# the laplace transform - 𝐿sin2 using laplace transform 𝐿sin2 = ... transform of derivatives ....

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The Laplace Transform

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INTRODUCTION

Definition

Let be a function defined for 0. Then the integral

=

0

Notation

Examples =

=

=

Let see some examples 1

Solve 1 using Laplace Transform

1 = (1)

0

=

0

0

=

0

=

0

= 0 +1

=

1

Let see some examples

Solve using Laplace Transform

= ()

0

=

0

0

=

2 0

=()()

()

2

0 0

0

2=

1

2

DIff Integr

+

- 1

+

0

2

Let see some examples 5

Solve 5 using Laplace Transform

5 = (5)

0

= (+5)

0

(+5)

0

=(+5)

( + 5) 0

=(+5)

( + 5)

+5 0

( + 5)=

1

+ 5

Let see some examples sin 2

Solve sin 2 using Laplace Transform

sin 2 = (sin 2)

0

= (sin 2)

0

(sin 2)

0

= sin 2

2 cos 2

2

4sin 4

2

0

1 +4

2 (sin 2)

0

= sin 2

2 cos 2

20

DIff Integr

+ sin 2

- 2 cos 2

+

4 sin 2

2

Let see some examples sin 2 cont

1 +4

2 (sin 2)

0

= sin 2

2 cos 2

20

0 0 2 =2

s2

Thus

1 +4

2 (sin 2)

0

=2

2

And

(sin 2)

0

=2

2 + 4

Let see some examples (piecewise)

Try this

cos 4

2

2 + sin

TRANSLATION THEOREM

Translation on the s-Axis

Examples

Translation on the t-Axis

Example

Solution

Evaluate

CONVOLUTION AND TRANSFORM OF PERIODIC FUNCTION

Transform of Derivatives

Transform of Derivatives

Contoh:- + 2 + = 0

Then *+ = 2 0 0

And *+ = 0

And *+ = ()

Thus ,2 0 0 - + 2 0 + = 0

Convolution

Convolution

Transform of Periodic Function

Example

Solution

Solution(cont.)

INVERSE LAPLACE TRANSFORM

Example

Inverse Laplace transform

Solve 12

1

3

2

First, try to expand it

14

2

4

4+

1

6

= 14

2 1

4

4+ 1

1

6

Inverse Laplace transform

Step 2nd

14

2 1

4

4+ 1

1

6

= 411

2 41

1

4+ 1

1

6

Now, let solve one by one

Inverse Laplace transform

Given 411

2, we need to find which theorem

in your laplace table match this Laplace Transform

So, we know that =!

+1

Then since we knew that +1 = 2

We can conclude that + 1 = 2, = 1

Inverse Laplace transform

Now, we know n=1

So,

1 =1!

2=

1

2

Eh!, the laplace that we are trying to solve is 4

2 , so we need to modify a bit.

Inverse Laplace transform

Try to match:-

=

2=

4

2

Wow, obviously, = 4,

So

411

2= 4

Inverse Laplace transform

Try to solve others:-

411

4

+ 1 = 4, = 3

3 =3!

4=

6

4

411

4=

6

4

=2

3

Thus

411

4=

2

33

Inverse Laplace transform

Try to solve others:-

11

6

+ 1 = 6, = 5

5 =5!

6=

120

6

11

6=

120

6

=1

120

Thus

411

4=

1

1205

Inverse Laplace transform

12

1

3

2

= 4 2

33 +

1

1205

Try this

Solution

Solution (cont)

PROPERTIES OF INVERSE LAPLACE TRANSFORM

Inverse of 1st and 2nd translation

Example of 1st translation inverse

Solution

Example of 2nd translation inverse

Inverse of

Example

Solution

SOLUTION OF INITIAL VALUE PROBLEM

Application

Example

Solution

Solving Linear ODE

Application

Figure 1

1

3

2

Application

Solve the system Figure 1 under the conditions E(t) = 60 V, L = 1 h, R = 50 Ohm, C = 104 f, and the currents 1 2 are initially zero.

Given that:-

1

+ 2 =

2

+ 2 1 = 0

Application

How to solve it?

1st step

1

+ 2 =

1

+ 502 = 60

And

2

+ 2 1 = 0

50 1042

+ 2 1 = 0

Application

How to solve it? 2nd Step Applying the Laplace transform to each equation of the system and simplifying gives

1

+ 502 = 60

=> ,1 1(0)- + 502 =60

50 1042

+ 2 1 = 0

0.005,2 2(0)- + 2 1 = 0

Application

How to solve it? 2nd Step (Cari 2 )

1 + 502 =60

0.0052 + 2 = 1

0.0052 + 2 + 502 =60

2 2 + 200 + 10000

200=

60

2 =12000

+ 100 2

Application

How to solve it?

2nd Step (Cari 1 ) 0.0052 + 2 = 1

0.00512000

+ 100 2+

12000

+ 100 2= 1

1 =60

+ 100 2+

12000

+ 100 2

1 =60 + 12000

+ 100 2

Application

How to solve it?

3rd Step

Solving the system for 1 and 2 and decomposing the results into partial fractions gives

1 =60 + 12000

+100 2=

6

5

6

5 +100

60

+100 2

2 =12000

+ 100 2=

6

5

6

5 + 100

120

+ 100 2

Application

How to solve it?

4th step

1 t =6

5

6

5e100t 60te100

2 t =6

5

6

5e100t 120te100