Meeting w3 chapter 2 part 1

Chapter 2 Analog Control SystemEddy Irwan Shah Bin SaadonDept. of Electrical EngineeringPPD, [email protected]

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Outline:1. Introduction2. Laplace Transform – Table/ Theorem/ Eg.3. Common Time Domain Input Function4. Transfer Function – Open/ Closed Loop & Eg.5. Electrical Elements Modelling – Table & Eg.6. Mechanical Elements Modelling - Table & Eg.7. Block Diagram Reduction - Table & Eg.8. System Response – Poles/ Zeros, Second

Order, Steady State Error, Stability Analysis

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1. Intro - Objective of this chapter

After completing this chapter you will be able to:

Describe the fundamental of Laplace transforms.

Apply the Laplace transform to solve linear ordinary differential equations.

Apply Mathematical model, called a transfer function for linear time-invariant electrical, mechanical and electromechanical systems.

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2. What is Laplace Transform?

Laplace transform is a method or techniques used to transform the time (t) domain to the Laplace/frequency (s) domain

What is algebra & calculus?

Time Domain Frequency Domain

Differential equations

Input q(t)

Output h(t)

Algebraic equations

Input Q(s)

Output H(s)

Calculus Algebra

Laplace Transformation

Inverse Laplace Transformation

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Laplace Transform (cont.)

The Laplace transform solution consists of the following three steps:

(1) the Laplace transformation of q1(t) and (r dhldt + h = Gq) to frequency domain

(2) the algebraic solution for H(s)

(3) the inverse Laplace transformation of H(s) to time domain h(t).

(4) The calculus solution is shown as step 4.

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Definition of the Laplace Transform

Laplace transform is defined as

Inverse Laplace transform is defined as

)(tf )()(0

sFdtetf st

L

L-1

j

j

st tfdsesFj

sF

)()(

2

1)]([

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Laplace Theorem

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Laplace Table

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Example 1

Find the Laplace transform for 1)( tfSolution:

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Example 2

Find the Laplace transform for atetf )(

Solution:

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Example 3

Find the inverse Laplace transform of

Solution: 21

32

10)(

ssssF

Expanding F(s) by partial fraction:

Where,

Then, taking the inverse Laplace transform

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Example 4

Given the ,solve for y(t) if all initial conditions are zero. Use the Laplace transform method.

Solution:Substitute the corresponding F(s) for each term:

Solving for the response:

Where, K1= 1 when s=0 K2=-2 when s=-4 K3= 1 when s=-8

Hence

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3. Common Time Domain Input Functions Unit Step Function

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Unit Ramp Function

cont.

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Unit Impulse Function

cont.

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4. Transfer Function

Definition:Ratio of the output to the input; with all initial conditions are zero

If the transformed input signal is X(s) and the transformed output signal is Y(s), then the transfer function M(s) is define as;

From this,

Therefore the output is

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TF of Linear Time Invariant Systems In practice, the input-output relation of lines time-invariant system

with continuous-data input is often described by a differential equation

The linear time-invariant system is described by the following nth-order differential equation with constant real coefficients;

c(t) is output

r(t) is input

).()(

....)()(

)()(

.....)(

)(

)(

011

1

1

011

1

1

trbdt

tdrb

dt

trdb

dt

trdb

tcadt

tdca

dt

tcda

td

tcda

m

m

mm

m

m

n

n

nn

n

n

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cont.

Taking the Laplace transform of both sides,

If we assume that all initial conditions are zero, hence

Now, form the ratio of output transform, C(s) divided by input transform. The ratio, G(s) is called transfer function.

).(___)(....)()(

)(___)(.....)()(

01

1

01

1

trofconditioninitialsRbsRsbsRsb

tcofconditioninitialsCasCsasCsam

mm

m

nn

nn

)().....()().....( 011

1011

1 sRbsbsbsbsCasasasa mm

mm

nn

nn

)....(

)....(

)(

)()(

011

1

011

1

asasasa

bsbsbsb

sR

sCsG

nn

nn

mm

mm

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cont.

The transfer function can be represented as a block diagram

General block diagram

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Block Diagram of Open Loop System

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Block Diagram of Closed Loop System

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Example 1

Problem: Find the transfer function represented by

Solution:

Taking the Laplace transform of both sides, assuming zero initial conditions, we have

The transfer function, G(s) is

)()(2)(

trtcdt

tdc

)()(2)( sRsCssC

2

1

)(

)()(

ssR

sCsG

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Example 2

Problem: Use the result of Example 1 to find the response, c(t), to an input, r(t)=u(t), a unit step and assuming zero initial conditions.

Solution:Since r(t)=u(t), R(s)=1/s, hence

Expanding by partial fractions, we get

Finally, taking the inverse Laplace transform of each term yields

)2(

1)()()(

sssGsRsC

2

2/12/1)(

ss

sC

tetc 2

2

1

2

1)(

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Example 3 Problem: Find the transfer function, G(s)=C(s)/R(s), corresponding to the

differential equation

Solution:

rdt

dr

dt

rdc

dt

dc

dt

cd

dt

cd34573

2

2

2

2

3

3

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Example 4

Problem: Find the differential equation corresponding to the transfer function,

Solution:

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12)(

2

ss

ssG

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Example 5

Problem: Find the ramp response for a system whose transfer function is,

Solution:

)8)(4()(

ss

ssG

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