Thompson - Computing for Scientists and Engineers (523188), страница 71
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Because the Lorentzian’s properties are almostexclusively used in connection with resonances, we express it in terms of the angular frequency,. This is the variable most often encountered in acoustics, optics,and engineering, whereas the energy, E, usually occurs as the variable of interest inatomic and subatomic physics.We define the Lorentzian function by(10.26)whereis called the resonance frequency and is called the Full Width at HalfMaximum (FWHM) of the resonance. You will understand the use of these termsby looking at Figure 10.5 and working Exercise 10.12.FIGURE 10.5 The Lorentzian function (10.26) as a function of frequency from the central freShown for(dotted curve),(solidcurve), and(dashed curve).quency, measured in units of the FWHM390FOURIER INTEGRAL TRANSFORMSExercise 10.12Show that the Lorentzian decreases to half its maximum value of unity at pointsabove and below the resonance frequencyas seen in Figure 10.5.
nBoth Gaussians and Lorentzians have convenient properties for describing data,so they are occasionally used as alternative line shapes. They also appear commingled in the Voigt function-- the convolution of a Lorentzian with a Gaussian —whose properties are discussed in Section 10.4. It is therefore interesting to compare and contrast the Gaussian and Lorentzian. In order to do this we need a common unit for the variables x and We choose the Lorentzian measured in units ofFWHM, relative toand a Gaussian with the same FWHM as the Lorentzian,which therefore sets the value ofYou may wish to derive the following comparative properties.Exercise 10.13(a) Use the definition of the Gaussian, (10.21), with the definition of itsFWHM to show that if the Gaussian FWHM is to be unity, then its standarddeviation0.4246.(b) Show that if a Gaussian and Lorentzian are to have the same FWHM, thenthe appropriate scale for the Gaussian relative to the Lorentzian scale chosenabove is= 1.665 x.(c) Show that when its argument is two half widths on either side of the maximum the Gaussian has decreased to 1/16 of its peak value, but the Lorentzianhas decreased to only 1/5 of its peak value.(d) Compare the second derivatives of Gaussian and Lorentzian at the peak totheir values at the peak to show that the Gaussian with unit FWHM is decreasingas -16 ln(2) = -11.09, somewhat faster than for the Lorentzian, for which theratio is decreasing as -8.
nFIGURE 10.6 Comparison of Lorentzian (dashed) with Gaussian (dotted) for the same FWHM.10.2EXAMPLES OF FOURIER TRANSFORMS391These analytic properties are illustrated in Figure 10.6, in which the Lorentzian andGaussian are plotted on a common scale in units of the FWHM. The most noticeable difference between them is the broader wings on the Lorentzian. When the twofunctions are convolved into the Voigt function (Section 10.4) a behavior intermediate between the Lorentzian and Gaussian is obtained when their arguments aremeasured in the units of this figure.Now that we have compared Gaussians and Lorentzians, is the Fourier integraltransform of the Lorentzian as easy to derive as for the Gaussian?Fourier integral transform of a LorentzianWe have now understand some properties of the Lorentzian function (10.26), so it istime to determine its Fourier transform.
Because we are using the variable forreasons described at the start of the preceding subsection), we denote the conjugatevariable in its Fourier transform by t, reminiscent of time.The Fourier integral transform of the Lorentzian, (10.26), we denote by P (t).From the definition (10.4) of the Fourier transform and by making some changes ofvariables, we can write compactly(10.27)where the integral(10.28)in terms of the variables(10.29)and the parameter(10.30)Although the integral in (10.28) can readily be evaluated by the method of contour integration in the complex plane, it is interesting and instructive to use anothermethod.Exercise 10.14(a) Verify the substitutions of variables made to express P (t).(b) By differentiating I in (10.28) under the integral sign twice with respect to s,then rearranging the integrand, show that I satisfies the differential equation392FOURIER INTEGRAL TRANSFORMS(10.31)(c) Assume that the integral in this expression can be set to its average value,namely zero, then verify that this differential equation is satisfied by(10.32)(d) Show that for solutions that don’t diverge asonly solutions withexponent -|s| are acceptable.(e) Determine the pre-exponentials in (10.32) by evaluating the integral (10.28)explicitly for s = 0 by making use of the substitution v = tan to show thatI(0) =nSo, after all this exercising we have the Fourier integral transform of the Lorentzianfunction (10.26) as(10.33)In order to emphasize the complementarity between the widths of functions andthe widths of their Fourier transforms, we have in Figure 10.7 the exponential partof P (t) for the three values of the Lorentzian FWHM used in Figure 10.6.
Notice that if the Lorentzian, L, is broad in the space, then its Fourier transform, P,is narrow in t space, and vice versa. This is similar to the behavior of the wedgeand Gaussian functions and their transforms discussed above.FIGURE 10.7 Magnitude of the Fourier integral transform of the Lorentzian, |P(t)|, expressedin units ofas a function of t. The exponential-decay curves are for = 1/2 (dotted), = 1(solid), and = 2 (dashed). Notice the inverse relation between the Lorentzians, Figure 10.5, andtheir transforms shown here.10.3CONVOLUTIONS AND FOURIER TRANSFORMS393A further interesting property of the Lorentzian and its transform is that althoughhas a very nonlinear dependence on the resonance frequencyand on theFWHMcan be simply transformed to produce a linear dependence onExercise 10.15(a) Show thatcan be determined from the oscillations in P (t ) .(b) Show that the graph of ln P(t) against t is a straight line with slope given byn10.3 CONVOLUTIONS AND FOURIER TRANSFORMSIn this section we introduce the convolution operation, which is of broad applicability in both mathematical analysis and numerical applications.
Convolutions becomeparticularly simple in the context of Fourier integral transforms, in that in the transform space, convolution is replaced by multiplication. We show how this resultgreatly simplifies the mathematics of computing transforms, and we culminate ouranalysis with the practical example, much used in optics and in astrophysics, of theVoigt profile, which is the convolution of a Gaussian with a Lorentzian. The numerics and a program for computing the Voigt profile are presented in Project 10 inSection 10.4.Convolutions: Definition and interpretationSuppose that we have two functions, y1 and y2, of the same variable x.
Let us construct the following integral of their product:(10.34)We use the special “star” notation to denote convolution. After the convolution theresult is a function of the variable x, which is kept fixed during the integration overx' under the integral. Some books show on the left-hand side an x argument afterboth y1 and y2. This is misleading, because it is the result of the convolution that isa function of x, rather than y1 and y2 separately.What is the interpretation of the convolution? Consider the integral of the product of two functions evaluated at the same argument. This is just the overlap between the two functions, including their relative signs.
The convolution, on theother hand, has the arguments of the two functions shifted by an amount dependingon the variable x. The particular choice of arguments is made clear by consideringy1 as a function peaked around x' = 0. Then the integral samples only values of y2near the particular choice of argument x.394FOURIER INTEGRAL TRANSFORMSIn spite of this viewpoint, the convolution is independent of the order in whichthe functions are written down.
That is,(10.35)Exercise 10.16In the definition of the convolution, (10.34), change the variable of integrationx' by setting x' = x - x". Thus show that the convolution definition is thensatisfied if the two functions are interchanged, that is, (10.35) holds. nThis commutation property of convolutions is useful in practical calculations.Convoluting a boxcar with a LorentzianAs an illustration of convolution in the context of resolution of a measurement, consider the “boxcar-averaging” function with unit area(10.36)Three examples of this boxcar are shown in Figure 10.8.
(For L = 0.1 this is, tocontinue our agricultural analogy, a grain silo rather than a boxcar.)FIGURE 10.8 The boxcar (or “window”) function for convolutions, shown for widths ofL = 0.1, 1, and 2.10.3CONVOLUTIONS AND FOURIER TRANSFORMS395The Lorentzian is given by, as discussed at the end of Section 10.2,(10.37)Lorentzians with various widths are shown in Figure 10.5. If the boxcar and Lorentzian functions are inserted in the convolution definition (10.34) and the change ofvariable to x - x' =is made, then the integration can be readily performed to produce the convolution of a boxcar with a Lorentzian(10.38)Ifthe inverse tangent is to be taken in the first quadrant, but otherwisemultiples of will need to be added to the values returned for the angle by your calculator, which is in radians in (10.38).Exercise 10.17(a) Perform the integration with the change of variables suggested in order toverify (10.38).(b) Show that ifthat is, if the width of the averaging function is muchless than the width of the Lorentzian, then the convolution just collapses to theoriginal Lorentzian, y2.
nFIGURE 10.9 Convolution of a boxcar (Figure 10.8) with a Lorentzian (Figure 10.5) for aLorentzian FWHMand 2.shown for the three boxcar widths in Figure 10.8. namely L = 0.1, 1,396FOURIER INTEGRAL TRANSFORMSThe property worked out in Exercise 10.17 (b) is shown in Figure 10.9 (whichhasin which the convolution for L = 0.1 is essentially the original Lorentzian, as one can see because its maximum value is essentially unity and its FWHM isnearly 2 units of x, that is,You can also see from this figure that the convolution becomes broader and squatter as L increases. The decrease of peak resolution as the boxcar width, L, increases can readily be seen by considering the peakheight as a function of L.Exercise 10.18(a) Show that the peak value of the convoluted distribution (10.38) is at x = 0,just as for the Lorentzian.(b) Use the formula relating tangents of doubled angles to the tangents of theangles to show that the peak value is at the origin, where(10.39)(c) From the first two terms of the Taylor expansion of the arctangent function,show that the peak value given by (10.39) can be approximated by( 10.40)in which the leading term is just the peak value of the original Lorentzian(10.37).
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