Mathematical Methods for Physics and Engineering - Matematica.NET

INTEGRAL TRANSFORMSgiven by∫1 ∞{∫ ∞}˜h(k) = √ dz e −ikz f(x)g(z − x) dx2π −∞−∞= √ 1 ∫ ∞{∫ ∞}dx f(x) g(z − x) e −ikz dz .2π −∞−∞If we let u = z − x in the second integral we have∫1 ∞{∫ ∞}˜h(k) = √ dx f(x) g(u) e −ik(u+x) du2π −∞−∞= √ 1 ∫ ∞∫ ∞f(x) e −ikx dx g(u) e −iku du2π−∞−∞= 1 √2π× √ 2π ˜f(k) × √ 2π˜g(k) = √ 2π ˜f(k)˜g(k). (13.38)Hence the Fourier transform of a convolution f ∗ g is equal to the product of theseparate Fourier transforms multiplied by √ 2π; this result is called the convolutiontheorem.It may be proved similarly that the converse is also true, namely that theFourier transform of the product f(x)g(x) is given byF[f(x)g(x)] = √ 1 ˜f(k) ∗ ˜g(k). (13.39)2π◮Find the Fourier transform of the function in figure 13.3 representing two wide slits byconsidering the Fourier transforms of (i) two δ-functions, at x = ±a, (ii) a rectangularfunction of height 1 and width 2b centred on x =0.(i) The Fourier transform of the two δ-functions is given by∫1 ∞˜f(q) = √ δ(x − a) e −iqx dx + 1 ∫ ∞√ δ(x + a) e −iqx dx2π −∞2π −∞= √ 1 (e −iqa + e iqa) = 2cosqa √ .2π 2π(ii) The Fourier transform of the broad slit is˜g(q) = √ 1 ∫ be −iqx dx = 1 [ ] e−iqx b√2π −b2π −iq−b= −1iq √ 2π (e−iqb − e iqb )= 2sinqbq √ 2π .We have already seen that the convolution of these functions is the required functionrepresenting two wide slits (see figure 13.6). So, using the convolution theorem, the Fouriertransform of the convolution is √ 2π times the product of the individual transforms, i.e.4cosqa sin qb/(q √ 2π). This is, of course, the same result as that obtained in the examplein subsection 13.1.2. ◭448