# Chapter 9: Fourier Transform

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The material covered in this chapter is also presented in Boas Chapter 7, Sections 12.

## 9.1 Fourier Transform

Can Fourier series be applied to functions $f(x)$ that are not periodic? Strictly speaking the answer is no. But we can generalize the approach to provide a positive answer. The trick is to take the periodicity length $2L$ to infinity, so that the function becomes periodic with an infinite period --- which is the same thing as not being periodic at all. A consequence of this limiting procedure is that the set of wave numbers implicated in the Fourier expansion will no longer be discrete, but will form a continuum. Discrete sums will therefore be replaced by integrals, and the standard Fourier series will become a Fourier transform. This is the topic of this chapter.

Recall from Chapter 7 --- refer back to Sec.7.7 --- that the Fourier series of a function $f(x)$, periodic with period $2L$, is given by

\begin{equation} f(x) = \sum_{n=-\infty}^\infty c_n e^{in\pi x/L}, \tag{9.1} \end{equation}

with coefficients

\begin{equation} c_n = \frac{1}{2L} \int_{-L}^L f(x) e^{-in\pi x/L}\, dx. \tag{9.2} \end{equation}

Writing $e^{in\pi x/L} = e^{i k_n x}$, we see that the set of wave numbers involved in the expansion is given by $k_n = n\pi/L$, so that the set of wavelengths is $\lambda_n = 2\pi/k_n = 2L/n$. With $n$ an integer ranging from $n=-\infty$ to $n = \infty$, these are infinite but discrete sets.

We wish to take the limit $L \to \infty$, and for this purpose it is useful to rewrite Eqs.(9.1) and (9.2) in terms of $k_n$. We have

\begin{align} f(x) &= \sum_{n=-\infty}^\infty c_n e^{i k_n x} \Delta n, \tag{9.3a} \\ c_n &= \frac{1}{2L} \int_{-L}^L f(x) e^{-i k_n x}\, dx, \tag{9.3b} \end{align}

where we inserted a trivial factor $\Delta n = 1$ in the Fourier series, to make it outrageously obvious that the sum is taken in unit steps of $n$. The reason for doing this is that we now wish to replace the summation variable with $k_n$, which increases in steps of $\Delta k_n = (\pi/L) \Delta n$. This gives us

\begin{align} f(x) &= \frac{L}{\pi} \sum_{k_n=-\infty}^\infty c_{k_n} e^{i k_n x} \Delta k_n, \tag{9.4a} \\ c_{k_n} &= \frac{1}{2L} \int_{-L}^L f(x) e^{-i k_n x}\, dx, \tag{9.4b} \end{align}

and these equations indicate that the limit $L \to \infty$ is problematic: the sum for $f(x)$ comes with an overall factor of $L$ that goes to infinity, and the integral for $c_{k_n}$ comes with a factor of $1/L$ that goes to zero.

These problems go away if we define $g_{k_n} := L c_{k_n}/\pi$ and assume that as $L \to \infty$, $c_{k_n} \to 0$ so that the product $L c_{k_n}$ stays finite. With this notation the equations become

\begin{align} f(x) &= \sum_{k_n=-\infty}^\infty g_{k_n} e^{i k_n x} \Delta k_n, \tag{9.5a} \\ g_{k_n} &= \frac{1}{2\pi} \int_{-L}^L f(x) e^{-i k_n x}\, dx, \tag{9.5b}

and taking the limit no longer seems problematic. With $L \to \infty$ we have that $\Delta k_n = (\pi/L) \Delta n$ becomes extremely small, and should therefore be interpreted as the infinitesimal $dk$. In the limit the discrete variable $k_n$ becomes the continuous variable $k$, the coefficients $g_{k_n}$ become the function $g(k)$, and the discrete sum over $k_n$ becomes an integral.

Taking the limit therefore takes us to the pair of equations

\begin{align} f(x) &= \int_{-\infty}^\infty g(k) e^{i k x}\, dk, \tag{9.6a}\\ g(k) &= \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-i k x}\, dx. \tag{9.6b} \end{align}

The function $g(k)$ is called the Fourier transform of $f(x)$. Equation (9.6a) is the generalization of the Fourier series to nonperiodic functions, and Eq.(9.6b) allows us to calculate $g(k)$ from $f(x)$; this complex function of $k$ is the generalization of the Fourier coefficients $c_n$.

The Fourier transform is often used to represent functions of time. In this context the variable $x$ is replaced by $t$, and the wave number $k$ is replaced by the angular frequency $\omega$. It is conventional to let $k \to -\omega$ when making this replacement. It is also conventional to change the notation of the Fourier transform, so that $g(k)$ is replaced by $g(\omega)$ instead of $g(-\omega)$. So the Fourier pair becomes

begin{align} f(t) &= \int_{-\infty}^\infty g(\omega) e^{-i \omega t}\, d\omega, \tag{9.7a} \\ g(\omega) &= \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{i \omega t}\, dt \tag{9.7b} \end{align}

when dealing with functions of time. In this new context the Fourier transform $g(\omega)$ is sometimes called the frequency spectrum of the function $f(t)$. It is important to understand that Eqs.(9.6) and (9.7) have precisely the same mathematical content; they differ in notation only.

## 9.2 Two-Step Function

As a first example we consider the two-step function defined by

\begin{equation} f(t) = \left\{ \begin{array}{lr} 1 & \quad -\tau < t < \tau \\ 0 & \quad \text{otherwise} \end{array} \right.. \tag{9.8} \end{equation}

Its Fourier transform is calculated with the help of Eq.(9.7b). We have

\begin{equation} g(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty f(t) e^{i\omega t}\, dt = \frac{1}{2\pi} \int_{-\tau}^\tau e^{i\omega t}\, dt = \frac{e^{i\omega t}}{2\pi i \omega} \biggr|^\tau_{-\tau} = \frac{1}{2\pi i \omega} (e^{i\omega\tau} - e^{-i\omega\tau}), \tag{9.9} \end{equation}

or

\begin{equation} g(\omega) = \frac{\sin(\omega\tau)}{\pi \omega}. \tag{9.10} \end{equation}

The function and its Fourier transform are shown in Fig.9.1. Figure 9.1: Two-step function and its Fourier transform. The plots were produced with $\tau = 1$.

This example (and the other ones below) reveals an interesting connection between the functions $f(t)$ and $g(\omega)$. Both functions can be characterized by a certain spread. In the case of $f(t)$, the function is nonzero for a finite interval of time only, and the half-width $\Delta t = \tau$ can be taken as a measure of its spread. In the case of $g(\omega)$, the function is nonzero over the entire interval $-\infty < \omega < \infty$, but it is sharply peaked around $\omega = 0$. A rough measure of its spread can be given by the half-width $\Delta \omega = \pi/(2\tau)$, at which point $g$ has decreased by an approximate factor of $0.6$. The product of the time and frequency spreads,

\begin{equation} \Delta \omega \Delta t = \frac{\pi}{2} \simeq 1.6, \tag{9.11} \end{equation}

is a number of order unity, independent of $\tau$. This is a universal feature of functions and their Fourier transforms: the product $\Delta\omega \Delta t$ is always of order unity. This means that a function $f(t)$ that has a wide spread in time will have a Fourier transform $g(\omega)$ that has a narrow spread in frequency; conversely, a function $f(t)$ that has a narrow spread in time will have a Fourier transform $g(\omega)$ that has a wide spread in frequency. This observation has far-reaching consequences in quantum mechanics, where frequency is related to energy by Planck's formula $E = \hbar \omega$. The relation $\Delta E \Delta t \simeq \hbar$ is one of the famous Heisenberg uncertainty relations.

## 9.3 Damped Wave

For a second example we consider the damped cosine wave described by

\begin{equation} f(t) = e^{-\kappa |t|} \cos(\Omega t), \tag{9.12} \end{equation}

where $\kappa$ is the damping constant, and $\Omega$ is the wave's angular frequency. To calculate the Fourier transform it is helpful to convert the cosine into complex exponentials. For $t > 0$ we write

\begin{equation} f(t) = e^{-\kappa t} \cos(\Omega t) = \frac{1}{2} e^{-\kappa t} ( e^{i\Omega t} + e^{-i\Omega t} ) = \frac{1}{2}\Bigl[ e^{i(\Omega + i\kappa) t} + e^{i(-\Omega + i\kappa)t} \Bigr], \tag{9.13} \end{equation}

while for $t < 0$,

\begin{equation} f(t) = e^{\kappa t} \cos(\Omega t) = \frac{1}{2} e^{\kappa t} ( e^{i\Omega t} + e^{-i\Omega t} ) = \frac{1}{2}\Bigl[ e^{i(\Omega - i\kappa) t} + e^{i(-\Omega - i\kappa)t} \Bigr]. \tag{9.14} \end{equation}

Inserting this in Eq.(9.7b) gives

\begin{align} g(\omega) &= \frac{1}{4\pi} \int_{-\infty}^0 \Bigl[ e^{i(\Omega - i\kappa) t} + e^{i(-\Omega - i\kappa)t} \Bigr] e^{i\omega t} \, dt \nonumber \\ & \quad \mbox{} + \frac{1}{4\pi} \int_0^\infty \Bigl[ e^{i(\Omega + i\kappa) t} + e^{i(-\Omega + i\kappa)t} \Bigr] e^{i\omega t} \, dt \nonumber \\ &= \frac{1}{4\pi i} \Biggl\{ \frac{e^{i(\omega+\Omega-i\kappa)t}}{\omega+\Omega-i\kappa} \biggr|^0_{-\infty} + \frac{e^{i(\omega-\Omega-i\kappa)t}}{\omega-\Omega-i\kappa} \biggr|^0_{-\infty} \nonumber \\ & \quad \mbox{} + \frac{e^{i(\omega+\Omega+i\kappa)t}}{\omega+\Omega+i\kappa} \biggr|^\infty_0 + \frac{e^{i(\omega-\Omega+i\kappa)t}}{\omega-\Omega+i\kappa} \biggr|^\infty_0 \Biggr\} \nonumber \\ &= \frac{1}{4\pi i} \biggl\{ \frac{1}{\omega+\Omega-i\kappa} + \frac{1}{\omega-\Omega-i\kappa} - \frac{1}{\omega+\Omega+i\kappa} - \frac{1}{\omega-\Omega+i\kappa} \biggr\}; \tag{9.15} \end{align}

the contributions from $t = \pm \infty$ vanish because of the factors $e^{\mp \kappa t}$ in the exponentials. Combining the first and third terms within the curly brackets gives

$\frac{2i\kappa}{(\omega+\Omega)^2 + \kappa^2}$

after simplification, while the second of fourth terms combine to give

$\frac{2i\kappa}{(\omega-\Omega)^2 + \kappa^2}.$

Our final expression for the Fourier transform is therefore

\begin{equation} g(\omega) = \frac{\kappa}{2\pi} \biggl[ \frac{1}{(\omega+\Omega)^2 + \kappa^2} + \frac{1}{(\omega-\Omega)^2 + \kappa^2} \biggr]. \tag{9.16} \end{equation}

The function and its Fourier transform are displayed in Fig.9.2. Figure 9.2: Damped cosine wave and its Fourier transform. The plots were produced with $\Omega = 1$ and $\kappa = 0.2$.

Exercise 9.1: Reproduce the calculations that lead to Eq.(9.16). This is a good exercise in complex algebra.

The damped cosine wave has a spread in time that is measured by the half-width $\Delta t = 1/\kappa$, which marks the point at which the wave's envelope has decreased by a factor of $1/e \simeq 0.4$. Its Fourier transform features two peaks at $\omega = \pm\Omega$, and the spread in frequency is measured by the half-width of each peak, given by $\Delta \omega = \kappa$, the point at which the peak has decreased by a factor of $1/2$. With these measures of spread we have that

\begin{equation} \Delta \omega \Delta t = 1, \tag{9.17} \end{equation}

and again we find that the right-hand side is a number of order unity, independent of $\kappa$ and $\Omega$.

## 9.4 Truncated Cosine Function

Our third example is concerned with a truncated cosine function described by

\begin{equation} f(t) = \left\{ \begin{array}{ll} \cos t & \quad -\frac{\pi}{2} < t < \frac{\pi}{2} \\ 0 & \text{otherwise} \end{array} \right. ./tag{9.18} \end{equation}

With $\cos t = \frac{1}{2} (e^{it} + e^{-it})$ the Fourier transform is calculated as

\begin{align} g(\omega) &= \frac{1}{2\pi} \int_{-\pi/2}^{\pi/2} \cos t\, e^{i\omega t}\, dt = \frac{1}{4\pi} \int_{-\pi/2}^{\pi/2} \bigl[ e^{i(\omega+1)t} + e^{i(\omega-1)t} \bigr]\, dt \nonumber \\ &= \frac{1}{4\pi i} \biggl[ \frac{ e^{i(\omega+1)\pi/2} - e^{-i(\omega+1)\pi/2}}{ \omega + 1 } + \frac{ e^{i(\omega-1)\pi/2} - e^{-i(\omega-1)\pi/2}}{ \omega - 1 } \biggr] \nonumber \\ &= \frac{1}{4\pi i} \biggl[ \frac{i}{\omega+1}( e^{i\omega\pi/2} + e^{-i\omega\pi/2} ) - \frac{i}{\omega-1}( e^{i\omega\pi/2} + e^{-i\omega\pi/2} ) \biggr], \tag{9.19} \end{align}

where we used the identity $e^{\pm i\pi/2} = \pm i$ in the last step. The complex exponentials can be converted to a cosine function, and we arrive at

\begin{equation} g(\omega) = \frac{\cos(\omega\pi/2)}{\pi(1-\omega^2)} \tag{9.20} \end{equation}

after further simplification. The function and its Fourier transform are displayed in Fig.9.3.

Exercise 9.2: Reproduce the calculations that lead to Eq.(9.20).

Exercise 9.3: What is the value of $g(\omega)$ at $\omega = \pm 1$?

The time spread of the truncated cosine function can again be measured by the half-width $\Delta t = \pi/3$, at which point the function has decreased by a factor of $1/2$. The frequency spread of its Fourier transform is measured by the half-width $\Delta \omega = \pi/2$, the point by which it has decreased by a factor of approximately $0.53$. This gives us

\begin{equation} \Delta \omega \Delta t = \frac{\pi^2}{6} \simeq 1.6, \tag{9.21} \end{equation}

## 9.5 Delta Function

Even the delta function can be given a Fourier transform. If we set

\begin{equation} f(t) = \delta(t-t') \tag{9.22} \end{equation}

in Eq.(9.7b) we quickly arrive at

\begin{equation} g(\omega) = \frac{e^{i\omega t'}}{2\pi} \tag{9.23} \end{equation}

for its Fourier transform. In this case the spread of $f(t)$ is $\Delta t = 0$, while the spread of $g(\omega)$ is $\Delta\omega = \infty$, because the function just keeps on oscillating. With a generous attitude we might interpret the product $\Delta \omega \Delta t = \infty \cdot 0$ as a number of order unity.

If we insert $g(\omega)$ back into Eq.(9.7a) we obtain an interesting representation of the delta function,

\begin{equation} \delta(t-t') = \frac{1}{2\pi} \int_{-\infty}^\infty e^{-i\omega(t-t')}\, d\omega. \tag{9.24} \end{equation}

Because the delta function is real, taking the complex conjugate of both sides of this equation yields

\begin{equation} \delta(t-t') = \frac{1}{2\pi} \int_{-\infty}^\infty e^{+i\omega(t-t')}\, d\omega. \tag{9.25} \end{equation}

These equations make intuitive sense. When $t \neq t'$, the integrand is an oscillating function of $\omega$, the integral adds up the positive and negative contributions, and we end up with a zero value; this is what we expect of $\delta(t-t')$ when $t \neq t'$. When $t = t'$, however, the complex exponential becomes equal to one, and the integral gives infinity, precisely what we expect of the delta function.

We can produce alternative representations of the delta function by adapting the notation. If, in Eqs.(9.24) or (9.25), we let $t \to \omega$, $t' \to \omega'$, $\omega \to t$, and $d\omega \to dt$, we obtain

\begin{equation} \delta(\omega-\omega') = \frac{1}{2\pi} \int_{-\infty}^\infty e^{\pm i(\omega-\omega')t}\, dt. \tag{9.26} \end{equation}

In this case the original function is $\delta(\omega-\omega')$, a function of $\omega$, and its Fourier transform is $e^{\mp i\omega' t}/(2\pi)$, a function of $t$. The spread in frequency is $\Delta\omega = 0$, while the spread in time is $\Delta t = \infty$.

These considerations allow us to define the Fourier transform of a undamped, and untruncated, cosine function. Let us therefore insert

\begin{equation} f(t) = \cos(\Omega t) \tag{9.27} \end{equation}

within Eq.(9.7b) to calculate its Fourier transform. We have

\begin{equation} g(\omega) = \frac{1}{2\pi} \int_{-\infty}^\infty \cos(\Omega t) e^{i\omega t}\, dt = \frac{1}{4\pi} \int_{-\infty}^\infty e^{i(\omega+\Omega)t}\, dt + \frac{1}{4\pi} \int_{-\infty}^\infty e^{i(\omega-\Omega)t}\, dt, \tag{9.28} \end{equation}

and Eq.(9.26) converts this to

\begin{equation} g(\omega) = \frac{1}{2} \bigl[ \delta(\omega + \Omega) + \delta(\omega - \Omega) \bigr]. \tag{9.29} \end{equation}

The Fourier transform of a pure cosine function is therefore the sum of two delta functions peaked at $\omega = \pm \Omega$. This result can be thought of as the limit of Eq.(9.16) when $\kappa \to 0$. In this case we are dealing with a function $f(t)$ with $\Delta t = \infty$ and a Fourier transform $g(\omega)$ with $\Delta \omega = 0$.

## 9.6 Practice Problems

1. (Boas Chapter 7, Section 12, Problem 6) Find the Fourier transform of the function defined by $f(t) = t$ for $|t|< 1$ and $f(t) = 0$ otherwise.

2. (Boas Chapter 7, Section 12, Problem 7) Find the Fourier transform of the function defined by $f(t) = |t|$ for $|t|< 1$ and $f(t) = 0$ otherwise.

3. (Boas Chapter 7, Section 12, Problem 12) Find the Fourier transform of the truncated sine function defined by $f(t) = \sin t$ for $|t|< \pi/2$ and $f(t) = 0$ otherwise.

4. (Boas Chapter 7, Section 12, Problem 21) Find the Fourier transform of the Gaussian function $f(t) = \exp[-t^2/(2\sigma^2)]$. Hint: consider the change of variables $y = t - i \sigma^2 \omega$ to evaluate the integral.

5. (Boas Chapter 7, Section 12, Problem 24a) Find the Fourier transform of $f(t) = \exp(-|x|)$.

6. (Boas Chapter 7, Section 12, Problem 24c) Find the Fourier transform of $f(t) = 1/(1+t^2)$. Hint: make sure to work through the previous problem first.

## 9.7 Challenge Problems

1. If $g(\omega)$ is the Fourier transform of $f(t)$, what is the Fourier

of $df/dt$?

2. Calculate the Fourier transform of the function $f(t) = J_0(t)$. Hint: Start with the integral representation for the Bessel function, and do the $t$-integral first.