Routh Hurwitz Stability Criterion – Stability Analysis in Time Domain – Control Systems
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Routh Hurwitz Stability Criterion – Stability Analysis in Time Domain – Control Systems


Do subscribe to Ekeeda Channel and press bell icon to get updates about latest engineering HSC and IIT JEE main and advanced videos Hello friends in this video we are going to study about the Routh Hurwitz criterion that how we can make the comment about the stability of a system that whether the system is stable or unstable using the Routh Hurwitz criteron so let’s start with our topic before starting with the Routh Hurwitz criteria first we will study that what is the criteria to determine the stability we know that for a control system the stability it is always defined with respect to the location of roots of the characteristic equation so the characteristic equation it plays a very important role for a control system so if we write the characteristic equation for an nth order system it will be like so this is the characteristic equation of an 1/8 order system now this scientist hobbits he says that for a system to be stable the hobbits determinants they should be greater than zero and these hobbits determinant they are the principal minors of a following arrangement so if we see that arrangement it is like so this is the how it’s in determinant okay this is the generalized form of the hobbits determinant now we are going to take all the hobbits in determinant inside in it like first we are going to take this e a one then we are going to take a 1 a not a 3 a 2 so all the principal minors of this arrangement they should be greater than 0 for a system to be stable so that was the hobbits criteria tricky right so that was the generalized arrangement of the habits determinant and in this we are going to take different different determinants so first we have taken d1 will be a 1 should be greater than 0 then D 2 is the arrangement or the determinant of a 1 a not a 3 a 2 so that should be greater than 0 just like this may have D 3 T 4 till D K ok so all these determinants they should be greater than 0 for a system to be stable now if we see this method this Hurwitz criteria this is very time requiring because you have to first write the characteristic equation then you have to write this arrangement then you have to determine all the determinants and then you will check that whether all the determinants are greater than 0 so determinant over 3 cross 3 matrix for cross flow matrix it is very time requiring ok so this how its criteria it was mathematically very complex so the scientists root he makes some modifications in this criteria and he proposed his own criteria for determining the stability and that is known as the root stability criteria so let’s study about the roots stability criteria we will see one by one all the steps of this criteria that how we can determine the stability using this method so first step is to determine the characteristic equation of the system so characteristic equation of the system is given by 1 plus GSHS equals to 0 now let’s take an example of a characteristic equation so we have taken an example of a characteristic equation in this we have the coefficients as a 1 a 2 a 3 a 4 and a 5 and we have the power of s the highest part is s to the power 4 then s cube s square s and H to the power 0 ok now our second step will be to build the root and how this area can be made let’s see in step number 2 so this is how the route that is me let’s see the highest power of s we have taken this as an example of the characteristic equation so in this characteristic equation the highest power of s is 4 so first we are writing s to the power 4 then all the powers s cube s square s 1 and s 0 we have written on the left hand side of this line on the right hand side we are going to write the coefficients because the highest power it is an even number so we are going to start with the even numbers ok even powers of s so here we have s to the power 4 coefficient is even so here we are going to write even then coefficient of s to the power square because highest power is even so we are going to start with the even couette numbers even powers of s that is s 4 s square s to the power 0 we are going to write their coefficients as the first row of this and a 1 a 3 and a 5 then these odd powers of s coefficients we are going to write 8 to a fool if any of the coefficient is missing in the added then we will write 0 for that coefficient okay now s to the power 2 we will write here X 1 and this X 1 will be determined by multiplying this coefficient a 2 with a 3 minus a 1 with a 4 and divided by a 2 so this is how X 1 is determined then you determined x2 witches we have to multiply the first column with this coefficient a to my into a five minus a 1 into 0 divided by this a 2 so this is how x2 is determined now for the coefficients are not present so we are will write here 0 this coefficient is missing so we are writing zero game now wipin wipin will be determined by multiplying this the coefficient just above it x1 with a 4 minus a 2 with x2 divided by X 1 then y2 will be if you have to multiply this X 1 bit 0-0 with a 2 so the answer will be 0 so here 0 is written then by 3 will be 0 similarly set one bility we have 2 x 1 with X 2 minus X 1 multiplied with 0 okay and divided by Y 1 so in this way all the coefficients you have to determine and you have to complete till power of s is equals to 0 so you have to build the root arrays in the second step okay so now one next step is step number three and in step number three we are going to determine that how many number of sign changes are there in the first column of the root so to check this stability we will determine the number of total number of sign changes in the first column of the suppose we have taken this example and in this this even coefficient it is positive a two is positive and x1 it is coming out to be negative vibin is again positive and z1 is again positive so this x1 it is negative so positive positive negative so there will be one sign change from positive to negative then again it is positive so from minus to positive again a sign change so there are two sign changes in the first column that is from positive to negative and then from negative to positive so we have so there are total two sign changes so these sign changes in the first column of the Routh array they are indicating the number of roots of the characteristic equation which are lying on the right half of the S plane and if the roots they lie on the right half of the S plane it means that the system is unstable so if there is any single sign change in the first column of the Routh array it means that the system is unstable so here I have written to check the stability determine the total number of sign changes in the first column of the array and then the number of sign changes they indicates the number of roots of the characteristic equation in the right half of the S plane that is having a positive real part okay so this indicates that if there is a sign change then the system is unstable if there is no sign change then system is stable so this is how the Routh stability criteria it is defining the stability or it is checking the stability of the system now let’s see the necessary and the sufficient conditions for this root stability criteria so these three are the necessary conditions for a system to be stable first is that none of the coefficients of the characteristic equation they should be missing or they should be zero okay so no coefficients should be missing like if we are starting we have taken the example of the characteristic equation in the roots method in this characteristic equation all the powers of s are present starting from s for we have s cube s square s 1 and s 0 so all the coefficients are pieces present no coefficient was missing and none of the coefficient was 0 so this is the necessary condition for a system to be stable that none of the coefficients should be missing or 0 in the characteristic equation now every coefficient it should be real and it should have same sign so no coefficient of the in the cactus teqi question it should be imaginary all should be real and also they have same sign if what dish either they should have all pause or they should have negative sign okay so the sign of the coefficient should be same they should be real and no coefficient should be missing and zero now third necessary condition is that if the characteristic equation of the system it contains only or and or only even power of cases like take an example if we have s cube plus 4 s 1 equals to 0 so we have only two coefficients like only odd power of s are present and if we take an example like 2 s 4 plus 3 s squared plus 1 equals to 0 so only even powers of s are present so if this is the case then it indicates that the roots of the characteristic equation they have no real part and they have only imaginary part ok so these type of characteristic equation or these types of systems they have no roots in the real part the roots will have no real part they will have only the imaginary part and it means that the output response of the system it has sustained oscillations so these are the three necessary conditions for a system to be stable using the root stability criteria now what is the sufficient condition for it so this efficient condition of defining the stability in the route of route stability criteria is that each term of the first column of routes added should be positive and it should have the same sign if all the coefficient or all the terms in the first column they will have the same sign then there will be no sign change in the first column and no sign change means that the system is stable so the sufficient condition for a system to be stable is that the first column each term of the first column of routes arrey it should be positive and it should have the same sign okay now there are some limitations in this route stability criteria so let’s see these two limitations that how these limitations can be solved while checking the stability for a system now first limitation in this routes stability criteria is that if in the first column we are getting a term which is zero then how we are going to solve such kind of problems so first limitation will be now the first limitation we have is when first element of any row is zero and the same remaining row contains at least one nonzero element so when we are building the root array we are encountering a situation when the first element of a row it is coming out to be zero and this row is have at least one non-zero element so how we are going to solve such problems let’s have in the Routh array which we have build up this was the example which we have taken and we have seen that how the Routh array is built now in this Routh array if suppose this x1 it is coming out to be 0 so this is the first element of this row and this x2 it is nonzero and this is again 0 so this row it is having at least one nonzero element and first element is 0 so this is the situation which we are taking in this first case okay now such type of situations they can be solved by two methods first method is we will use s equals to 1 by 0 so this is our first method in which we will use s equals to 1 by Z we were having the characteristic equation in terms of s so in that characteristic equation we are going to replace s by 1 by Z we will get the characteristic equation in terms of Z now we will check determine if Z and we will repeat the same route alle calculations and then the sign change what we were doing for F of s okay so we were having the characteristic equation earlier in terms of is here now if we are having that situation that first element is zero then what we are going to do we will replace this s by one by Z so here we will get a 1 1 by Z to the power 4 plus a 2 1 by Z to the part 3 plus a 3 1 by Z to the part 2 plus a 4 by Z plus a 5 equals to 0 then we will simplify that equation and we will get F Z and then we will repeat the same steps we will build the root array then we will check the number of sign changes in the first column so same procedure will be repeated for F of Z so this is the first method to resolve this problem second method is 2 we have got 0 as the first element in the row so that 0 will be replaced by a small number epsilon ok so our second step will be replace 0 so zero will be replaced by the small positive number epsilon and then we will complete the array route array we are going to complete with this number epsilon so we are going to replace zero by epsilon then we will complete the array by using this number epsilon that it will be completed and then we are going to check the sign change by taking the limits epsilon tends to zero so let’s take an example so that this will be clear to you so this is the characteristic equation of the system now we are going to build the Routh array from it so Routh array will be the highest power of s and this is five so we are going to start with H to the power 5 10 4 3 2 1 and 0 now the coefficient of H to the power 5 is 2 then the next odd power is s cube so we will write its coefficient so 2 6 and then 1 now s to the power 4 so we will write s to the power 4 coefficient s to the power square coefficient and s to the power 0 coefficient so it will be 1 3 and 1 now we will do the mind s to the power 3 coefficient by this X 1 we will determine 1 into 6 minus 3 into 6 divided by 1 so this is 6 minus 6 divided by 1 so it will be 0 now next we have 1 into 1 that is 1 my two divided by 1 so it will be minus 1 here and next we will have 0 so this we have encountered the condition that first element it is coming out to be 0 and this row has one nonzero element so what we are going to do we will replace this 0 by Epsilon a positive number then we are going to melt the rest of the Routh array by using this positive number so epsilon into 3 3 epsilon minus 1 into minus 1 so it will be plus 1 divided by epsilon so this will be 3 epsilon plus 1 upon epsilon then next we will have epsilon into 1 and 1 into 0 upon epsilon so it will be 1 next we will have 0 then X to the power 1 we will multiply this with minus 1 epsilon with 1 divided by this so we are going to get this number as minus of so what we have done we have multiplied this three epsilon plus 1 upon epsilon with minus 1 so this is the term then minus of epsilon multiplied with 1 it will be epsilon divided by this term so this is s to the power 1 then rest we will get as 0 and 0 again we will have this term multiplied with 1 this term multiplied with 0 divided by this so we will get it as 1 0 and 0 so we have obtained the root arrayed it ok so now we are going to check the sign changes in this first column of the Routh array how we are going to check the sign change you may will take the limits epsilon tends to 0 so if we apply the here the limits epsilon tends to 0 then its value will come out to be limits epsilon tends to 0 for 3 epsilon plus 1 upon epsilon so here it will be 3 into 0 plus 1 upon 0 so it will tend towards positive infinity okay then we are going to apply the limits on the term minus okay now if here we will put epsilon equals to zero tends to zero then it will become 1 upon zero minus zero upon this is again 1 upon zero so this will come out to be a negative number now in this root area if we take two is positive one is positive then it is a positive number epsilon it is a small positive number then three epsilon plus 1 upon epsilon it is coming out to be positive infinity and then this term it is coming out to be negative so from positive infinity to negative it will be a sign change and again from negative to a positive one it will be a sign change so here we are having two sign changes so to sign changes means the system is unstable so we are getting the system as unstable because there are two sign changes so two roots are lying on the right half of the S plane so the system will be unstable so in this way when you encounter the situation that the first element of any low it is coming out to be zero and the rest of the elements at least one of the element is nonzero then either you can put s equals to one by Z and you can follow the same procedure or you can replace zero by a small positive number epsilon and then you can follow this method now the second limitation is in this Routh Hurwitz criteria is that when all the elements of any of the row it is coming out to be zero then what you are going to do you are going to form a subsidiary or you can say auxilary equation okay then you are going to differentiate that auxilary equation and then equate that auxiliary equation to 0 and then you will get the value of the is okay so let’s take an example so this is the Routh array we have s 4 s 3 s 2 we have the coefficients here when we are calculating the coefficient then this s 2 s square row this row all the elements are coming out to be 0 so we are going to form the subsidiary or the auxilary equation this auxilary equation is a s equals tune we have a 2 s cube plus a 4 s to the power 1 because here we are having the coefficient of s to the power 4 s square and s to the power 0 here we are having s to the power cube s to the power 1 so this a 2 multiplied with s cube plus a 4 multiplied with s 1 so this is the auxilary equation now we are going to differentiate this auxilary equation with respect to s so d a s by D s it will be here we are having s cube so it will be 3 a 2 s Square and here we are having a 4 s 1 so it will be a food so now we have got the coefficient of s Square and s to the power 0 s square coefficient is 3 a 2 and s 0 coefficient is a 4 so we are going to use this Co if these coefficients we will place it here where we have obtained x1 equals to 0 and x2 equals to 0 and then we will follow the same procedure we will build the rest of the Routh array and then we will check the number of sign changes in the first column of the Routh array so when you got the situation that all the elements of the row are coming out to be 0 then form the subsidiary or the auxilary equation and then differentiate with respect to s and then these coefficients use it in the Routh array where you have obtained 0 and then repeat the last procedure so in this video we have studied about the root stability criteria that how we can make a comment about the stability of a control system whether it is stable or not stable using the Routh Hurwitz criteria you have to just use the characteristic equation then form the Routh array and then check the number of sign changes in the first column of the roots area so I hope that this video is clear to you thank you

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