5. Use the Routh stability criterion to solve the following...

5. Use the Routh stability criterion to solve the following questions. (a) Determine the stability of the following systems with the characteristic equations: (i) s^5 + s^4 + 4s^3 + 4s^2 + 8s + 1 = 0 (ii) s^5 + 6s^4 + 2s^3 + 12s^2 + s + 6 = 0 (b) Consider the following general transfer function G(s) = 1 / (s^4 + a_3s^3 + a_2s^2 + a_1s + a_0). (i) Determine the conditions in terms of a_0, a_1, a_2 and a_3 for the existence of stable roots. (ii) By using the result from part (b)(i), determine the stability of the following systems with the characteristic equations: • s^4 + 5s^3 + 50s^2 + 3s + 2 = 0 • s^4 + 4s^3 - 10s^2 - 50s + 15 = 0

Transcript

00:01 So here in this question, we are considering about the fifth question where we have to use the routh stability where we have to use the routh stability criteria to solve the questions.

00:13 So in the first part we have to determine the stability of the following system with the characteristic equation.

00:19 First system is given that is s raised to the power 5 plus s raised to the power 4 plus 4 of s raised to the power 3 plus 8 of 4 of s raised to the power 3 plus 4 of s raised to the power 2 plus 8 of s plus 1 that is equals to 0.

00:37 So this is the term which we are given here.

00:40 So here we can make the table that is s raised to the power 5, s raised to the power 4, s raised to the power 3, s raised to the power 2, s raised to the power 1, s raised to the power 0.

00:49 This from here is 1, 4, 8, 1, 4, 1, 4 minus 4 divided by 1, x, 8 minus 1 divided by 1 that is 7 that from here is 0.

01:03 This from here is x minus 4 divided by x.

01:07 So according to the routh's criteria, we can say that 4 of x minus 7 divided by x become equals to 4 minus 7 divided by x.

01:15 That from here is equals to 4 minus 7 and x from here is equals to 0 and something divided by 0 become equals to infinity.

01:23 So answer from here is minus infinity.

01:25 Hence the answer to the first part.

01:27 Now in the second part, we are given the equation that is s raised to the power 5 plus 6 of s raised to the power 4 plus 2 of s raised to the power 3 plus 12 of s raised to the power 2 plus 4 of s plus 6 that is equals to 0.

01:41 So this from here is equals to s raised to the power 5, s raised to the power 4, s raised to the power 3, s raised to the power 2, s raised to the power 1 and s raised to the power 0.

01:51 This from here is 1, 2, 1, 6, 12, 6, 18 minus 12 divided by 6 that from here is equals to 6 divided by 6 that become equals to 1 and this from here is 6 minus 6 divided by 6 that is equals to 0.

02:07 So we can say that it is marginally stable.

02:11 From here we can say that this is marginally stable due to the row completion 0.

02:16 So the auxiliary equation from here will be equals to 6 of s raised to the power 4 plus 12 of s raised to the power 2 plus 6 and the value of da divided by ds that is differentiation of the auxiliary equation will be 24 of s raised to the power 3 plus 74 of s that plus 0.

02:34 So this is the value of d of a which is divided by ds that is equals to 24 of s raised to the power 3 plus 24 of s.

02:44 So this is the equation from here.

02:47 Hence the answer to the first part.

02:49 Now we are considering about the part b where we have to consider the general function where the general function ds is given that is equals to 1 which is divided by s raised to the power plus a3 s raised to the power 3 plus a2 s raised to the power 2 plus a1 of s plus a0.

03:07 So in the first part we have to determine the condition in terms of a0, a1, a2, a3 here...

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