(2) laplace transform

7/31/2019 (2) Laplace Transform

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Assistance lecturer

Helwan University

Faculty Of Engineering At Mattaria


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We need an easy method for solving linear differentialequations with initial conditions

This method will transform linear differential equations tosimple algebraic equations

Example:

y+ 2y 3y = g (t)

with the initial conditionsy (0) =y(0) = 0,

is transformed to: s2Y + 2sY 3Y = G, where Y and G are thetransforms ofy and g, respectively.

Thus, we easily get: Y = G/(s2 + 2s 3),

Finally, we apply the inverse transform to gety from Y.


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Letf(t) be a function defined and piecewise continuous on the

interval (0 t< ). We define its Laplace transform F(s) by:

0

)()]([)( dttfetfLsF st

Note thatf(t) is a function oft, while its transform F(s ) is a

function ofs .

F(s ) may or may not exist.


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f(t) = 1

f(t) = t

f(t) = t2

f(t) = tn

f(t) = eat

f(t) = cos (t)

f(t) = sin (t)

Find the Laplace transform of the following functions:


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f (t) f (s )

e at

tn

cos (t)

sin (t)

cosh (t)

sinh (t)

as

1

1

!

ns

n

22

s

22s

s

22s

s

22

s


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otherwise

tifktf

0

52)(

Find the Laplace transform of the following function:

Solution:

Applying the definition we get:


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Let F(s) = L[f(t)], and G (s) = L [ g (t) ], and let a andb be constants. Then:

L [a f(t) + b g(t)] = a F(s) + b G (s)

Proof :

Follows easily from the integral definition of the Laplacetransform.

(Note that integration is linear.)


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3cos)(

ttf

Find the Laplace transform of :Solution:Note that:

Now apply linearity to get:


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Let F(s) = L [f(t)], and let a be a constant. Then:

L [e atf(t)] = F(s a ) = F(s )|s sa

Proof:

Also follows easily from the integral definition:

0

)(

0

)()())(( dttfedttfeetfeL tasatstat


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Solution:Using First Shifting, we get:

2

2

2)2(

2

)2(

11

)()(

sstLteL ssss

t

Find the Laplace transform of:

f(t) = t e 2t


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Given a function F(s ), we seek a functionf(t), for

which F(s ) = L[f(t)].

f(t) is then called the inverse Laplace transformofF(s ), writtenf(t) = L1[F(s )].

We can findf(t) by consulting the table of basic

transforms and/or apply the properties of theLaplace transform.


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Inverting the rules for the Laplace transform, we get:

)!1(

11

1

n

t

sL

n

n

)sin(11

22

1t

s

L

)]([)(11

sFLesFLat

ass


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3

2 32)(

s

sssF

!2321

13

12

1)]([

2

32

11 tt

sssLsFL

Find the inverse Laplace transform of :

We break the fraction into 3 parts, and apply linearity to get :

Solution:


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52)(

2

ss

ssF

)2sin(2

1)2cos(

4

1

4)1()]([

2

1

2

11

tte

s

sLe

s

sLsFL

t

t


We first complete the square, apply first shifting and linearity

to get:

Solution:


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82)(

2

ss

ssF

)3sinh(3

1)3cosh(

9

1

9)1(

)]([2

1

2

11

tte

s

sLe

s

sLsFL

t

t


Complete the square first, and apply first shifting and linearity:Solution 1:


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24)2)(4()(

s

B

s

A

ss

ssF

tt eesFL241

3

1

3

2)]([

82)(

2

ss

ssF

Now multiply both sides by (s + 4)(s 2), substitute s = 4, 2 to get

A = 2/3 andB = 1/3. Finally, apply linearity to get:


Factorize the bottom, and split the fraction:

Solution 2:


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The previous method, called the method ofPartial Fractions,is

useful when we need to find the inverse Laplace transform of a

rational function,

i.e. a fraction of polynomials of the formf(s )/ g (s ).

It is also useful in integrating rational functions.

The method allows us to write any rational functionsf(s )/g (s ) ,

wherefhas degree less than g, as a sum of simple fractions.


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)1)(2(

1)(

22

sss

ssF

12)1)(2(

12222

s

EDs

s

C

s

B

s

A

sss

s

Now multiply both sides by s 2 (s 2)(s 2+ 1), and substitute s = 0, 2 to

get:A = 1/2 and C= 1/20. Also, comparing the coefficients ofs, s 4,

and s 3, we get:B = 1/4,D = 1/5, andE= 3/5.

Find the partial fraction expansion of :

Solution:We can write:


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)1)(2(

1)(

22

sss

ssF

1

)5/3()5/1(

2

20/14/12/1

)1)(2(

12222

s

s

ssssss

s

ttetsFLt

sin5

3cos

5

1

20

1

4

1

2

1)]([

21


Solution:Since

We apply linearity to get:


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Let F(s ) = L [f(t)], then:

L [f(t)] = s F(s ) f(0)

Proof: Using integration by parts, we get:

)0()(

)()0()(lim

)()()()()]([

0

0

0

0

0

fssF

dttfesfetfe

dttfestfedttfetfL

stst

t

ststst


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Let F(s) = L [f(t)], then:

L [f(t)] = s 2F(s) s f(0) f(0)

In General:

L (f (n) (t)) = s n F(s) s n1f(0) s n2f(0) f (n1)(0)

Proof:L[f(t)] = s L [f(t)] f(0)

= s [s F(s ) f(0)] f(0), etc..


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Let F(s ) = L [f(t)], then:

)(1

)(0 sFsdfL

t

Proof: Perform Integration by parts in the integral

definition.

We thus have:

t

dfsFs

L0

1)()(

1


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Starting with an initial value problem of the form

a y+ b y+ c y = g (t),

withy (0) andy (0) given,

we apply the Laplace Transform on both sides to get:

A (s 2 Ys y (0) y(0)) + b (s Yy (0)) + c Y = G (s),

Now solve the algebraic equation for Y. Then get the inverse transformy = L1 (Y).


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Taking the Laplace Transform on both sides,

we get:

s Y1 2Y = 5/(s 2)

Solving for Y, we get:

Y = 1/(s 2) + 5/(s 2)2

Taking the inverse transform, we get:

y = e 2t+ 5t e 2t= (1 + 5t)e 2t

Solvey2y = 5e 2t,y (0) = 1


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Taking the Laplace Transform on both sides, we get:[s 2 Ys(3) 0] + [s Y3] 2Y = 6/s

Solving for Y, we get:Y = (3s 2 + 3s + 6)/[s (s 1)(s + 2)]

Breaking Y into its partial fractions, we get:

Y = 3/s + 4/(s 1) + 2/(s + 2)

Taking the inverse transform, we get:

y = 3 + 4e t+ 2e 2t

Solvey+ y2y = 6, withy (0) = 3,y(0) = 0


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(2) laplace transform

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