Ppt in Powerpoint 2017 For GTU or GU Engineering Students.

1. Fraction
Decomposition
for Inverse
Laplace
Transform
AIET Computer Engineering (Enroll 51 to 62)
Faculty Name :- Prof. Chhaya V. Kariya.

2. EFFORTS BY…
 Rathod Aman (161210107051)
 Rohit Dwivedi (161210107052)
 Sakhiya Vivek (161210107053)
 Gajendra Sisodiya (161210107055)
 Thakur Riya (161210107056)
 Tiwari Vishalsagar (161210107057)
 Vaghasiya Dwarkesha (161210107058)
 Yadav Kalpesh (161210107062)

3.  What Is Partial Fraction Decomposition?
 Why We Use Partial Fraction Method?
 Benefits Of Using Partial Fraction Method
 Partial Fraction For Inverse Laplace
 Cases In Partial Fraction Expansion Of Inverse Laplace Transform
 P(s) Is A Quadratic With 2 Real Roots
 P(s) Is A Quadratic With A Double Root
 P(s) Is A Quadratic And Has Complex Conjugate Roots
 P(s) Is Of Degree 3 And Has 1 Real Root And 2 Complex Conjugate Roots
SESSIONS OUTCOME

4. What Is Partial Fraction Decomposition?
In algebra, the partial fraction decomposition or partial
fraction expansion of a rational function (that is,
a fraction such that the numerator and the denominator
are both polynomials) is the operation that consists in
expressing the fraction as a sum of a polynomial (possibly
zero) and one or several fractions with a simpler
denominator.

5. The importance of the partial fraction decomposition
lies in the fact that it provides an algorithm for
computing the antiderivative of a rational function. In
symbols, one can use partial fraction expansion to
change a rational fraction in the form f(x)/g(x).

6.  The partial fraction decomposition may be seen as the
inverse procedure of the more elementary operation of
addition of rational fraction, j which produces a single
rational fraction with a numerator and denominator
usually of high degree. The full decomposition pushes
the reduction as far as it can go.

7. Why We Use Partial Fraction Method?
Remember, the whole idea is to
make the rational function easier to
integrate…

8. BENIFITS OF USING PARTIAL FRACTION METHOD
 Usually partial fractions method starts with polynomial long division in order to
represent a fraction as a sum of a polynomial and an another fraction, where the
degree of the polynomial in the numerator of the new fraction is less than the
degree of the polynomial.
 We, however, never have to do this polynomial long division, when Partial
Fraction Decomposition is applied to problems.

9. PARTIAL FRACTION FOR INVERSE LAPLACE
Consider the Laplace Transform
Some manipulations must be done
before Y(s) can be inverted since it
does not appear directly in our
table of Laplace transforms.

10. As we will show :
 Now, we can invert Y(s). From the table, we see that the inverse of 1/(s-2) is
exp(2t) and that inverse of 1/(s-3) is exp(3t). Using the linearity of the inverse
transform, we have
 The method of partial fractions is a technique for decomposing functions like Y(s)
above so that the inverse transform can be determined in a straightforward
manner. It is applicable to functions of the form
 Where Q(s) and P(s) are polynomials and the degree of Q is less than the degree
of P.

11. CASES IN PARTIAL FRACTION EXPANSION OF INVERSE LAPLACE
TRANSFORM
 FOR SIMPLICITY WE ASSUME THAT Q AND P HAVE REAL COEFFICIENTS. WE CONSIDER THE
FOLLOWING CASES:
1. P(s) is a QUADRATIC with 2 REAL ROOTS
2. P(s) is a QUADRATIC with a DOUBLE ROOT
3. P(s) is a QUADRATIC and has COMPLEX CONJUGATE ROOTS
4. P(s) is of DEGREE 3 and has 1 REAL ROOT and 2 COMPLEX CONJUGATE ROOTS

12. P(S) IS A QUADRATIC WITH 2 REAL ROOTS
Consider the initial example on this page:
 The denominator can be factored: s^2-5s+6=(s-2)(s-3). The denominator has 2
real roots. In this case we can write:
 Here A and B are numbers. This is always possible if the numerator is of degree 1
or is a constant. The goal now is to find A and B. Once we find A and B, we can
invert the Laplace transform:

13.  If we add the two terms on the RHS, we get:
 Comparing LHS and RHS we have:
 The denominators are equal. The two sides will be equal if the numerators are
equal. The numerators are equal for all s if the coefficients of s and the constant
term are equal. Hence, we get the following equations:

14.  This is a system of 2 linear equations with 2 unknowns. It can be solved by
standard methods. We obtain A=4 and B=-2. Hence:
 If the denominator has n distinct roots:
then the decomposition has the form:
 Where A_1, A_2, ..., A_n are unknown numbers. The inverse transform is

15. THE UNKNOWN COEFFICIENTS CAN BE DETERMINED USING
THE SAME AS IN THE CASE OF ONLY 2 FACTORS, AS SHOWN
ABOVE.

16. P(S) IS A QUADRATIC WITH A DOUBLE ROOT
 Consider the example:
 The denominator has double root -2. The appropriate decomposition in this case
is
 Here A and B are numbers. From the table the inverse transform of 1/(s+2) is
exp(-2t) and the inverse transform of 1/(s+2)^2 is texp(-2t). Hence, the inverse
transform of Y(s) is

17. NOTE
USING THE SAME TECHNIQUE AS IN THE CASE OF
DISTINCT ROOTS ABOVE, IT CAN BE SHOWN THAT A=1
AND B=-2.

18. P(S) IS A QUADRATIC AND HAS COMPLEX CONJUGATE ROOTS
Consider the example:
The roots of the denominator are -2+/-i. We can complete the
square for the denominator. We have:
Hence, we have

19. Note the denominator (s+2)^2+1 is similar to that for Laplace
transforms of exp(-2t)cos(t) and exp(-2t)sin(t). We need to
manipulate the numerator. Note that in the formula in the
table, we have a=-2 and b=1. We look for a decomposition of
the form:
If we can find A and B, then:

20. The inverse transform is:
We can determine A and B by equating numerators in the expression:
Comparing coefficients of s in the numerator we conclude 3=A.
Comparing the constant terms we conclude 2A+B=9. Hence A=3 and
B=3.

21. P(S) IS OF DEGREE 3 AND HAS 1 REAL ROOT AND 2 COMPLEX
CONJUGATE ROOTS
Consider the example:
The roots of the denominator are 1 and -2+/-i. In this case we look for
a decomposition of the form:
where A, B and C are unknowns. The inverse transform of A/(s-1) is
Aexp(t). Once B and C are determined the second term can be
manipulated as in the previous example.

22. †HA∏K
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