3. Forward path of a unity-feedback system has the transfer...

3. Forward path of a unity-feedback system has the transfer function. egin{equation} G(s) = frac{K}{s(s+1)(1+3s)} end{equation} (a) Using Routh-Hurwitz method, judge the system stability when K=2 and find the condition that constant K must satisfy for the system to be stable. [10 marks] (b) If a system with a specified closed-loop transfer function T(s) is required to be stable, and that all the poles of the transfer function are at least at the distance x from the imaginary axis (i.e. have real parts less than -x), explain how you can test if this is fulfilled by using Routh- Hurwitz method. [6 marks]

Transcript

00:01 Here in the first part of this question we are given a forward pathway of a unity feedback which is represented by gs that is equals to k divided by the s multiplied by the s plus 1 which is further multiplied by the 1 plus 3s.

00:15 So here in the first part we are using the truth hurwitz method to judge the system stability where the k 2 when k is equals to 2 is here and after that we have to find out the condition that constant k must satisfy the system to be stable.

00:39 So here transfer function is given as transfer function is given as ts which is equals to gs divided by the 1 plus gs multiplied by the hs.

00:56 So for a unity forward path we are having the value of hs that is equals to 1.

01:03 So ts from here becomes equals to k divided by the s multiplied by the s plus 1 multiplied by the 3s plus 1 which is divided by the 1 plus k which is divided by k which is divided by the s multiplied by the s plus 1 multiplied by the 3s plus 1 which is further multiplied by the 1.

01:22 Solving the term from here we get the value of ts that becomes equals to k divided by 3 of s raised to the power 3 plus 4 of s raised to the power 2 plus s plus k.

01:32 Now the chart of the equation will be somewhat like this that is 3 of s raised to the power 3 plus 4 of s plus 4 of s raised to the power 2 plus s plus k that is equals to 0 it's 8.

01:48 Sorry it's s plus s plus k is equals to 8.

01:51 Now we are considering out about a root hits array.

01:55 So here we are writing the equation that is h3 here s2 s1 s0 then s3 for s3 there is 3 1 and here it is 4 then here it is k cross multiplying this term from here this become equals to 4 multiplied by the 1 minus 3 multiplied by the k which is divided by 4 and this from here is k.

02:17 So the stability of the system element of the first column of the array must be greater than 0.

02:24 So 4 minus 3k divided by the 4 is greater than 0 which means that k from here is less than 4 divided by 3 and k must be greater than 0.

02:35 So the value of k lies between 0 and 4 divided by 3.

02:40 In this case this will be completely stable.

02:48 So this will be completely stable and at the value that is k is equals to 0 and k is equals to 4 divided by 3.

02:55 This is marginally stable at that value.

03:00 So this is what the k is equals to 1.

03:03 Now we are considering about 4k is equals to 2.

03:06 The system is unstable.

03:12 As k is equals to 2 then this system from here 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 3 1 4 2 minus 1 divided by 2 and this is 2.

03:27 This is negative value and its negative value that's why it is unstable.

03:32 So in the first column number of charges of the signal of the element is equals to 2.

03:40 So 2 poles of a system 2 pole of the system is in the right side of the s plane.

03:54 So this is the answer to the part a of the question.

04:00 Now we are considering about the part b.

04:03 So for part b what we are assuming here that system has 3 pole and all are left then minus x and the negative x is for this plane.

04:13 So p1 is equals to minus x minus 1 p2 is equals to minus x minus 2 and p3 is equals to minus x minus 3...

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