Module 21 - Laplace Transforms (self Study)

SCHOOL OF MATHEMATICS MATHEMATICS FOR PART I ENGINEERING Self Study Course

MODULE 21

LAPLACE TRANSFORMS

Module Topics 1. Definition 2. Laplace transform of standard functions 3. Inverse Laplace transforms 4. Solution of second order linear ordinary differential equations with constant coefficients

A:

Work Scheme based on JAMES (THIRD EDITION)

1. Laplace transforms prove a very useful tool in analysing engineering systems and an introduction to the technique can be found in chapter 11 of J. Turn to p.767 and read the introductory section 11.1. 2. Study section 11.2.1 on p.769 which includes the definition of a Laplace transform, formula (11.2). Note that the transform variable s is in general complex, not real. The notation f (s) is used by some authors instead of F (s), as stated by J. just below equation (11.2), and the transform variable s is sometimes replaced by p. Comments (a) and (b) in section 11.2.1 are important – you should recall that improper integrals were studied in Module 10. The Heaviside unit step function, introduced in comment (c) on p.770, is widely used in more complicated examples, but it will not be considered further in this introductory module. Comment (d) is also not relevant for the material in this module. 3. Laplace transform pairs are usually quoted from tables. You should know how the simplest results are derived, however, so work through Examples 11.1 and 11.2. Note that the transforms exist only for particular values of s. Read through Examples 11.3 and 11.4 which supply additional standard results. The table of Laplace transforms on your Formula Sheet and in the Data Book contains the results 11.4 to 11.8. Note from J. that the transforms exist only for certain values of s, although these restrictions are not stated in the Data Book. 4. Since the definition of a Laplace transform involves an improper integral, the transform of some functions f (t) will not exist. Section 11.2.3, which it is not necessary to work through, considers the existence of Laplace transforms and Theorem 11.1 states necessary conditions for existence. In general terms the Laplace transforms of most functions exist, provided the functions are not too discontinuous and do not increase with 2 t faster than ekt , for some k. One function for which the transform does not exist is f (t) = et . 5. Turn to p.776 and section 11.2.4 on the properties of the Laplace transform. You must understand and be able to use the shaded results in this section, but the proofs are less important. Read property 11.1, the linearity property, and then work through Examples 11.7 and 11.8, which use the derived results (11.5) and (11.6). –1–

Theorem 11.2, the first shift theorem on p.778, is particularly important in applications. Note how it is written on the Formula Sheet and in the Data Book. Work through Examples 11.9 and 11.10. You should observe that in Example 11.9 the transform F (s) exists when Re(s) > 0. It follows that F (s + 2) exists when Re(s + 2) > 0, i.e. Re(s) + 2 > 0 or Re(s) > −2, as stated in the solution. The range of validity in Example 11.10 is found in a similar way. The derivative of transform result, Theorem 11.3, can be useful in some situations and this is also on the Formula Sheet (and in the Data Book for the special case t = 1). Study Examples 11.11, 11.12 and 11.13. Note that the answer to 11.12 is found with the aid of the derivative theorem, although it is easier 2 to proceed as follows. The table of Laplace transforms shows {t2 } = 3 (Re(s) > 0), and the first s 2 {t2 et } = (Re(s) > 1). This example illustrates the fact that there shift theorem then implies (s − 1)3 is often more than one way of calculating a transform and you can often save time by spotting the simplest method.

L

L

***Using the Formula Sheet (or the Data Book) do Exercises 3(a),(c),(g),(i),(k),(o) on p.782***

L

−1

6. Study section 11.2.7 on the inverse Laplace transform, denoted by , up to equation (11.11). If the transform F (s) appears in the tables the corresponding function f (t) can easily be written down. Study Examples 11.14 and 11.15, which are very straightforward. The linearity property at the end of the section is important in applications, since it extends the range of transforms which can be easily inverted. 7. Study section 11.2.8 including Examples 11.16 and 11.17. The use of partial fractions is crucial in many situations. If you cannot remember the general formulae for partial fractions then look back at section 14 of Module 1 or pp.101–107 in J. In Example 11.16 the original denominator is a product of two linear factors. The split into partial fractions is then relatively simple, the associated constants can be calculated and the function f (t) is easily written down using the standard results in the table. Example 11.17 has a more complicated denominator which must be expressed as the sum of three fractions. Again the constants must be found by the standard procedure, before using the linearity property and the table to carry out the inversion. As you know from earlier work, when a quotient is split into a linear combination of fractions the determination of the constants can be a lengthy process, and it this stage of the solution procedure which takes the most time. 8. Study the comments on inversion using the first shift theorem at the beginning of section 11.2.9. This is an extremely important procedure, but it is not straightforward and so it may take you a little time to understand it fully. Let us briefly summarize the method. A given transform may not appear in the table you are using and so cannot be inverted directly. However, the transform can often be expressed as a function of the variable s − a, say F (s − a). If the function F (s), which is obtained from the latter when the variable s − a is replaced by s, does appear in the table then its inverse f (t) can be written down, and the inverse of the original expression is eat f (t). The above procedure is now illustrated with a number of Examples. Study Example 11.18. This is already written as F (s + 2), so the shift theorem can be invoked immediately. Consider the steps in the solution in relation to the comments in the paragraph above. Study Example 11.19. It is necessary here to complete the square in the denominator, after which the shift theorem is again easy to use. Study Example 11.20. The first step is again to complete the square in the denominator, which this time becomes a function of s + 1. The whole fraction must depend on this variable, however, so you must next write the numerator in terms of s + 1. Thus write s + 7 = (s + 1) + 6 and then split the single fraction –2–

into two parts, which can be treated separately using the shift theorem and the table. The solution to this Example represents damped oscillatory motion and hence this type of Example often occurs when solving differential equations which model mechanical and electrical oscillations. Finally study Example 11.21. Here the transform is expressed as a combination of three separate fractions using the standard method. The shift theorem is needed to deal with one of these fractions. ***Do Exercises 4(a),(c),(f ),(h) on p.786*** 9. The most important use of Laplace transforms is in solving differential equations. Study section 11.3.1 on deriving the general formulae for Laplace transforms of ordinary derivatives. The formulae (11.12) and (11.13) display the results for the Laplace transforms of first and second derivatives whereas (11.14) gives the general formula for the transform of the nth derivative. Note that these formulae involve the transform F (s), and the initial values of the function f (t) and its derivatives. Results (11.12) and (11.13) appear on the Formula Sheet and in the Data Book. 10. Omit section 11.3.2 and move on to the following section, 11.3.3, which you should study. Taking the transform of equation (11.16) and solving for X(s) leads to the formula (11.17). The latter should not be remembered since when dealing with specific Examples one works through from the beginning, as shown in the Examples discussed below. Read the observations (a)–(d) on p.790. Study Example 11.23 which uses the Laplace transform to obtain the solution of an inhomgeneous second order linear ordinary differential equation subject to initial conditions. After taking Laplace transforms and rearranging, the solution for the transform X(s) is stated near the middle of p.791 as a sum of two terms which are each split up using partial fractions before inverting. Move on to p.792 and study Example 11.24, which gives the solution of another second order equation. Having determined X(s), the expression is split into a sum of three fractions, one of which has the general A + Bs 2 3 1 s form 2 . After finding A = 2/25 and B = −3/50, the fraction can be written − , s +1 25 s2 + 1 50 s2 + 1 which appear as the final two terms in the expanded form of X(s). 11. In this module only the basic ideas of Laplace transforms have been presented. The method is widely used in many Engineering applications, however, and most of you will study them again in more detail in next year’s Mathematics course. ***Do Exercises 5(a),(f ) on p.794***

B:

Work Scheme based on STROUD (FIFTH EDITION)

In this new edition of S. some basic material on Laplace transforms is in Programme 26 starting on p.1098. Note that S. uses the independent variable x whereas J. uses t, and it is t which is used in the module tests and examination papers. Study frames 1–30. S. concentrates on finding solutions to first order ordinary differential equations with constant coefficients and does not consider the shift theorem. Some simple second order ODEs are solved at the end of the Programme, but the omission of the shift theorem means that some of the difficulties that can arise are not discussed. Therefore, to complete the module you should study section 8 in A: Work scheme based on JAMES (THIRD EDITION).

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Specimen Test 21

1.

Obtain the Laplace transforms of the following functions: (i)

2.

(ii)

e−t cos 2t,

(iii)

t sin 3t.

Obtain the inverse Laplace transforms of the following functions: (i)

3.

3 + t3 ,

1 1 , + 3 s s+1

(ii)

s2

1 , − 2s + 5

(iii)

2 s2 (s2

+ 1)

.

Given that the function x(t) satisfies the differential equation dx d2 x + x = te−t +2 dt2 dt and the initial conditions x = 1 and

dx = −2 when t = 0, show that dt

L {x(t)} ≡ X(s) = (s + 1) 1

4

and hence find the solution for x(t).

–4–

+

s (s + 1)2