in the previous lecture we discussed the introduction to stability analysis now in this presentation we are going to discuss the rao thirvids criteria or the rh criteria which is a very famous method to determine the stability of a control system so let's get started so before moving on to routh orbits criteria firstly we will discuss the necessary conditions of stability in the previous lecture we discussed that if we know the poles of closed loop transfer function we can comment on the stability of the system moreover we also discussed that the closed loop poles are the roots of characteristic equation so if we are given a characteristic equation we have to find out the roots of that characteristic equation in order to have the closed loop poles and if we know the closed loop poles we can comment on the stability of the system so this is what we discussed in the previous lecture but there is a problem here if the characteristic equation is of higher order then it is not possible for us to calculate its roots to determine the closed loop poles so if the characteristic equation is of second order that is quadratic equation or a cubic equation then we can solve that characteristic equation in order to find out its roots so that we can have the closed loop poles and we can comment on the stability but if the characteristic equation is of higher order for example if it is of eighth order then it is not possible for us to find the roots of that characteristic equation because in that case it will have eight different roots that is eight closed loop poles so it will be difficult for us to comment on the stability of that system so to make this point more clear to you let us take the generalized form of transfer function the transfer function of any linear closed loop system can be represented as cs over rs equal to b naught multiplied with s to the power m plus b 1 multiplied with s to the power m minus 1 plus so on up to b m divided by a naught multiplied with s to the power n plus a 1 multiplied with s to the power n minus 1 plus so on up to a n so this is the generalized form of transfer function of any closed loop system this is we have discussed in the previous lectures we can write this transfer function in the form of ns over ds where n s represent the numerator part of transfer function and d of s represents the denominator part of transfer function from the previous lecture we know that if we equate the denominator part of transfer function equal to 0 we will have the characteristic equation of that system so the characteristic equation for this closed loop system will be ds equal to 0 that is a naught multiplied with s to the power n plus a 1 multiplied with s to the power n minus 1 plus so on up to a n equal to 0 this is the characteristic equation for this closed loop transfer function now we can see here this is a nth order characteristic equation that is it will have n different roots that is there are n number of closed loop poles present in this transfer function now for example if the value of n is equal to 8 then it will be eighth order characteristic equation so we will have to calculate eight different roots in order to comment on the stability and it is very difficult for us to calculate the roots of our eighth order equation but don't worry we don't have to do this if we have the rh criteria if we apply the routher bits criteria we can comment on the stability of the system without even solving this characteristic equation to find out its roots rh criteria gives us the necessary and sufficient conditions that this characteristic equation must follow in order to be a stable system and that is what we are going to discuss in this lecture so firstly let us discuss the necessary conditions that this characteristic equation must follow in order to be a stable system so the first necessary condition is all the coefficients of the polynomial must have the same sign so it says that if this is the characteristic polynomial then all the coefficient of this polynomial must have the same sign so if the coefficients are positive then all the coefficients must be positive or if the coefficients are negative then all the coefficients of this characteristic polynomial must be negative so this is the first necessary condition for stability let us now move on to the second condition which is none of the coefficient vanishes that is all the powers of s must be present in the characteristic equation now it means that no power of s must be absent in the characteristic equation for example this is a nth order equation so we are having n different powers of s s power n s power n minus 1 and so on up to s power 0. so coefficients of all powers of s must be present in this characteristic equation let me explain this with the help of an example if the value of n is equal to 4 then it will be a fourth order equation so we will have s power 4 s power 4 minus 1 that is s power 3 we will have a term of s squared s power 1 and s power 0. so coefficients of all the 5 powers of s from s to the power 4 to s to the power 0 must be present in this characteristic equation and none of the coefficients should vanish if any of the coefficient is missing in the characteristic equation we can say that the system is unstable and that is why we call this condition as the necessary condition for stability so i hope you got these two conditions so we can say that these two conditions are necessary for a system to be stable but these conditions are not sufficient conditions that is these two conditions are the necessary conditions for the stability of a system if any of these two conditions is not satisfied we can directly say that the system is unstable but if both the conditions are satisfied we need to further check it for the stability and for that sake we will apply the routh test which we are going to discuss in the later section of this lecture now moving on to the next point if any polynomial satisfies the above to necessary conditions then it is called as the hervitz polynomial so if any characteristic polynomial satisfies these two necessary conditions then we will call it as the hervitz polynomial for example if this characteristic equation satisfies these two necessary conditions of stability then we will call this polynomial that is a naught s power n plus a 1 s to the power n minus 1 plus so on up to a n as the hervitz polynomial and we can note one more important point if any polynomial is not a hervitz polynomial then in that case we can directly comment on the stability that is we can say that the system is unstable and if the characteristic equation satisfies both the necessary conditions that is it is a herbits polynomial then we need to further check for the sufficient conditions of stability by applying the routh test so i hope you got the necessary conditions of stability let us now move on to discuss the routh orbits criteria in which we will discuss the necessary and sufficient conditions of stability so moving on to rauth herbert's criteria this method is a combination of independent researchers of two different mathematicians edward routh and adolf herwitz edward john routh was an english mathematician and in 1876 he proposed an efficient recursive algorithm to determine whether all the roots of the characteristic polynomial of a linear system lie in the left half plane in 1895 german mathematician adolf hervetz independently proposed a method to arrange the coefficient of characteristic polynomial into a square matrix called the hervitz matrix and he showed that the polynomial is stable if and only if the sequence of determinants of its principal sub matrices are all positive so these two are the independent procedures given by edward john routh in 1876 and by adolf herves in 1895 and we study the rao tharwitz criteria as a combination of researchers of both these mathematicians so we can say that rh criteria is a mathematical test that is a necessary and sufficient condition for the stability of an lti system and it states that all the roots of a characteristic equation lie in the left half plane that is the system will be stable if and only if a certain set of algebraic combinations of its coefficients have the same sign so this method involves some steps which include tabulating the coefficients of characteristic equation in a particular way and the tabulation of coefficients gives an array which is called as the routh array moreover we need to interpret the routh table to determine the number of poles in the right half of s plane so firstly whenever we have the characteristic equation in front of us we have to tabulate the coefficients of characteristic equation in a particular way and the tabulation of coefficients gives an array which is called as routh array we will discuss the method of forming the route array in a while then after that we need to interpret the routh table to determine the number of poles in the right half of s plane so as we have the route array we have to check for the sign of coefficients and then we can comment on the stability of the system so i hope you understood the outlines of rh criteria let us now move on to discuss the method by which we can form the routh array from a given characteristic equation so moving on to the method of forming the routes array so for that sake let us consider the characteristic equation which is a naught multiplied with s to the power n plus a 1 multiplied with s power n minus 1 plus a 2 multiplied with s power n minus 2 plus so on up to a n equal to 0. so this is the characteristic equation of a given control system and we have assumed that this is a hervids polynomial that is it follows the necessary conditions of stability what are the necessary conditions of stability yes all the coefficients of characteristic polynomial must have the same sign and none of the coefficient vanishes so if we assume that all the coefficients a naught a 1 a 2 and so on up to a n are having the same sign then the first condition will be satisfied moreover we can see that we are having all the powers of s from s to the power n to s to the power 0 so all the powers of s are present in this characteristic polynomial so we can say that this characteristic polynomial follows both the necessary conditions and hence it is a herbits polynomial and hence we can check for the necessary and sufficient conditions by applying the routes test and in order to apply the routes test firstly we need to form the routes array so let us now move on to discuss the method of forming the route array for that sake firstly we need to create a table like this and here we have the decreasing powers of s for example we will have the s power n then we will have s power n minus 1 and in the same way we will fill all the decreasing powers of s in a vertical manner and inside this table we fill the coefficients of characteristic equation and the first two rows can be directly written from the characteristic equation itself so let's see how we can do it firstly we will fill the first coefficient a naught which is the coefficient of s power n in front of s power n now we need to fill all the coefficients in a horizontal manner in an alternate fashion for example if we have written a naught here we have to skip a1 here we can write a2 so here a2 will come after that we need to skip a3 so after this a4 will come then we need to skip a5 after this a6 welcome so in this way in this first row we have filled the coefficients of all the even powers of s now what are we left with yes the coefficients of odd powers of s so now if we move on to the second row we will have a 1 which is the coefficient of s power n minus 1 so we will place it in front of s power n minus 1 now again we have to fill it in the horizontal manner in an alternate fashion so if we have filled a1 now we have to skip a2 and here a3 will come now again we have to skip a4 here a5 will come and in the same way here a7 will come i hope you got this in the first row we have a naught then we need to skip a1 here a2 will come then we need to skip a3 here a4 will come and after that a6 will come and we have to continue this till the last even coefficient in the same way in the second row we need to fill the odd coefficients so here a1 will come which is the coefficient of s power n minus 1 then we need to skip a 2 because we have already filled a2 here a3 will come then we need to skip a4 here a5 will come and in the same way here a7 will come this thing will be more clear to you when we will see some examples now moving on in order to fill the third row we need to perform certain calculations so let us see how we can do it so the third row will be the row of s power n minus 2 and let us consider that the first coefficient that is needed to be placed here is equal to b1 now we need to calculate b1 and we will do this by using these two rows so what will be b1 b1 will be equal to a1 multiplied with a2 minus a naught multiplied with a3 over a1 so in order to calculate b1 we need to multiply a1 with a2 minus a naught multiplied with a3 over a1 it is very similar to calculating the determinant of these four coefficients and then after that taking the negative sign as common so i hope you got this b1 is equal to a1 multiplied with a2 minus a naught multiplied with a3 over a1 let us now move on to fill the second coefficient b2 now what will be b2 b2 will be equal to a1 multiplied with a4 minus a naught multiplied with a5 over a1 that is b2 will be equal to a1 multiplied with a4 minus a naught multiplied with a5 over a1 i hope you are getting this let us now move on to the next coefficient consider the next coefficient to be b3 so b3 will be equal to a1 multiplied with a6 minus a naught multiplied with a7 over a1 what will be b3 b3 will be equal to a1 multiplied with a6 minus a naught multiplied with a7 over a1 so in this way the next row can be obtained by doing such calculations in the previous rows i hope you are getting this let us now move on to the next row which is the fourth row which is the row of s power n minus 3. let us consider the first coefficient that is to be placed here equal to c1 now we need to calculate c1 in the same way we calculated b1 so c1 will be equal to b1 multiplied with a3 minus a1 multiplied with b2 over b1 see here it is similar to calculating the determinant of these four coefficients and then taking minus sign as common note that let me repeat this one more times c1 will be equal to b1 multiplied with a3 minus a1 multiplied with b2 over b1 i hope you are getting this this is the standard method to form the routes array from a given characteristic equation and believe me it is very simple it will be more clear to you when we will see the examples so let us now move on to the next coefficient let us consider the next coefficient equal to c2 so c2 will be equal to b1 multiplied with a5 minus a1 multiplied with b3 over b1 what will be c2 c2 will be b1 multiplied with a5 minus a1 multiplied with b3 over b1 all of the time we need to divide with the first coefficient itself note that in the same way we need to proceed till s power 0 and if we proceed till s power 0 the last coefficient we will get here is the constant term which is a n so the last coefficient will be a n and this is called as the routh array for this characteristic equation i hope you got the method of forming the routes array let us now move on to the routes stability criteria so the route stability criteria says that all the terms in the first column of the routes array must have the same sign now we can see this is the route array and this is the first column of row 3. what are the elements present in the first column a naught a 1 b 1 c 1 and so on up to a n these are the coefficients of first column of route sorry and according to route's stability criteria in order to have the stable system all the terms in the first column that is all these terms must have the same sign moreover there should not be any sign change in the first column of route sorry that is what we discussed so after forming the route array if all the terms in the first column have the same sign then we can say that this system is a stable system so i hope you got the method of forming the routes array from a given characteristic equation and the route stability criteria let us now move on to discuss some examples based on this method examine the stability of given equations using routes method so two characteristic equations are given to us and we need to examine the stability of these systems by using the routes method so we know that if we are given a characteristic equation we need to find out the roots of characteristic equation and the roots of characteristic equation are the poles of closed loop transfer function and if we know the poles of closed loop transfer function we can comment on the stability that is what we discussed in the previous lecture but in this case we don't have to find out the roots of characteristic equation we just have to apply the rh criteria and then we can easily comment on the stability of the system let us check this out so firstly let us consider this first equation which is s cube plus 6 s squared plus 11 s plus 6 equal to 0. so before moving on to form the routes array for this characteristic equation firstly we need to check for the necessary conditions what are the necessary conditions of stability yes the first condition is all the coefficients of s must have the same sign so we can see here the first coefficient is a naught equal to 1 a naught equal to 1 a 1 equal to 6 a 1 equal to 6 a 2 equal to 11 a 2 equal to 11 and a 3 which is the coefficient of s power 0 is equal to 6. so we can see here all the coefficients have the same sign so the first condition is satisfied now we need to check for the second condition what is the second condition yes none of the coefficient vanishes that is all the powers of s must be present in the characteristic equation so we can see here this is a third order characteristic equation and all the powers of s are present in this case s power 3 is there s power 2 is present s to the power 1 is present and s power 0 is also present and hence we can say that this characteristic polynomial is also a hervitz polynomial as it satisfies both the necessary conditions of stability if any of the condition is not satisfied we could have directly declared the system as unstable but in this case both the necessary conditions of stability are satisfied and hence in order to comment on the stability we need to form the routes array for this characteristic equation so let us now move on to form the routes array and for that sake we need to form a table in which we have the decreasing powers of s in this place and inside this table we fill the coefficients and we know that we can fill the first two rows directly from the characteristic equation so considering the two rows first row is the s power 3 which is the maximum power of s then we have s power 2. now we need to fill these two rows by looking into the coefficients so the first coefficient will be the coefficient of s power 3 which will be equal to 1. now we need to fill this in the alternate fashion so we need to skip this coefficient and we will move on to the third coefficient which is equal to 11 so the second entry here will be 11 and after that if we skip this coefficient we are not left with any coefficient so we will have only two entries in the first row let us now move on to the second row in which we will have coefficient of s power two in the first column so the first entry will be equal to six now we need to skip this coefficient so we will have 6 in this place moving on to the next row which is the row of s power 1 and in order to fill this row we need to make certain calculations so this entry will be equal to 6 multiplied with 11 minus 1 multiplied with 6 over 6 we can see here 6 multiplied with 11 minus 6 multiplied with 1 over 6. if we simplify this we will have the coefficient equal to 10. now what will be the coefficient that should be present here yes it should be 6 multiplied with 0 as no coefficient is present here minus 1 multiplied with 0 over 6 so it will be equal to 0 i hope you got this in this way we are done with third row let us now move on to the fourth row which is the row of s power 0 and this is the last row so if we want to calculate the coefficient of s power 0 it will be equal to 10 multiplied with 6 minus 6 multiplied with 0 over 10 so it will be 10 multiplied with 6 minus 6 multiplied with 0 over 10 and if we simplify this we will have 6 and we can see here this last coefficient is the constant term of characteristic equation i hope you got this now this is the routes array for this characteristic equation now if we check for the route stability criteria for the stability of system all the coefficients of first column in the routes array must have the same sign we can see here all the coefficients in the first column are positive as the number of sign changes equal to zero and hence we can say that all the coefficients are having same sign and that's why it satisfies the route stability criteria that's why this system is a stable system let us now move on to the second example in which we are having the characteristic equation as sq plus 4 s squared plus s plus 16 equal to 0. now again we can see that this characteristic equation is satisfying both the necessary conditions of stability and hence we can say that this characteristic polynomial is a herwitz polynomial now in order to comment on the stability again we need to form the routes array so i want you all to pause this video and form the routes array for this characteristic polynomial and let's see how many of you would do it correctly i hope you're done so firstly let's check for the coefficients coefficient of s power 3 is equal to 1 coefficient of s power 2 is equal to 4 coefficient of s is equal to 1 and the coefficient of s power 0 is equal to 16. let us now form the routes array so we will have the highest power of s here s cube and a decreasing power of s as squared and we know we can fill these two rows directly by using these four coefficients and we need to fill these four coefficients in alternate fashion so we will have one now we have to skip this coefficient so the second coefficient will be one so in this way we are done with the first row let us now move on to the second row the first coefficient to be filled in the second row is equal to 4 so it will be 4 and now we need to skip this coefficient so the second coefficient will be equal to 16. let's check this from the characteristic equation in the first row the first coefficient is equal to 1 then we need to skip this coefficient so the second coefficient is also equal to 1. now in the second row we need to fill this coefficient which is equal to 4 and then we need to skip this coefficient so the alternate coefficient is equal to 16. so the second coefficient in the second row is 16. let me give you one simpler method to fill coefficient in these two rows consider these four coefficients 1 4 1 and 16. fill these four coefficients in this manner 1 4 1 and 16. fill this in the vertical manner and in that case you'll not have to consider the alternate coefficients you can fill it directly let me repeat this one more time if we have the coefficients 1 4 1 16 so we can fill it in this way 1 4 1 16. if we fill the coefficients in this way we don't have to take the alternate coefficients but remember these two methods are independent from each other if you are filling the coefficients in a horizontal manner you have to take the coefficients in an alternate fashion and if you are filling the coefficient in a vertical manner then we don't have to consider the alternate fashion remember this so i hope you got this let us now move on to the third row which is the row of s power one so the first coefficient will be four multiplied with one minus one multiplied with 16 over four so it will be minus three i hope you got this now what will be the second coefficient of third row yes it will be 0 because we are not having any coefficients here so it will be 4 multiplied with 0 minus 1 multiplied with 0 over 4 which will be equal to 0 i hope you got this let us now move on to the last row which is s power 0 and directly we can say it will be equal to the constant value which is 16 right but let's calculate it it will be minus 3 multiplied with 16 minus 4 multiplied with 0 over minus 3 and if we solve this we will have 16 which is the constant value in the characteristic equation so in this way we are done with the forming of routes array for this characteristic equation let us now check for the route stability criteria and we know that for the system to be stable number of sign changes in the first column of routes array should be equal to zero but in this case we can see that there are two sign changes if we move from this coefficient to this coefficient there is no sign change but if we move from this coefficient to this coefficient there is a sign change this is positive and this is negative so up till now there is one sign change now if we move from this coefficient to this coefficient there is one more sign change this coefficient is negative and this one is positive so from this coefficient to this coefficient we are having two sign changes so we can say that number of sign changes in this routes array is equal to 2 and that's why we can say that all the coefficients in the first column of routes array are not having same sign and that's why this system will be unstable moreover number of sign changes in the first column of routes array tells us the number of poles in the right half plane so if the number of sign changes is equal to 2 we can say that this system is having two poles in the right half plane so we can say that number of poles for this system in the right half plane is equal to 2. so i hope you got this in this lecture we discussed the necessary conditions of stability the method of forming the routes array and the route stability criteria and based on that we discussed these two examples in which we saw that in this first example the number of sign changes in the first column is equal to 0 and hence this system is stable and in the second example the number of sign changes in the first column is equal to 2 and hence this system is unstable number of sign changes is equal to number of poles in the right half plane and that's why we said that the number of poles for this system in the right half plane is equal to 2 so i hope you got this in the next lecture we will discuss some more examples and the special cases of rh criteria as of now we are done with this lecture thank you for watching this lecture i'll end this one here see you in the next 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