The Fourier Transform

Section 11.4 The Fourier Transform

Let \(U(t)\) be a real-valued function with period \(2\pi\) which is piecewise continuous such that \(U\,'(t)\) also exists and is piecewise continuous. Then \(U(t)\) has the complex Fourier series representation

\begin{equation*} U(t) = \sum_{n=-\infty}^{\infty}c_ne^{int}, \text{ where } \end{equation*}

\begin{equation*} c_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}U(t) e^{-int}\,dt \text{ for all } n\text{.} \end{equation*}

The coefficients \(\{c_nt\}\) are complex numbers. Previously, we expressed \(U(t)\) as a real trigonometric series:

\begin{equation} U(t) =\frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos nt+b_n\sin nt)\text{.}\tag{11.4.1} \end{equation}

Hence, a relationship between the coefficients is

\begin{align*} a_n=c_n+c_{-n} \amp \text{ for } n=0,1,2, \ldots , \text{ and }\\ b_n=i(c_n-c_{-n}) \amp \text{ for } n=1,2, \ldots \text{,} \end{align*}

These relations are easy to establish. We start by writing

\begin{align} U(t) \amp = c_0 + \sum_{n=1}^{\infty}c_ne^{int} + \sum_{n=1}^{\infty}c_{-n}e^{-int}\tag{11.4.2}\\ \amp = c_0 + \sum_{n=1}^{\infty}c_n(\cos nt+i\sin nt) + \sum_{n=1}^{\infty}c_{-n}(\cos nt-i\sin nt)\notag\\ \amp = c_0 + \sum_{n=1}^{\infty}[(c_n+c_{-n}) \cos nt+i(c_n-c_{-n}) \sin nt]\text{.}\notag \end{align}

Comparing Equation (11.4.2) with Equation (11.4.1), we see that \(a_0=2c_0\text{,}\) \(a_n=c_n+c_{-n}\text{,}\) and \(b_n=i( c_n-c_{-n})\text{.}\)

If \(U(t)\) and \(U\,'(t)\) are piecewise continuous and have period \(2L\text{,}\) then \(U(t)\) has the complex Fourier series representation

\begin{equation} U(t) = \sum_{n=-\infty}^{\infty}c_ne^{i\frac{\pi nt}{L}}, \text{ where }\tag{11.4.3} \end{equation}

\begin{equation} c_n = \frac{1}{2L}\int_{-L}^{L}U(t) e^{-i\frac{\pi nt}{L}}\,dt \text{ for all } n\text{.}\tag{11.4.4} \end{equation}

We have seen how periodic functions are represented by trigonometric series. Many practical problems involve nonperiodic functions. A representation analogous to Fourier series for a nonperiodic function \(U(t)\) is obtained by considering the Fourier series of \(U(t)\) for \(-L\lt t\lt L\) and then taking the limit as \(L \to \infty\text{.}\) The result is known as the Fourier transform of \(U(t)\text{.}\)

Start with a nonperiodic function \(U(t)\) and consider the periodic function \(U_L(t)\) with period \(2L\text{,}\) where

\begin{align*} U_L(t) \amp = U(t) \amp \amp \text{ for } -L\lt t \le L, \text{ and }\\ U_L(t) \amp =U_L(t+2L) \amp \amp \text{ for all } t\text{.} \end{align*}

Then \(U_L(t)\) has the complex Fourier series representation

\begin{equation} U_L(t)=\sum_{n=-\infty}^{\infty}c_ne^{i\frac{\pi nt}{L}}\text{.}\tag{11.4.5} \end{equation}

We introduce some terminology to discuss the terms in Equation (11.4.5), first

\begin{equation} w_n=\frac{\pi n}{L} \text{ is called the frequency. }\tag{11.4.6} \end{equation}

If \(t\) denotes time, then the units for \(w_n\) are radians per unit time. The set of all possible frequencies is called the frequency spectrum, i.e.,

\begin{equation*} \left\{\ldots ,\frac{-3\pi}{L},\frac{-2\pi}{L},\frac{-\pi}{L},\frac{\pi}{L}, \frac{2\pi}{L},\frac{3\pi}{L}, \ldots\right\}\text{.} \end{equation*}

It is important to note that as \(L\) increases, the spectrum becomes finer and approaches a continuous spectrum of frequencies. It is reasonable to expect that the summation in the Fourier series for \(U_L(t)\) will give rise to an integral over \([-\infty .\infty ]\text{.}\) This is stated in the following important theorem.

Theorem 11.4.1. Fourier Transform.

Let \(U(t)\) and \(U\,'(t)\) be piecewise continuous, and

\begin{equation*} \int_{-\infty}^{\infty}\left| U(t) \right| dt\lt M\text{,} \end{equation*}

for some positive constant \(M\text{.}\) The Fourier transform \(F(w)\) of \(U(t)\) is defined as

\begin{equation} F(w) = \frac{1}{2\pi}\int\nolimits_{-\infty}^{\infty}U(t)e^{-iwt}\,dt\text{.}\tag{11.4.7} \end{equation}

At points of continuity, \(U(t)\) has the integral representation

\begin{equation*} U(t) = \int\nolimits_{-\infty}^{\infty}F(w)e^{iwt}\,dw\text{.} \end{equation*}

and at a point \(t=a\) of discontinuity of \(U\text{,}\) the integral converges to \(\frac{U(a^-)+U(a^+)}{2}\text{.}\)

Note: It is common to express the fact that \(U\) is transformed into \(F\) by using the operator notation

\begin{equation*} \mathrm{F}\big(U(t)\big)=F(w) \end{equation*}

Proof.

Set \(\Delta w_n=w_{n+1}-w_n=\frac{\pi}{L}\) and \(\frac{1}{2L}=\frac{1}{2\pi}\Delta w_n\text{.}\) These quantities are used in conjunction with Equations (11.4.3), (11.4.4), (11.4.5) and the frequency in (11.4.6) to obtain

\begin{align} U_L(t) \amp = \sum_{n=-\infty}^{\infty}\left[\frac{1}{2L} \int_{-L}^{L}U(t) e^{-iw_nt}\,dt\right] e^{iw_nt}\tag{11.4.8}\\ \amp = \sum\limits_{n=-\infty}^{\infty}\left[\frac{1}{2\pi}\int_{-L}^{L}U(t) e^{-iw_nt}\,dt\right] e^{iw_nt}\Delta w_n\text{.}\notag \end{align}

If we define \(F_L(w)\) by

\begin{equation*} F_L(w) = \frac{1}{2\pi}\int_{-L}^{L}U(t)e^{-iwt}\,dt\text{,} \end{equation*}

then Equation (11.4.8) can be written as

\begin{equation} U_L(t) = \sum_{n=-\infty}^{\infty}F_L(w_n)e^{iw_nt}\Delta w_n\text{.}\tag{11.4.9} \end{equation}

As \(L\) gets large, \(F_L(w_n)\) approaches \(F(w_n)\) and \(\Delta w_n\) tends to zero. Thus the limit on the right-hand side of Equation (11.4.9) can be viewed as an integral. This substantiates the Fourier integral representation

\begin{equation*} U(t) = \int_{-\infty}^{\infty}F(w)e^{iwt}dw\text{.} \end{equation*}

A more rigorous proof of this fact can be found in advanced texts.

The following table depicts some important properties of the Fourier transform.

Table 11.4.2. Important Properties of the Fourier Transform


Linearity	\(\mathrm{F}\big(aU_1(t) + bU_2(t)\big) =a\,\mathrm{F}(U_1(t)) +b\,\mathrm{F}\big(U_2(t)\big)\text{;}\)
\hdashline Symmetry	If \(\mathrm{F}\big(U(t)\big)=F(w)\text{,}\) then \(\mathrm{F}\big(F(t)\big) = \frac{1}{2\pi}U(-w)\text{;}\)
\hdashline Time scaling	\(\mathrm{F}(U(at)) = \frac{1}{\|a\|}F(\frac{w}{a})\text{;}\)
\hdashline Time shifting	\(\mathrm{F}(U(t-t_0)) = e^{-it_0w}F(w)\text{;}\)
\hdashline Frequency shifting	\(\mathrm{F}\big(e^{-iw_0t}U(t)\big) = F(w-w_0)\text{;}\)
\hdashline Time differentiation	\(\mathrm{F}\big(U\,'(t)\big) = iwF(w)\text{;}\)
\hdashline Frequency differentiation	\(\frac{d^nF(w)}{dw^n} = \mathrm{F}\big((-it)^nU(t)\big)\text{;}\)
\hdashline Moment Theorem	If \(M_n = \int_{-\infty}^{\infty}t^nU(t)\,dt\text{,}\) then \((-i)^nM_n = 2\pi F^{(n)}(0)\text{.}\)

Example 11.4.3.

Show that \(F\big(e^{-|t|}\big) = \frac{1}{\pi(1+w^2)}\text{.}\)

Solution.

Using formula (11.4.7) we obtain

\begin{align*} F(w) \amp =\frac{1}{2\pi}\int\nolimits_{-\infty}^{\infty}e^{-\left| t\right| }e^{-iwt}\,dt\\ \amp = \frac{1}{2\pi}\int\nolimits_{-\infty}^{0}e^{(1-iw)t}\,dt + \frac{1}{2\pi}\int_0^{\infty}e^{(-1-iw) t}\,dt\\ \amp = \frac{1}{2\pi (1-iw)}e^{(1-iw) t}\bigg|_{t=-\infty}^{t=0}+\frac{1}{2\pi (-1-iw)}e^{(-1-iw) t} \bigg|_{t=0}^{t=\infty}.\\ \amp = \frac{1}{2\pi (1-iw)}+\frac{1}{2\pi (1+iw)}\\ \amp = \frac{1}{\pi (1+w^2)}\text{.} \end{align*}

Example 11.4.4.

Show that \(\mathrm{F}\big(\frac{1}{1+t^2}\big) = \frac{1}{2}e^{-|w|}\text{.}\)

Solution.

Using the result of Example 11.4.3 and the symmetry property and the symmetry property, we obtain

\begin{equation*} \mathrm{F}\left(\frac{1}{\pi (1+t^2)}\right) = \frac{1}{2\pi}e^{-|-w|} = \frac{1}{2\pi}e^{-|w|}\text{.} \end{equation*}

By the linearity property, with each term is multiplied by \(\pi\text{,}\) we get

\begin{equation*} \mathrm{F}\left(\frac{1}{1+t^2}\right) = \frac{1}{2}e^{-|w|} \end{equation*}

Exercises Exercises

1.

Find \(\mathfrak{F}\big(U(t)\big)\) for each of the following:

(a)

\(U(t) = \begin{cases}1 \amp \text{ for } |t| \lt 1,\\ 0 \amp \text{ for } |t| > 1. \end{cases}\)

Solution.

\(\mathfrak{F}\big(U(t)\big) =\frac{\sin(w)}{\pi w}\text{.}\)

(b)

\(U(t) = \begin{cases}1-|t| \amp \text{ for } |t| \le 1,\\ 0 \amp \text{ for } |t|>1. \end{cases}\)

Solution.

\(\mathfrak{F}\big(U(t)\big) =\frac{1-\cos(w)}{\pi w^2}=\frac{2\sin^2(\frac{w}{2})}{\pi w^2}\text{.}\)

2.

Let \(U(t) = \begin{cases}\sin t \amp \text{ for } |t| \le \pi,\\ \;\;\, 0 \amp \text{ for } |t| > \pi. \end{cases}\) Show that \(\mathfrak{F}\big(U(t)\big) =\frac{i\sin \pi w}{\pi (1-w^2)}\text{.}\)

3.

Use the symmetry and linearity properties and the result of Exercise a to show that

\begin{equation*} \mathfrak{F}\left(\frac{\sin t}{t}\right) = \begin{cases}\frac{1}{2} \amp \text{ for } |w| \lt 1,\\ 0 \amp \text{ for } |w| > 1. \end{cases} \end{equation*}

4.

Let \(U(t) =e^{-\frac{t^2}{2}}\text{.}\) Show that \(\mathfrak{F}\big(U(t)\big) = \frac{1}{\sqrt{2\pi}}e^{-\frac{w^2}{2}}\text{.}\) \hint{Use the integral definition and combine the terms in the exponent, then complete the square and use the fact that \(\int\nolimits_{-\infty}^{\infty}e^{-\frac{r^2}{2}}\,dt=\sqrt{2\pi}\text{.}\)}

5.

Use the time scaling property and the example in the text to show that

\begin{equation*} \mathfrak{F}(e^{-a|t|}) =\frac{|a|}{\pi(a^2+w^2)} \end{equation*}

6.

Use the symmetry and linearity properties and the result of Exercise 11.4.2 to show that

\begin{equation*} \mathfrak{F}\left(\frac{i\sin \pi t}{1-t^2}\right) = \begin{cases}\frac{i\sin w}{2} \amp \text{ for } |w| \le \pi,\\ 0 \amp \text{ for } |w| > \pi. \end{cases} \end{equation*}

7.

Use the time differentiation property and the result of Exercise 11.4.4 to show that

\begin{equation*} \mathfrak{F}\left(te^{-\frac{r^2}{2}}\right) =\frac{-iwe^{-\frac{w^2}{2}}}{\sqrt{2\pi}} \end{equation*}

8.

Use the symmetry and linearity properties and the results of Exercise b to show that

\begin{equation*} F\left(\frac{\sin^2\frac{t}{2}}{t^2}\right) = \begin{cases}\frac{1-|w|}{4\pi} \amp \text{ for } |w| \le 1, \\ 0 \amp \text{ for } |w| >1. \end{cases} \end{equation*}

9.

Write a report on the Fourier transform. Discuss some of the ideas you found in the literature that are not mentioned in the text.

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