Root Locus Analysis (3)

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# Root Locus Analysis (3) - PowerPoint PPT Presentation

Root Locus Analysis (3). Hany Ferdinando Dept. of Electrical Eng. Petra Christian University. General Overview. This section explain how to use the Root Locus method to design compensator The compensator is lead and lag compensator Students need some basic geometry for precision drawing

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## Root Locus Analysis (3)

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### Root Locus Analysis (3)

Hany Ferdinando

Dept. of Electrical Eng.

Petra Christian University

General Overview
• This section explain how to use the Root Locus method to design compensator
• The compensator is lead and lag compensator
• Students need some basic geometry for precision drawing
• Pre-requisite: student has to understand how to plot Root Locus in complex plane
What is compensator
• Compensation is the modification of the system dynamic to satisfy the given specification
• The lag-lead compensator will not be discussed
• The compensator adds zero and pole to the system
• It pulls the root locus to the right
• It tends to lower the system’s relative stability
• It slows down the settling time
• It pulls the root locus to the left
• It tends to make the system more stable
• It speeds up the settling time
Proof

Proof that

If R1C1 > R2C2 then it is lag network, otherwise it is lead network

Consider

s = -1 ± j√3

Based on the closed-loop poles, z = 0.5, wn = 2 rad/s and Static velocity error (Kv) = 2s-1

Proof it!!

Problem:

It is desired to modify the closed-loop poles so that the wn = 4 rad/s without changing the damping ratio z

For the wn = 4 rad/s without changing the damping ratio z, the desired closed-loop poles are inline with the line between the origin and the closed-loop poles of the system

s = -2 ± j2√3

If G(s) is the open-loop transfer function, then the open-loop of compensated system is

How to find a and T????

Desired closed-loop pole

A

P

/2

/2

C

B

D

O

PB is the bisector between PA and PO

f is the desired angle

Angle of

at s = -2+j√3 is -210o

Therefore the f is 30o

From the previous slide method, the point C and D is approximately -5.4 and -2.9

You can try it by yourself!!

From the previous slide, you can calculate the T, aT and all Rs and Cs until you get:

where K = 4Kc

K = 18.7916

Compensated open-loop transfer function is

with Gc(s) is

Find the value for all Rs and Cs in the circuit

Lag Compensator (1)

Consider

s = -0.3307 ± j0.5864

Based on the closed-loop poles, z = 0.491, wn = 0.673 rad/s and Static velocity error (Kv) = 0.53s-1

Proof it!!

Lag Compensator (2)

Problem:

It is desired to increase the Kv to about 5s-1 without appreciably changing the location of the dominant closed-loop poles

Lag Compensator (3)

The new Kv is about 10 times of the old Kv, therefore the b (a in the Lead Compensator) is set to 10.

The zero of the compensator is also 10 times of the pole.

We get

K = 1.06Kc

Lag Compensator (4)

The Root Locus plot of the compensated system is very close to the uncompensated one. Therefore, we need Matlab to help us.

The new dominant closed-loop pole with the same z is -0.31±j0.55 (from Matlab). The gain K is 1.0235

Find the pole in complex plane

To find the desired dominant complex pole in the Root Locus plot, one can use rlocfind(num,den) function. Output of this function is the selected pole and the gain at that pole.

When this function is called, there is a pointer one can use to select a pole in the complex plane. Usually, the Root Locus is drawn first.

Next…

Bode diagram as a tool for frequency response analysis is the next topic.

Prepare yourself for this…