Question

Consider the transfer function

\(G(s)H(s)=\frac{K}{(s^3+4s^2+s-6) }\)

The root-locus plot of the system passes through s = 0. The value of K at this point will be

1. 10
2. 0
3. 6
4. 8

Answer 1

Correct Answer - Option 3 : 6

Analysis:

Method-1:

\(G(s)H(s)=\frac{K}{(s^3+4s^2+s-6) }\)

For root locus we use magnitude criteria to find gain.

i.e \(\left| {Gain} \right| = 1\) at some 's' value.

at s = 0

\(\left| {\frac{k}{{0 + 0 + 0 - 6}}} \right| = 1\)

k = 6

Method-2:

Given, the open-loop transfer function as:

\(G(s)H(s)=\frac{K}{(s^3+4s^2+s-6) }\)

The root-Locus will intersect the jω-axis at s = 0 and can be found using Routh Array.

Now, the characteristic equation is given as:

1 + G(s)H(s) = 0

s³ + 4s² + s - 6 = 0

For the characteristic equation, we form the Routh’s array a

s3	1	1
s2	4	4 - K
s1	\(\frac{4-K+6}{4}\)	0
s0	K - 6

The root-locus plot of the system passes through s = 0

So, put s⁰ row  = 0

k - 6 = 0

K = 6