The Fourier Transform
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The Fourier Transform As we have seen, any (sufficiently smooth) function f(t) that is periodic can be built out of sin’s and cos’s. We have also seen that complex exponentials may be used in place of sin’s and cos’s. We shall now use complex exponentials because they lead to less writing and simpler computations, but yet can easily be converted into sin’s and cos’s. If f(t) has period 2ℓ, its (complex) Fourier series expansion is ℓ ∞ π π ik ℓ t 1 ik ℓ t f(t)= cke with ck = 2ℓ f(t)e− dt (1) k= ℓ X−∞ Z− ik π t ik π (t+2ℓ) Not surprisingly, each term cke ℓ in this expansion also has period 2ℓ, because cke ℓ = ik π t i2kπ ik π t cke ℓ e = cke ℓ . We now develop an expansion for non-periodic functions, by allowing complex exponentials (or equivalently sin’s and cos’s) of all possible periods, not just 2ℓ, for some fixed ℓ. So, from now on, do not assume that f(t) is periodic. For simplicity we’ll only develop the expansions for functions that are zero for all sufficiently large t . | | With a little more work, one can show that our conclusions apply to a much broader class of functions. Let L > 0 be sufficiently large that f(t) = 0 for all t L. We can get a Fourier series expansion for the part | | ≥ of f(t) with L<t<L by using the periodic extension trick. Define F (t) to be the unique function − L determined by the requirements that i) F (t)= f(t) for L<t L L − ≤ ii) FL(t) is periodic of period 2L Then, for L<t<L, − L ∞ π π ik L t 1 ik L t f(t)= FL(t)= ck(L)e where ck(L)= 2L f(t)e− dt (2) k= L X−∞ Z− If we can somehow take the limit L , we will get a representation of f that is is valid for all t’s , not → ∞ just those in some finite interval L<t<L. This is exactly what we shall do, by the simple expedient − of interpreting the sum in (2) as a Riemann sum approximation to a certain integral. For each integer k, th π π define the k frequency to be ωk = k L and denote by ∆ω = L the spacing, ωk+1 ωk, between any two iωt − successive frequencies. Also define fˆ(ω)= ∞ f(t)e− dt. Since f(t) = 0 for all t L −∞ | |≥ L π ∞ R π ∞ 1 ik L t 1 i(k L )t 1 iωk t 1 ˆ 1 ˆ ck(L)= 2L f(t)e− dt = 2L f(t)e− dt = 2L f(t)e− dt = 2L f(ωk)= 2π f(ωk) ∆ω L Z− Z−∞ Z−∞ In this notation, ∞ 1 ˆ iωkt f(t)= FL(t)= 2π f(ωk)e ∆ω (3) k= X−∞ for any L<t<L. As we let L , the restriction L<t<L disappears, and the right hand − → ∞ − 1 ∞ ˆ iωt side converges exactly to the integral 2π f(ω)e dω. To see this, cut the domain of integration into −∞ y R 1 ˆ iωt y = 2π f(ω)e ω 0 ω1 ω2 ω3 ∆ω c Joel Feldman. 2007. All rights reserved. March 1, 2007 The Fourier Transform 1 small slices of width ∆ω and approximate, as in the above figure, the area under the part of the graph of 1 ˆ iωt 1 ˆ iωk t 2π f(ω)e with ω between ωk and ωk + ∆ω by the area of a rectangle of base ∆ω and height 2π f(ωk)e . 1 iωt 1 iωk t ∞ ˆ ∞ ˆ This approximates the integral 2π f(ω)e dω by the sum 2π k= f(ωk)e ∆ω. As L the −∞ → ∞ approximation gets better and better−∞ so that the sum approaches the integral. So taking the limit of (3) as R P L gives → ∞ 1 ∞ ˆ iωt ˆ ∞ iωt f(t)= 2π f(ω)e dω where f(ω)= f(t)e− dt (4) Z−∞ Z−∞ The function fˆ is called the Fourier transform of f. It is to be thought of as the frequency profile of the signal f(t). Example 1 Suppose that a signal gets turned on at t = 0 and then decays exponentially, so that at f(t)= e− if t 0 0 if t<≥ 0 for some a> 0. The Fourier transform of this signal is t(a+iω) ∞ iωt ∞ at iωt ∞ t(a+iω) e− ∞ 1 fˆ(ω)= f(t)e− dt = e− e− dt = e− dt = = 0 0 (a + iω) 0 a + iω Z−∞ Z Z − √ 2 2 1 ω ˆ iφ(ω) Since a + iω has modulus a + ω and phase tan− a , we have that f(ω) = A(ω)e with A(ω) = 1 = 1 and φ(ω)= tan 1 ω . The amplitude A(ω) and phase φ(ω) are plotted below. a+iω √a2+ω2 − a | | − A(ω) a a ω ϕ(ω) a a ω Example 2 Suppose that a signal consists of a single rectangular pulse of width 1 and height 1. Let’s say that it gets turned on at t = 1 and turned off at t = 1 . The standard name for this “normalized” − 2 2 rectangular pulse is 1 1 if 1 <t< 1 rect(t)= 2 2 0 otherwise− 1 1 t − 2 2 It is also called a normalized boxcar function. The Fourier transform of this signal is 1/2 iωt 1/2 iω/2 iω/2 ∞ iωt iωt e− e e− 2 ω rect(ω)= rect(t) e− dt = e− dt = = − = ω sin 2 1/2 iω 1/2 iω Z−∞ Z− − − d c Joel Feldman. 2007. All rights reserved. March 1, 2007 The Fourier Transform 2 1/2 when ω = 0. When ω = 0, rect(0) = 1/2 dt = 1. By l’Hˆopital’s rule 6 − R d sin ω 1 cos ω lim rect(ω) = lim 2 2 = lim 2 2 2 =1= rect(0) ω 0 ω 0 ω 0 → → ω → 1 d d so rect(ω) is continuous at ω = 0. There is a standard function called “sinc” that is defined(1) by sinc ω = sin ω ω ω . In this notation rect(ω) = sinc 2 . Here is a graph of rect(ω). d d 1 d rect(ω) d ω 2π 2π − Properties of the Fourier Transform Linearity If α and β are any constants and we build a new function h(t)= αf(t)+ βg(t) as a linear combination of two old functions f(t) and g(t), then the Fourier transform of h is ∞ iωt ∞ iωt ∞ iωt ∞ iωt hˆ(ω)= h(t)e− dt = αf(t)+ βg(t) e− dt = α f(t)e− dt + β g(t)e− dt (L) Z−∞ Z−∞ Z−∞ Z−∞ = αfˆ(ω)+ βgˆ(ω) Time Shifting Suppose that we build a new function h(t)= f(t t ) by time shifting a function f(t) by t . The easy − 0 0 way to check the direction of the shift is to note that if the original signal f(t) has a jump when its argument t = a, then the new signal h(t) = f(t t ) has a jump when t t = a, which is at t = a + t , t units to − 0 − 0 0 0 the right of the original jump. t0 f(t) h(t)= f(t t ) − 0 t a a + t0 The Fourier transform of h is ′ ∞ iωt ∞ iωt ∞ iω(t +t0) hˆ(ω)= h(t)e− dt = f(t t )e− dt = f(t′)e− dt′ where t = t′ + t , dt = dt′ − 0 0 Z−∞ Z−∞ Z−∞ (T) ′ iωt0 ∞ iωt iωt0 = e− f(t′)e− dt′ = e− fˆ(ω) Z−∞ (1) sin ω There are actually two different commonly used definitions. The first, which we shall use, is sinc ω = ω . The second is sin πω sinc ω = πω . It is sometimes called the normalized sinc function. c Joel Feldman. 2007. All rights reserved. March 1, 2007 The Fourier Transform 3 Scaling t If we build a new function h(t)= f α by scaling time by a factor of α> 0, then the Fourier transform of h is ′ ∞ iωt ∞ t iωt ∞ iωαt hˆ(ω)= h(t)e− dt = f e− dt = α f(t′)e− dt′ where t = αt′, dt = αdt′ α (S) Z−∞ Z−∞ Z−∞ = αfˆ(αω) Example 3 Now consider a signal that consists of a single rectangular pulse of height H, width W and centre C. y H y = rHWC (t) t C W C + W − 2 2 t C The function rHWC (t) = H rect −W (with rect the normalized rectangular pulse of Example 2) has height H and jumps when t C = 1 , i.e. t = C 1 W and so is the specified signal. By combining properties (L), −W 2 2 ± ± t C (T) and (S), we can determine the Fourier transform of rHWC (t) = H rect −W for any H, C and W . We build it up in three steps. First we consider the special case in which H = 1 and C = 0. Then we have R (t) = rect t . So, ◦ 1 W according to the scaling property (S) with t t t f(t) = rect(t), h(t)= R1(t) = rect W = f W = f α with α = W we have ˆ ˆ ˆ 2 Wω 2 Wω R1(ω)= h(ω)= αf(αω)= W rect(Wω)= W Wω sin 2 = ω sin 2 t C Next we allow a nonzero C too and consider R d(t) = rect − = R (t C).