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*}

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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}

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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*}

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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}

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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 \(b_n=i( c_n-c_{-n})\text{.}\)

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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}

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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 \(U(t)\text{.}\)

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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*}

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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}

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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}

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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*}

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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.

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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}

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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 \(\frac{U(a^-)+U(a^+)}{2}\text{.}\)

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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*}

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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}

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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}

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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.

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

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Table 11.4.2. Important Properties of the Fourier Transform

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

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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*}

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Example 11.4.4.

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

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Solution.

Using the result of Example 11.4.3, 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|}. \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*}

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Exercises Exercises

1.

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

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(a)

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

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Solution.

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

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(b)

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

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Solution.

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

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2.

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

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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*}

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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 \(\int\nolimits_{-\infty}^{\infty}e^{-\frac{r^2}{2}}\,dt=\sqrt{2\pi}\text{.}\)}

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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*}

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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*}

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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*}

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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*}

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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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