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Methods of finding Laplace transforms and inverse transforms. Properties of inverse Laplace transforms. Convolution of two functions. Convolution theorem. Heaviside expansion formulas.




Methods of finding Laplace transforms


1. Direct method. Direct use of definition.

 

2. Use of tables.


3. Series method. If F(t) has a power series expansion given by


ole.gif


one can obtain its Laplace transform by taking the sum of the Laplace transforms of each term in the series. Thus the Laplace transform of 1) is given by


ole1.gif



4. Method of differential equations. A method involving finding a differential equation satisfied by F(t) and then applying the various rules and theorems pertaining to Laplace transforms to obtain the desired transform f(s).


Example


5. Differentiation with respect to a parameter. This method employs Leibnitz’s Rule for differentiating under an integral sign along with the various rules and theorems pertaining to Laplace transforms to arrive at the desired transform. See “Spiegel. Laplace Transforms (Schaum)” for examples.


6. Miscellaneous methods employing various devices and techniques.




Uniqueness of inverse Laplace transforms. The Laplace transform of a null function N(t) is zero. A consequence of this fact is that if L[F(t)] = f(s) then also L[F(t) + N(t)] = f(s). From this it follows that we can have two different functions with the same Laplace transform.


Example. The two different functions F1(t) = e-4t and


             ole2.gif


have the same Laplace transform i.e. 1/(s + 4)


Thus if we allow null functions, we can see that the inverse Laplace transform is not unique. It is unique, however, if we disallow null functions (which do not in general arise in cases of physical interest).


Lerch’s theorem. If a function F(t) defined on the positive real axis, t ≥0, is piecewise regular and of exponential order then the inverse Laplace transform is unique.



Inverse Laplace transforms


 

            f(s)

   L-1[f(s)] = F(t)

1.

             ole3.gif


            1

2.

             ole4.gif  

            t

3.

  ole5.gif

             ole6.gif  

4.

               ole7.gif

             ole8.gif  

5.

            ole9.gif

          ole10.gif

6.

             ole11.gif  

            cos at

7.

           ole12.gif   

        ole13.gif

8.

           ole14.gif

        cosh at




































Properties of inverse Laplace transforms


1. Linearity property. Let c1 and c2 be any constants and F1(t) and F2(t) be functions with Laplace transforms f1(s) and f2(s) respectively. Then


3)      L-1[c1 f1(s) + c2 f2(s)] = c1 L-1 [f1(s)] + c2 L-1 [f2(s)] = c1 F1(t) + c2 F2(t)


The inverse Laplace transform thus effects a linear transformation and is a linear operator.


Example.


             ole15.gif




2. First translation (or shifting) property. If L-1[f(s)] = F(t), then


4)        L-1[f(s - a)] = eat F(t)



Example 1. Since


             ole16.gif


we obtain


             ole17.gif



We can generalize on this example. Let a and b be arbitrary constants. Since


             ole18.gif


we can employ the method of completing the square to obtain the general result


             ole19.gif


where

 

             ole20.gif



Remark. From item 3 in the above Inverse Laplace Transforms table we have


             ole21.gif


Applying 4) to this formula gives the following much used formula:


             ole22.gif




3. Second translation (or shifting) property. If L-1[f(s)] = F(t), then


ole23.gif



Example. Since


             ole24.gif


we obtain


             ole25.gif




4. Change of scale property. If L-1[f(s)] = F(t), then


ole26.gif




Example. Since


             ole27.gif


we obtain


             ole28.gif




5. Inverse Laplace transform of derivatives. If L-1[f(s)] = F(t), then


  ole29.gif



Example. Since


             ole30.gif


and


             ole31.gif



we obtain


             ole32.gif



6. Inverse Laplace transform of integrals. If L-1[f(s)] = F(t), then



ole33.gif



Example. Let


             ole34.gif


Then


             ole35.gif


Now it can be shown that


             ole36.gif



Consequently


             ole37.gif

 

 

 

7. Multiplication by sn. If L-1 [f(s)] = F(t) and F(0) = 0, then

 

9)      L-1 [sf(s)] = F'(t)


Thus multiplication by s has the effect of differentiating F(t). If F(0) ≠ 0, then

 

10)      L-1 [sf(s) - F(0)] = F'(t).


Since, for any constant c, L [cδ(t)] = c it follows that L-1 [c] = cδ(t) where δ(t) is the Dirac delta function or impulse function. Thus 10) can be written

 

11)    L-1 [sf(s)] = F'(t) + F(0)δ(t)


Generalizations of these results can be made for L-1 [sn f(s)], n = 2, 3, ......



Example. Since


             ole38.gif


and sin 0 = 0, we obtain


             ole39.gif


 

8. Division by s. If L-1 [f(s)] = F(t), then


ole40.gif


Example. Since


             ole41.gif


we obtain


             ole42.gif




Def. Convolution of two functions. The function h defined by


             ole43.gif


is called the convolution of f and g and often denoted by f*g. It can be shown that f*g = g*f. Proof



9. Convolution theorem. Let L-1[f(s)] = F(t) and L-1[g(s)] = G(t). Then


ole44.gif



Example. Since


             ole45.gif                                      


and


             ole46.gif                                                  


we obtain


             ole47.gif

 





Methods of finding inverse Laplace transforms


1. Use of Tables


2. Method of partial fractions. Any rational function of the form p(s)/q(s) where p(s) and q(s) are polynomials in which p(s) is of lesser degree than q(s) can be written as a sum of fractions of the types


             ole48.gif


where n is a positive integer and all coefficients are real if all coefficients in the original polynomials were real. Fractions of these types are called partial fractions. The method consists of reducing a rational function p(s)/q(s) to a sum of partial fractions and then finding the inverse transform of that sum of partial fractions. For information on partial fractions and reducing a rational function p(s)/q(s) to a sum of partial fractions see Partial Fractions. A method closely related to this one uses the Heaviside expansion formula.


3. Heaviside expansion formulas.


Fundamental Theorem of Algebra. If g(x) is a polynomial with real coefficients, then g(x) can be written as the product of linear and quadratic factors with real coefficients:


            g(x) = c(x - α1)(x - α2) ... (x2 + b1x + c1) (x2 + b2x + c2) ...


where bi2- 4ci < 0.


Theorem 1. Let y = L-1[p(s)/q(s)], where p(s) and q(s) are polynomials and the degree of q(s) is greater than the degree of p(s). Then the term in y corresponding to an unrepeated linear factor s - a of q(s) is given by


             ole49.gif


where Q(s) is the product of all the factors of q(s) except s - a.



Corollary. Let y = L-1[p(s)/q(s)]. Let q(s) be completely factorable into unrepeated linear factors


            (s - a1), (s - a2), ........ , (s - an)


Then


ole50.gif    

 

where Qi(s) is the product of all the factors of q(s) except the factor s - ai. 



Theorem 2. Let y = L-1[p(s)/q(s)], where p(s) and q(s) are polynomials and the degree of q(s) is greater than the degree of p(s). Then the terms in y corresponding to a repeated linear factor (s - a)r in q(s) are


             ole51.gif


where f(s) is the quotient of p(s) and all factors of q(s) except (s - a)r.


Theorem 3. Let y = L-1[p(s)/q(s)], where p(s) and q(s) are polynomials and the degree of q(s) is greater than the degree of p(s). Then the terms in y corresponding to an unrepeated, irreducible quadratic factor (s + a)2 + b2 of q(s) are


             ole52.gif

s

where fi and fr are, respectively, the real and imaginary parts of f(-a + ib) and f(s) is the quotient of p(s) and all the factors of q(s) except (s + a)2 + b2 .


There is a fourth theorem dealing with repeated, irreducible quadratic factors but because of its complexity and limited usefulness we will not present it.



Problem. Find


             ole53.gif


Solution. We have


            p(s) = 2s2 - 4

            q(s) = (s + 1)(s - 2)(s - 3) = s3 - 4s2 + s + 6

            q'(s) = 3s2 - 8s + 1

            a1 = -1, a2 = 2, a3 = 3


Then by 14) above the required inverse y = L-1[p(s)/q(s)] is given by


             ole54.gif


                                                                                     ole55.gif  




4. Series methods. If f(s) has a series expansion in inverse powers of s given by


ole56.gif


then under suitable circumstances we can invert term by term to obtain



ole57.gif



Problem. Find


             ole58.gif


Solution. Expanding e-1/ s as an infinite series, we obtain


             ole59.gif



                         ole60.gif



Inverting term by term, using Rule 3 in the above table of inverse Laplace transforms, we get


             ole61.gif



                                     ole62.gif



                                     ole63.gif



                                     ole64.gif





5. Method of differential equations. A method involving finding a differential equation satisfied by f(s) and then applying the various rules and theorems pertaining to Laplace transforms to arrive at the desired function F(t).


6. Differentiation with respect to a parameter. This method employs Leibnitz’s Rule for differentiating under an integral sign along with the various rules and theorems pertaining to Laplace transforms to arrive at the desired function F(t). See “Spiegel. Laplace Transforms (Schaum)” for examples.


7. The Complex Inversion formula. A method employing complex variable theory to evaluate the Complex Inversion formula.


8. Miscellaneous methods employing various devices and techniques.





References

  Murray R. Spiegel. Laplace Transforms. (Schaum)

  C. R. Wylie, Jr. Advanced Engineering Mathematics.



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