Thompson - Computing for Scientists and Engineers (523188), страница 70
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Dothese properties also hold for the Fourier integral transform, in which k varies continuously?Exercise 10.6(a) Use (10.16) to show that when k is an even integer, including zero,Y (k) = 0, just as for the series.(b) Show that the modulus of the Fourier transform, |Y ( k )| l/ k when k isan odd integer, just as for the Fourier series. n384FOURIER INTEGRAL TRANSFORMSFIGURE 10.2 Fourier integral transform of the square pulse, as given by (10.16). but plottedwith the phase factor omitted. The solid points are the discrete Fourier transform values at integerfrequencies.You will have noticed something peculiar about the k-space Fourier amplitude,Y (k), for k near zero.
As long as k is infinitesimally removed from zero (10.16)applies and Y (k) varies smoothly with k. But right at k = 0 the amplitude must bezero. It is, however, usual to ignore this fact and to extend the definition of Y (k)given by (10.16) into the origin. A Maclaurin expansion of the sine function followed by division by the denominator then produces(10.17)On this basis, we may now graph the Fourier integral transform, apart from thecomplex factor in (10.16), as shown in Figure 10.2.
The values that correspond inmagnitude to the Fourier series coefficients, bk, for the square pulse are also indicated for the odd integers.If you have some experience with diffraction of waves, you will notice thatY (k) varies with k just like the amplitude of a wave diffracted through an aperture.Indeed, detailed study (presented, for example, in optics texts) will show you thatthis is exactly so.Now that we have seen the similarities and differences of Fourier series and integral transforms for the square pulse, it is interesting to consider another example toreveal other facets of integral transforms.Fourier transform of the wedge functionThe wedge function, considered for the discrete transform in Section 9.5, illustrateshow the behavior of the integral transform varies as the input function is scaled in10.2EXAMPLES OF FOURIER TRANSFORMS385shape.
We therefore have a wedge whose shape is adjustable, namely(10.18)Although the wedge always has unit area, its width (and therefore its height) is controlled by the value of L. This property makes it convenient for later investigation.Exercise 10.7Derive the Fourier integral transform of the wedge function by inserting the definition of the wedge function, (10.18), in the formula (10.4) for the transform,then carrying out the indicated integration.
This is straightforward but tedious.Show, after several carefully executed steps, that(10.19)Unlike the square-pulse example above, no special treatment is necessary fork = 0, except that the limit as k approaches zero must be taken carefully. nYou may surmise, correctly, that the behavior of the Fourier integral transform neark = 0 is related to the Wilbraham-Gibbs phenomenon discussed in Section 9.6, inthat functions that have discontinuities in their values need special treatment.Note that there is dimensional consistency between the arguments in the originalfunction and the arguments in the transform, as follows.
Suppose, for example, thatx is a distance, then so must be L. Then in the transform (10.19) k has dimensionsof reciprocal distance, but the product kL, being the argument of the sine function,must be dimensionless, as required for any mathematical function.Exercise 10.8Suppose that x = t, where t represents time, and thatShow that the Fourier transform (10.19) becomesalso a time.(10.20)where represents angular frequency.
Is the productquired? ndimension-free, as re-The Fourier integral transform of the wedge function is shown in Figure 10.3for two values of L, namely L = 1 (in the same units as x) and L = 2.386FOURIER INTEGRAL TRANSFORMS0FIGURE 10.3 Fourier integral transform of the wedge function (10.18), as given by (10.19).Shown for a wedge of width L = 1 (solid curve) and for a wider wedge (L = 2, dashed curve) whichproduces a narrower transform.Notice immediately in Figure 10.3 that as the wedge function becomes broaderin the x space (although always keeping unit area) it becomes proportionally narrower in the k space. Indeed, the Fourier integral transform as a function of L is obtained merely by scaling the k values so that the product kL is constant, because (10.19)shows that Y(k) depends only upon this product.
For example, in Figure 10.3 thezeros of the transform occur with twice the frequency for L = 2 as for L = 1.Exercise 10.9Verify the following properties of the Fourier integral transform of the wedgefunction:(a) Y (k) is symmetric about k = 0.(b) Y (k) falls to half its maximum value at k = 2.78/L.(c) The zeros of Y(k) occur atwith n a nonzero integer.(d) A secondary maximum of Y (k) occurs nearwhere its height isabout 0.045 that of the first maximum. nThe results from this exercise are important in the theory of wave diffraction fromapertures. See, for example, the books on vibrations and waves by Ingard and byPippard.Gaussian functions and Fourier transformsWe investigated the Gaussian function in the context of maximum likelihood andleast squares in Section 6.1. We repeat its definition here as10.2EXAMPLES OF FOURIER TRANSFORMS387(10.21)We have renamed the function, commemorating Gauss, and we have included thestandard deviation,as a parameter in the function definition to emphasize its importance.
Also, this Gaussian is centered on x = 0, rather than on an average valueas in Section 6.1. Such a shift of origin is not important for our current use and interpretation of the Gaussian. You probably already derived some properties of theGaussian function by working Exercise 6.2. The Gaussian is of interest in Fouriertransforms because of its simple behavior under the integral transform. Also, it is asmoothly varying function that is often suitable for use as a windowing or dampingfunction, of the kind discussed in Section 9.8.The process of deriving the Fourier integral transform of a Gaussian is quitetricky.
It is therefore interesting to begin with some numerical explorations by Fourier transforming some discretized Gaussian data using the FFT program developedin Section 9.7.Exercise 10.10(a) Write a simple main program that generates an array of data from the Gaussian function (10.21), given the user’s choice ofx range, and the index vthat determines the number of points, N, in the FFT through N = 2 v. Includean option that makes a file to output the FFT results or plotting. The programshould then use the FFT for these data.(b) Run the program for a range of parameter values. Suitable values are= 1/2, 1, 2, and -2 < x < 2, and v = 7 so that N = 128.2(c) Plot the FFT output from each run. By plotting ln(FFT) versus k , showthat, through about k 1 such a plot is linear and its slope is aboutnYou will notice after the calculations in this exercise that the function obtained in kspace is not very smooth when k increases beyond unity.
The limiting process thatwe carried out analytically to obtain the Fourier integral from the Fourier series istherefore not so easy numerically.This exercise has probably suggested to you that the Gaussian and its integraltransform have the same functional dependence on their arguments, x or k, and thattheir width (standard-deviation) parameters are the inverse of each other. Let us demonstrate this analytically.Exercise 10.11(a) Insert the Gaussian function (10.21), which is normalized to unity when integrated over all x space, into formula (10.4) for its Fourier integral transform,which we will callChange variables toin order toshow that(10.22)388FOURIER INTEGRAL TRANSFORMSHere the variableis given by(10.23)(b) Use the value for the integral over y in (10.22) from Abramowitz and Stegun, equations (7.1.1) and (7.1.16), namelyto show that(10.24)where in k space(10.25)is the standard deviation of the Gaussian in k space.
nThis result confirms the numerical surmise made in Exercise 10.10.Now that we have the Gaussian in both x and k spaces, we can examine some oftheir properties. First, we can display both on the same graph, which is very economical for the author and very enlightening for the student. Figure 10.4 showsthree general-purpose Gaussians that may be used for both x and k spaces. Becauseof the reciprocal relations between the standard deviations, the plot foris thesame for both. Further, if in x space we choosethe broadest of thethree curves, thenin k space, and vice versa.The reciprocity between widths in x and k spaces has important interpretations inoptics and quantum physics, since a narrow pulse in x space (well-localized) is necessarily broad in k space (delocalized).
When appropriately interpreted in quantummechanics, this provides an example of the Heisenberg Uncertainty Relations.FIGURE 10.4 Gaussian distributions and their Fourier integral transforms. The distributionwith magnitudein x space transforms into that with magnitudethe transforms are identical.in k space. For10.2EXAMPLES OF FOURIER TRANSFORMS389A further property of the Gaussian function may be introduced, namely its relation to Dirac delta distributions discussed at the end of Section 10.1. If the Gaussian width is steadily decreased, the function steadily becomes more and moreconcentrated towards the origin. Thus you can imagine that the limiting behavior asof the Gaussian is a Dirac delta distribution, because if it occurs inside anintegral with another function only the value of that function near the origin will survive in the integration, so that eventuallywill be obtained.Lorentzian functions and their propertiesWe previously encountered the Lorentzian function in Section 8.1 when investigating forced motion and resonances.
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