CONTROL SYSTEM - ROOT LOCUS TECHNIQUE MCQs

This section of Control Systems Multiple Choice Questions and Answers (MCQs) focuses on “Root Locus Techniques ”. These MCQs are very helpful for the preparation of University exams, Engineering exams, and various competitive exams.

CONTROL SYSTEM – ROOT LOCUS TECHNIQUE MCQs

1.The root locii starts from …….. with k=0.

A) an open loop zero

B) an open loop poles

C) closeloop zero

D) close loop pole

ANSWER= (B) an open-loop poles
Explanation:- The root locus starts from an open-loop pole with k=0 and terminates either on open-loop zero or infinity with k=∞.

2. Any point on the real axis is a part of the root locus if and only if :

A) The number of poles to its left is even.

B) The number of poles to its left is odd.

C) The number of poles and zeros to its right is even.

D) The number of poles and zeros to its right is odd.

ANSWER= (D) The number of poles and zeros to its right is odd.
Explanation:- Any point on the real axis is a part of the root locus if and only if the sum of a number of poles and zeros to the right of this point is odd.

3. The number of branches of root locus terminating at infinity is equal to:

A) Number of open-loop poles minus zeros.

B) Number of open-loop poles plus zeros.

C) Number of close loop poles minus zeros.

D) Number of close loop poles plus zeros.

ANSWER= (A) Number of open-loop poles minus zeros.
Explanation:- The root locus terminates either on open-loop zero or infinity. The number of branches of root locus terminating at infinity is equal to number of open-loop poles minus zeros(P-Z).

4. The forward path transfer function of a unity feedback system is given by G(s)=K/s(s+4)(s+5). Determine the angle of Asymptotes.

A) 60º, 150º, 300º

B) 60º, 180º, 300º

C) 80º, 150º, 300º

D) 60º, 120º, 300º

ANSWER= (B) 60º , 180º, 300º
Explanation:-The branches of root locus tend to infinity along a straight line called asymptotes. The angle of asymptotes is given by ∅=(2K+1)180º/(P-Z).The total number of asymptotes = P-Z.

5. The centroid of asymptotes of the unity feedback system having transfer function G(s)=K/s(s+2)(s²+ 6s+25) is

A) -1

B) 2

C) -3

D) -2

ANSWER= (D) -2
Explanation:- The transfer function has no zeros and 4 poles i.e 0,-2,-3+j4 ,-3-j4. so the centroid of asymptotes is calculated by using formula σ = (sum of poles – sum of zeros)/(P-Z) =(0-2-3+j4 -3-j4)/(4-0) =-2

6. If root locus lies between two adjacent open-loop poles on the real axis then there will be :

A) At least one break-in point.

B) At least one break-away point.

C) No break-away or break-in point.

D) Maybe both break-away and break-in point occurs.

ANSWER= (B) At least one breakaway point.
Explanation:- If the root locus lies between two adjacent open-loop poles on the real axis then there will be at least one breakaway point. If the root locus lies between two adjacent open-loop zeros on the real axis then there will be at least one break-in point. If the root locus lies between open-loop poles and zeros, then there will be no breakaway or break-in point or maybe both occur.

7. Determine the break away point of the root locus for the openloop transfer function G(s)H(s)=K/s(s+2)(s+4) is:

A) -3.15

B) 3.15

C) -0.85

D) 0.85

ANSWER= (C) -0.85
Explanation:- The break away points are calculated by the roots of dk/ds =0. The characteristic equation of the given transfer function is s(s+2)(s+4)+k=0 then we get k= -(s³+ 6s² +8s) ,so dk/ds = – (3s³+ 12s +8)=0 .Hence s = -0.85 and -3.15. Since -3.15 is not the part of the root locus therefore break away point is -0.85.

8. The point of intersection of root locus branch of the unity feedback open loop transfer function G(s)=K/s(s+4)(s+5) with the imaginary axis is:

A) +j4.47 and -j4.47

B) +j3.16 and -j3.16

C) +j2.5 and -j2.5

D) +j2 and -j2

ANSWER= (A) +j4.47 and -j4.47
Explanation:- The intersection of root locus branch with jw axis can be determined through the Routh Hurwitz criterion. For the given transfer function , in the Routh table for K=180 ,the auxiliary equation A(s)=9s²+ k =0 .so s=+j4.47 and -j4.47.

9.The point of intersection of root locus with the imaginary axis may be calculated by using:

A) Nyquist criterion

B) Bode plot

C) Polar plot

D) Routh criterion

ANSWER= (D)Routh criterion
Explanation:- The point of intersection of root locus with the imaginary axis may be calculated by using the Routh criterion and obtain auxiliary equation A(s) from the Routh table and equate it equals to zero i.e A(s)=0.

10. The end point of root locus is:

A) open loop pole

B) closed loop pole

C) open loop zero

D) closed loop zero

ANSWER= (C) open loop zero
Explanation:- The root locus terminates either on open-loop zero or infinity.

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