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If a feedback control system has its open-loop transfer function

\(G(s)H(s)=\frac{K}{s(s+2)(s^2+2s+5)}\)

The coordinates of the centroid of the asymptotes of its root-locus are


1. -1 and 0
2.

1 and 0


3. 0 and -1
4. 0 and 1

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Correct Answer - Option 1 : -1 and 0

Concept:

1. Every branch of a root locus diagram starts at a pole (K = 0) and terminates at a zero (K = ∞) of the open-loop transfer function.

2. Root locus diagram is symmetrical with respect to the real axis.

3. Number of branches of the root locus diagram are:

N = P if P ≥ Z

= Z, if P ≤ Z

4. Number of asymptotes in a root locus diagram = |P – Z|

5. Centroid: It is the intersection of the asymptotes and always lies on the real axis. It is denoted by σ.

\(\sigma = \frac{{\sum {P_i} - \sum {Z_i}}}{{\left| {P - Z} \right|}}\)

ΣPi is the sum of real parts of finite poles of G(s)H(s)

ΣZi is the sum of real parts of finite zeros of G(s)H(s)

6. Angle of asymptotes: \({\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}\)

l = 0, 1, 2, … |P – Z| – 1

7. On the real axis to the right side of any section, if the sum of total number of poles and zeros are odd, root locus diagram exists in that section.

8. Break-in/away points: These exist when there are multiple roots on the root locus diagram.

At the breakpoints gain K is either maximum and/or minimum.

So, the roots of \(\frac{{dK}}{{ds}}\) are the breakpoints.

Application:

The loop transfer function of a system is given by: 

\(G(s)H(s)=\frac{K}{s(s+2)(s^2+2s+5)}\)

Number of open-loop poles = 4

The real part of open-loop poles: 0, -2, -1, -1 

Number of open-loop zeros = 0

Number of Asymptotes = |4 – 0| = 4

Centroid, \(\sigma = \frac{{\left( {-2-1-1} \right) - \left( { 0} \right)}}{4} = - 1\)

Centroid \( = \left( { - 1,0} \right)\)

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