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A unity feedback control system has an open loop transfer function which is given as \(G(s) =\frac{K} {{s(s + 4)}}\). Find the angle of asymptotes.
1. 55°, 56°
2. 90°, 270°
3. 45°, 115°
4. 109°, 34°

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Correct Answer - Option 2 : 90°, 270°


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 the total number of poles and zeros are odd, the 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 break points.


\(G\left( s \right) = \frac{K}{{s\left( {s + 4} \right)}}\)

Number of poles (P) = 2

Number of zeros (Z) = 0

P – Z = 2

Angle of asymptotes,\({\theta _l} = \frac{{\left( {2l + 1} \right)\pi }}{{P - Z}}\), l = 0, 1

θl = 90°, 270°

So option (2) is the correct answer.

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