The Fourier Transform of a Gaussian Function

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 transform 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 function. 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