Distributions And Their Fourier Transforms
Chapter 4Distributions and Their FourierTransforms4.1The Day of ReckoningWe’ve been playing a little fast and loose with the Fourier transform — applying Fourier inversion, appealing to duality, and all that. “Fast and loose” is an understatement if ever there was one, but it’s also truethat we haven’t done anything “wrong”. All of our formulas and all of our applications have been correct,if not fully justified. Nevertheless, we have to come to terms with some fundamental questions. It will takeus some time, but in the end we will have settled on a very wide class of signals with these properties: The allowed signals include δ’s, unit steps, ramps, sines, cosines, and all other standard signals thatthe world’s economy depends on. The Fourier transform and its inverse are defined for all of these signals. Fourier inversion works.These are the three most important features of the development to come, but we’ll also reestablish someof our specific results and as an added benefit we’ll even finish off differential calculus!4.1.1A too simple criterion and an exampleIt’s not hard to write down an assumption on a function that guarantees the existence of its Fouriertransform and even implies a little more than existence. IfZ f (t) dt then F f and F 1f exist and are continuous. Existence follows from F f (s) ZZ e 2πist f (t) dt e 2πist f (t) dt Z f (t) dt .
138Chapter 4 Distributions and Their Fourier TransformsHere we’ve used that the magnitude of the integral is less that the integral of the magnitude.1 There’sactually something to say here, but while it’s not complicated, I’d just as soon defer this and othercomments on “general facts on integrals” to Section 4.3; read it if only lightly — it provides some additionalorientation.Continuity is the little extra information we get beyond existence. Continuity follows as follows. For anys and s0 we haveZ Z 00 2πist F f (s) F f (s ) ef (t) dt e 2πis t f (t) dt Z Z 00 (e 2πist e 2πis t )f (t) dt e 2πist e 2πis t f (t) dt R As a consequence of f (t) dt we can take the limit as s0 s inside the integral. If we do that0then e 2πist e 2πis t 0, that is, F f (s) F f (s0) 0 as s0 swhich says that F f (s) is continuous. The same argument works to show that F 1f is continuous.2We haven’t said anything here about Fourier inversion — no such statement appears in the criterion. Let’slook right away at an example.The very first example we computed, and still an important one, is the Fourier transform of Π. We founddirectly thatZ Z 1/2 2πistF Π(s) eΠ(t) dt e 2πist dt sinc s . 1/2No problem there, no problem whatsoever. The criterion even applies; Π is in L1 (R) sinceZ Z 1/2 Π(t) dt 1 dt 1 . 1/2Furthermore, the transform F Π(s) sinc s is continuous. That’s worth remarking on: Although the signaljumps (Π has a discontinuity) the Fourier transform does not, just as guaranteed by the preceding result— make this part of your intuition on the Fourier transform vis à vis the signal.Appealing to the Fourier inversion theorem and what we called duality, we then saidZ F sinc(t) e 2πist sinc t dt Π(s) . Here we have a problem. The sinc function does not satisfy the integrability criterion. It is my sad dutyto inform you thatZ sinc t dt . I’llR give you two ways of seeing the failure of sinc t to be integrable. First, if sinc did satisfy the criterion sinc t dt then its Fourier transform would be continuous. But its Fourier transform, which has12Magnitude, not absolute value, because the integral is complex number.So another general fact we’ve used here is that we can take the limit inside the integral. Save yourself for other things andlet some of these “general facts” ride without insisting on complete justifications — they’re everywhere once you let the rigorpolice back on the beat.
4.1 The Day of Reckoning139to come out to be Π, is not continuous. Or, if you don’t like that, here’s a direct argument. We canfind infinitely many intervals where sin πt 1/2; this happens when t is between 1/6 and 5/6, and thatrepeats for infinitely many intervals, for example on In [ 16 2n, 56 2n], n 0, 1, 2, . . . , because sin πtis periodic of period 2. The In all have length 2/3. On In we have t 56 2n, so11 t 5/6 2nandZInThenZ sin πt 11dt π t 2π 5/6 2nX sin πt dt π t nZInZdt In1 21.2π 3 5/6 2n sin πt 11 Xdt .π t 3π n 1 5/6 2nIt’s true that sinc t sin πt/πt tends to 0 as t — the 1/t factor makes that happen — but not“fast enough” to make the integral of sinc t converge.This is the most basic example in the theory! It’s not clear that the integral defining the Fourier transformof sinc exists, at least it doesn’t follow from the criterion. Doesn’t this bother you? Isn’t it a littleembarrassing that multibillion dollar industries seem to depend on integrals that don’t converge?In fact, there isn’t so much of a problem with either Π or sinc. It is true that(Z 1 t 12e 2πist sinc s ds 0 t 12 However showing this — evaluating the improper integral that defines the Fourier transform — requiresspecial arguments and techniques. The sinc function oscillates, as do the real and imaginary parts of thecomplex exponential, and integrating e 2πist sinc s involves enough cancellation for the limitlima b Zbe 2πist sinc s dsato exist.Thus Fourier inversion, and duality, can be pushed through in this case. At least almost. You’ll noticethat I didn’t say anything about the points t 1/2, where there’s a jump in Π in the time domain. Inthose cases the improper integral does not exist, but with some additional interpretations one might beable to convince a sympathetic friend thatZ e 2πi( 1/2)s sinc s ds 12 in the appropriate sense (invoking “principle value integrals” — more on this in a later lecture). At bestthis is post hoc and needs some fast talking.3The truth is that cancellations that occur in the sinc integral or in its Fourier transform are a very subtleand dicey thing. Such risky encounters are to be avoided. We’d like a more robust, trustworthy theory.3One might also then argue that defining Π( 1/2) 1/2 is the best choice. I don’t want to get into it.
140Chapter 4 Distributions and Their Fourier TransformsThe news so farwhenHere’s a quick summary of the situation. The Fourier transform of f (t) is definedZ f (t) dt . We allow f to be complex valued in this definition. The collection of all functions on R satisfying thiscondition is denoted by L1(R), the superscript 1 indicating that we integrate f (t) to the first power.4The L1 -norm of F is defined byZ kf k1 f (t) dt . Many of the examples we worked with are L1 -functions — the rect function, the triangle function, theexponential decay (one or two-sided), Gaussians — so our computations of the Fourier transforms in thosecases were perfectly justifiable (and correct). Note that L1 -functions can have discontinuities, as in therect function.The criterion says that if f L1(R) then F f exists. We can also sayZ Z F f (s) e 2πist f (t) dt f (t) dt kf k1 . That is: The magnitude of the Fourier transform is bounded by the L1 -norm of the function.This is a handy estimate to be able to write down — we’ll use it shortly. However, to issue a warning:Fourier transforms of L1 (R) functions may themselves not be in L1, like for the sinc function, so wedon’t know without further work what more can be done, if anything.The conclusion is that L1 -integrability of a signal is just too simple a criterion on which to build a reallyhelpful theory. This is a serious issue for us to understand. Its resolution will greatly extend the usefulnessof the methods we have come to rely on.There are other problems, too. Take, for example, the signal f (t) cos 2πt. As it stands now, this signaldoes not even have a Fourier transform — does not have a spectrum! — for the integralZ e 2πist cos 2πt dt does not converge, no way, no how. This is no good.Before we bury L1(R) as too restrictive for our needs, here’s one more good thing about it. There’s actuallya stronger consequence for F f than just continuity. IfZ f (t) dt then F f (s) 0 as s . 4And the letter “L” indicating that it’s really the Lebesgue integral that should be employed.
4.1 The Day of Reckoning141This is called the Riemann-Lebesgue lemma and it’s more difficult to prove than showing simply that F fis continuous. I’ll comment on it later; see Section 4.19. One might view the result as saying that F f (s) isat least trying to be integrable. It’s continuous and it tends to zero as s . Unfortunately, the factthat F f (s) 0 does not imply that it’s integrable (think of sinc, again).5 If we knew something, or couldinsist on something about the rate at which a signal or its transform tends to zero at then perhapswe could push on further.4.1.2The path, the wayTo repeat, we want our theory to encompass the following three points: The allowed signals include δ’s, unit steps, ramps, sines, cosines, and all other standard signals thatthe world’s economy depends on. The Fourier transform and its inverse are defined for all of these signals. Fourier inversion works.Fiddling around with L1 (R) or substitutes, putting extra conditions on jumps — all have been used. Thepath to success lies elsewhere. It is well marked and firmly established, but it involves a break with theclassical point of view. The outline of how all this is settled goes like this:1. We single out a collection of functions S for which convergence of the Fourier integrals is assured,for which a function and its Fourier transform are both in S, and for which Fourier inversion works.Furthermore, Parseval’s identity holds:Z Z 2 f (x) dx F f (s) 2 ds . This much is classical; new ideas with new intentions, yes, but not new objects. Perhaps surprisinglyit’s not so hard to find a suitable collection S, at least if one knows what one is looking for. Butwhat comes next is definitely not “classical”. It had been first anticipated and used effectively in anearly form by O. Heaviside, developed, somewhat, and dismissed, mostly, soon after by less talentedpeople, then cultivated by and often associated with the work of P. Dirac, and finally refined byL. Schwartz.2. S forms a class of test functions which, in turn, serve to define a larger class of generalized functions ordistributions, called, for this class of test functions the tempered distributions, T . Precisely becauseS was chosen to be the ideal Fourier friendly space of classical signals, the tempered distributionsare likewise well suited for Fourier methods. The collection of tempered distributions includes, forexample, L1 and L2-functions (which can be wildly discontinuous), the sinc function, and complexexponentials (hence periodic functions). But it includes much more, like the delta functions andrelated objects.3. The Fourier transform and its inverse will be defined so as to operate on these tempered distributions,and they operate to produce distributions of the same type. Thus the inverse Fourier transform canbe applied, and the Fourier inversion theorem holds in this setting.4. In the case when a tempered distributions “comes from a function” — in a way we’ll make precise— the Fourier transform reduces to the usual definition as an integral, when the integral makessense. However, tempered distributions are more general than functions, so we really will have donesomething new and we won’t have lost anything in the process.5For that matter, a function in L1 (R) need not tend to zero at ; that’s also discussed in Appendix 1.
142Chapter 4 Distributions and Their Fourier TransformsOur goal is to hit the relatively few main ideas in the outline above, suppressing the considerable massof details. In practical terms this will enable us to introduce delta functions and the like as tools forcomputation, and to feel a greater measure of confidence in the range of applicability of the formulas.We’re taking this path because it works, it’s very interesting, and it’s easy to compute with. I especiallywant you to believe the last point.We’ll touch on some other approaches to defining distributions and generalized Fourier transforms, butas far as I’m concerned they are the equivalent of vacuum tube technology. You can do distributions inother ways, and some people really love building things with vacuum tubes, but wouldn’t you rather learnsomething a little more up to date?4.2The Right Functions for Fourier Transforms: Rapidly DecreasingFunctionsMathematics progresses more by making intelligent definitions than by proving theorems. The hardestwork is often in formulating the fundamental concepts in the right way, a way that will then make thedeductions from those definitions (relatively) easy and natural. This can take awhile to sort out, anda subject might be reworked several times as it matures; when new discoveries are made and one seeswhere things end up, there’s a tendency to go back and change the starting point so that the trip becomeseasier. Mathematicians may be more self-conscious about this process, but there are certainly examplesin engineering where close attention to the basic definitions has shaped a field — think of Shannon’s workon Information Theory, for a particularly striking example.Nevertheless, engineers, in particular, often find this tiresome, wanting to do something and not “just talkabout it”: “Devices don’t have hypotheses”, as one of my colleagues put it. One can also have too muchof a good thing — too many trips back to the starting point to rewrite the rules can make it hard tofollow the game, especially if one has already played by the earlier rules. I’m sympathetic to both of thesecriticisms, and for our present work on the Fourier transform I’ll try to steer a course that makes thedefinitions reasonable and lets us make steady forward progress.4.2.1Smoothness and decayTo ask “how fast” F f (s) might tend to zero, depending on what additional assumptions we might makeabout the function f (x) beyond integrability, will lead to our defining “rapidly decreasing functions”,and this is the key. Integrability is too weak a condition on the signal f , but it does imply that F f (s) iscontinuous and tends to 0 at . What we’re going to do is study the relationship between the smoothnessof a function — not just continuity, but how many times it can be differentiated — and the rate at whichits Fourier transform decays at infinity.We’ll always assume that f (x) is absolutely integrable, and so has a Fourier transform. Let’s suppose,more stringently, that xf (x) is integrable, i.e.,Z xf (x) dx .
4.2 The Right Functions for Fourier Transforms: Rapidly Decreasing Functions143Then xf (x) has a Fourier transform, and so does 2πixf (x) and its Fourier transform isZ F ( 2πixf (x)) ( 2πix)e 2πisxf (x) dx Z Z d 2πisxdee 2πisx f (x) dx f (x) dx dsds (switching d/ds and the integral is justified by the integrability of xf (x) ) d(F f )(s)dsThis says that the Fourier transform F f (s) is differentiable and that its derivative is F ( 2πixf (x)). Whenf (x) is merely integrable we know that F f (s) is merely continuous, but with the extra assumption on theintegrability of xf (x) we conclude that F f (s) is actually differentiable. (And its derivative is continuous.Why?)For one more go-round in this direction, what if x2 f (x) is integrable? Then, by the same argument,Z F (( 2πix)2f (x)) ( 2πix)2e 2πisx f (x) dx Z 2Z d 2πisxd2d2 2πisx f(x)dx eef(x)dx (F f )(s) ,222 dsds dsand we see that F f is twice differentiable. (And its second derivative is continuous.)Clearly we can proceed like this, and as a somewhat imprecise headline we might then announce: Faster decay of f (x) at infinity leads to a greater smoothness of the Fourier transform.Now let’s take this in another direction, with an assumption on the smoothness of the signal. Supposef (x) is differentiable, that its derivative is integrable, and that f (x) 0 as x . I’ve thrown in allthe assumptions I need to justify the following calculation:Z F f (s) e 2πisx f (x) dx x Z 2πisxe 2πisxe f (x) f 0 (x) dx 2πis x 2πis (integration by parts with u f (x), dv e 2πisx dx)Z 1 e 2πisx f 0 (x) dx (using f (x) 0 as x )2πis 1 (F f 0)(s)2πisWe then have F f (s) 11 (F f 0)(s) kf 0k1 .2πs2πsThe last inequality follows from the result: “The Fourier transform is bounded by the L1-norm of thefunction”. This says that F f (s) tends to 0 at like 1/s. (Remember that kf 0k1 is some fixed numberhere, independent of s.) Earlier we commented (without proof) that if f is integrable then F f tends to 0at , but here with the stronger assumptions we get a stronger conclusion, that F f tends to zero at acertain rate.
144Chapter 4 Distributions and Their Fourier TransformsLet’s go one step further in this direction. Suppose f (x) is twice differentiable, that its first and secondderivatives are integrable, and that f (x) and f 0 (x) tend to 0 as x . The same argument givesZ F f (s) e 2πisx f (x) dx Z 1 e 2πisx f 0 (x) dx (picking up on where we were before)2πis x Z 2πisxee 2πisx1000 f (x) f (x) dx2πis 2πis x 2πis(integration by parts with u f 0 (x), dv e 2πisx dx)Z 1 e 2πisx f 00(x) dx (using f 0 (x) 0 as x )2(2πis) 1 (F f 00)(s)(2πis)2Thus F f (s) 1kf 00k1 2πs 2and we see that F f (s) tends to 0 like 1/s2.The headline: Greater smoothness of f (x), plus integrability, leads to faster decay of the Fourier transform at .Remark on the derivative formula for the Fourier transform The astute reader will have noticedthat in the course of our work we rederived the derivative formulaF f 0(s) 2πisF f (s)which we’ve used before, but here we needed the assumption that f (x) 0, which we didn’t mentionbefore. What’s up? With the technology we have available to us now, the derivation we gave, above, isthe correct derivation. That is, it proceeds via integration by parts, and requires some assumption likef (x) 0 as x . In homework (and in the solutions to the homework) you may have given aderivation that used duality. That only works if Fourier inversion is known to hold. This was OK when therigor police were off duty, but not now, on this day of reckoning. Later, when we develop a generalizationof the Fourier transform, we’ll see that the derivative formula again holds without what seem now to beextraneous conditions.We could go on as we did above, comparing the consequences of higher differentiability, integrability,smoothness and decay, bouncing back and forth between the function and its Fourier transform. The greatinsight in making use of these observations is that the simplest and most useful way to coordinate all thesephenomena is to allow for arbitrarily great smoothness and arbitrarily fast decay. We would like to haveboth phenomena in play. Here is the crucial definition.Rapidly decreasing functionsA function f (x) is said to be rapidly decreasing at if1. It is infinitely differentiable.
4.2 The Right Functions for Fourier Transforms: Rapidly Decreasing Functions1452. For all positive integers m and n,xmdnf (x) 0dxnasx In words, any positive power of x times any order derivative of f tends to zero at infinity.Note that m and n are independent in this definition. That is, we insist that, say, the 5th power of x timesthe 17th derivative of f (x) tends to zero, and that the 100th power of x times the first derivative of f (x)tends to zero; and whatever you want.Are there any such functions? Any infinitely differentiable function that is identically zero outside somefinite interval is one example, and I’ll even write down a formula for one of these later. Another example is2f (x) e x . You may already be familiar with the phrase “the exponential grows faster than any power2of x”, and likewise with the phrase “e x decays faster than any power of x.”6 In fact, any derivative of2e x decays faster than any power of x as x , as you can check with L’Hopital’s rule, for example.We can express this exactly as in the definition:xmdn x2 0edxnas x There are plenty of other rapidly decreasing functions. We also remark that if f (x) is rapidly decreasingthen it is in L1(R) and in L2(R); check that yourself.An alternative definition An equivalent definition for a function to be rapidly decreasing is to assumethat for any positive integers m and n there is a constant Cmn such thatxmdnf (x) Cmndxnasx .In words, the mth power of x times the nth derivative of f remains bounded for all m and n, though theconstant will depend on which m and n we take. This condition implies the “tends to zero” condition,above. Convince yourself of that, the key being that m and n are arbitrary and independent. We’ll usethis second, equivalent condition often, and it’s a matter of taste which one takes as a definition.Let us now praise famous men It was the French mathematician Laurent Schwartz who singled outthis relatively simple condition to use in the service of the Fourier transform. In his honor the set of rapidlydecreasing functions is usually denoted by S (a script S) and called the Schwartz class of functions.Let’s start to see why this was such a good idea.1. The Fourier transform of a rapidly decreasing function is rapidly decreasing. Let f (x) bea function in S. We want to show that F f (s) is also in S. The condition involves derivatives of F f , sowhat comes in is the derivative formula for the Fourier transform and the version of that formula for higherderivatives. As we’ve already seen 2πisF f (s) F62df (s) .dxI used e x as an example instead of e x (for which the statement is true as x ) because I wanted to include x ,2and I used e x instead of e x because I wanted the example to be smooth. e x has a corner at x 0.
146Chapter 4 Distributions and Their Fourier TransformsAs we also noted,dF f (s) F ( 2πixf (x)) .dsBecause f (x) is rapidly decreasing, the higher order versions of these formulas are valid; the derivationsrequire either integration by parts or differentiating under the integral sign, both of which are justified.That is, dn (2πis)nF f (s) F n f (s)dx dnF f (s) F ( 2πix)nf (x) .nds(We follow the convention that the zeroth order derivative leaves the function alone.)Combining these formulas one can show, inductively, that for all nonnegative integers m and n, n ddmmF( 2πix)f(x) (2πis)n m F f (s) .ndxdsNote how m and n enter in the two sides of the equation.We use this last identity together with the estimate for the Fourier transform in terms of the L1 -norm ofthe function. Namely, s n dmdn mF f (s) (2π)m n F(x f (x))mdsdxn (2π)m ndn m(x f (x))dxn1The L1 -norm on the right hand side is finite because f is rapidly decreasing. Since the right hand sidedepends on m and n, we have shown that there is a constant Cmn withdmF f (s) Cmn .dsmsnThis implies that F f is rapidly decreasing. Done.2. Fourier inversion works on S. We first establish the inversion theorem for a timelimited functionin S. Suppose that f (t) is smooth and for some T is identically zero for t T /2, rather than just tendingto zero at . In this case we can periodize f (t) to get a smooth, periodic function of period T . Expandthe periodic function as a converging Fourier series. Then for T /2 t T /2,f (t) X n X n Xn cn e2πint/Tee2πint/T2πint/T 1T 1TZ T /2e 2πinx/Tf (x) dx T /2Z e 2πinx/T f (x) dx Xn e2πint/T F f nT1.TOur intention is to let T get larger and larger. What we see is a Riemann sum for the integralZ e2πist F f (s) ds F 1 F f (t) ,
4.2 The Right Functions for Fourier Transforms: Rapidly Decreasing Functions147and the Riemann sum converges to the integral because of the smoothness of f . (I have not slippedanything past you here, but I don’t want to quote the precise results that make all this legitimate.) Thusf (t) F 1 F f (t) ,and the Fourier inversion theorem is established for timelimited functions in S.When f is not timelimited we use “windowing”. The idea is to cut f (t) off smoothly.7 The interestingthing in the present context — for theoretical rather than practical use — is to make the window so smooththat the “windowed” function is still in S. Some of the details are in Section 4.20, but here’s the setup.We take a function c(t) that is identically 1 for 1/2 t 1/2, that goes smoothly (infinitely differentiable)down to zero as t goes from 1/2 to 1 and from 1/2 to 1, and is then identically 0 for t 1 and t 1.This is a smoothed version of the rectangle function Π(t); instead of cutting off sharply at 1/2 we bringthe function smoothly down to zero. You can certainly imagine drawing such a function:In Section 4.20 I’ll give an explicit formula for this.Now scale c(t) to cn (t) c(t/n). That is, cn (t) is 1 for t between n/2 and n/2, goes smoothly downto 0 between n/2 and n and is then identically 0 for t n. Next, the function fn (t) cn (t) · f (t) isa timelimited function in S. Hence the earlier reasoning shows that the Fourier inversion theorem holdsfor fn and F fn . The window eventually moves past every t, that is, fn (t) f (t) as n . Someestimates based on the properties of the cut-off function — which I won’t go through — show that theFourier inversion theorem also holds in the limit.3. Parseval holds in S. We’ll actually derive a more general result than Parseval’s identity, namely:If f (x) and g(x) are complex valued functions in S thenZ Z f (x)g(x) dx F f (s)F g(s)ds . As a special case, if we take f g then f (x)f (x) f (x) 2 and the identity becomesZ Z 2 f (x) dx F f (s) 2 ds . 7 The design of windows, like the design of filters, is as much an art as a science.
148Chapter 4 Distributions and Their Fourier TransformsTo get the first result we’ll use the fact that we can recover g from its Fourier transform via the inversiontheorem. That is,Z F g(s)e2πisx ds .g(x) The complex conjugate of the integral is the integral of the complex conjugate, henceZ g(x) F g(s)e 2πisx ds . The derivation is straightforward, using one of our favorite tricks of interchanging the order of integration: Z Z Z 2πisxf (x)g(x) dx f (x)F g(s)eds dx Z Z f (x)F g(s)e 2πisx ds ds Z Z f (x)F g(s)e 2πisx dx dx Z Z f (x)e 2πisx dx F g(s) ds Z F f (s)F g(s)ds All of this works perfectly — the initial appeal to the Fourier inversion theorem, switching the order ofintegration — if f and g are rapidly decreasing.4.3A Very Little on IntegralsThis section on integrals, more of a mid-chapter appendix, is not a short course on integration. It’s hereto provide a little, but only a little, background explanation for some of the statements made earlier. Thestar of this section is you. Here you go.Integrals are first defined for positive functions In the general approach to integration (of realvalued functions) you first set out to define the integral for nonnegative functions. Why? Because howevergeneral a theory you’re constructing, an integral is going to be some kind of limit of sums and you’ll want toknow when that kind of limit exists. If you work with positive (or at least nonnegative) functions then theissues for limits will be about how big the function gets, or about how big the sets are where the functionis or isn’t big. You feel better able to analyze accumulations than to control conspiratorial cancellations.So you first define your integral for functions f (x) with f (x) 0. This works fine. However, you knowfull well that your definition won’t be too useful if you can’t extend it to functions which are both positiveand negative. Here’s how you do this. For any function f (x) you let f (x) be its positive part:f (x) max{f (x), 0}Likewise, you letf (x) max{ f (x), 0}be its negative part.8 (Tricky: the “negative part” as you’ve defined it is actually a positive function; taking f (x) flips over the places where f (x) is negative to be positive. You like that kind of thing.) Thenf f f 8A different use of the notation f than we had before, but we’ll never use this one again.
4.3 A Very Little on Integrals149while f f f .You now say that f is integrable if both f and f are integrable — a condition which makes sense sincef and f are both nonnegative functions — and by definition you setZf ZZf f .(For complex-valued functions you apply this to the real and imaginary parts.) You follow this approachfor integrating functions on a finite interval or on the whole real line. Moreover, according to this definition f is integrable if f is because thenZ f Z(f f ) ZZf f and f and f are each integrable.9 It’s also true, conversely, that if f is integrable then so is f . Youshow this by observing thatf f and f f and this implies that both f and f are integrable. You now know where the implicationZ f (t) dt F f exists comes from. You get an easy inequality out of this development:Zf Z f .In words, “theRabsolute value of the integral is at most the integral of the absolute value”. RAnd sure that’strue, because f may involve cancellations of the positive and negative values of f while f won’t havesuch cancellations. You don’t shirk from a more formal argument:Zf ZZZ(f f ) f ZZf (f f ) ZZZf f Zf f (since f and f are both nonnegative) f . You now know where the second inequality inZZ 0 2πist 2πis0 t F f (s) F f (s ) ef (t) dt e 0e 2πist e 2πis t f (t) dtcomes from; this came up in showing that F f is continuous.RRRSome authorsreserve“summable” for the case when f , i.e., for when both f and f are finite. TheyRR theR term Rstill define f f f but they allow the possibility that one of the integrals on the right may be , i
Distributions and Their Fourier Transforms 4.1 The Day of Reckoning We’ve been playing a little fast and loose with the Fourier transform — applying Fourier inversion, appeal-ing to duality, and all that. “Fast and loose” is an understatement if ever there was one,
and Laplace transforms F(s) Z 0 f(t)e st dt. Laplace transforms are useful in solving initial value problems in differen-tial equations and can be used to relate the input to the output of a linear system. Both transforms provide an introduction to a more general theory of transforms, which are u
1 0 e iux (u) du Reference: Kendall’s Advanced Theory of Statistics, Volume I, chapter 4 Liuren Wu (Baruch) Fourier Transforms Option Pricing 8 / 22. Fourier transforms and inversions of European options Take a European call option as an example. We perform the following . Di usions, Econometrica, 68(6), 1343{1376.
Laplace vs. Fourier Transform Laplace transform: Fourier transform Laplace transforms often depend on the initial value of the function Fourier transforms are independent of the initial value. The transforms are only the same if the function is the same both sides of the y-axis (so the unit step function is different). 0 F(s) f (t)e stdt f ′(t) sF(s)
Malus Lagrange Legendre Laplace The committee examining his paper had expressed skepticism, in part due to not so rigorous proofs. Amusing aside . The inverse Fourier transform of the product of two Fourier transforms is the convolution of the two inverse Fourier transforms: What do we use convolution for?
Alternatively, the 3D Fourier slice theorem makes it pos-sible to compute the 3D Fourier transform of the unknown radioactive distribution from the set of 2D Fourier transforms of the projection data[20 25]. These direct Fourier meth-ods (DFM) have the potential to substantially speed up the reconstruction process (when a simple 3D .
Gambar 5. Koefisien Deret Fourier untuk isyarat kotak diskret dengan (2N1 1) 5, dan (a) N 10, (b) N 20, dan (c) N 40. 1.2 Transformasi Fourier 1.2.1 Transformasi Fourier untuk isyarat kontinyu Sebagaimana pada uraian tentang Deret Fourier, fungsi periodis yang memenuhi persamaan (1) dapat dinyatakan dengan superposisi fungsi sinus dan kosinus.File Size: 568KB
Fourier transform of functions that diff using definition of Fourier transformations. Keywords: fourier transforms, power series, taylor's and maclaurin series and gamma function. GJSFR-F Classification: FOR Code: infinitely terms. Hence, the method is useful to find the icult to obtain their
2020 Sutherland, Alister Peasant seals and sealing practices in eastern England, c. 1200-1500 Ph.D. . 2015 Harris, Maureen ‘A schismatical people’: conflict between ministers and their parishioners in Warwickshire between 1660 and 1714. Ph.D. 2015 Harvey, Ben Pauper narratives in the Welsh borders, 1790 - 1840. Ph.D. 2015 Heaton, Michael English interwar farming: a study of the financial .
