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## How to prepare Root Locus for GATE ECE 2017 || Solving Root Locus Simplified

Root locus is important topic from GATE point of view. One or two questions are always expected from this topic. Experts at GATE- Prepladder shares  tips and tricks to solve questions on Root Locus

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The Root Locus is defined as locus of closed loop poles obtained when system gain K which varies from 0 to ∞.

It determines the relative stability of system w.r.t. variance in system gain K. How to prepare root locus from gate point of view is discussed.

## Tips and Tricks To solve questions on Root Locus

#Tip 1

Angle criterion is used to find whether any point in plane lies on root locus or not. When it is confirmed that the point is lying on root locus then using magnitude criterion we can find the value of K at that point.

< G(s) H(s) = ± (2K+1)180

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#Tip2

The denominator tells the number of poles and numerator tells the number of open loop zero.

G(s) H(s) = K (s+3)/(s+2)

One open loop pole s= -2 . One open loop zero at  s= -3

#Tip3

The number of branches of root locus are

N= P, if P > Z

= Z, if Z > P

= P = Z, if P = Z

Where

P = Number of finite poles

Z = Number of finite zeros

N = Number of separate root loci

#Tip4

The angle of asymptotes is given by

Φ = (2K+1)180º/(P-Z)

Where K= 0,1,2,...........

The total number of asymptotes = P – Z

#Tip5

Break away points are given by dK/ds

#Tip6

Generally we plot the root locus for different values of K where

In Standard Form

G(s) H(s) = K (s+a)/(s+b)(s+c)

But if it is given that

G(s) H(s) = 5 (s+8)/s (s+K)

Then for the plot, we have to find the equivalent transfer function of standard forward means.

For this characteristic equation becomes

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

1+ 5 (s+8) /s (s+ K) = 0

s (s+K) + 5 (s+8) = 0

s2 +5s + 40 + Ks = 0

Dividing by (s2+ 5s + 40) both side

1+ Ks/s2 + 5s + 40 = 0

Let new transfer function become G1(s) H1(s), then

G1(s) H1(s) = Ks/ s2 + 5s + 40

Which is standard form.

## Solved Example

Q1     If G(s) H(s) = K/s (s2 + 6s +10), find break away point.

Soln.      1 + G(s) H(s) = 0

s(s2 + 6s + 10) + K = 0

K = -s3 – 6s2 – 10s

dK/ds = -(3s2 + 12s+ 10) = 0

Therefore,  s1 = -1.1835 and s2 = -2.815 are the break away points

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## Examples From GATE

Question : The feedback configuration and the pole-zero locations of

G(s) = s2- 2s+2/( s2+ 2s+2)

are shown below. The root locus for negative values of k, i.e. for k –infinity<k<3 , has breakaway/break-in points and angle of departure at pole P (with respect to the positive real axis) equal to

Solution :

must read : Most Important Topics For GATE-2017 (ECE)

Question :

Solution :

Any point on real axis of s- is part of root locus if number of OL poles and zeros to right of that point is even. Thus (B) and (C) are possible option. The characteristics equation is

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

Or 1+k(1-s)/(s(s+3) =0

Or k = s2+3s/(s(s+3)

For break away & break in point

dk/ds =0 ,

-s2 + 2s + 3 =0

Hence  1 - must be the break away point and 3 must be the break in point.

Best Wishes !!

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