Download presentation

Presentation is loading. Please wait.

Published byNasir Alden Modified over 2 years ago

1
LAPLACE TRANSFORMS

2
INTRODUCTION

3
The Laplace Transformation Differential equations Input excitation e(t) Output response r(t) Time DomainFrequency Domain Algebraic equations Input excitation E(s) Output response R(s) Laplace Transform Inverse Laplace Transform

4
THE LAPLACE TRANSFORM

5

6
THE INVERSE LAPLACE TRANSFORM

7

8
Functional Laplace Transform Pairs

9
Operational Laplace Transform Pairs

10
Inverse Laplace Transform The inverse Laplace transform is usually more difficult than a simple table conversion.

11
Partial Fraction Expansion If we can break the right-hand side of the equation into a sum of terms and each term is in a table of Laplace transforms, we can get the inverse transform of the equation (partial fraction expansion).

12
Repeated Roots In general, there will be a term on the right- hand side for each root of the polynomial in the denominator of the left-hand side. Multiple roots for factors such as (s+2) n will have a term for each power of the factor from 1 to n.

13
Complex Roots Complex roots are common, and they always occur in conjugate pairs. The two constants in the numerator of the complex conjugate terms are also complex conjugates. where K * is the complex conjugate of K.

14
Solution of Partial Fraction Expansion The solution of each distinct (non-multiple) root, real or complex uses a two step process. The first step in evaluating the constant is to multiply both sides of the equation by the factor in the denominator of the constant you wish to find. The second step is to replace s on both sides of the equation by the root of the factor by which you multiplied in step 1

15

16
The partial fraction expansion is:

17
The inverse Laplace transform is found from the functional table pairs to be:

18
Repeated Roots Any unrepeated roots are found as before. The constants of the repeated roots (s-a) m are found by first breaking the quotient into a partial fraction expansion with descending powers from m to 0:

19
The constants are found using one of the following:

20

21
The partial fraction expansion yields:

22
The inverse Laplace transform derived from the functional table pairs yields:

23
A Second Method for Repeated Roots Equating like terms:

24
Thus

25
Another Method for Repeated Roots As before, we can solve for K 2 in the usual manner.

26

27
Unrepeated Complex Roots Unrepeated complex roots are solved similar to the process for unrepeated real roots. That is you multiply by one of the denominator terms in the partial fraction and solve for the appropriate constant. Once you have found one of the constants, the other constant is simply the complex conjugate.

28
Complex Unrepeated Roots

29

30
Case 1: Functions with repeated linear roots Consider the following example: F(s) should be decomposed for Partial Fraction Expansion as follows:

31
Using the residue method:

32
so and f(t) = [-6e-t + (6 + 12t)e-2t ]u(t)

33
Case 2: Functions with complex roots If a function F(s) has a complex pole (i.e., a complex root in the denominator), it can be handled in two ways: 1) By keeping the complex roots in the form of a quadratic 2) By finding the complex roots and using complex numbers to evaluate the coefficients

34
Example: Both methods will be illustrated using the following example. Note that the quadratic terms has complex roots.

35
Method 1: Quadratic factors in F(s) F(s) should be decomposed for Partial Fraction Expansion as follows:

36
A)Find A, B, and C by hand (for the quadratic factor method): Combining the terms on the right with a common denominator and then equating numerators yields:

37
so now manipulating the quadratic term into the form for decaying cosine and sine terms:

38
so The two sinusoidal terms may be combined if desired using the following identity:

39
so

40
Method 2: Complex roots in F(s) Note that the roots of are so

41
A)Find A, B, and C by hand (for the complex root method): F(s) should be decomposed for Partial Fraction Expansion as follows:

42
The inverse transform of the two terms with complex roots will yield a single time-domain term of the form Using the Residue Theorem:

43

44
So, This can be broken up into separate sine and cosine terms using

45
so

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google