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5 Fourier and Laplace Transforms

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5 Fourier and Laplace Transforms 5 Fourier and Laplace Transforms “There is no branch of mathematics, however abstract, which may not some day be applied to phenomena of the real world.”, Nikolai Lobatchevsky (1792-1856) 5.1 Introduction In this chapter we turn to the study of Fourier transforms, which provide integral representations of functions defined on the entire real line. Such functions can represent analog signals. Recall that analog signals are continuous signals which are sums over a continuous set of fre- quencies. Our starting point will be to rewrite Fourier trigonometric series as Fourier exponential series. The sums over discrete frequencies will lead to a sum (integral) over continuous frequencies. The resulting integrals will be complex integrals, which can be evaluated using contour methods. We will investigate the properties of these Fourier transforms and get prepared to ask how the analog signal representations are related to the Fourier se- ries expansions over discrete frequencies which we had seen in Chapter 2. Fourier series represented functions which were defined over finite do- mains such as x 2 [0, L]. Our explorations will lead us into a discussion of the sampling of signals in the next chapter. We will also discuss a related integral transform, the Laplace transform. In this chapter we will explore the use Laplace transforms are useful in solving initial value problems in differen- of integral transforms. Given a function f (x), we define an integral transform to tial equations and can be used to relate the input to the output of a linear a new function F(k) as system. Both transforms provide an introduction to a more general theory Z b F(k) = f (x)K(x, k) dx. of transforms, which are used to transform specific problems to simpler a ones. Here K(x, k) is called the kernel of the In Figure 5.1 we summarize the transform scheme for solving an initial transform. We will concentrate specifi- cally on Fourier transforms, value problem. One can solve the differential equation directly, evolving the Z ¥ initial condition y(0) into the solution y(t) at a later time. fˆ(k) = f (x)eikx dx, −¥ However, the transform method can be used to solve the problem indi- and Laplace transforms rectly. Starting with the differential equation and an initial condition, one Z ¥ F(s) = f (t)e−st dt. computes its Transform (T) using 0 Z ¥ Y(s) = y(t)e−st dt. 0 170 fourier and complex analysis Figure 5.1: Schematic of using trans- ODE, y(0) T Alg Eqn forms to solve a linear ordinary differ- ential equation. solve solve y(t) IT Y(s) Applying the transform to the differential equation, one obtains a simpler (algebraic) equation satisfied by Y(s), which is simpler to solve than the original differential equation. Once Y(s) has been found, then one applies the Inverse Transform (IT) to Y(s) in order to get the desired solution, y(t). We will see how all of this plays out by the end of the chapter. We will begin by introducing the Fourier transform. First, we need to see how one can rewrite a trigonometric Fourier series as complex exponential series. Then we can extend the new representation of such series to ana- log signals, which typically have infinite periods. In later chapters we will highlight the connection between these analog signals and their associated digital signals. 5.2 Complex Exponential Fourier Series Before deriving the Fourier transform, we will need to rewrite the trigonometric Fourier series representation as a complex exponential Fourier series. We first recall from Chapter 2 the trigonometric Fourier se- ries representation of a function defined on [−p, p] with period 2p. The Fourier series is given by ¥ a0 f (x) ∼ + ∑ (an cos nx + bn sin nx) ,( 5.1) 2 n=1 where the Fourier coefficients were found as 1 Z p an = f (x) cos nx dx, n = 0, 1, . , p −p 1 Z p bn = f (x) sin nx dx, n = 1, 2, . .( 5.2) p −p In order to derive the exponential Fourier series, we replace the trigono- metric functions with exponential functions and collect like exponential terms. This gives ¥ inx −inx inx −inx a0 e + e e − e f (x) ∼ + ∑ an + bn 2 n=1 2 2i a ¥ a − ib ¥ a + ib = 0 + ∑ n n einx + ∑ n n e−inx.( 5.3) 2 n=1 2 n=1 2 fourier and laplace transforms 171 The coefficients of the complex exponentials can be rewritten by defining 1 c = (a + ib ), n = 1, 2, . .( 5.4) n 2 n n This implies that 1 c¯ = (a − ib ), n = 1, 2, . .( 5.5) n 2 n n So far, the representation is rewritten as ¥ ¥ a0 inx −inx f (x) ∼ + ∑ c¯ne + ∑ cne . 2 n=1 n=1 Re-indexing the first sum, by introducing k = −n, we can write −¥ ¥ a0 −ikx −inx f (x) ∼ + ∑ c¯−ke + ∑ cne . 2 k=−1 n=1 Since k is a dummy index, we replace it with a new n as −¥ ¥ a0 −inx −inx f (x) ∼ + ∑ c¯−ne + ∑ cne . 2 n=−1 n=1 We can now combine all the terms into a simple sum. We first define cn for negative n’s by cn = c¯−n, n = −1, −2, . a0 Letting c0 = 2 , we can write the complex exponential Fourier series repre- sentation as ¥ −inx f (x) ∼ ∑ cne ,( 5.6) n=−¥ where 1 c = (a + ib ), n = 1, 2, . , n 2 n n 1 c = (a− − ib− ), n = −1, −2, . , n 2 n n a c = 0 .( 5.7) 0 2 Given such a representation, we would like to write out the integral forms of the coefficients, cn. So, we replace the an’s and bn’s with their integral representations and replace the trigonometric functions with complex expo- nential functions. Doing this, we have for n = 1, 2, . , 1 c = (a + ib ) n 2 n n 1 1 Z p i Z p = f (x) cos nx dx + f (x) sin nx dx 2 p −p p −p 1 Z p einx + e−inx i Z p einx − e−inx = f (x) dx + f (x) dx 2p −p 2 2p −p 2i 1 Z p = f (x)einx dx.( 5.8) 2p −p 172 fourier and complex analysis It is a simple matter to determine the cn’s for other values of n. For n = 0, we have that Z p a0 1 c0 = = f (x) dx. 2 2p −p For n = −1, −2, . ., we find that Z p Z p 1 −inx 1 inx cn = c¯n = f (x)e dx = f (x)e dx. 2p −p 2p −p Therefore, we have obtained the complex exponential Fourier series coeffi- cients for all n. Now we can define the complex exponential Fourier series for the function f (x) defined on [−p, p] as shown below. Complex Exponential Series for f (x) Defined on [−p, p] ¥ −inx f (x) ∼ ∑ cne ,( 5.9) n=−¥ Z p 1 inx cn = f (x)e dx.( 5.10) 2p −p We can easily extend the above analysis to other intervals. For example, for x 2 [−L, L] the Fourier trigonometric series is ¥ a0 npx npx f (x) ∼ + ∑ an cos + bn sin 2 n=1 L L with Fourier coefficients 1 Z L npx an = f (x) cos dx, n = 0, 1, . , L −L L 1 Z L npx bn = f (x) sin dx, n = 1, 2, . L −L L This can be rewritten as an exponential Fourier series of the form Complex Exponential Series for f (x) Defined on [−L, L] ¥ −inpx/L f (x) ∼ ∑ cne ,( 5.11) n=−¥ Z L 1 inpx/L cn = f (x)e dx.( 5.12) 2L −L We can now use this complex exponential Fourier series for function de- fined on [−L, L] to derive the Fourier transform by letting L get large. This will lead to a sum over a continuous set of frequencies, as opposed to the sum over discrete frequencies, which Fourier series represent. 5.3 Exponential Fourier Transform Both the trigonometric and complex exponential Fourier series provide us with representations of a class of functions of finite period in fourier and laplace transforms 173 terms of sums over a discrete set of frequencies. In particular, for functions defined on x 2 [−L, L], the period of the Fourier series representation is 2L. We can write the arguments in the exponentials, e−inpx/L, in terms of −iw x the angular frequency, wn = np/L, as e n . We note that the frequencies, np nn, are then defined through wn = 2pnn = L . Therefore, the complex exponential series is seen to be a sum over a discrete, or countable, set of frequencies. We would now like to extend the finite interval to an infinite interval, x 2 (−¥, ¥), and to extend the discrete set of (angular) frequencies to a continuous range of frequencies, w 2 (−¥, ¥). One can do this rigorously. n It amounts to letting L and n get large and keeping L fixed. p We first define Dw = L , so that wn = nDw. Inserting the Fourier coeffi- cients (5.12) into Equation (5.11), we have ¥ −inpx/L f (x) ∼ ∑ cne n=−¥ ¥ 1 Z L = ∑ f (x)einpx/L dx e−inpx/L n=−¥ 2L −L ¥ Dw Z L = ∑ f (x)eiwnx dx e−iwn x.( 5.13) n=−¥ 2p −L Now, we let L get large, so that Dw becomes small and wn approaches the angular frequency w. Then, 1 ¥ Z L f (x) ∼ lim ∑ f (x)eiwnx dx e−iwn xDw Dw!0,L!¥ 2p n=−¥ −L 1 Z ¥ Z ¥ = f (x)eiwx dx e−iwx dw.( 5.14) 2p −¥ −¥ Looking at this last result, we formally arrive at the definition of the Definitions of the Fourier transform and Fourier transform.
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