[PDF] Section 7.4: Inverse Laplace Transform A natural question to ask

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Section 7.4: Inverse Laplace Transform

A natural question to ask about any function is whether it has an inverse function. We now ask this question about the Laplace transform: given a functionF(s), will there be a functionf(t) such thatF(s) =Lffg(s)? It turns out that there is at most one continuous functionf(t) which satises this property (there could be innitely many discontinuous functions with the same Laplace transform, but we prefer to work with continuous functions). Denition.Given a functionF(s), if there is a functionf(t) that is continuous on [0;1) and satisesLffg=F, then we say thatf(t) is the inverse Laplace transformofF(s) and employ the notationf=L1fFg. This idea has more than theoretical interest, however; we'll see in the next section that nding inverse Laplace transforms is a critical step in solving initial value problems. To determine the inverse Laplace transform of a function, we try to match it with the form of an entry in the right-hand column of a Laplace table.

Example 1.DetermineL1fFgfor

(a)F(s) =2s

3, (b)F(s) =3s

2+ 9, (c)F(s) =s1s

22s+ 5.

Solution.(a)L12s

3 (t) =L12!s 3 (t) =t2 (b)L13s 2+ 9 (t) =L13s 2+ 32 (t) = sin(3t) (c)L1s1s

22s+ 5

(t) =L1s1(s1)2+ 22 (t) =etcos(2t). Of course, very often the transform we are given will not correspond exactly to an entry in the Laplace table. One tool we can use in handling more complicated functions is the linearity of the inverse Laplace transform, a property it inherits from the original Laplace transform. Theorem 1.Assume thatL1fFg;L1fF1g, andL1fF2gexist and are continuous on[0;1)and letcbe any constant. Then L

1fF1+F2g=L1fF1g+L1fF2g;

L

1fcFg=cL1fFg:

Example 2.DetermineL15s66ss

2+ 9+32s2+ 8s+ 10

Solution.By linearity, we have

L

15s66ss

2+ 9+32s2+ 8s+ 10

= 5L11s6 6L1ss 2+ 9 +32
L11s

2+ 4s+ 5

1 2 Rewriting the nal term using completing the square, it becomesL11(s+ 2)2+ 12

Therefore, by the Laplace table we see that

L

15s66ss

2+ 9+32s2+ 8s+ 10

(t) = 5e6t6cos(3t) +32 e2tsint:

Example 3.DetermineL15(s+ 2)4

Solution.The fourth power in the denominator suggests that the inverse Laplace trans- form is of the form L

1n!(sa)n+1

(t) =eattn:

In this case,a=2;n= 3, so by linearity we have

L

15(s+ 2)4

(t) =56

L13!(s+ 2)4

(t) =56 e2tt3:

Example 4.DetermineL13s+ 2s

2+ 2s+ 10

Solution.Using completing the square, the denominator can be rewritten as s

2+ 2s+ 10 =s2+ 2s+ 1 + 9 = (s+ 1)2+ 32:

Therefore, the form ofF(s) suggests the following two formulas from the Laplace table: L

1sa(sa)2+b2

(t) =eatcos(bt); L

1b(sa)2+b2

(t) =eatsin(bt); where we havea=1;b= 3. Thus, we want to write

3s+ 2s

2+ 2s+ 10=As+ 1(s+ 1)2+ 32+B3(s+ 1)2+ 32

for an appropriate choice of constantsA;B. By clearing the denominator, we have the equation

3s+ 2 =A(s+ 1) + 3B=As+ (A+ 3B):

Equating coecients gives usA= 3;B=1=3, so by linearity, we have L

13s+ 2s

2+ 2s+ 10

(t) = 3L1s+ 1(s+ 1)2+ 32 (t)13

L13(s+ 1)2+ 32

(t) = 3etcos(3t)13 etsin(3t): 3 In Example 4, we found it easier to take the Laplace transform of the broken up fraction than of the original combined fraction. The procedure we used above, which you may recall from Calculus 2 and which will prove very useful in simplifying these functions, is called themethod of partial fractions. Because of its importance, we now review this method.

Review of Partial FractionsA rational functionP(s)=Q(s), whereP(s);Q(s) are polynomials with the degree ofP

less than the degree ofQ, has a partial fraction expansion based on the factorization of Q. There are three main cases to consider, the easiest being that of nonrepeated linear factors. IfQ(s) is a product of distinct linear factors,

Q(s) = (sr1)(sr2)(srn);

then the partial fraction expansion has the form

P(s)Q(s)=A1sr1+A2sr2++Ansrn:

The constantsA1;:::;Ancan be derived by clearing denominators and either equating coecients of the resulting polynomials as in Example 4, or by careful substitution. We demonstrate the second way in the following examples. Example 5.DetermineL1fFgforF(s) =7s1(s+ 1)(s+ 2)(s3). Solution.Since the denominator has three distinct linear factors, the partial fraction expansion has the form

7s1(s+ 1)(s+ 2)(s3)=As+ 1+Bs+ 2+Cs3;

and upon clearing denominators (multiplying through by (s+ 1)(s+ 2)(s3)), we have (1) 7s1 =A(s+ 2)(s3) +B(s+ 1)(s3) +C(s+ 1)(s+ 2): If we substitutes-values which make the linear factors zero, then each time two terms will drop out and it will be simple to solve for each coecient. First lets=1; then (1) becomes

8 =4A)A= 2:

Similarly, lettings=2 gives

15 = 5B)B=3;

and lettings= 3 yields

20 = 20C)C= 1:

4 Therefore, we can now use linearity to calculate the inverse Laplace transform: L

17s1(s+ 1)(s+ 2)(s3)

(t) =L12s+ 13s+ 2+1s3 (t) = 2L11s+ 1 (t)3L11s+ 2 (t) +L11s3 (t) = 2et3e2t+e3t: The second scenario involves repeated linear factors. If the highest power ofsrthat dividesQ(s) is (sr)m, then the portion of the partial fraction expansion corresponding to (sr)mis A

1sr+A2(sr)2++Am(sr)m:

Example 6.DetermineL1s2+ 9s+ 2(s1)2(s+ 3)

Solution.The partial fraction expansion has the form s

2+ 9s+ 2(s1)2(s+ 3)=As1+B(s1)2+Cs+ 3:

Clearing denominators gives

(2)s2+ 9s+ 2 =A(s1)(s+ 3) +B(s+ 3) +C(s1)2: Again, we substitute values which make the linear factors vanish. Lettings= 1 in (2) gives

12 = 4B)B= 3;

and lettings=3 yields

16 = 16C)C=1:

To nd the third constant, we can plug in any other value fors; a convenient choice is s= 0:

2 =3A+ 3B+C=3A+ 91

6 =3A)A= 2:

Therefore, the inverse Laplace transform is

L

1s2+ 9s+ 2(s1)2(s+ 3)

(t) =L12s1+3(s1)21s+ 3 (t) = 2L11s1 (t) + 3L11(s1)2 (t)L11s+ 3 (t) = 2et+ 3tete3t: The most dicult case is that of a quadratic factor. Ifmis the highest power of (s)2+2that dividesQ(s), then the partial fraction expansion for that term is A

1s+B1(s)2+2+A2s+B2[(s)2+2]2++Ams+Bm[(s)2+2]m:

5 However, for looking up Laplace transforms it is more convenient to have numerators of the formAi(s) +Bi, so we can equivalently write A

1(s) +B1(s)2+2+A2(s) +B2[(s)2+2]++Am(s) +Bm[(s)2+2]m:

Example 7.DetermineL12s2+ 10s(s22s+ 5)(s+ 1)

Solution.Since the quadratic factor in the denominator is irreducible, we rewrite using completing the square:s22s+ 5 = (s1)2+ 22. Then the partial fraction expansion has the form2s2+ 10s(s22s+ 5)(s+ 1)=A(s1) + 2B(s1)2+ 22+Cs+ 1:

Clearing denominators gives

(3) 2s2+ 10s= [A(s1) + 2B](s+ 1) +C(s22s+ 5): There are two linear terms involved in this expression, so pick the twos-values which make them vanish. Whens=1, we get

8 = 8C)C=1:

Whens= 1, we get

12 = 4B+ 4C= 4B4)4B= 16)B= 4:

For the nal constant, sets= 0:

0 =A+ 2B+ 5C)A= 2(4) + 5(1) = 3:

Therefore, we nd the inverse Laplace transform to be L

12s2+ 10s(s22s+ 5)(s+ 1)

=L13(s1) + 2(4)(s1)2+ 221s+ 1 = 3L1s1(s1)2+ 22 + 4L12(s1)2+ 22

L11s+ 1

= 3etcos(2t) + 4etsin(2t)et:

Homework: pp. 374-375, #1-29 odd.

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