root locus 1

Root Locus Analysis (1)Root Locus Analysis (1)Hany FerdinandoHany Ferdinando

Dept. of Electrical Eng.Dept. of Electrical Eng.

Petra Christian UniversityPetra Christian University

General OverviewGeneral Overview

This section discusses how to plot the This section discusses how to plot the Root Locus methodRoot Locus method

Step by step procedure is used inline with Step by step procedure is used inline with an examplean example

Finally, some comments are given as the Finally, some comments are given as the complement for this sectioncomplement for this section

Why Root LocusWhy Root Locus

Closed-loop poles’ location determine the Closed-loop poles’ location determine the stability of the systemstability of the system

Closed-loop poles’ location is influenced Closed-loop poles’ location is influenced as the gain is variedas the gain is varied

Root locus plot gives designer information Root locus plot gives designer information how the gain variation influences the how the gain variation influences the stability of the systemstability of the system

Plot ExamplePlot Example

-6 -5 -4 -3 -2 -1 0 1 2-4

-3

-2

-1

0

1

2

3

4Root Locus

Real Axis

Imag

inar

y Ax

is

Important Notes:Important Notes:

Poles are drawn as ‘x’ while zeros are Poles are drawn as ‘x’ while zeros are drawn as ‘o’drawn as ‘o’

Gain at poles is zero, while gain at zeros Gain at poles is zero, while gain at zeros is infinityis infinity

Pole is the starting point and it Pole is the starting point and it mustmust finish finish at zero; therefore, for every pole there at zero; therefore, for every pole there should be corresponding zeroshould be corresponding zero

Root locus is plot on the s planeRoot locus is plot on the s plane

StandardizationStandardization

G(s)

H(s)

+-

R(s) C(s)

)()(1

)(

)(

)(

sHsG

sG

sR

sC

Find Characteristic Equation!!Find Characteristic Equation!!

1 + G(s)H(s) = 01 + G(s)H(s) = 0

How to make it?How to make it?

1.1. Start from the characteristic equationStart from the characteristic equation

2.2. Locate the poles and zeros on the s planeLocate the poles and zeros on the s plane

3.3. Determine the root loci on the real axisDetermine the root loci on the real axis

4.4. Determine the asymptotes of the root lociDetermine the asymptotes of the root loci

5.5. Find the breakaway and break-in pointsFind the breakaway and break-in points

6.6. Determines the angle of departure (angle of Determines the angle of departure (angle of arrival) from complex poles (zeros)arrival) from complex poles (zeros)

7.7. Find the points where the root loci may cross Find the points where the root loci may cross the imaginary axisthe imaginary axis

Example (1)Example (1)

)2)(1()(

sss

KsG 1)( sH

1. Start from the characteristic equation1. Start from the characteristic equation

0)2)(1(

1

sss

K

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

Imag

inar

y Pa

rt

Example (2)Example (2)

2. Locate the poles and zeros on the s plane2. Locate the poles and zeros on the s plane

Example (3)Example (3)

3. Determine the root loci on the real axis3. Determine the root loci on the real axis

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

Imag

inar

y Pa

rt

Example (4a)Example (4a)

4. Determine the asymptotes of the root loci4. Determine the asymptotes of the root loci

03

)12(180

ko

103

]0[)]2()1(0[

Example (4b)Example (4b)

4. Determine the asymptotes of the root loci4. Determine the asymptotes of the root loci

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Real Part

Imag

inar

y Pa

rt

Example (5)Example (5)

5. Find the breakaway and break-in point5. Find the breakaway and break-in point

From the characteristic equation, findFrom the characteristic equation, find

)23( 23 sssK

then calculate…then calculate…

0)263( 2 ssds

dK

s = -0.4266 and s = -1.5744s = -0.4266 and s = -1.5744

Example (6)Example (6)

6. Determine the angle from complex pole/zero6. Determine the angle from complex pole/zero

This example has no complex This example has no complex poles and zeros, therefore, poles and zeros, therefore, this step can be skipped!!!this step can be skipped!!!

polepole = 0 – sum from pole + sum from zero= 0 – sum from pole + sum from zero

zerozero = 0 – sum from zero + sum from pole = 0 – sum from zero + sum from pole

Example (7)Example (7)

7. Find the points where the root loci may cross the 7. Find the points where the root loci may cross the imaginary axisimaginary axis

Do this part by substituting jDo this part by substituting j for all s in the for all s in the characteristic equationcharacteristic equation

0)2()3(

0)(2)(3)(32

23

jK

Kjjj

= = ±√2, K = 6 or ±√2, K = 6 or = 0, K = 0 = 0, K = 0

Example (8)Example (8)

-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5-1.5

-1

-0.5

0

0.5

1

1.5

Root Locus

Real Axis

Imag

inar

y Ax

is

CommentsComments

nnthth degree algebraic equation in s degree algebraic equation in s If n-mIf n-m≥2 then a1 is negative sum of the roots of ≥2 then a1 is negative sum of the roots of

the equation and is independent of Kthe equation and is independent of K It means if some roots move on the locus toward the It means if some roots move on the locus toward the

left as K increased then the other roots left as K increased then the other roots mustmust move move towards the right as K is increasedtowards the right as K is increased

n1n

1n

m1m

1m

a...sas

)b...sbK(sH(s)G(s)

Root Locus in MatlabRoot Locus in Matlab

Function rlocus(num,den) draws the Root Locus Function rlocus(num,den) draws the Root Locus of a system. Another version in state space is of a system. Another version in state space is rlocus(A,B,C,D)rlocus(A,B,C,D)

01 den

numThe characteristic equationThe characteristic equation

Those functions are for negative Those functions are for negative feedback (normal transfer function)feedback (normal transfer function)

Next…Next…

Topic for the next meeting is Root Locus Topic for the next meeting is Root Locus in positive feedbackin positive feedback