 Namaste, welcome to the session Routh Criteria with Special Cases. At the end of this session, students will be able to apply Routh Criteria for special cases and analyze the stability of system. Now in this session we are going to see Routh Criteria, Routh Criteria with special case 1 and then Routh Criteria with special case 2. Now before proceeding further, watch the video here and recall Routh Criteria and how to write Routh table for this equation. So I think you have recalled how to write Routh Criteria. Now for the given equation we can write the Routh table or it is also called as Routh array as for s raise to n rho the coefficients are a0, a2, a4, a6 and then for s raise to n minus 1, a1, a3, a5, a7. The b1, b2, b3 calculated as then next rho s raise to n minus 3 is calculated for c1, c2, c3 as again take a pause here and recall what is the stability condition of Routh's array. So I think you are recalled the stability condition of Routh's array or Routh's table. So the necessary and sufficient condition for a system to be stable is that all the terms in the first column of Routh's array must have same sign and there should not be any sign changes in the first column of Routh's array. System is unstable if any sign change occurs in the first column and the number of sign changes in the first column indicates number of roots in the right half of the s plane. Now let us see Routh criteria with special case 1. So what is special case 1? So in the Routh table if there is a 0 only in the first column of a row or first element of any row of Routh array is 0 and remaining row contains at least one non-zero element. Then it is known as a special case 1. Now due to this what happens is Routh test fails as a term in the new row becomes infinite so we cannot find the stability of the system. So to overcome this in the special case of Routh criteria we have to replace 0 by epsilon and complete the Routh table and then let epsilon tends to 0 and again analyze the stability from the number of sign changes in the first column of Routh table. Now let us have one example on Routh criteria with special case 1. So for the given equation below find out stability of the system. So for this equation we can write the Routh table as for s raised to 5 row we have the coefficients here 1, 2 and 6. So for s raised to 4 row we have the values as 3 and 6 and 9. So we can divide this row by 3 so we will get 1, 2, 3 the new values after dividing s raised to 4 row by 3. Then for calculation of s raised to 3 values so here 1 into 2, 2 minus 1 into 2 so 2 minus 2 is 0 divided by 1 is 0. So here in the first column of Routh array or Routh table we have 0 and then in next value is 1 into 6 minus 1 into 3 that is 3 divided by 1 will get 3 here. So according to the special case 1 if we have a 0 in the first column only and only 1 0 is there then what we have to do is we have to consider 0 as epsilon and 3 as it is for the row s raised to 2 we have here 2 into epsilon minus 3 divided by epsilon and 3 into epsilon minus 0 because there is no value here. So we have 2 divided by epsilon then we have 3 here for the s raised to 1 row the next value is 3 minus 3 epsilon square divided by 2 epsilon minus 3. So after calculating this we will get this value. The next value will be 0 for the s raised to 0 through we will get 3 value here. So once table is over so limit epsilon tends to 0 2 epsilon minus 3 divided by epsilon will give us minus infinity. And for the s raised to 1th row limit epsilon tends to 0 3 minus 3 epsilon square divided by 2 epsilon minus 3 will give us plus 3 here. So what we can observe from the first column of Routh table we have the values as 1 1 then epsilon minus infinity plus 3 and 3. So if we observe here we have first sign change from positive to negative then second sign change negative to positive. So what we can conclude here is there are 2 sign changes so the system is unstable with 2 poles in the right half of the s plane. So this is how Routh criteria with special case 1 is solved. Now let us see Routh criteria with special case 2. The Routh table gate fills when there is an entire row that consists of zeros. So to overcome Routh criteria with special case 2 we have to form the auxiliary equation from the above row of the row which consists of all the zeros. Then we have to differentiate the polynomial with respect to s and replace the row of zeros by its coefficient. Then we have to continue the construction of the Routh table and then analyze the stability from the number of sign changes in the first column. So let us have example on Routh criteria with special case 2. So for the given equation is the system stable. So we have this equation and from this equation the Routh table is written as for the first row s raised to 6 we have the values 1, 8, 20 and 16. For the s raised to 5 we have the values as 2, 12, 16 and there is no further so 0 here. So in the second row that is s raised to 5 we can divide each value by 2 so the new values comes here 1, 6, 8. So to find out the next row values that is s raised to 4 so it is nothing but 8 minus 6 divided by 1 gives us 2. Then 1 into 20, 20 minus 8, 12 divided by 1, 12 then further value is 1 into 16 minus 0 that is 16 here. So we have s raised to 4 row values as 2, 12, 16 so again dividing this row by 2 will give us new values as 1, 6, 8. Then for the next row s raised to 3 the values will be 1 into 6, 6 minus 1 into 6 that will give us 0 so here we get 0. Then further value 1 into 8 minus 1 into 8 that will be 0 again. So if you observe here s raised to 3 row consist of all the 0s. So according to the Routh criteria with special case 2 we have to write the auxiliary equation from the above row of the row which is consisting of all the 0s. So here equation becomes s raised to 4 plus 6x square plus 8 so further moving to the solution we have to differentiate this equation. So that will give us 4s cube plus 12s and then replacing these 0s by the coefficients of differential equation as 4 and 12. So here for s raised to 3 row we have to consider new values as 4 and 12. Now further simplifying this we can divide this row by 4 that will give us 1 and 3. So let us calculate the values for s square that will give us 1 into 6 minus 1 into 3 divided by 1 that will be 3 and next value will be 1 into 8 minus 0 divided by 1 that will give us 8. So further calculating values of s raised to 1 and s raised to 0 will give us 1 by 3 and 8. Now if you observe this column first column of the Routh array we have all the values positive 1 1 1 then 1 3 1 by 3 8. We have to solve this auxiliary equation then that will give us s is equal to plus minus j root 2 and s is equal to plus minus j 2. So by observing these values we can conclude that the system is marginally stable. So in this way we can solve the Routh criteria for the special case 1 and 2. These are references. Thank you.