$$\require{cancel}$$

# 10.6: Common Fourier Transforms

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

To accumulate more intuition about Fourier transforms, let us examine the Fourier transforms of some interesting functions. We will just state the results; the calculations are left as exercises.

## Damped waves

We saw in Section 10.2 that an exponentially decay function with decay constant $$\eta \in \mathbb{R}^+$$ has the following Fourier transform: $f(x) = \left\{\begin{array}{cl}e^{-\eta x}, & x \ge 0 \\ 0, & x < 0,\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{-i}{k-i\eta}.$ Observe that $$F(k)$$ is given by a simple algebraic formula. If we “extend” the domain of $$k$$ to complex values, $$F(k)$$ corresponds to an analytic function with a simple pole in the upper half of the complex plane, at $$k = i\eta$$.

Next, consider a decaying wave with wave-number $$q \in \mathbb{R}$$ and decay constant $$\eta \in \mathbb{R}^+$$. The Fourier transform is a function with a simple pole at $$q + i \eta$$: $f(x) = \left\{\begin{array}{cl}e^{i (q + i\eta) x}, & x \ge 0 \\ 0, & x < 0.\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{-i}{k-(q + i\eta)}.$

On the other hand, consider a wave that grows exponentially with $$x$$ for $$x < 0$$, and is zero for $$x > 0$$. The Fourier transform is a function with a simple pole in the lower half-plane: $f(x) = \left\{\begin{array}{cl}0, & x \ge 0 \\ e^{i (q - i\eta) x}, & x < 0.\end{array}\right. \;\; \overset{\mathrm{FT}}{\longrightarrow} \;\; F(k) = \frac{i}{k-(q - i\eta)}.$ From these examples, we see that oscillations and amplification/decay in $$f(x)$$ are related to the existence of poles in the algebraic expression for $$F(k)$$. The real part of the pole position gives the wave-number of the oscillation, and the distance from the pole to the real axis gives the amplification or decay constant. A decaying signal produces a pole in the upper half-plane, while a signal that is increasing exponentially with $$x$$ produces a pole in the lower half-plane. In both cases, if we plot the Fourier spectrum of $$|F(k)|^2$$ versus real $$k$$, the result is a Lorentzian peak centered at $$k = q$$, with width $$2\eta$$.

## Gaussian wave-packets

Consider a function with a decay envelope given by a Gaussian function: $f(x) = e^{iq x} \, e^{-\gamma x^2}, \;\;\;\mathrm{where}\; q \in \mathbb{C},\; \gamma \in \mathbb{R}.$ This is called a Gaussian wave-packet. The width of the envelope is usually characterized by the Gaussian function’s standard deviation, which is where the curve reaches $$e^{-1/2}$$ times its peak value. In this case, the standard deviation is $$\Delta x = 1/\sqrt{2\gamma}$$.

We will show that $$f(x)$$ has the following Fourier transform: $F(k) = \sqrt{\frac{\pi}{\gamma}} \, e^{-\frac{(k-q)^2}{4\gamma}}.$

To derive this result, we perform the Fourier integral as follows: \begin{align} F(k) &= \int_{-\infty}^\infty dx \, e^{-ikx}\, f(x) \\ &= \int_{-\infty}^\infty dx \, \exp\left\{-i(k-q)x -\gamma x^2\right\}.\end{align} In the integrand, the expression inside the exponential is quadratic in $$x$$. We complete the square: \begin{align} F(k) &= \int_{-\infty}^\infty dx \, \exp\left\{-\gamma\left(x + \frac{i(k-q)}{2\gamma}\right)^2 + \gamma\left(\frac{i(k-q)}{2\gamma}\right)^2\right\} \\ &= \exp\left\{ - \frac{(k-q)^2}{4\gamma}\right\}\; \int_{-\infty}^\infty dx \, \exp\left\{-\gamma\left(x + \frac{i(k-q)}{2\gamma}\right)^2\right\}.\end{align} The remaining integral is the Gaussian integral with a constant imaginary shift in $$x$$. By shifting the integration variable, one can show that this is equal the standard Gaussian integral, $$\sqrt{\pi/\gamma}$$; the details are left as an exercise. We thus arrive at the result stated above.

The Fourier spectrum, $$|F(k)|^2$$, is a Gaussian function with standard deviation $\Delta k = \frac{1}{\sqrt{2(1/2\gamma)}} = \sqrt{\gamma}.$ Figure $$\PageIndex{1}$$

Once again, the Fourier spectrum is peaked at a value of $$k$$ corresponding to the wave-number of the underlying sinusoidal wave in $$f(x)$$, and a stronger (weaker) decay in $$f(x)$$ leads to a broader (narrower) Fourier spectrum. These features can be observed in the plot above.

This page titled 10.6: Common Fourier Transforms is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Y. D. Chong via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.