Calculate the Fourier Transform of a Function

The Fourier transform is an integral transform widely used in physics and engineering. They are widely used in signal analysis and are well-equipped to solve certain partial differential equations.

The convergence properties of the Fourier transform are quite severe due to the lack of the exponential decay term as seen in the Laplace transform, and it means that functions like polynomials, exponentials, and trigonometric functions all do not have Fourier transforms in the usual sense. However, we can make use of the Dirac delta function to assign these functions Fourier transforms in a way that makes sense.

Because even the simplest functions that are encountered may need this type of treatment, it is recommended that you be familiar with the properties of the Laplace transform before moving on.

Preliminaries

• We define the Fourier transform of <math>f(t)</math> as the following function, provided the integral converges.
  • <math>\hat{f}(\omega) = \mathcal{F}\{f(t)\} = \int_{-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t</math>
• The inverse Fourier transform is defined in a similar manner. Notice the symmetry present between the Fourier transform and its inverse, a symmetry that is not present in the Laplace transform.
  • <math>f(t) = \mathcal{F}^{-1}\{\hat{f}(\omega)\} = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}\mathrm{d}\omega</math>
• There are many other definitions of the Fourier transform. The above definition making use of angular frequency is one of them, and we will use this convention in this article. See the tips for two other commonly used definitions.
• The Fourier transform and its inverse are linear operators.
  • <math>\int_{-\infty}^{\infty}[af(t) + bg(t)]e^{-i\omega t}\mathrm{d}t = a\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t + b\int_{-\infty}^{\infty}g(t)e^{-i\omega t}\mathrm{d}t</math>

Steps

The Basics

1. Substitute the function into the definition of the Fourier transform. As with the Laplace transform, calculating the Fourier transform of a function can be done directly by using the definition. We will use the example function <math>f(t) = \frac{1}{t^{2}+1},</math> which definitely satisfies our convergence criteria.
   • <math>\mathcal{F}\left\{\frac{1}{t^{2}+1}\right\} = \int_{-\infty}^{\infty}\frac{e^{-i\omega t}}{t^{2}+1}\mathrm{d}t</math>
2. Evaluate the integral using any means possible. This integral resists the techniques of elementary calculus, but we can make use of Integrate Using Residue Theory instead. The absolute value stems from the fact that we close the contour in the lower half plane when <math>\omega > 0</math> and in the upper half plane when <math>\omega < 0.</math>
   • <math>\mathcal{F}\left\{\frac{1}{t^{2}+1}\right\} = \pi e^{-|\omega|}</math>
3. Evaluate the Fourier transform of the rectangular function. The rectangular function <math>\operatorname{rect}(t),</math> or the unit pulse, is defined as a piecewise function that equals 1 if <math>-\frac{1}{2} < t < \frac{1}{2},</math> and 0 everywhere else. As such, we can evaluate the integral over just these bounds. The result is the cardinal sine function.
   • <math>\mathcal{F}\{\operatorname{rect}(t)\} = \int_{-1/2}^{1/2}e^{-i\omega t}\mathrm{d}t = \frac{2}{\omega}\sin\frac{\omega}{2}</math>
   • If the unit pulse is shifted such that the bounds are 0 and 1, then there exists an imaginary component as well, as seen by the graph above.
     • <math>\int_{0}^{1}e^{-i\omega t}\mathrm{d}t = \frac{\sin\omega}{\omega} + i\left(\frac{\cos\omega-1}{\omega}\right)</math>
4. Evaluate the Fourier transform of the Integrate Gaussian Functions The Gaussian function is one of the few functions that is its own Fourier transform. We integrate by completing the square.
   • <math>\begin{align}\mathcal{F}\{e^{-t^{2}}\} &= \int_{-\infty}^{\infty}e^{-t^{2}}e^{-i\omega t}\mathrm{d}t \\

&= \int_{-\infty}^{\infty}e^{-(t^{2}+i\omega t - \omega^{2}/4 + \omega^{2}/4)}\mathrm{d}t \\ &= e^{-\omega^{2}/4}\int_{-\infty}^{\infty}e^{-(t+i\omega/2)^{2}}\mathrm{d}t \\ &= \sqrt{\pi}e^{-\omega^{2}/4}\end{align}</math>

Properties of the Fourier Transform

1. Determine the Fourier transform of a derivative. A simple integration by parts, coupled with the observation that <math>f(t)</math> must vanish at both infinities, yields the answer below.
   • <math>\begin{align}\mathcal{F}\{f^{\prime}(t)\} &= \int_{-\infty}^{\infty}f^{\prime}(t)e^{-i\omega t}\mathrm{d}t,\ \ u = e^{-i\omega t},\ v = f^{\prime}(t)\mathrm{d}t \\

&= i\omega\hat{f}(\omega)\end{align}</math>

1. • In general, we can take <math>n</math> derivatives.
     • <math>\mathcal{F}\{f^{(n)}(t)\} = (i\omega)^{n}\hat{f}(\omega)</math>
   • This yields the interesting property, stated below, which may be familiar in quantum mechanics as the form that the momentum operator takes in position space (on the left) and momentum space (on the right).
     • <math>-i\frac{\mathrm{d}}{\mathrm{d}t} \to \omega</math>
2. Determine the Fourier transform of a a function multiplied by <math>t^{n}</math>. The symmetry of the Fourier transform gives the analogous property in frequency space. We will first work with <math>n = 1</math> and then generalize.
   • <math>\begin{align}\mathcal{F}\{tf(t)\} &= \int_{-\infty}^{\infty}tf(t)e^{-i\omega t}\mathrm{d}t \\

&= \int_{-\infty}^{\infty}i\frac{\partial}{\partial\omega}(e^{-i\omega t})f(t)\mathrm{d}t \\ &= i\frac{\mathrm{d}}{\mathrm{d}\omega}\hat{f}(\omega)\end{align}</math>

1. • In general, we can multiply by <math>t^{n}.</math>
     • <math>\mathcal{F}\{t^{n}f(t)\} = i^{n}\frac{\mathrm{d}^{n}}{\mathrm{d}\omega^{n}}\hat{f}(\omega)</math>
   • We immediately obtain the below result. This is a symmetry that is not fully realized with the Laplace transforms between the variables <math>t</math> and <math>s.</math>
     • <math>i\frac{\mathrm{d}}{\mathrm{d}\omega} \to t</math>
2. Determine the Fourier transform of a function multiplied by <math>e^{iat}</math>. Multiplication by <math>e^{iat}</math> in the time domain corresponds to a shift in the frequency domain.
   • <math>\mathcal{F}\{e^{iat}f(t)\} = \int_{-\infty}^{\infty}f(t)e^{-i(\omega - a)t}\mathrm{d}t = \hat{f}(\omega - a)</math>
3. Determine the Fourier transform of a shifted function <math>f(t-c)</math>. A shift in the time domain corresponds to multiplication by <math>e^{-i\omega c}</math> in the frequency domain, which again illustrates the symmetry between <math>t</math> and <math>\omega.</math> We can easily evaluate this using a simple substitution.
   • <math>\begin{align}\mathcal{F}\{f(t-c)\} &= \int_{-\infty}^{\infty}f(t-c)e^{-i\omega t}\mathrm{d}t \\

&= \int_{-\infty}^{\infty}f(t)e^{-i\omega(t+c)}\mathrm{d}t \\ &= e^{-i\omega c}\hat{f}(\omega)\end{align}</math>

1. Determine the Fourier transform of a stretched function <math>f(ct)</math>. The stretch property seen in the Laplace transform also has an analogue in the Fourier transform.
   • <math>\begin{align}\mathcal{F}\{f(ct)\} &= \int_{-\infty}^{\infty}f(ct)e^{-i\omega t}\mathrm{d}t,\ \ u = ct \\

&= \frac{1}{c}\int_{-\infty}^{\infty}f(u)e^{-i\omega u/c}\mathrm{d}u \\ &= \frac{1}{c}\hat{f}\left(\frac{\omega}{c}\right)\end{align}</math>

Distributions

1. Evaluate the Fourier transform of <math>e^{iat}</math>. If you have had some exposure to Laplace transforms before, you know that the exponential function is the "simplest" function that has a Laplace transform. In the case of the Fourier transform, this function is not well-behaved because the modulus of this function does not tend to 0 as <math>t\to\infty.</math> Nevertheless, its Fourier transform is given as the delta function.
   • <math>\mathcal{F}\{e^{iat}\} = 2\pi\delta(\omega - a)</math>
   • The imaginary exponential oscillates around the unit circle, except when <math>t = 0,</math> where the exponential equals 1. You can think of the contributions by the oscillations as canceling themselves out for all <math>t \neq 0.</math> At <math>t = 0,</math> the integral of the function then diverges. The delta function is then used to model this behavior.
   • This result gives us the Fourier transform of three other functions for "free." The Fourier transform of the constant function is obtained when we set <math>a = 0.</math>
     • <math>\mathcal{F}\{1\} = 2\pi\delta(\omega)</math>
   • Using Euler's formula, we get the Fourier transforms of the cosine and sine functions.
     • <math>\mathcal{F}\{\cos at\} = \pi(\delta(\omega - a) + \delta(\omega + a))</math>
     • <math>\mathcal{F}\{\sin at\} = -i\pi(\delta(\omega - a) - \delta(\omega + a))</math>
2. Evaluate the Fourier transform of <math>t^{n}e^{iat}</math>. We can use the shift property to compute Fourier transforms of powers.
   • <math>\mathcal{F}\{t^{n}e^{iat}\} = 2\pi i^{n}\frac{\mathrm{d}^{n}}{\mathrm{d}\omega^{n}}\delta(\omega - a)</math>

Video

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Tips

• There are two other commonly used conventions for the Fourier transform.
  • Some authors define the Fourier transform to split the factor of <math>2\pi</math> evenly between the integrals.
  • The result is a greater symmetry between the transforms.
    • <math>\hat{f}(\omega) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-i\omega t}\mathrm{d}t</math>
    • <math>f(t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\hat{f}(\omega)e^{i\omega t}\mathrm{d}\omega</math>
  • Others use the normal frequency variable <math>\xi,</math> which is related to angular frequency by <math>\omega = 2\pi\xi.</math>
    • <math>\hat{f}(\xi) = \int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t}\mathrm{d}t</math>
    • <math>f(t) = \int_{-\infty}^{\infty}\hat{f}(\xi)e^{i2\pi\xi t}\mathrm{d}\xi</math>

Related Articles

• Calculate the Laplace Transform of a Function
• Perform an Inverse Discrete Fourier Transform by Hand Using the Shortcut Method

See also

• Use Calculus to Rotate Curves Around an Axis
• Integrate by Substitution
• Evaluate Complete Elliptic Integrals
• Integrate
• Take Derivatives
• Understand Calculus
• Differentiate Exponential Functions
• Calculate the Laplace Transform of a Function
• Pass Calculus
• Do Implicit Differentiation

Read more ...