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Section 18.5 Properties of the Fourier Transform

In this section, you will find activities that encourage you to prove two important properties of the Fourier transform. These properties are relatively easy to prove. Each activity has a hint if you get stuck and the full solution if you are short on time.

Activity 18.5. The Fourier Transform of a Shifted Function.

Show that the Fourier transform of a function that is shifted by an amount \(x_0\) is related to the Fourier transform of the original function by a constant phase related to the amount of the shift, i.e.

\begin{equation} {\cal{F}}(f(x-x_0)) =e^{-ikx_0}\, {\cal{F}}(f(x)).\tag{18.5.1} \end{equation}
Hint.

Use the substitution \(y=x-x_0\) in the definition of the Fourier transform.

Solution.

Plug \(f(x-x_0)\) into the definition of the inverse Fourier transform and use the substitution \(y=x-x_0\text{,}\) \(dy=dx\text{.}\) Notice that the (infinite) limits of integration don't change for this substitution.

\begin{align} {\cal{F}}(f(x-x_0) \amp =\tilde{f}_{x_0}(k)\tag{18.5.2}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-ikx}\,f(x-x_0)\, dx\tag{18.5.3}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-ik(y+x_0)}\, f(y)\, dy\tag{18.5.4}\\ \amp = e^{-ikx_0}\left(\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-iky}\, f(y)\, dy\right)\tag{18.5.5}\\ \amp = e^{-ikx_0}\tilde{f}(k)\tag{18.5.6}\\ \amp = e^{-ikx_0}{\cal{F}}(f)\tag{18.5.7} \end{align}

Activity 18.6. The Fourier Transform of a Derivative.

Show that the Fourier Transform of the derivative of a function is simply related to the Fourier transform of the original function, as given by the following formula

\begin{equation} {\cal{F}}\left(\frac{d}{dx}\, f(x)\right) =\tilde{f}(k)\, ik.\tag{18.5.8} \end{equation}
Hint.

Take the derivative of both sides of the definition of the inverse Fourier transform with respect to \(x\) and simplify. Interpret your simplified expression as the inverse Fourier transform of something.

Solution.

The definition of the inverse Fourier transform is given in (18.3.2). Take the derivative of both sides of this equation with respect to \(x\) and simplify. Then interpret the expression as an inverse Fourier transform, again using (18.3.2)

\begin{align} f(x)\amp = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \tilde{f}(k)\, e^{ikx}\, dk\tag{18.5.9}\\ \frac{d}{dx}\, f(x)\amp =\frac{d}{dx}\,\left[ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \tilde{f}(k)\, e^{ikx}\, dk \right]\tag{18.5.10}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \tilde{f}(k)\, \left(\frac{d}{dx}e^{ikx}\right)\, dk\tag{18.5.11}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \tilde{f}(k)\, \left(ik\, e^{ikx}\right)\, dk\tag{18.5.12}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \left(\tilde{f}(k)\, ik\right)\, e^{ikx}\, dk\tag{18.5.13}\\ \amp ={\cal{F}}^{-1}\left(\tilde{f}(k)\, ik\right)\tag{18.5.14}\\ \Rightarrow {\cal{F}}\left(\frac{d}{dx}\, f(x)\right) \amp =\tilde{f}(k)\, ik\tag{18.5.15} \end{align}