The Fourier transform of a gaussian function Kalle Rutanen 25.3.2007, 25.5.2007 1 1 Abstract In this paper I derive the Fourier transform of a family of functions of the form 2 f(x) = ae−bx . I thank ”Michael”, Randy Poe and ”porky_pig_jr” from the newsgroup sci.math for giving me the techniques to achieve this. The intent of this particular Fourier transform function is to give information about the frequency space behaviour of a Gaussian filter. 2 Integral of a gaussian function 2.1 Derivation Let 2 f(x) = ae−bx with a > 0, b > 0 Note that f(x) is positive everywhere. What is the integral I of f(x) over R for particular a and b? Z ∞ I = f(x)dx −∞ To solve this 1-dimensional integral, we will start by computing its square. By the separability property of the exponential function, it follows that we’ll get a 2-dimensional integral over a 2-dimensional gaussian. If we can compute that, the integral is given by the positive square root of this integral. Z ∞ Z ∞ I2 = f(x)dx f(y)dy −∞ −∞ Z ∞ Z ∞ = f(x)f(y)dydx −∞ −∞ ∞ ∞ Z Z 2 2 = ae−bx ae−by dydx −∞ −∞ ∞ ∞ Z Z 2 2 = a2e−b(x +y )dydx −∞ −∞ ∞ ∞ Z Z 2 2 = a2 e−b(x +y )dydx −∞ −∞ Now we will make a change of variables from (x, y) to polar coordinates (α, r). The determinant of the Jacobian of this transform is r. Therefore: 2 2π ∞ Z Z 2 I2 = a2 re−br drdα 0 0 2π ∞ Z 1 Z 2 = a2 −2bre−br drdα 0 −2b 0 2 2π ∞ a Z 2 = e−br drdα −2b 0 0 a2 Z 2π = −1dα −2b 0 −2πa2 = −2b πa2 = b Taking the positive square root gives: rπ I = a b 2.2 Example Requiring f(x) to integrate to 1 over R gives the equation: rπ I = a = 1 b ⇔ r b a = π And substitution of: 1 b = 2σ2 Gives the Gaussian distribution g(x) with zero mean and σ variance: 2 1 − x g(x) = √ e 2σ2 σ 2π 3 3 The Fourier transform We will continue to evaluate the bilateral Laplace transform B(s) of f(x) by using the intermediate result derived in the previous section. The Fourier trans- form is then given by F (w) = B(iw). 1 Z ∞ B(s) = √ f(x)e−sxdx 2π −∞ ∞ 1 Z 2 = √ ae−bx e−sxdx 2π −∞ ∞ 1 Z 2 = √ ae−(bx +sx)dx 2π −∞ Next we will complete the square in the exponent. b(x + k)2 = bx2 + 2bkx + bk2 s By comparing factors of x we see that 2bk = s and thus k = 2b . Now: s s2 b(x + )2 − = bx2 + sx 2b 4b Z ∞ 2 1 −(b(x+ s )2− s ) B(s) = √ ae 2b 4b dx 2π −∞ 2 Z ∞ 1 s −b(x+ s )2 = √ e 4b ae 2b dx 2π −∞ s Next we will make a change of variables by x(u) = u − 2b . The determinant of the Jacobian of this transformation is 1. Thus: 2 Z ∞ 1 s −bu2 B(s) = √ e 4b ae du 2π −∞ By using the result from the previous section, the integral is solved as: r 1 s2 π B(s) = √ e 4b a 2π b a s2 = √ e 4b 2b The associated Fourier transform is then: 4 F (w) = B(iw) 2 a − w = √ e 4b 2b Thus the Fourier transform of a Gaussian function is another Gaussian func- tion. Requiring f(x) to integrate to 1 over R gives: 1 s2 B1(s) = √ e 4b 2π F1(w) = B1(iw) 2 1 − w = √ e 4b 2π 5.