LAPLACE TRANSFORM · JABATAN KEJURUTERAAN ELETRIK, POLITEKNIK MERLIMAU 27.11.2014 . Laplace...

LAPLACE TRANSFORM EE602 : CIRCUIT ANALYSIS

SARIATI DALIB JABATAN KEJURUTERAAN ELETRIK,

POLITEKNIK MERLIMAU 27.11.2014

Laplace Transform

SARIATI DALIB Page 1

Chapter 2: Laplace Transform

2.1 Understand the use of Laplace Transform and Inverse Laplace Transform in

solving network analysis

2.1.1 Define the Laplace Transform of an expression by using the integral

definition.

𝐹(𝑠) = ∫ 𝑒−𝑠𝑡∞

0𝑓(𝑡) 𝑑𝑡

2.1.2 Determine the Laplace Transform of a function by using table.

Table Laplace Transform

2.1.3 Use the linearity property. a) 𝐿𝑓 + 𝑠 = 𝐿𝑓 + 𝐿𝑠

b) 𝐿 𝑘𝑓 = 𝑘 𝐿 𝑓 if k is constant

2.1.4 Use the first shift theorem. 𝑙𝑒𝑎𝑡 𝑓(𝑡)𝑠 = 𝐹(𝑠 − 𝑎)

2.1.5 Use the Laplace Transform of derivatives.

𝐿 𝑡𝑛 𝑓(𝑡)𝑠 = (−1)𝑛

𝐹𝑛(𝑠)

2.1.6 Use the Laplace Transform of integrals. 𝐿 𝑡 𝑓(𝑡)𝑠 = 𝐹′ (s)

2.1.7 Determine the Inverse Laplace Transforms of some standard functions.

Table Inverse Laplace Transform

𝐿−1

𝐹(𝑠) = 𝑓(𝑡)

Linearity Property : 𝐿𝑓 + 𝑠 = 𝐿𝑓 + 𝐿𝑠 𝐿 𝑘𝑓 = 𝑘 𝐿 𝑓

First Shift Theorem:

𝐿−1𝐹 (𝑠 − 𝑎) = 𝑒𝑎𝑡 𝑓(𝑡)

2.1.8 Compute the Inverse Laplace Transforms using partial fractions.

2.1.9 Compute the Inverse Laplace Transforms by completing the square.

2.1.10 Compute the Laplace Transform of first and second derivatives.

2.1.11 Apply the Laplace Transform to solve differential equations.

2.1.12 Apply the Laplace Transform in RLC circuit analysis:

a. RL Series circuit

b. RC Series circuit

c. LC Series circuit

d. RLC Series circuit

e. RLC Parallel circuit

Laplace Transform


Chapter 2 : Laplace Transform

Introduction:

The main idea behind the Laplace Transformation is that we can solve an

equation (or system of equations) containing differential and integral terms

by transforming the equation in "t-space" to one in "s-space". This makes the

problem much easier to solve. The kinds of problems where the Laplace

Transform is invaluable occur in electronics..

If needed we can find the inverse Laplace transform, which gives us the

solution back in "t-space".

Definition of Laplace Transform

Let be a given function which is defined for . If there exists a

function so that

,

Then is called the Laplace Transform of , and will be denoted by

. Notice the integrator where is a parameter which may be

real or complex.

Thus,

The symbol which transform into is called the Laplace transform

operator.

Laplace Transform


Laplace Transform by Direct Integration

To get the Laplace transform of the given function , multiply by

and integrate with respect to from zero to infinity. In symbol,

Example 1 : Find the Laplace transform of when by using Direct

Integration Method

Solution 01

ℒ𝑓(𝑡) = ∫ 𝑒−𝑠𝑡𝑓(𝑡)𝑑𝑡∞

0

ℒ1 = ∫ 𝑒−𝑠𝑡(1)𝑑𝑡∞

0

ℒ1 = ∫ 𝑒−𝑠𝑡𝑑𝑡∞

0

ℒ1 = −1

𝑠[ 𝑒−𝑠𝑡]

∞

0

= −1

𝑠[

1

𝑒−𝑠𝑡]

= −1

𝑠 [

1

∞−

1

𝑒0]

= −1

𝑠(0 − 1) =

1

𝑠

Thus,

Laplace Transform


Example 2: Find the Laplace transform of by using Direct Integration

Method

Solution 02

Thus,

Laplace Transform


Example 3: Find the Laplace transform of . by using Direct

Integration Method

Solution 03

For .

Using integration by parts: . Let

Using integration by parts again. Let

http://www.mathalino.com/reviewer/integral-calculus/integration-by-parts-techniques-of-integration

Laplace Transform


Thus,

Therefore,

Find the Laplace transform of the following when by using Direct

Integration Method

1. 𝑓(𝑡) = 4

2. 𝑓(𝑡) = 𝑒2𝑡

3. 𝑓(𝑡) = 𝑒−4𝑡

Exercise 1

Laplace Transform


Table of Laplace Transforms of Elementary Functions

Below are some functions and their Laplace transforms .

e-at 1

𝑠 + 𝑎

Laplace Transform


Use table to find the laplace transform of the following;

a) 𝑓(𝑡) = 4 d) 𝑓(𝑡) = 𝑐𝑜𝑠 6𝑡

b) 𝑓(𝑡) = 𝑡3 e) 𝑓(𝑡) = 𝑠𝑖𝑛 3𝑡

c) 𝑓(𝑡) = 𝑒2𝑡 f)𝑓(𝑡) = 𝑒−3𝑡

Exercise 2

Laplace Transform


Properties of Laplace Transform

1. Constant Multiple

If is a constant and is a function of , then

Example:

1. 𝑓(𝑡) = 4 𝑐𝑜𝑠𝑡

𝐹(𝑠) = 4 [𝑠

𝑠2 + 12]

=4𝑠

𝑠2 + 1

2. 𝑓(𝑡) = 5𝑡

𝐹(𝑠)= 5 ℒ t

= 5 [1

𝑠2]

= 5

𝑠2

Linearity Property

2. Linearity Property

If and are constants while and are functions of whose

Laplace transform exists, then

Proof of Linearity Property

Laplace Transform


This property can be easily extended to more than two functions as shown

from the above proof. With the linearity property, Laplace transform can also

be called the linear operator.

Example 01: Linearity

Find the Laplace transform of .

Solution 01

𝑇ℎ𝑢𝑠 𝐹(𝑠) = 5 − 2𝑠

𝑠2

Exercise : Linearity

Determine the Laplace Transform of following functions by using table and

theorems

1. 𝑓(𝑡) = 6 𝑒−5𝑡 + 𝑒3𝑡 + 5𝑡3 − 9

2. 𝑓(𝑡) = 4𝑐𝑜𝑠4𝑡 − 9𝑠𝑖𝑛4𝑡 + 2𝑐𝑜𝑠10𝑡

3. 𝑓(𝑡) = 3 sin3

2 𝑡

4. 𝑓(𝑡) = 5 𝑒2𝑡 + 4 sin 3𝑡

5. 𝑓(𝑡) = 3 − 3𝑒−2𝑡

6. 𝑓(𝑡) = 6𝑡 + 5𝑡3

7. 𝑓(𝑡) = 2𝑠𝑖𝑛4𝑡 – 7𝑐𝑜𝑠 3𝑡

8. 𝑓(𝑡) = 3 + 2𝑡 − 6𝑡2

Laplace Transform


3. First Shifting Property

If , when then,

In words, the substitution for in the transform corresponds to the

multiplication of the original function by .

Proof of First Shifting Property

Example 01:First Shifting Property


Solution 01

………(1)

ℒ𝑒2𝑡 =1

𝑠 − 2… … … . (2)

Replace s in (1) with (s-2) Thus,

Laplace Transform


Example 02: First Shifting Property

Find the Laplace transform of 𝑓(𝑡) = 𝑒−5𝑡𝑠𝑖𝑛3𝑡

Solution 02

Using table : ……..(1)

ℒ𝑒−5𝑡 =1

𝑠 + 5… … … . (2)

Replace S in (1) with (S+5)

Thus,

ℒ(𝑒−5𝑡 sin 3𝑡) =3

(𝑠2 + 10𝑠 + 25) + 9=

3

𝑠2 + 10𝑠 + 34

Example 03 : First Shifting Property


Solution 03

Using table : ℒ𝑐𝑜𝑠𝑡 𝑡 =𝑠

𝑠2+ 12 ……..(1)

ℒ𝑒−3𝑡 =1

𝑠 + 3… … … . (2)

Replace S in (1) with (S+3)

Thus,

Laplace Transform


Exercise: First Shifting Property

Determine the Laplace Transform of following functions by using table and

theorems

1. 𝑓(𝑡) = 𝑒2𝑡𝑐𝑜𝑠2𝑡

2. 𝑓(𝑡) = 𝑡 𝑒2𝑡

3. 𝑓(𝑡) = 𝑒3𝑡(2𝑡 + 3)

4. 𝑓(𝑡) = 𝑒−𝑡𝑡

5. 𝑓(𝑡) = 𝑒𝑡 sin 2𝑡.

Laplace Transform


Laplace Transform of Derivatives

For first-order derivative:

For second-order derivative:

For third-order derivative:

For nth order derivative:

Proof of Laplace Transform of Derivatives

Using integration by parts,

Thus,

Apply the limits from 0 to ∞:

Laplace Transform


Example 01: Laplace Transform of Derivatives

Find the Laplace transform of using the transform of derivatives.

Solution 01

..........

..........

..........

Example 02: Laplace Transform of Derivatives


Solution 02

..........

Laplace Transform


Example 03 : Laplace Transform of Derivatives


Solution 03

..........

Laplace Transform


Laplace Transform of Intergrals

Theorem

If , then

Proof

Let

then, and

Taking the Laplace transform of both sides,

From Laplace transform of derivative, and from the

Theorem above,

Thus,

http://www.mathalino.com/reviewer/advance-engineering-mathematics/laplace-transforms-derivatives

Laplace Transform


Example 01: Laplace Transform of Intergrals


Solution 01

Hence,

Laplace Transform


Summary of Laplace Transform Properties

Properties of Laplace Transform

1. Linearity Property: Constant Multiple

If is a constant and is a function of , then

2. Linearity Property- addition/subtraction of function.

If and are constants while and are functions of whose

Laplace transform exists, then

Example:

3. First Shifting Property

If , then,

4. Transforms of Derivatives

The Laplace transform of the derivative exists when , and

In general, the Laplace transform of nth derivative is

5. Transforms of Integrals

Theorem

If , then

http://www.mathalino.com/reviewer/advance-engineering-mathematics/linearity-property-laplace-transform

http://www.mathalino.com/reviewer/advance-engineering-mathematics/first-shifting-property-laplace-transform

Laplace Transform


The Inverse

Laplace Transform

Laplace Transform


The Inverse Laplace Transform

Definition

From , the value is called the inverse Laplace transform of

.

In symbol,

where is called the inverse Laplace transform operator.

Table of Inverse Laplace Transforms of Elementary Functions

𝐹(𝑠) = ℒ𝑓(𝑡) 𝑓(𝑡) = ℒ−1F(s)

Laplace Transform


Exercise : Inverse Laplace Transform

Find the inverse Laplace transform of the following;

a) 𝐹(𝑠) = 1

𝑠

b) 𝐹(𝑠) = 2

𝑠

c) 𝐹(𝑠) = 2

3𝑠

d) 𝐹(𝑠) = 1

𝑠2

e) 𝐹(𝑠) = 4

𝑠2

f) 𝐹(𝑠) = − 1

𝑠

g) 𝐹(𝑠) = 5

𝑠3

h) 𝐹(𝑠) = 12

𝑠3

i) 𝐹(𝑠) = 3

𝑠+4

j) 𝐹(𝑠) = 2

2𝑠−6

k) 𝐹(𝑠) = 8

9 +𝑠2

l) 𝐹(𝑠) = 3

𝑠+6

m) 𝐹(𝑠) = 𝑠

𝑠2+16

n) 𝐹(𝑠) = 3𝑠

𝑠2+9

Laplace Transform


Theorems on Inverse Laplace Transformation

Theorem 1: Linearity Theorem

If a and b are constants,

Example 01: : Linearity Theorem

Find the inverse transform of 𝐹(𝑠) = 8

𝑠3+

3

𝑠2+

1

𝑠

Solution 01

𝑓(𝑡) = ℒ−1 8

𝑠3+

3

𝑠2+

1

𝑠

= 8ℒ−1 1

𝑠3+ 3ℒ−1

1

𝑠2+ ℒ−1

1

𝑠

= 8 [𝑡3−1

(3 − 1)!] + 3 [

𝑡2−1

(2 − 1)!] + [𝑡]

= 8 𝑡2

2+ 3

𝑡

1+ 1

= 4 𝑡2 + 3𝑡 + 1


Find the inverse transform of 𝐹(𝑠) =6

𝑠2+9

Solution 02

𝑓(𝑡) = ℒ−16

𝑠2 + 9

ℒ−1 3

𝑠2 + 9 = 6 ℒ−1

1

𝑠2 + 32

=6

3ℒ−1

3

𝑠2 + 32

= 2𝑠𝑖𝑛3𝑡

Laplace Transform



Find the inverse transform of .

Solution 03

𝑓(𝑡) = ℒ−1 5

𝑠 − 2−

4𝑠

𝑠2 + 9

= 5ℒ−1 [1

𝑠 − 2] − 4ℒ−1 [

𝑠

𝑠2 + 32]

= 5𝑒2𝑡 − 4𝑐𝑜𝑠3𝑡

Exercise : : Linearity Theorem

Using table ,find the inverse Laplace transform of :

a) 𝐹(𝑠) = 2

𝑠3 − 5

𝑠

b) 𝐹(𝑠) =3

𝑠+

1

𝑠+3

c) 𝐹(𝑠) = 𝑠+3

𝑠2+9

Laplace Transform


Theorem 2: First Shift Theorem

Example 1: First Shift Theorem

Find the inverse transform of 4𝑠+4

(𝑠+1)2+9

Solution 1:

𝑓(𝑡) = ℒ−1 4𝑠 + 4

(𝑠 + 1)2 + 9

= ℒ−1 4(𝑠 + 1)

(𝑠 + 1)2 + 9

= 4𝑒−𝑡ℒ−1 𝑠

𝑠2 + 32

= 4𝑒−𝑡𝑐𝑜𝑠3𝑡


Find the inverse transform of 2

(𝑠+3)4

Solution 2:

𝑓(𝑡) = ℒ−1 2

(𝑠 + 3)4 = 2𝑒−3𝑡ℒ−1

1

𝑠4

= 2𝑒−3𝑡ℒ−1 3!

𝑠3+1 𝑥

13!

= 2𝑒−3𝑡𝑥1

6ℒ−1

3!

𝑠3+1

= 𝑒−3𝑡

3 𝑡3

Laplace Transform



Find the inverse transform of 𝑠+4

(𝑠+2)2+4

Solution 3:

𝑓(𝑡) = ℒ−1 𝑠 + 4

(𝑠 + 2)2 + 4 = ℒ−1

𝑠 + 2 + 2

(𝑠 + 2)2 + 22

= ℒ−1 𝑠 + 2

(𝑠 + 2)2 + 22 +

2

(𝑠 + 2)2 + 22

= 𝑒−2𝑡ℒ−1𝑠

𝑠2 + 22+ 𝑒−2𝑡 ℒ−1

2

𝑠2 + 22

= 𝑒−2𝑡𝑐𝑜𝑠2𝑡 + 𝑒−2𝑡𝑠𝑖𝑛2𝑡

Exercise: First Shift Theorem

Find the inverse laplace transform of the following by using table and

theorems :

𝑎) 𝐹(𝑠) =3

𝑠 + 1

𝑏) 𝐹(𝑠) =3

(𝑠 + 1)6

𝑐) 𝐹(𝑠) =3

(𝑠 + 1)6 + 9

𝑑) 𝐹(𝑠) =𝑠 + 2

(𝑠 + 2)6 + 16

Laplace Transform


Inverse Laplace Transform by Partial Fraction Expansion

This technique uses Partial Fraction Expansion to split up a complicated

fraction into forms that are in the Laplace Transform table. As you read

through this section, you may find it helpful to refer to the section on partial

fraction expansion techniques.

1. Distinct Real Roots

Consider first an example with distinct real roots.

Example: Distinct Real Roots

Q: Find the inverse Laplace Transform of:

Solution:

We can find the two unknown coefficients using the "cover-up" method or

residue method:

𝐴 =𝑠 + 1

(𝑠 + 2)|

𝑠= 0

=1

2

𝐵 =𝑠 + 1

𝑠|

𝑠= −2=

−1

−2=

1

2

So

𝐹(𝑠) = 1/2

𝑆+

1/2

𝑠 + 2

And the inverse laplace

𝑓(𝑡) =1

2 +

1

2𝑒−2𝑡

http://lpsa.swarthmore.edu/BackGround/PartialFraction/PartialFraction.html

http://lpsa.swarthmore.edu/LaplaceZTable/LaplaceZFuncTable.html



http://lpsa.swarthmore.edu/BackGround/PartialFraction/PartialFraction.html#cover-up

Laplace Transform


2. Repeated Real Roots

Consider next an example with repeated real roots (in this case at the origin,

s=0).

Example: Repeated Real Roots

Q: Find the inverse Laplace Transform of the function F(s).

Solution 1:

We can find two of the unknown coefficients ( A and C)using the "cover-

up" method.

𝐴 =𝑠2 + 1

𝑠2|

𝑠= −2

=5

4

𝐶 =𝑠2 + 1

(𝑠 + 2)|

𝑠= 0

=1

2

We find B using cross-multiplication: 𝑠2 + 1 = 𝑠2𝐴 + 𝑠(𝑠 + 2)𝐵 + (𝑠 + 2)𝐶

𝑠2 + 1 = 𝑠2𝐴 + 𝐵𝑠2 + 2𝐵𝑠 + 𝐶𝑠 + 2𝐶

Equating like powers of "s" gives us:

power of "s" left side

coefficient

right side

coefficient

s2 1 A + B

s1 0 2B + C

s0 1 2C

𝐴 + 𝐵 = 1

𝐵 = 1 − 𝐴 = 1 −5

4= −

1

4

𝐹(𝑠) =5/4

𝑠 + 2−

1/4

𝑠+

1/2

𝑠2

And inverse laplace : 𝑓(𝑡) =5

4 𝑒−2𝑡 −

1

4+

1

2𝑡



http://lpsa.swarthmore.edu/BackGround/PartialFraction/PartialFraction.html#cross-multiplication

Laplace Transform


Solution 2:

We can find two of the unknown coefficients ( A and C)using cross –

multiplication.

𝑠2 + 1 = 𝑠2 ( 𝑠 + 2) (𝐴

𝑠 + 2+

𝐵

𝑠+

𝐶

𝑠2 )

𝑠2 + 1 = 𝑠2𝐴 + 𝑠(𝑠 + 2)𝐵 + (𝑠 + 2)𝐶

𝐿𝑒𝑡 𝑠 = −2 , 4 + 1 = 4𝐴

𝑇ℎ𝑢𝑠 , 𝐴 =5

4

𝐿𝑒𝑡 𝑠 = 0 , 0 + 1 = 2𝐶

𝑇ℎ𝑢𝑠 , 𝑐 =1

2

𝐿𝑒𝑡 𝑠 𝑏𝑒 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟, 𝑠 = 1 , 1 + 1 = 𝐴 + 3𝐵 + 3𝐶

2 =5

4+ 3𝑏 +

1

2𝑥3

3𝐵 = 2 −3

2−

5

4

3𝐵 = −3

4

𝑇ℎ𝑢𝑠 , 𝐵 = −1

4

𝐹(𝑠) = 𝑠2 + 1

𝑠2(𝑠 + 2)=

54

𝑠 + 2 +

−14

𝑠+

12

𝑠2

𝑓(𝑡) =5

4𝑒−2𝑡 −

1

4+

1

2𝑡

Many texts use a method based upon differentiation of the fraction when

there are repeated roots.

Laplace Transform


3. Complex Roots

Another case that often comes up is that of complex conjugate

roots. Consider the fraction:

The second term in the denominator cannot be factored into real terms. This

leaves us with two possibilities - either accept the complex roots, or find a

way to include the second order term.

Simplify the function F(s) so that it can be looked up in the Laplace

Transform table.

Solution:

Another way to expand the fraction without resorting to complex numbers

is to perform the expansion as follows.

𝐴 =𝑠 + 3

𝑠2 + 4𝑠 + 5|

𝑠= −5

= −5 + 3

(−5)2 + 4(−5) + 5

= −2

25 − 20 + 5

= −2

10= −

1

5

We can find the quantities B and C from cross-multiplication.

𝑠 + 3 = 𝐴(𝑠2 + 4𝑠 + 5) + (𝐵𝑠 + 𝐶)(𝑠 + 5)



Laplace Transform


𝑠 + 3 = (𝐴𝑠2 + 4𝐴𝑠 + 5𝐴) + (𝐵𝑠2 + 5𝐵𝑠 + 𝐶𝑠 + 5𝐶)

𝑠 + 3 = 𝑠2(𝐴 + 𝐵) + 𝑠(4𝐴 + 5𝐵 + 𝐶) + (5𝐴 + 5𝐶)

If we equate like powers of "s" we get

order of

coefficient

left side

coefficient

right side

coefficient

2nd (s2) 0 A+B

1st (s1) 1 4A+5B+C

0th (s0) 3 5A+5C

𝐸𝑞𝑛1. 𝐴 + 𝐵 = 0

𝐴 = −𝐵 𝑇ℎ𝑢𝑠 𝐵 = −𝐴 , 𝐵 = 1/5

Eqn 3. 5A + 5C = 3

5𝑐 = 3 − 5𝐴

𝐶 =3 − 5 (−

15

)

5=

3 + 1

5=

4

5

Thus

𝐹(𝑠) = −

15

𝑆 + 5 +

15

𝑠 +45

𝑠2 + 4𝑠 + 5

Finally, we get

𝐹(𝑠) =−

15

𝑠 + 5 +

(15

) 𝑠 +45

𝑠2 + 4𝑠 + 5

The inverse Laplace Transform is given below :

𝐹(𝑠) =−

15

𝑠 + 5 + (

1

5)𝑥

𝑠 + 4

𝑠2 + 4𝑠 + 5

http://lpsa.swarthmore.edu/LaplaceXform/InvLaplace/InvLaplaceXformPFE.html#Method_2

Laplace Transform


Then use method : Completing the square

𝐹(𝑠) =−

15

𝑠 + 5 + (

1

5)𝑥

𝑠 + 2 + 2

(𝑠 + 2)2 + 1

𝐹(𝑠) =−

15

𝑠 + 5 + (

1

5) [

𝑠 + 2

(𝑠 + 2)2 + 1] + [

2

(𝑠 + 2)2 + 1]

The inverse laplace transform :

𝑓(𝑡) = −1

5𝑒−5𝑡 +

1

5𝑒−2𝑡[𝑐𝑜𝑠𝑡 + 2𝑠𝑖𝑛𝑡 ]

𝑓(𝑡) = −0.2𝑒−5𝑡 + 0.2𝑒−2𝑡[𝑐𝑜𝑠𝑡 + 2𝑠𝑖𝑛𝑡 ]

Laplace Transform


How To Complete The Square

Given a term of the form

αs2+bs+c

it is often useful to express it as

(s + d)2+e

This is often useful when solving certain quadratic equations. In our case it

lets us more easily use Laplace Transform Tables (in the table the form

(s+a)2+ω02 comes up frequently).

Without loss of generality, we will only consider the case where α=1. If we

divide the original equation by through by α and let b=β/α and c=γ/α, we

get

s2+bs+c

and our task is to express it as

(s+d)2+e

We start by setting the two terms to be equal to each other

x2 + bx + c = (x+d)2 + e

Expand the right hand side

x2 + bx + c = x2 + 2dx + d2 + e

Equating the coefficients of like powers of x we get

b = 2d,

c = d2 + e or d = b/2, and e = c - d2


Laplace Transform


Example 1: Completing the square

Complete the square for the expression: s2+2s+10

Solution:

The original function is of the form "s2 + bs + c", so b=2, c=10, and

d = b/2 = 1

e = c - d2 = 10 - 1 = 9.

The desired expression is "(s+d)2 + e" or (s+1)2+9

Thus : s2+2s+10 = (s+1)2+9

Example 2: Completing the square

Complete the square for the expression: x2+4x+29

Solution:

The original function is of the form "x2 + bx + c", so b=4, c=29, and

d = b/2 = 2

e = c - d2 = 29 - 4 = 25.

The desired expression is "(x+d)2 + e" or (x+2)2+25

Thus , x2+4x+29 = (x+2)2+25

Laplace Transform


Exercise 1:

Perform the indicated operation:

Solution 1

For

𝐴 =𝑠 − 5

𝑠 − 2|

𝑠= −3=

−3 − 5

−3 − 2=

8

5

𝐵 =𝑠 − 5

𝑠 + 3|

𝑠= 2=

2 − 5

2 + 3= −

3

5

Thus,

Laplace Transform


Exercise 2:

Find the inverse transform of

Solution 2

For

𝐴 =2𝑠2 + 5𝑠 − 6

(𝑠 + 3)(𝑠 − 5)|

𝑠= 1

=2 + 5 − 6

4(−4)=

1

−16

𝐵 =2𝑠2 + 5𝑠 − 6

(𝑠 − 1)(𝑠 − 5)|

𝑠= −3

=2(−3)2 + 5(−3) − 6

(−3 − 1)(−3 − 5)=

18 − 15 − 6

(−4 )(−8)= −

3

32

𝑐 =2𝑠2 + 5𝑠 − 6

(𝑠 − 1)(𝑠 + 3)|

𝑠= 5

= 2(5)2 + 5(5) − 6

(5 − 1)(5 + 3)=

50 + 25 − 6

(4)(8)=

69

32

Therefore,

Laplace Transform


Using Inverse Laplace Transforms to Solve Differential Equations

The Laplace Transform Of A Derivatives

Let f(t) be a function of f and let F(s) be the laplace transform of f. The value

of f and its derivatives when t=0 are denoted by f(0) , f’(0) , f’’’(0) and so on.

The nth derivatives of f is denoted by fn(t). Then it can be shown that the

laplace transform of fn(t) is given by:

fn(t). = Sn F(s) – Sn-1 f(0) – Sn-2 f’(0) …….f(n-1)(0)

Two common cases are when n=1 and n=2:

f1 (t) = SF(s) – f(0)

f2 (t) = S2 F(s) – S1 f(0) – f ’(0)

Example 1: Derivatives

Q: The laplace transform of f(t) is F(s). Given f(0) = 2 and f’(0) = -3. Write

expression for the laplace transform of :

a) f ‘ (t)

b) f 11 (t)

Solution:

a) 𝐹(𝑠) = 𝑠𝐹(𝑠) − 𝑓(0) = 𝑠𝐹(𝑠) − 2

b) 𝐹(𝑠) = 𝑠2𝐹(𝑠) − 𝑠𝑓(0) − 𝑓′(0) = 𝑠2𝐹(𝑠) − 2𝑠 + 3

Laplace Transform


Example 2: Derivatives

Q; If the initial value, f(0)=y(0) = 2 and f’(0)=y’(0) = -3. Write expression for the

laplace transform of :

a) 2f’’ – 3f’ + f b) 𝒇′′

𝟐+ 𝟐𝒇′ − 𝒇 c)

𝟑𝒅𝟐𝒚

𝒅𝒕𝟐+

𝒅𝒚

𝒅𝒕

Solution;

𝑎) 2𝑓′′ − 3𝑓′ + 𝑓 = 2 [𝑠2𝐹(𝑠) − 𝑠𝑓(0) − 𝑓′(0)] − 3[𝑠𝐹(𝑠) − 𝑓(0)] + 𝐹(𝑠)

= 2[𝑠2𝐹(𝑠) − 2𝑠 + 3] − 3[𝑠𝐹(𝑠) − 2] + 𝐹(𝑠)

= 2𝑠2 𝐹(𝑠) − 4𝑠 + 6 − 3𝑠𝐹(𝑠) + 6 + 𝐹(𝑠)

= 𝐹(𝑠)[ 2𝑠2 − 3𝑠 + 1] − 4𝑠 + 12

𝑏)1

2𝑓′′ + 2𝑓′ − 𝑓 =

1

2[𝑠2𝐹(𝑠) − 𝑠𝑓(0) − 𝑓′

(0)] + 2[𝑠𝐹(𝑠) − 𝑓(0)] − 𝐹(𝑠)

= 1

2[𝑠2𝐹(𝑠) − 2𝑠 + 3] + 2[𝑠𝐹(𝑠) − 2] − 𝐹(𝑠)

= [1

2𝑠2𝐹(𝑠)) − 𝑠 +

3

2] + [2𝑠𝐹(𝑠) − 4] − 𝐹(𝑠)

= 𝐹(𝑠) [1

2𝑠2 + 2𝑠 − 1] − 𝑠 +

3

2− 4

𝐹(𝑠) [1

2𝑠2 + 2𝑠 − 1] − 𝑠 +

5

2

𝑐) 3𝑑2𝑦

𝑑𝑡2+

𝑑𝑦

𝑑𝑡= 3𝑦′′ + 𝑦

= 3[𝑠2𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0)] + 𝑌(𝑠)

= 3[𝑠2𝑌(𝑠) − 2𝑠 + 3] + 𝑌(𝑠)

= 3𝑌(𝑠)(𝑠2 + 1) − 6𝑠 + 9

Laplace Transform


Solving Differential Equation using laplace Transform:

Example 1: Solve DE

Q: Consider the initial value problem.

Solve 𝑑𝑦

𝑑𝑡− 3𝑦 = 0 , 𝑦(0) = 4

Solution:

𝑑𝑦

𝑑𝑡− 3𝑦 = 0

𝑠𝑌(𝑠) − 𝑦(0) − 3𝑌(𝑠 ) = 0

𝑠𝑌(𝑠) − 4 − 3𝑌(𝑠 ) = 0

𝑌(𝑠)[𝑠 − 3] = 4

𝑌(𝑠) =4

𝑠 − 3

Inverse laplace

𝑦(𝑡) = 4𝑒3𝑡

Exercise: Solve DE

1. Solve 𝑑𝑥

𝑑𝑡− 2𝑥 = 2𝑒3𝑡 , 𝑥(0) = 2

2. Solve 𝑑2𝑦

𝑑𝑡2 − 𝑦 = 2 , 𝑦(0) = 𝑦′(0) = 0

3. Solve 𝑦′′ + 𝑦′ − 2𝑦 = 4 , 𝑦(0) = 2, 𝑦′(0) = 1

Laplace Transform


Example 2:Solve DE

Write down the subsidiary equations for the following differential equations

and hence solve them.

𝑑𝑦

𝑑𝑡+ 𝑦 = 𝑠𝑖𝑛3𝑡, given that y = 0 when t = 0.

Solution:

Taking Laplace transform of both sides gives:

𝑠𝑌(𝑠) − 𝑦(0) + 𝑌(𝑠) =3

𝑠2 + 32

𝑌(𝑠)[ 𝑠 + 1] − 𝑦(0) =3

𝑠2 + 9

𝑌(𝑠)[ 𝑠 + 1] =3

𝑠2 + 9 𝑠𝑖𝑛𝑐𝑒 𝑦(0) = 0

Solving for Y(s) and finding the partial fraction decomposition gives:

𝑌(𝑠) =3

(𝑠 + 1)(𝑠2 + 9)=

𝐴

𝑠 + 1+

𝐵𝑠 + 𝐶

(𝑠2 + 9)

𝐴 =3

𝑠2 + 9|

𝑠= −1=

3

10

Apply ‘cross multiplication’:

3 = 𝐴(𝑠2 + 9) + (𝐵𝑠 + 𝐶)(𝑠 + 1)

3 = 𝐴𝑠2 + 9𝐴 + 𝐵𝑠2 + 𝐵𝑠 + 𝐶𝑠 + 𝐶

power of "s" left side

coefficient

right side

coefficient

s2 0 A + B

s1 0 B +C

s0 3 9A + C

http://www.intmath.com/methods-integration/11-integration-partial-fractions.php

Laplace Transform


𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1 ∶ 𝐴 + 𝐵 = 0

𝐵 = −𝐴

= −3

10

𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2; 𝐵 + 𝐶 = 0

𝐶 = −𝐵

= − [−3

10] =

3

10

So

𝑌(𝑠) =3

(𝑠 + 1)(𝑠2 + 9)=

𝐴

𝑠 + 1+

𝐵𝑠 + 𝐶

(𝑠2 + 9)

=

310

𝑠 + 1+

(−3

10) 𝑠 +3

10(𝑠2 + 9)

=3

10[

1

𝑠 + 1+

−𝑠 + 1

(𝑠2 + 32]

= 3

10[

1

𝑠 + 1+

−𝑠

(𝑠2 + 32)+

1

(𝑠2 + 32)]

Finding the inverse Laplace tranform gives us the solution for y as a function

of t:

𝑦(𝑡) =3

10𝑒−𝑡 −

3

10𝑐𝑜𝑠3𝑡 +

1

10𝑠𝑖𝑛3𝑡

Laplace Transform


Example 5 : Solve DE

Solve 𝑑2𝑦

𝑑𝑡2+ 2

𝑑𝑦

𝑑𝑡+ 5𝑦 = 0 , given that 𝑦′(0) = 0, 𝑦(0) = 1when t = 0

Solution:

Taking Laplace transform of both sides and applying initial conditions of y(0) =

1 and y'(0) = 0 gives:

𝑠2𝑌(𝑠) − 𝑠𝑦(0) − 𝑦′(0) + 2(𝑠𝑌(𝑠) − 𝑦(0) + 5𝑌(𝑠) = 0

(𝑠2𝑌(𝑠) − 𝑠) + 2(𝑠𝑌(𝑠) − 1) + 5𝑌(𝑠) = 0

(𝑠2 + 2𝑠 + 5)𝑌(𝑠) = 𝑠 + 2

Solving for Y and completing the square on the denominator gives:

𝑌(𝑠) = 𝑠 + 2

𝑠2 + 2𝑠 + 5=

𝑠 + 2

(𝑠 + 1)2 + 4

=𝑠 + 1 + 1

(𝑠 + 1)2 + 4

=𝑠 + 1

(𝑠 + 1)2 + 22+

1

(𝑠 + 1)2 + 22

=𝑠 + 1

(𝑠 + 1)2 + 22+

1

2

2

(𝑠 + 1)2 + 22

Now, finding the inverse Laplace Transform gives us the solution for y as a

function of t:

𝑦(𝑡) = 𝑒−𝑡𝑐𝑜𝑠2𝑡 +1

2𝑒−𝑡𝑠𝑖𝑛2𝑡

http://www.intmath.com/quadratic-equations/2-solving-quadratic-equations-completing-square.php

Laplace Transform


Example 6: Solve DE

Solve

y'' + y' - 2y = 4 y(0) = 2 y'(0) = 1

Solution

[ y’’ + y’ – 2y] = 4

A table of Laplace transforms is useful here. We get

(s2 y - 2s - 1) + (s y - 2) - 2 y = 4/s

Next, combine like terms to get

(s2 + s - 2) y = 4/s + 2s + 3

Putting under a common denominator, dividing and factoring we get

𝐿𝑌(𝑠) =2𝑠2 + 3𝑠 + 4

(𝑠)(𝑠 − 1)(𝑠 + 2)

Use partial fractions. We write

𝐿𝑌(𝑠) =2𝑠2 + 3𝑠 + 4

(𝑠)(𝑠 − 1)(𝑠 + 2)=

𝐴

𝑠+

𝐵

𝑠 − 1+

𝐶

(𝑠 + 2)

which gives

A(s - 1)(s + 2) + Bs(s + 2) + Cs(s - 1) = 2s2 + 3s + 4

Letting s = 0 gives -2A = 4 A = -2

Letting s = 1 gives 3B = 9 B = 3

Letting s = -2 gives 6C = 6 C = 1

Now we solve

𝐿𝑌(𝑠) =−2

𝑠+

3

𝑠 − 1+

1

(𝑠 + 2)

Now we can use the table to get

𝑦(𝑡) = −2 + 3𝑒𝑡 + 𝑒−2𝑡

http://ltcconline.net/greenl/courses/106/Techniques/parfrac1.htm

Laplace Transform


Apply Laplace Transform In RLC

Circuit Analysis

Laplace Transform


Laplace Transform


Laplace Transform in RL Circuit

Example 1: RL circuit

By using laplace transform, find i(t) when switch is close at t= 0. The initial

current value i(0) = 0.

Solution:

Derive the circuit equation in t-domain then transform these differential

equation in s-domain.

By KVL:

In t-domain; 𝑉 = 𝑖𝑅 + 𝐿 𝑑𝑖/𝑑𝑡

100 = 25𝑖 + 0.1𝑑𝑖

𝑑𝑡

Applying laplace transform:

In s-domain:

100

𝑠= 25𝐼(𝑠) + 0.1[𝑠𝐼(𝑠) − 𝑖(0)]

= 25𝐼(𝑠) + 0.1[𝑠𝐼(𝑠)] = 𝐼(𝑠)[25 + 0.1𝑠]

Divide both sides with 0.1:

1000

𝑠= 𝐼(𝑠)[250 + 𝑠]

𝐼(𝑠) =1000

𝑠(250+𝑠)

By partial Fraction :

𝐼(𝑠) =1000

𝑠(𝑠 + 250)=

𝐴

𝑠+

𝐵

𝑠 + 250

Laplace Transform


𝐴 =1000

(𝑠 + 250)|

𝑠= 0

=1000

0 + 250= 4

𝐵 =1000

𝑠|𝑠= −250

=1000

−250= −4

we get : A = 4 and B = -4,

Thus :

𝐹(𝑠) = 4

𝑠−

4

𝑠 + 250

By inverse laplace transformation, we get: 𝑓(𝑡) = 4 − 4𝑒−250𝑡

Example 2: RL circuit

Find the current i(t) in the RL circuit shown below when switch is close at t= 0.

Solution:

Derive the circuit equation in t-domain then transform these differential

equation in s-domain.

By KVL:

In t-domain; 𝑉 = 𝑖𝑅 + 𝐿 𝑑𝑖/𝑑𝑡

4 = 2𝑖 + 2𝑑𝑖/𝑑𝑡

Laplace Transform


Applying laplace transform both sides:

4

𝑠= 2𝐼(𝑠) + 2[𝑠𝐼(𝑠) − 𝑖(0)]

4

𝑠= 2𝐼(𝑠) + 2[𝑠𝐼(𝑠) − 2]

4

𝑠= 𝐼(𝑠)(2𝑠 + 2) − 4

4

𝑠+ 4 = 𝐼(𝑠)(2𝑠 + 2)

4 + 4𝑠

𝑠= 𝐼(𝑠)(2𝑠 + 2)

𝐼(𝑠)(2𝑠 + 2) = 4(1 + 𝑠)

2𝑠(𝑠 + 1)

𝐼(𝑠) = 2(𝑠 + 1)

𝑠(𝑠 + 1)

𝐼(𝑠) = 2

𝑠

Then Inverse the laplace transform: 𝑖(𝑡) = 2 𝐴𝑚𝑝

Example 3: RL circuits

A series RL circuit with R = 60Ω and L = 10H connected in series with 120V DC

source. At t= 0 , the switch is closed. Find :

a) The current in the circuit, voltage across the resistor and inductor at any

instant of time after t=0.

b) The current at t = 0.65s.

Laplace Transform


Solution;

At t=0 , the switch S is closed. By KVL,

𝑉𝑠 = 𝑖(𝑡)𝑅 + 𝐿𝑑𝑖(𝑡)

𝑑𝑡

120 = 𝑖(𝑡)60 + 10𝑑𝑖(𝑡)

𝑑𝑡

Applying Laplace transformation both sides:

120

𝑠= 60𝐼(𝑠) + 10 [𝑠𝐼(𝑠) − 𝑖(0)]

As initial current is not given, it can be assumed zero. Thus

𝐼(𝑠)[ 60 + 10𝑠] =120

𝑠

𝐼(𝑠)[ 𝑠 + 6] =12

𝑠

𝐼(𝑠) =12

𝑠(𝑠 + 6)

By partial fraction expansion,

𝐼(𝑠) =12

𝑠(𝑠 + 6)=

𝐴

𝑠+

𝐵

𝑠 + 6

𝐴 =12

(𝑠 + 6)|

𝑠= 0

=12

6= 2

𝐵 =12

𝑠|

𝑠= −6=

12

−6= −2

Thus

𝑰(𝒔) =𝟐

𝒔−

𝟐

𝒔 + 𝟔

By inverse laplace transform:

𝒊(𝒕) = 𝟐 − 𝟐𝒆−𝟔𝒕

Laplace Transform


The voltage across the resistor is calculated by

𝑉𝑅 = 𝑅𝑖(𝑡)

𝑽𝑹 = 𝟔𝟎𝒊(𝒕) = 𝟏𝟐𝟎(𝟏 − 𝒆−𝟔𝒕)

The voltage across the inductor is determined from

𝑽𝑳 = 𝑳𝒅𝒊

𝒅𝒕= 𝟏𝟎

𝒅

𝒅𝒕[ 𝟐(𝟏 − 𝒆−𝟔𝒕)] = 𝟏𝟐𝟎𝒆−𝟔𝒕

Current at t= 0.65s

𝒊(𝒕) = 𝟐(𝟏 − 𝒆−𝟔(𝟎.𝟔𝟓)) = 𝟏. 𝟗𝟗𝟔𝑨

Exercise: RL circuit

By using laplace transform, find i(t) when switch is close at t= 0. The initial

current value

i(0) = 2A.

Laplace Transform


Laplace Transform in RC Circuit

Consider the series RC circuit shown below. Let the witch S is closed at t = 0.

Applying KVL around the loop, we get:

−𝑉(𝑠) + 𝑖(𝑡)𝑅 +1

𝐶∫ 𝑖(𝑡)𝑑𝑡 = 0

Applying Laplace Transform , we get:

𝐼(𝑠)𝑅 +1

𝐶[𝐼(𝑠)

𝑠+

𝑞(0)

𝑠] =

𝑉(𝑠)

𝑠

Or

𝐼(𝑠)𝑅 +1

𝐶[𝐼(𝑠)

𝑠+

𝑣(0)

𝑠] =

𝑉(𝑠)

𝑠

Where

𝑣(0) =𝑞(0)

𝑐

Assume zero initial voltage, we get :

𝐼(𝑠) [𝑅 +1

𝐶𝑠] =

𝑉𝑠

𝑠

𝐼(𝑠) [𝑅𝐶𝑠 + 1

𝐶𝑠] =

𝑉𝑠

𝑠

𝐼(𝑠) =

𝑉𝑠𝑅

𝑠 +1

𝑅𝐶

By inverse laplace transform, 𝑖(𝑡) =𝑉𝑠

𝑅 𝑒−

1

𝑅𝐶

Laplace Transform


Example : RC circuit

The capacitor in the circuit is shown below has initial charge qo = 600µC with

polarity as shown. If the switch is closed at t=0, obtain the current for t>0.

Solution:

The switch s is closed at t = 0. The initial charge qo across the capacitor does

not change instantaneously.

By KVL,

−𝑉(𝑠) + 𝑖(𝑡)𝑅 +1

𝐶∫ 𝑖(𝑡)𝑑𝑡 = 0

−50 + 𝑖(𝑡)10 +1

100𝑥10−6∫ 𝑖(𝑡)𝑑𝑡 = 0

𝐼(𝑠)10 +1

100𝑥10−6[𝐼(𝑠)

𝑠+

𝑞(0)

𝑠] =

50

𝑠

Since 𝑣(0) =𝑞(0)

𝑐

𝐼(𝑠)10 +1

100𝑥10−6[𝐼(𝑠)

𝑠+

600 𝑥 10−6

𝑠] =

50

𝑠

𝐼(𝑠) [10 +104

𝑠] =

50

𝑠−

6

𝑠

𝐼(𝑠) [10𝑠 + 104

𝑠] =

44

𝑠

𝐼(𝑠) =44

10(𝑠 + 103)

By inverse laplace transform 𝑖(𝑡) = 4.4𝑒−103𝑡

Example 1: RC circuit

Laplace Transform


Given v(t) = ½ cost. Find Vc(t) when switch is close at t= 0. Assume all initial

value is zero.

Vc(t)

Solution:

Applying KVL around the loop, we get:

−𝑉(𝑠) + 𝑖(𝑡)𝑅 +1

𝐶∫ 𝑖(𝑡)𝑑𝑡 = 0

𝑖(𝑡)2 +1

1/4∫ 𝑖(𝑡)𝑑𝑡 =

1

2𝑐𝑜𝑠𝑡

Applying Laplace Transform , we get:

𝐼(𝑠)2 + 4 [𝐼(𝑠)

𝑠] =

1

2(

𝑠

𝑠2 + 1)

𝐼(𝑠) [2 + 4

𝑠] =

1

2(

𝑠

𝑠2 + 1)

𝐼(𝑠) [2𝑠 + 4

𝑠] =

1

2(

𝑠

𝑠2 + 1)

𝐼(𝑠) [2𝑠 + 4

𝑠] =

1

2(

𝑠

𝑠2 + 1)

𝐼(𝑠) [2(𝑠 + 2)

𝑠] =

1

2(

𝑠

𝑠2 + 1)

𝐼(𝑠) =1

2(

𝑠

𝑠2 + 1) 𝑥

𝑠

2(𝑠 + 2)

𝐼(𝑠) =1

4[

𝑠2

(𝑠2 + 1)(𝑠 + 2)]

𝑉𝑐(𝑠) = 𝐼(𝑠)𝑍𝑐(𝑠)

Laplace Transform


𝑉𝑐(𝑠) = 1

4[

𝑠2

(𝑠2 + 1)(𝑠 + 2)] 𝑥

4

𝑠

𝑉𝑐(𝑠) = [𝑠

(𝑠2 + 1)(𝑠 + 2)]

By partial fraction expansion:

𝑉𝑐(𝑠) = [𝑠

(𝑠2 + 1)(𝑠 + 2)] =

𝐴

𝑠 + 2+

𝐵𝑠 + 𝐶

𝑠2 + 1

[𝑠

(𝑠2 + 1)(𝑠 + 2)] =

𝐴(𝑠2 + 1)

𝑠 + 2+

(𝐵𝑠 + 𝐶)(𝑠 + 2)

𝑠2 + 1

𝑠 = 𝐴(𝑠2 + 1) + (𝐵𝑠 + 𝐶)(𝑠 + 2)

Solve it as usual then we get: A= -2/5, B = 2/5 and C = 1/5 and

𝑉𝑐(𝑠) =−

25

𝑠 + 2 +

25

𝑠 +15

𝑠2 + 1

𝑉𝑐(𝑡) = −2

5𝑒−2𝑡 +

2

5 𝑐𝑜𝑠𝑡 +

1

5𝑠𝑖𝑛𝑡

Laplace Transform


Exercise: RC circuit

1. Consider the circuit below when the switch is closed at t=0 with Vc(0) =

1.0V. Solve the current i(t) in the circuit

2. The capacitor in the circuit shown in the circuit below has initial charge

qo = 600µF with polarity shown. If the switch is closed at t=0, obtain the

current and charge for t>0.

1µF

100µF

Ω

Laplace Transform


Example : RL circuits

In the circuit shown below , the switch is in position (a) for along time and is

moved to position (b) at t= 0. Find the current i(t) whn t>0.

Solution:

When the switch is at position (a) , the 12Ω resistor and 0.4H inductor in series

are excited by 10V source. Now , the inductor stores electrical energy in the

form of current.

When the switch is at position(a)

As the switch is at position a for long time, the circuit will be in a steady state

and the inductor behaves as a short circuit for DC. Hence , current in the

circuit is

𝑖(𝑡) =𝑉

𝑅=

10

12

𝑖(𝑡) =5

6

The switch is moved tp position b at t=0. The circuit is now energized by the

60V source, and the response of the circuit is due to the initial current in the

inductor and the 60V source

Laplace Transform


At t= 0, the switch is moved to position b and that instant, the current in the

inductor due to the previous operation is

At t = 0 , the

𝐼𝑜 = 𝑖(0) =5

6𝐴

The current is known as initial current and its direction is in the same direction

of assumed current i9t) due to 60V source as shown in figure b. by applying

KVL in figure b we get:

60 = 12𝑖(𝑡) + 0.4𝑑𝑖

𝑑𝑡

By laplace transformation both sides:

12𝐼(𝑠) + 0.4[𝑠𝐼(𝑠) − 𝑖(0)] =60

𝑠

Substituting 𝑖(0) =5

6𝐴 ,

𝐼(𝑠)[ 0.4𝑠 + 12] −0.4𝑥5

6=

60

𝑠

𝐼(𝑠)[ 𝑠 + 30) −5

6=

150

𝑠

𝐼(𝑠)[ 𝑠 + 30] =150

𝑠+

5

6

𝐼(𝑠) =(

56) (𝑠 + 180)

𝑠(𝑠 + 30)

By partial fraction:

𝐼(𝑠) =(

56) (𝑠 + 180)

𝑠(𝑠 + 30)=

𝐴

𝑠+

𝐵

𝑠 + 30

𝐴 =(

56) (𝑠 + 180)

𝑠 + 30|

𝑠= 0

=(

56) (0 + 180)

0 + 30=

150

30= 5

Laplace Transform


𝐵 =(

56) (𝑠 + 180)

𝑠|

𝑠= −30

=(

56) (−30 + 180)

−30=

125

−30= −

25

6

Hence 𝐼(𝑠) =5

𝑠−

25/6

𝑠+30

By inverse laplace transform;

𝑖(𝑡) = 5 −25

6𝑒−30𝑡

Laplace Transform


Example : RLC circuit

Refer to the circuit shown below. Assume that the current is initially relaxed.

Let R = 4Ω, L= 1H and C = 0.2F

Solution:

Applying KVL around the loop, we get

𝑉𝑠 = 𝑅𝑖(𝑡) + 𝐿𝑑𝑖

𝑑𝑡+

1

𝑐 ∫ 𝑖(𝑡)𝑑𝑡

10 = 4𝑖(𝑡) + 1𝑑𝑖

𝑑𝑡+

1

0.2 ∫ 𝑖(𝑡)𝑑𝑡

Applying Laplace transform both sides, we get:

10

𝑠= 4𝐼(𝑠) + 1[𝑠𝐼(𝑠) − 𝑖(0)] + 5 [

𝐼(𝑠)

𝑠+

𝑞(0)

𝑠]

As the circuit is initially relaxed, i(0) = 0 and q(0) = 0

10

𝑠= 4𝐼(𝑠) + 𝑠𝐼(𝑠) + 5 [

𝐼(𝑠)

𝑠]

10

𝑠= 𝐼(𝑠) [4 + 𝑠 +

5

𝑠]

𝐼(𝑠) [ 4 + 𝑠 +5

𝑠] =

10

𝑠

𝐼(𝑠) [4𝑠 + 𝑠2 + 5

𝑠] =

10

𝑠

𝐼(𝑠) = 10

𝑠 𝑥

𝑠

𝑠2 + 4𝑠 + 5

Laplace Transform


𝐼(𝑠) = 10

𝑠2 + 4𝑠 + 5

𝐼(𝑠) = 10

(𝑠 + 2)2 + 1

By inverse laplace transformation

𝒊(𝒕) = 𝟏𝟎𝒆−𝟐𝒕𝒔𝒊𝒏𝒕

Exercise:

1. Consider the circuit when the switch is closed at t=0, VC(0)=1.0 V. Solve

for the current i(t) in the circuit.

2. Solve for i(t) for the circuit, given that V(t) = 10 sin5t V, R = 4 W and L = 2

H.

Laplace Transform


References:

1. Alexander Sadiku (2007), Fundamental of Electric Circuits (3rd), Mc Graw

Hill. (ISBN: 9780071105828)

2. Nilsson, James W & Riedel, Susan A (2008), Electric Circuits (8th), Pearson

Prentice Hall. (ISBN-13: 9780131989252)

3. William H. Hayt Jr, Jack E. Kennedy and Steven M. Durbin (2007),

Engineering Circuit Analysis (7th), Mc Graw Hill. (ISBN-13: 9780072866117)

4. http://www.mathalino.com/

5. http://lpsa.swarthmore.edu/LaplaceXform/FwdLaplace/LaplaceDiffEq.h

tml

6. http://math.fullerton.edu/mathews/c2003/LaplaceTransformMod.html

http://www.mathalino.com/

http://lpsa.swarthmore.edu/LaplaceXform/FwdLaplace/LaplaceDiffEq.html

http://lpsa.swarthmore.edu/LaplaceXform/FwdLaplace/LaplaceDiffEq.html

http://math.fullerton.edu/mathews/c2003/LaplaceTransformMod.html

LAPLACE TRANSFORM · JABATAN KEJURUTERAAN ELETRIK, POLITEKNIK MERLIMAU 27.11.2014 . Laplace...

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