Section 18.4 Fourier Transform of the Delta Function
The easiest and one of the most important examples of a Fourier Transform is the delta function!
Activity 18.3. 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.
Solution.
\begin{equation}
\tilde{\delta}_{x_0}(k)
=\frac{1}{\sqrt{2\pi}}\, e^{-ikx_0}\tag{18.4.1}
\end{equation}
\begin{align}
\tilde{\delta}_{x_0}(k)
\amp =\frac{1}{\sqrt{2\pi}}
\int_{-\infty}^{\infty}e^{-ikx}\delta(x-x_0)\, dx\tag{18.4.2}\\
\amp =\frac{1}{\sqrt{2\pi}}\, e^{-ikx_0}\tag{18.4.3}
\end{align}
Now, we can use the inverse Fourier transform to derive the important exponential representation of the delta function, (17.6.2).
Activity 18.4. 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{18.4.4}
\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{18.4.5}\\
\amp =\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty}
\left(\frac{1}{\sqrt{2\pi}}e^{-ikx_0}\right) \, e^{ikx}\, dk\tag{18.4.6}\\
\amp =\frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ik(x-x_0)}\, dk\tag{18.4.7}
\end{align}