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CHAPTER 2LAPLACE TRANSFORM

AND TRANSFER FUNCTION

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LAPLACE TRANSFORM AND TRANSFER FUNCTION

2.1 Explain the concept of Laplace Transform2.2 Understand the concept of transfer

function2.3 Understand block diagram representation2.4 Explain Signal Flow Graph representationIdentify the Mason’s gain formula

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Definition of Laplace Transform

The transform method is used to solve certain problems, that are difficult to solve directly.

In this method the original problems is first transformed and solved.

Laplace transform is one of the tools for solving ordinary linear differential equations.

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First : Convert the given differential equation from time domain to complex frequency domain by taking Laplace transform of the equation

From this equation, determine the Laplace transform of the unknown variable

Finally, convert this expression into time domain by taking inverse Laplace transform

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Laplace transform method of solving differential equations offers two distinct advantages over classical method of problem solving

From this equation, determine the Laplace transform of the unknown variable

Finally, convert this expression into time domain by taking inverse Laplace transform

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The Laplace transform is defined as below: Let f(t) be a real function of a real variable t

defined for t>0, then

Where F(s) is called Laplace transform of f(t). And the variable ‘s’ which appears in F(s) is frequency dependent complex variable

It is given by, where = Real part of complex variable ‘s’ = Imaginary part of complex

variable ‘s’

dt e.(t) f f(t) LF(s) st-

0

jω σ s

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Example 1

Find the Laplace transform of e-at and 1 for t ≥ 0

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Find the Laplace transform of e-at and 1 for t ≥ 0

Solution : i) f(t) = e-at

ii) f(t) = 1

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Inverse Laplace Transform

The operation of finding out time domain function f(t) from Laplace transform F(s) is called inverse Laplace transform and denoted as L-1

Thus,

The time function f(t) and its Laplace transform F(s) is called transform pair

f(t) (f(t)) LL F(s)L -1-1

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Properties of Laplace Transforms

The properties of Laplace transform enable us to find out Laplace transform without having to compute them directly from the definition.

The properties are given:

A) The Linear PropertyThe Laplace transformation is a linear

operation – for functions f(t) and g(t), whose Laplace transforms exists, and constant a and b, the equation is : bLg(t) aLf(t) g(t) b f(t) a L

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B) DifferentiationAccording to this property, It means that inverse Laplace transform of

a Laplace transform multiplied by s will give derivative of the function if initial conditions are zero.

C) n-fold differentiationAccording to this property,

f(0) - sLf(t) dt

df(t) L

(0)f ... - (0)fs- f(0)s- Lf(t)s dt

f(t)d L 1-n'2-n1-nn

n

n

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D) Integration Property

In general, the Laplace transform of an order n is

Laplace transform exists if f(t) does not grow too fast as

s

)0(f Lf(t)

s

1 )( L

1

0

t

f

s

)0(f...

s

)0(f

s

)0(f Lf(t)

s

1

f(t)dt....L

n

1-n

2-n

n

1-n

n

n

t

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E) Time Shift

The Laplace transform of f(t) delayed by time T is equal to the Laplace transform of f(t) multiplied by e-sT ; that is

L[f ( t – T ) u( t – T )] = e-sT F(s), where u (t – T) denotes the unit step function, which is shifted to the right in time by T.

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F) Convolution IntegralThe Laplace transform of the product of

two functions F1(s) and F2(s) is given by the convolution integrals

where L-1F1(s) = f1(t) and L-1F2(s) = f2(t)

d)(f)-(tf

)d-t(ft)(f s)(s)F(FL

2

t

0

1

2

t

0

1211-

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G) Product TransformationThe Laplace transform of the product of two

functions f1(t) and f2(t) is given by the complex convolution integral

H) Frequency Scaling

The inverse Laplace transform of the functions

)d(F)(Fj 2

1 t)(t)f(fL 2

jc

j-c

1211-

f(t) F(s)L whereaf(at), a

sF 1-

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I) Time ScalingThe Laplace transform of a functions

Lf(t) F(s) whereaF(as) a

tfL

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J) Complex TranslationIf F(s) is the Laplace transform of f(t) then

by the complex translation property,

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K) Initial Value Theorem

The Laplace transform is very useful to find the initial value of the time function f(t). Thus if F(s) is the Laplace transform of f(t) then,

The only restriction is that f(t) must be continuous or at the most , a step discontinuity at t=0.

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L) Final Value Theorem

Similar to the initial value, the Laplace transform is also useful to find the final value of the time function f(t).

Thus if F(s) is the Laplace transform of f(t) then the final value theorem states that,

The only restriction is that the roots of the denominator polynomial of F(s) i.e poles of F(s) have negative or zero real parts

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Example 2

Find the Laplace transform of sin t

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Find the Laplace transform of sin t

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f(t) F(s)

1 1/s

Constant K K/s

K f(t), K is constant K F(s)

t 1/s2

tn n/sn+1

e-at 1/s+a

eat 1/s-a

e-at tnn/((s+a)n+1 )

sin t /(s2 + 2)

cos t s/(s2 + 2)

e-at sin t /((s+a)2 + 2)

Table of Laplace Transforms:Table 1 : Standard Laplace Transform pairs

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Table 1 Contd….

f(t) F(s)

e-at cos t (s+a)/((s+a)2 + 2)

sinh t /(s2 - 2)

cosh t s/(s2 - 2)

t e-at 1/(s+a)2

1 - e-at a/s(s+a)

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Table 2 : Laplace transforms of standard time functions

Function f(t) Laplace Transform F(s)

Unit step = u(t) 1/s

A u(t) A/s

Delayed unit step = u(t-T) e-Ts/s

A u(t-T) Ae-Ts /s

Unit ramp = r(t) = t u(t) 1/s2

At u(t) A/s2

Delayed unit ramp = r(t-T) = (t-T) u(t-T)

e-Ts /s2

A(t-T) u(t-T) Ae-Ts /s 2

Unit impulse = (t) 1

Delayed unit impulse = (t-T) e-Ts

Impulse of strength K i.e K (t) K

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Inverse Laplace Transform

Let F(s) is the Laplace transform of f(t) then the inverse Laplace transform is denoted as,

The F(s), in partial fraction method, is written in the form as,

Where N(s) = Numerator polynomial in s D(s) = Denominator polynomial in s

F(s)L f(t) -1

D(s)

N(s) F(s)

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Simple and Real Roots

The roots of D(s) are simple and realThe function F(s) can be expressed as,

where a, b, c… are the simple and real roots of D(s).

The degree of N(s) should be always less than D(s)

c)...-b)(s-a)(s-(s

N(s)

D(s)

N(s) F(s)

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This can be further expressed as,

where K1, K2, K3 … are called partial fraction coefficients

The values of K1, K2, K3 … can be obtained as,

.... c)-(s

K

b)-(s

K

a)-(s

K

c)...-b)(s-a)(s-(s

N(s) F(s) 321

cs

bs

as

F(s) c).-(s K

F(s) b).-(s K

F(s) a).-(s K

3

2

1

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In general, and so on

Where sn = nth root of D(s)

Is standard Laplace transform pair.Once F(s) is expressed in terms of partial

fractions, with coefficients K1, K2 … Kn, the inverse Laplace transform can be easily obtained

nss F(s) .)s-(s K nn

a)(s

1 eL at

... eK eK eK F(s)L f(t) ct3

bt2

at1

-1

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Example 3

Find the inverse Laplace transform of given F(s)

4)3)(ss(s

2)(s F(s)

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Solution : The degree of N(s) is less than D(s). Hence F(s) can be expressed as,

)4s(2

1 -

3)(s3

1 6

1 F(s)

2

1-

)3(-4)x(-4

2)(-4

4)3)(ss(s

2)(s).4(s F(s) 4).(s K

3

1

4)(-3)x(-3

2)(-3

4)3)(ss(s

2)(s3).(s F(s) 3).(s K

6

1

3x4

2

4)3)(ss(s

2)(ss. F(s) s. K where

4)(s

K

3)(s

K

s

K F(s)

44s3

33s2

00s1

321

s

s

s

s

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Taking inverse Laplace transform,

4t-3t- e 2

1 - e

3

1

6

1 f(t)

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Multiple Roots

The given function is of the form,

There is multiple root of the order ‘n’ existing at s=a.

The method of writing the partial fraction expansion for such multiple roots is,

where represents remaining terms of the expansion of F(s)

D(s) a)-(s

N(s) F(s)

n

(s)D'

(s)N'

a)-(s

K ...

a)-(s

K

a)-(s

K

a)-(s

K F(s) 1-n

2-n2

1-n1

n0

(s)D'

(s)N'

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A separate coefficient is assumed for each power of repetative root, starting from its highest power n to 1

For ease of solving simultaneous equations, find the coefficient K0 by the same method for simple roots

Finding the Laplace inverse transform of expanded F(s) refer to standard transform pairs,

as F(s) .a)-(s K n

0

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Example 4

Obtain the inverse Laplace transform of given f(s)

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Complex Conjugate Roots

If there exists a quadratic term in D(s) of F(s) whose roots are complex conjugates then the F(s) is expressed with a first order polynomial in s in the numerator as,

Where (s2 + s + ) is the quadratic whose roots are complex conjugates while represents remaining terms of the expansion.

The A and B are partial fraction coefficients.

(s)D'

(s)N'

) s (s

B As F(s)

2

(s)D'

(s)N'

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The method of finding the coefficients is same as the multiple roots.

Once A and B are known, then use the following method for calculating inverse Laplace transform.

Consider A and B are know.

Now complete the square in the denominator by calculating last term as,

Where L.T = Last term M.T =

Middle term F.T = First term

) s (s

B As (s)F

21

4(F.T.)

(M.T.) L.T.

2

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4 -

2

s

B As

4

4

s s

B As s)(F

4 L.T.

2

2222

21

2

where

Now adjust the numerator As + B in such a way that it is of the form

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Example 5

Find the inverse Laplace transform of

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Application of Laplace Transform in Control System

The control system can be classified as electrical, mechanical, hydraulic, thermal and so on.

All system can be described by integrodifferential equations of various orders

While the o/p of such systems for any i/p can be obtained by solving such integrodifferential equations

Mathematically, it is very difficult to solve such equations in time domain

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The Laplace transform of such integrodifferential equations converts them into simple algebraic equations

All the complicated computations then can be easily performed in s domain as the equations to be handled are algebraic in nature.

Such transformed equations are known as equations in frequency domain

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By eliminating unwanted variable, the required variable in s domain can be obtained

By using technique of Laplace inverse, time domain function for the required variable can be obtained

Hence making the computations easy by converting the integrodifferential equations into algebraic is the main essence of the Laplace transform