GATE Electrical Engineering (EE) Test: The Root Locus Technique- 2 Free

Here, P = 4,Z = 1, P - Z = 3
No. of branches of RL terminating at zero = 1.
No. of branches of RL terminatina at infinity = 3

Since out of the given option, real axis between s = -2 and s = -4 is only a part of root locus, therefore breakaway point will lie between -2 and -4.

Given, s² + 2s + 10 + K(s² + 6s + 10) = 0

Here, number of open loop poles, P = 2. Number of open loop zero, Z = 2
∴ P - Z = 0.
Angle of asymptotes are given by:

Sines P - Z = 0, therefore there are no angle of asymptotes for the root locus of the given system.

Given, 

1+ G(s) = 0
or, s (s²+ 6 s + 12) + K = 0
or, K = - (s³ + 6s² + 12s)
For B.A. points, 

or, s² + 4s + 4 = 0
or, (s + 2)² = 0
or, s = - 2 , - 2
For given O.L.T.F., P = 3, Z = 0, P - Z = 3

Poles are at s = 0 and s = -3 ± j√3 = -3 ± j1.732
Since entire -ve real axis is a part of root locus, therefore s = -2 is a valid breakaway point.

Statement 1 is true. The Routh-Hurwitz criterion is used to detect when closed-loop characteristic polynomial roots lie on the imaginary axis by forming the Routh array; setting an entire row to zero (or examining sign changes) yields the frequencies and gain values at which imaginary-axis crossings occur.

Statement 2 is true. A breakaway or break-in point on the real axis corresponds to a point where two or more root-locus branches meet or separate. At that point the characteristic equation has a repeated (multiple-order) root, e.g., a double root for two branches meeting.

Statement 3 is true. Let L(s)=G(s)H(s). From 1+K L(s)=0 we get K = -1/L(s). A breakaway/break-in on the real axis occurs where dK/ds = 0 (extremum of K versus s). Differentiating gives:

K = -1/L(s)

dK/ds = 0 ⇒ (1/L(s)^2)·dL/ds = 0 ⇒ dL/ds = 0.

Thus the breakaway points satisfy d/ds [G(s)H(s)] = 0.

Statement 4 is false. For a root of multiplicity n the angles of arrival/departure are given by θ = (2q+1)π/n (i.e. θ = (2q+1)×180°/n) for q = 0,1,...,n-1. Adjacent branches are separated by 360°/n. The single value "180°/n" alone is not the general correct statement; it is incomplete and therefore the given statement is incorrect.

Conclusion: Statements 1, 2 and 3 are correct and statement 4 is incorrect; hence Option A is the correct choice.

From the root sensitivity, standpoint, a system should not be operated at the breakaway points because root sensitivity at breakaway point is infinite.

at breakaway points

Here, α = closed loop poles(s)
and β = system gain (K)

There is possibility of complex B.A. point. Also when a pole is added to OLTF, root locus pushes to right and reverse happens when a zero is added.

Here, P = 3, Z = 1, P - Z = 2
No. of branches of RL terminating at zero = 1.
No. of branches of RL termination at infinity = 2

Here, two branches of RL has to terminate at infinity.
Thus, the two imaginary poles will terminate at infinity.
Hence, we need to find angle of departure for which we join all poles and zeros with the imaainarv Dole at s = j1.

Thus, φ = (90° -135°) = - 45°
∴  Angle of departure is,
φ_D = 180° + φ
= 180° - 45° = 135°

- Root Locus Concept: The root locus is a graphical representation that shows how the roots of a system change with variations in a system parameter, typically the gain.

- Breakaway Point: This is a point on the real axis between two poles of the system where root locus branches diverge or converge.

- Statement C Analysis: A breakaway point cannot exist to the left of a zero on the real axis, because that portion is not influenced by the zero for root locus. Therefore, statement C is incorrect.

Here,
P = 2, Z = 3 and Z > P
∴ No. of branches of RL terminating at infinity
= Z - P = 3 - 2 = 1

To find whether root locus intersects the imaginary axis or not, we construct Routh’s array. Characteristic equation is
s³ + βs² + Ks + Kα = 0
Routh's array is:

For β > α > 0, root-loci doesn't intersect the jω-axis and is stable for all positive values of K (no sign change in first column for β > α > 0 and K> 0).
Also, P = 3, Z = 1, P - Z = 2
Thus, two branches of RL will terminate at infinity while one branch will terminate at a zero.
Therefore, possible root locus diagram will be represented by option (a).