Thompson - Computing for Scientists and Engineers (523188), страница 69
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Now one lets L and N tend to infinity carefully (formathematical rigor, see Churchill’s text), so that the summation in (10.1) becomesan integral and the variablebecomes continuous. Also, we can revert to the notation of k for the variable complementary to x.
Now, however, k varies continuously rather than discretely. Finally, we have the result for the Fourier integraltransform pair(10.3)and the inverse transformation(10.4)Both integrals in this transform pair are on an equal footing, and there is nomathematical distinction between the variables x and k. For this reason, one will10.1FROM FOURIER SERIES TO FOURIER INTEGRALS379often see the expressions with the signs of the exponents reversed. As long as onemaintains complete consistency of sign throughout the use of a Fourier integraltransform pair, such a sign change is not significant.Exercise 10.1Verify all the steps between (10.1) and (10.4) for the derivation of the Fourierintegral transform. nFor the scientist and engineer, x and k describe physical measurements, so thattheir dimensions must be reciprocal in order that the argument of the complex exponential be dimensionless.
If x is position, then k is the wavenumber, given in termsof the wavelength,and having dimensions of reciprocal length.When x represents time, t, then k is replaced by the complementary variable the angular frequency,where T is the period and f is the frequency.The dimensions of are radian per unit time.Waves and Fourier transformsSince, as just described, we usually have the transform combination kx or U.it isinteresting to consider briefly the extension of the Fourier integral transform to twovariables, namely(10.5)Here the complex exponential represents a monochromatic (single wavelength andfrequency) plane wave travelling along the x direction with wavenumber k at angularfrequencyas we discussed in Section 2.4.
Its associated amplitude is Y (k ,Thus, the double integral over wavenumbers and frequencies, y (x,t), represents aphysical wave, called a wave packet. Suppose that the speed of the wavelet of givenfrequency, given byis fixed, as it is for a light wave in vacuum,then the double integral must collapse to a single integral, because when one of ork is given, the other is determined. This is called a nondispersive wave.
How tohandle the mathematics of this is considered below in discussing the Dirac delta distribution.Dirac delta distributionsA rather strange integral property can be deduced from the Fourier integral transformpair (10.3) and (10.4). Indeed, when first used by the physicist Dirac it disquietedmathematicians so much that it took a few years to provide a sound basis for its use.Suppose that we substitute (10.3) into (10.4), taking care to use a different variableunder the integral sign in (10.3), say k’, since it is a dummy variable of integration.One then finds, by rearranging the integrals, that380FOURIER INTEGRAL TRANSFORMS(10.6)where the Dirac delta distribution is(10.7)Exercise 10.2Work through the steps leading from (10.3) and (10.4) to the last two equations.nTo the extent that in (10.6) Y is any reasonably well-behaved function that allowsconstruction of its Fourier integral transform and rearrangement of the integrals thatlead to (10.7), we have a remarkable identity.
The Dirac delta distribution must besuch that it can reproduce a function right at the point k, even though its values atmany other points k’ appear under the integral. Clearly, must be distributed withrespect to k - k' such that it is concentrated near k' = k.Recollection of a property of the discrete Fourier transform is illustrative at thispoint, and it makes an analogy between the Kronecker delta and the Dirac delta. Referring to (9.9), we see that summation over the discrete data corresponds to integration over the continuous variable in (10.7).
The summation leads to selection bythe Kronecker delta of discrete matching k and l values, corresponding to matchingof k and k' values that is implied for the continuous variables in the Dirac delta. Ourexamples of Fourier transforms in Section 10.2, especially the integral transform ofthe Gaussian function, provide examples of Dirac delta distributions.10.2 EXAMPLES OF FOURIER TRANSFORMSWe now explore several examples of Fourier integral transforms, initially to understand their relations to other Fourier expansions and to be able to use them in applications such as convolutions and calculation of spectral profiles.Exponential decay and harmonic oscillationWe discussed in Section 9.2 the discrete Fourier transforms for exponential decayand harmonic oscillation, and we obtained analytic expressions for both of them.Suppose that the function to be transformed is(10.8)The transition to the Fourier integral transform can be made from the expression forthe general complex exponential discrete transform (9.18) by lettingas10.2and assuming that Redenominator to lowest order inEXAMPLES OF FOURIER TRANSFORMS381to guarantee convergence.
By expanding thethen lettingwe readily find that(10.9)where k is now a continuous variable. The transition from the sum in the discretetransform to the integral in the integral transform is a good example of the limit processes involved in the Riemann definition of an integral.Exercise 10.3Show all the steps between (9.18) and (10.9), paying particular attention to theorder in which the limits on N and h are taken. nThis integral transform of the complex exponential can now be applied to the twolimiting behaviors for Ifis purely real and positive, then we have exponentialdecay, while for purely imaginary we have harmonic oscillation.
These are justthe analytical examples considered in Section 9.2 for the discrete transform.Exponential decay, integral transforms can be scaled, as we discussed in Section 9.2, so thatand thus k is measured in units ofTo understand thetransition from discrete to integral transform it is interesting to consider the case of Nlarge and h small but finite. You can easily calculate from (9.18) and (10.9) the ratioof the midpoint discrete transform to the integral transform, where the midpoint frequency is calculated from (9.15) asAswe find that(10.10)For zero frequency (k = 0) and in the limit of large N, one can show simply from(9.18) and (10.9) withthat the DFT overpredicts the transform comparedwith the FIT by an amount that increases as h increases.Exercise 10.4(a) Make graphs of the real and imaginary parts of the integral transform (10.9)and compare them with calculations of the discrete transform for large N, sayN = 128 for h = 1.
A small computer program, such as you may have writtenfor Exercise 9.6 (b), would be helpful for this.(b) For the midpoint ratio of discrete to integral transforms fill in the steps in thederivation of the result (10.10).(c) For the zero-frequency difference between discrete and integral transformsshow that for Nh >> 1 the difference is given by(10.11)where the last approximation is from a Maclaurin expansion in h. n382FOURIER INTEGRAL TRANSFORMSFIGURE 10.1 Ratios of the discrete Fourier transform (9.20) at the midpoint frequency k = N/2to the Fourier integral transform at the corresponding frequency p/h, shown for discrete transformsof lengths N = 4, 8, and 16.
The dashed line is the asymptotic ratio given by (10.10).Numerical comparisons between discrete and integral transforms of the dampedexponential at the midpoint value k = N /2 are shown in Figure 10.1 for small values of N. The dashed line in Figure 10.1 is the result (10.10). Notice that for Nand h both small the discrete transform actually gives a negative real transform atk = N /2 that is quite different from the real part of the integral transform, which iseverywhere positive. This emphasizes the need for care in the limit process of deriving the integral transform from the discrete transform, as described above (10.3) andworked out by the conscientious reader in Exercise 10.3.Harmonic oscillution integral transforms are included in (10.9), as in Section 9.2 for the discrete transform, by settinga pure oscillator with asingle frequency k0.
In order to consider this rigorously, a small positive real part,has to be included in to produce convergence of the geometric series for Nlarge, as discussed above (9.21). Then one may letto obtain the followingresult for the Fourier integral transform of the harmonic oscillator:(10.12)This shows a simple pole at the resonance frequency k0, whereas the discrete transform has a distribution of values that are peaked in magnitude near k0, as shown inFigure 9.4.In the next examples we do not have nice functions that readily allow the discreteand integral Fourier transforms to be compared directly and analytically. For mostof these examples the discrete transform can be computed numerically, for exampleby using the FFT algorithm described in Sections 9.3 and 9.7, whereas it is the integral transform that can usually be expressed in analytical form.10.2EXAMPLES OF FOURIER TRANSFORMS383The square-pulse functionFor the first example of a Fourier integral transform we choose the first examplefrom the Fourier series in Section 9.5, the square pulse.
It has the same definitionas in (9.50), namely(10.13)In order to compute the integral transform in k space you need only plug this expression into (10.4). Because the integration is over all x, we have to supplement thedefinition (10.13) with the constraint that(10.14)This function can now be used in the defining equation for the Fourier integral transform, (10.4).Exercise 10.5Perform the indicated integration in (10.4) to show that the Fourier integraltransform of the square pulse is given by(10.15)and(10.16)fornThe correspondence of these results with those for the Fourier series coefficientsof the square pulse can be clarified as follows. For the series we used an expansionin terms of cosines and sines, and we found in Section 9.5 that only the sine termsare nonzero. Equation (9.52) and Figure 9.8 show that the envelope of the seriescoefficients, b k l/k, and that the coefficients vanish for k an even integer.
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