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Section 12.4 Fourier Transform of the Delta Function

The easiest and one of the most important examples of a Fourier Transform is the delta function!

Activity 12.3. The Fourier Transform of a Delta Function.

Show that the Fourier Transform of the delta function \(f(x)=\delta(x-x_0)\) is a constant phase that depends on \(x_0\text{,}\) where the peak of the delta function is.

Answer.
\begin{equation} \tilde{\delta}_{x_0}(k) =\frac{1}{\sqrt{2\pi}}\, e^{-ikx_0}\tag{12.4.1} \end{equation}
Solution.
\begin{align} \tilde{\delta}_{x_0}(k) \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}e^{-ikx}\delta(x-x_0)\, dx\tag{12.4.2}\\ \amp =\frac{1}{\sqrt{2\pi}}\, e^{-ikx_0}\tag{12.4.3} \end{align}

Now, we can use the inverse Fourier transform to derive the important exponential representation of the delta function, (11.6.2).

Activity 12.4. The inverse Fourier Transform of a Delta Function.

Use the inverse Fourier Transform of the delta function to derive the exponential representation of the delta function.

\begin{equation} \delta(x-x_0) = \frac{1}{2\pi}\, \int_{-\infty}^{\infty} e^{ik(x-x_0)}\, dk\tag{12.4.4} \end{equation}
Solution.

We just showed that the Fourier transform of the delta function is a constant phase. This means that the inverse Fourier transform of the phase is a delta function. So, just write down this statement:

\begin{align} \delta(x-x_0) \amp ={\cal{F}}^{-1} \left(\frac{1}{\sqrt{2\pi}}\, e^{-ikx_0}\right)\tag{12.4.5}\\ \amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \left(\frac{1}{\sqrt{2\pi}}e^{-ikx_0}\right) \, e^{ikx}\, dk\tag{12.4.6}\\ \amp =\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ik(x-x_0)}\, dk\tag{12.4.7} \end{align}