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

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.

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°

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).