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)(sY sdt 

 yd  L

  n

n

n

OBTAINING TRANSFER FUNCTION BY USING LAPLACE

TRANSFORMATION

Up until now we have been obtaining transfer function using BlockDiagram and Reduction method. However, another simple way to obtain

the transfer function of a system is by using Laplace Transformation

property.

From the definition of Transfer Function itself:

“ Ratio of Laplace transform of the output to the Laplace transform of the input

under all initial condition are set to zero.”

Then the

Therefore:

  )(

)(

)(

)()(

s X 

sY 

t  x

t  y L

input  L

output  LsG  

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Example :

Find the transfer function G(s) for the following differential equations if the

input is u(t) and output is y(t).

)(4)(

5)(

8)(

42

2

t udt 

t du y

dt 

t dy

dt 

t  yd 

Solution :

  )(4)0()()(5)0()(8)0()0()(4   2 sU U ssU sY  yssY  ysysY s  

 All initial condition are set to zero, thus y(0) and U(0) are zero

)(4)()(5)(8)(4  2

sU ssU ssY ssY sY s     )(4584)(   2 sU ssssY   

584

4

)(

)()(

2

ss

s

sU 

sY sG

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OK Now lets go back to solving Transfer Function for Electrical Network

system…..

The relationship between voltage and current and charge under zero initialcondition are as table below;

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For electrical system impedance s I sV 

s Z    )(

 As an example :

Show that the transfer function G(s)= V(s)/I(s) of a pure capacitor.

dt 

t dvC t i  )(

)(  

Taking Laplace both sides;  

dt 

t dvC  Lt i L

  )()(

)()(   ssCV s I   

  sC s I sV 

sG  1

)(  

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In solving electrical system, there are 2 important laws normally used;

1. Kirchoff Voltage Law – Sum of voltage around a closed loop is equal

to zero

2. Kirchoff Current Law – Sum of current flowing into a node is equal to zero.

Example 1 :

Find the transfer function G(s) for the electrical circuit if

Output :   )(t vc Input : )(t vi

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Solution :

Transform the circuit into this circuit

a) Using Voltage Divider method

s

s

s

sV 

sV 

input  L

output  LsG

i

c

211

1

)(

)()(

12

1)(

2

sssG

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 Analogy between Electrical system with Mechanical system networks

It is already known that there are 3 passive components in mechanical

system that is

a. Spring b. Mass c. Damper  

For mechanical system, impedance, Z(s) is defined;

nt  Displaceme

Force

s X 

sF 

s Z    )(

)(

)(

Mechanical properties are analogous to electrical properties that is;

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Force, Velocity and Displacement translation for spring mass and damper 

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The basic concept are;

1. Number of equation of motion required is equal to the number of linearly

independent motion2. Linear independence is defined as a point of motion in a system that can

still move if all other points of motion are held still.

3. For mechanical system, equation that describes the motion is known as

equation of motion.

4. Linear independence is known as degree of freedom.5. In solving mechanical system, one important law must be used that is

Newton’s Law that states that the sum of forces on a mass is equals to

zero.

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Example 1 :

Find the transfer function)(

)()(

sF 

s X sG   for the system shown below;

Using Newton Law , we obtain

)(

)(

t  f kx x f  x M 

 x M t  f kx x f 

v

v

 Applying Laplace transform both sides, we obtain

  sF k sf  M ss X sF skX s X sf s MX s

v

v

2

2

  k s f  MssF s X 

sG

v

 

2

1)(

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Example 2:

[ Sum of impedance at x1]X1(s) – [Sum of impedance between x1 and x2]X2(s)

= [Sum of applied force at x1]

Transfer Function for mechanical motion that has more than 1 DOF

Rule 1 :

 – [Sum of impedance between x1 and x2]X1(s)+ [ Sum of impedance at x2 ]X2(s)

= [Sum of applied force at x2]

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Solution :

Transformed the system

Sum of impedance at x1

34

21221

2

2

21211

2

ss

sss

K K sf sf  M s vv

Sum of impedance between x1 and x2

s22

Sum of applied force at x1

0

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Sum of impedance at x2

2322

2K sf sf  M s vv  

Sum of applied force at x2

)(sF 

Put all into Rule 1 to obtain

    022)(34 212   s X ss X ss

  sF s X sss X s     )(24222 22

1

Put into Cramer’s rule

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Class Exercise;

  )157(

13:

23

2

ssss

s

sF 

s X  Ans

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