number of zeros and number of branches which are going to approach infinity. In step number two, draw the pole zero plot, identify where the root locus exists, identify by visual method, the existence of breakaway point. In step number three, calculate angle of asymptote. Step number four, determine the centroid. Step number five, draw a rough sketch based on the above steps. Step number six, calculate the breakaway point. Step number seven, calculate angle of departure for complex conjugate poles and angle of arrival for complex conjugate zeros. Step number eight, draw a final plot of the above step. Commend regarding the stability and performance. Example one, plot the root locus using MATLAB and conventional method. Using MATLAB, we will try to solve the problem and compare the root locus plot with the conventional method. Construct the root locus for the given system having open loop transfer function g of s h of s is equal to k divided by s into s plus 5 into s plus 10. So if you want to use the MATLAB, we need to modify the open loop transfer function equation that is given to us. We need to write it in polynomial form. So at numerator, we have only k whereas at denominator, we have s into s plus 5 into s plus 10. We will try to convert the simplified form into a polynomial equation. So we are going to have k divided by s cube plus 15 s square plus 50 s. Now you need to convert this polynomial equation into a row matrix. At the numerator, we have constant term and at the denominator, we have polynomial equation. So from the denominator, we have to see which is the highest coefficient and accordingly, we are going to construct the row matrix. So at the denominator, we have 1, 15, 50 and 0 whereas at numerator, we are going to have 0, 0, 0 and 1. That represents the row matrix of given open loop transfer function. So we are going to use the command prompt and write num that represents numerator is equal to 0, 0, 0 and 1. Then represents denominator and we are going to write 1, 15, 50 and 0 in the square brackets. And we are going to define a variable named as g. We are going to call the in-built function tf that represents the transfer function. And for that, we are going to pass the arguments as num and den. So g is equal to tf num and den. And at last, we are going to call the function for plotting of the root locus that is r locus of g. We are going to solve first problem, construct root locus for given open loop transfer function g of s h of s is equal to k divided by s into s plus 5 into s plus 10. So first step is count the number of zeros and number of poles and number of branches which are going to approach infinity. So count of z is equal to 0 count of pole is 3 and it is going to lie at 0 minus 5 and minus 10. And number of branches which are approaching infinity is p minus z. So 3 minus 0 that is 3. So 3 branches will approach infinity. Step number 2. So in this step, you need to find the existence of root locus and possibility of existence of breakaway point. We have poles which are existing at 0 minus 5 and minus 10. So if you try to draw a section in between 0 and minus 5 and observe the number of roots which are present on its right hand side, the count is 1. So that is odd. So root locus is going to exist between 0 and minus 5. So if you take one more section in between minus 5 and minus 10 and observe the number of roots which are there on its right hand side, the count goes to be 2. So that is even and root locus is not going to exist between minus 5 and minus 10. If you take one more section towards the left hand side of minus 10 and observe number of roots on its right hand side, the count goes to 3. So that is odd. So root locus is going to exist from minus 10 up to minus infinity. So root locus exists between 0 and minus 5 and it also exists from minus 10 up to minus infinity. So apart from this, we are going to see whether there is a possibility of existence of breakaway point. So if we see the prediction of breakaway point, we can see that minimum one breakaway point exists in between 0 and minus 5 because these are two adjacently placed poles and in between that we have root locus. So from that there is chance that minimum one breakaway point exists between two adjacently placed poles. So for that we can say that breakaway point exists between 0 and minus 5 and you can write by visual method. So we have step number 3. Here we are going to find asymptote. So theta is equal to plus or minus 2q plus 1 into 180 degrees divided by p minus z. So in this case the p minus z value is 3. So you are supposed to select the value of q as 0, 1 and 2. So if you place the value of q as 0, you are going to get theta 1 as 60 degrees. If you place the value of q as 1, you are going to get theta 2 value as 180. And if you place the value of q as 2, you are going to get theta 3 value as 300 degrees. So we have step number 4 wherein we are going to find centroid that is represented by sigma. Sigma is equal to summation of real part of poles of open loop transfer function minus summation of real part of zeros of open loop transfer function divided by p minus z. So centroid equal to real part of poles are from here we can find 0 minus 5 minus 10. We don't have 0 so minus 0 p minus z is 3. So it is minus 15 by 3. Therefore the centroid is located at minus 5. So you need to show this centroid on this plot. So as you can see over here the centroid is going to lie at this minus 5. So I am representing at centroid is equal to minus 5. Apart from that we can see that there are 3 asymptotes which are supposed to pass through these asymptotes with angle of 60, 180 and 300. So you need to draw it from these centroid points theta 1 is equal to 60, theta 2 is equal to 180 degree and theta 3 is equal to 300 degrees. These are my references. Thank you.