SCHOOL OF ENGINEERING & BUILT ENVIRONMENT Mathematics Laplace Transforms 1. Introduction ……………………………………. 1 2. Definition ………………………………………. 2 3. Properties of the Laplace Transform …………… 4 4. Inverse Laplace Transforms …………………… 6 5. Using Laplace Transforms to Solve ODEs ……. 9 6. The Step Function ……………………………... 11 7. The Dirac Delta Function ……………………… 20 8. Convolution ……………………………………. 25 9. A Note on Control Theory …………………….. 26 Tutorial Exercises ……………………………… 27 Answers to Tutorial Exercises …………………. 33 1 Laplace Transforms 1) Introduction The idea of transforming a “difficult” problem into an “easier” problem is one that is used widely in mathematics. Diagrammatically we have Problem Transform Transformed Problem Difficult Solve Solution Transformed Inverse Solution Transform There are many types of transforms available to mathematicians, engineers and scientists. We are going to examine one such transformation, the Laplace transform, which can be used to solve certain types of differential equations and also has applications in control theory. 2) Definition The Laplace transform operates on functions of t . Given a function f (t ) we define its Laplace transform as ∞ L[ f (t ) ] = f (t ) e − s t dt 0 where s is termed the Laplace variable. We sometimes denote the Laplace transform by F ( s ) or f . We start off with a function of t and end up with a function of s ; s is, in fact, a complex variable, but this need not concern us too much. Because the function of t is often some form of time signal, we often talk about moving from the time domain to the Laplace domain when we perform a Laplace transformation. Note: Laplace transforms are only concerned with functions where t ≥ 0. 2 Examples (1) f (t) = 1: ∞ L[1] = 1.e −st dt 0 t = ∞ = − 1 −st e s t = 0 Strictly speaking, we can’t set t equal to ∞, but we can “take the limit” as t heads towards infinity. Providing s > 0 , thereby ensuring that we have a negative exponential, the limit of the inside of the square brackets as t tends to infinity will be zero. Also, since e 0 = 1, this leaves us with 1 1 L[1] = 0 − − = . s s 1 So f (t) = 1 → L[1] = s (2) f (:t) = t ∞ − s t L[t ] = t .e dt 0 This integral requires integration by parts to complete the process, but with the same assumptions regarding s as before, it can readily be shown that 1 L[t ] = . s 2 1 So f (t) = t → L[t ] = s 2 Very quickly the integrations required to complete the Laplace transformation become difficult and messy. For this reason, we generally work from a table of pre-determined Laplace transforms (see Appendix). 3 Examples Using Table of Laplace Transforms (3) (a) Determine L[t 3 ]. t 3 is not in the table explicitly, but t n is: n ! L[t n ] = s n + 1 For t 3 we require n = 3 : 3! 3×2×1 6 L[t 3 ] = = = s 4 s 4 s 4 (b) Determine L[ e − 2 t ]. From table: −α 1 L[ e t ] = . s + α Set α = 2 : − 1 L[ e 2 t ] = s + 2 (c) Determine L[sin ( 4t ) ]. From table: ω L[sin (ω t ) ] = . s 2 + ω 2 Set ω = 4 : 4 4 L[sin ( 4t ) ] = = s 2 + 4 2 s 2 + 16 4 3) Properties of the Laplace Transform The Laplace transform has several special properties that make it a useful mathematical tool. We consider some of these now. a) Linearity Suppose we have two functions along with their respective Laplace transforms: L[ f (t ) ] = F ( s ) L[ g ( t ) ] = G ( s ) . The property of linearity means that L[ a f (t ) + b g (t ) ] = a F ( s ) + b G ( s ) providing a and b are constants. This makes the transformation of a string of functions straightforward. Example s ω 2 s + 3 ω (4) L[ 2 cos (ω t ) + 3 sin (ω t ) ] = 2 + 3 = s 2 + ω 2 s 2 + ω 2 s 2 + ω 2 Warning: The Laplace transform of a product is NOT EQUAL TO the product of the individual Laplace transforms. We have to invoke other properties of the Laplace transform to deal with such. b) The First Shifting Theorem Suppose a function f (t ) has the Laplace transform F ( s ) . It is easily demonstrated that L[ e −α t f ( t ) ] = F ( s + α ) . Example (5) From tables and Example (3)(a) we have 6 L[t 3 ] = . [1] s 4 By the first shifting property − 6 L[ e 2 t t 3 ] = . [2] ( s + 2) 4 To obtain [2] from [1] we merely replace s by ( s + 2) . 5 c) Transformation of Derivatives As before, denote the Laplace transform (LT) of f (t ) by F ( s ) . Now consider the LT of the derivative of f (t ) , denoted by f ( t ) : ∞ − s t L[ f (t ) ] = f (t ) e dt . 0 Integrate by parts (integrating f ( t ) and differentiating e − s t ): ∞ ∞ = []f ( t ) e − s t − f ( t ) ( − s ) e − s t dt 0 0 ∞ − s t = 0 − f ( 0) + s f ( t ) e dt 0 = − f ( 0 ) + s F ( s ) . So L[ f (t ) ] = s F ( s ) − f ( 0 ) . We have expressed the Laplace transform of a derivative in terms of the Laplace transform of the undifferentiated function. In effect, the Laplace transform has converted the operation of differentiation into the simpler operation of multiplication by s . In a similar fashion, using repeated integration by parts, we can show that L[ f (t ) ] = s 2 F ( s ) − s f ( 0) − f ( 0 ) . This is one of the most important properties of the Laplace transform. The Laplace transform “gets rid of” derivatives; just the thing for solving differential equations! When we come to solve differential equations using Laplace transforms we shall use the following alternative notation: L[x] = x L[ x ] = s x − x(0) 2 L[ x] = s x − s x(0) − x (0) . However, before we can solve differential equations, we need to look at the reverse process of finding functions of t from given Laplace transforms. 6 4) Inverse Laplace Transforms So far, we have looked at how to determine the LT of a function of t , ending up with a function of s . The table of Laplace transforms collects together the results we have considered, and more. When we apply Laplace transforms to solve problems we will have to invoke the inverse transformation. That is, given a Laplace transform F ( s ) we will want to determine the corresponding f (t ) . In general we have γ + ∞ − 1 j L 1 [ F ( s ) ] = F ( s ) e s t ds , 2π j γ − j ∞ where the evaluation of the integral requires a knowledge of complex analysis, which is too difficult to consider here. Instead, we shall rely on the table of Laplace transforms used in reverse to provide inverse Laplace transforms. This will mean manipulating a given Laplace transform until it looks like one or more entries in the right of the table. The inverse is then determined from the left of the table. The following examples illustrate the main algebraic techniques required. These include completing the square, factorisation and the formation of partial fractions. See separate documents for the details of completing the square and partial fractions. Examples 3 (6) Invert the Laplace transform . s 6 n! The closest entry in the table is which is inverted as: sn + 1 n! → t n . sn + 1 Setting n = 5 : 5! 120 → t 5 . [Note: 5! = 5×4×3×2×1 = 120 ], so → t 5 . s 6 s 6 In the given Laplace transform there is a 3 in the numerator but we would like there to be a 120 to match the table entry. We can re-write the transform providing we do not alter its “net value”: 3 1 3 120 1 120 = 3 = = . 6 6 6 6 s s 120 s 40 s The term in the square brackets is now exactly the table entry so we can invert that and simply multiply by the fraction in front: 3 1 → t 5 . s 6 40 7 1 (7) Invert . s 2 + 2 s + 5 This requires the technique of “completing the square” and a little bit of fine tuning to re-write it in a form that can be inverted from tables. Note that the closest entry in the table gives: ω → e −α t sin (ω t ) . ( s + α ) 2 + ω 2 Now complete the square in the denominator: s 2 + 2 s + 5 = ( s + 1) 2 − 1 + 5 = ( s + 1) 2 + 4 . The given transform becomes: 1 1 1 = = . s 2 + 2 s + 5 ( s + 1) 2 + 4 ( s + 1) 2 + 2 2 This is now very close to the table entry with α = 1 and ω = 2 . We would like there to be a 2 on top so “fine tuning” gives 1 = 1 2 2 2 2 2 , ( s + 1) + 2 2 ( s + 1) + 2 where the term in the square brackets is exactly the table entry with α = 1 and ω = 2 . Inverting this and multiplying by the fraction gives 1 2 − → 1 t 2 2 2 e sin ( 2t ) . 2 ( s + 1) + 2 s + 5 (8)(a) Invert . s 2 − 2 s − 3 Factorising the denominator and splitting the result into its partial fractions deals with this one. Note that the details of the partial fraction expansion have been omitted. s + 5 s + 5 2 1 = = − s 2 − 2 s − 3 ( s − 3) ( s + 1) s − 3 s + 1 1 −α Use → e t twice with α = − 3 and α = +1 to give ( s + α ) 2 1 − − → 2e 3t − e t s − 3 s + 1 8 s + 5 (8)(b) Invert by completing the square.