In the root locus plot shown in the figure, the pole/zero marks a

SOLUTION

Given root locus diagram is:

Note:

This pole-zero diagram has no pole-zero marked and also the arrow is not marked in the diagram.

We can use some of the properties of the root-locus plot.

1) Root locus plot always originates from open-loop poles and terminates at either open-loop zeroes or infinity.

2) The point at which multiple roots of the characteristic equation are present is known as breakaway/break in or saddle point.

Now, analyzing the given options:

1) From the root locus diagram, we can see that branches are tending to infinity. This indicates that the number of poles ≠ No. of zeroes.

Hence option D is eliminated.

Now, analyzing options A, B and C we can see that s = -1, -2, -4, and -7 are the possible location of poles and zeroes.

• From the root locus diagram, we can see that very close to two origin two poles are approaching each other, so the possible location of poles may be s = -1 & s = -2.
• Here, we have not considered two zeroes because it is actually not possible, as one branch end is at infinity.
• In this root-locus, near the origin, two-zeroes are possible only in the case when one branch end must be at some poles. But here, it is not the case.
• Also in none of the options A, B, and C two zeroes are given and option D is already ruled out.
• (s + 1) & (s + 2) are poles. Now our target is to determine whether s = -7 & s = -4 are zero or pole.
• It is confirmed from root locus that one of them is a pole and one of them is a zero.

Let us try to find out centroid.

In our root locus plot, centroid must be in LHS of s-plane

Option A is ruled out since s + 1 is a pole.

Option B:

\(\frac{{s + 4}}{{\left( {s + 1} \right)\left( {s + 2} \right)\left( {s + 7} \right)}}\)

The centroid is calculated as:

\(\frac{{\sum real\;part\;of\;poles - \sum real\;part\;of\;zeroes}}{{P - z}}\)

P = No. of poles

z = No. of zeroes

Centroid will be:

\(= \frac{{\left( { - 1 - 2 - 7} \right) - \left( { - 4} \right)}}{2}\)

\(= \frac{{ - 10 + 4}}{2}\)

= - 3

Centroid (-3, 0) in LHS of the s-plane

Option C \( = \frac{{s + 7}}{{\left( {s + 1} \right)\left( {s + 2} \right)\left( {s + 4} \right)}}\) 

\(= \frac{{\sum real\;part\;of\;poles - \sum real\;part\;of\;zeroes}}{{P - z}}\)

\(= \frac{{ - 1 - 2 - 4 - \left( { - 7} \right)}}{2}\)

= 0

Hence option B is correct.

And the root locus plot will be:

\(G\left( s \right) = \frac{{\left( {s + 4} \right)}}{{\left( {s + 1} \right)\left( {s + 2} \right)\left( {s + 7} \right)}}\)

Angle of Asymptotes is:

\({\phi _A} = \frac{{\left( {2k + 1} \right)180^\circ }}{{P - z}}\) 

P – z = 3 – 1 = 2

When \(k = 0 = \frac{{180^\circ }}{2} = 90^\circ \)

\(k = 1 = \frac{{3\; \times \;180^\circ }}{2} = 270^\circ \) 

Option B is correct.