The Fourier Transform

(1)

Department of Mathematics and Statistics, Villanova University, Villanova, PA, USA

5.1 Definition and examples

For a given function f such that $\int _{-\infty }^{\infty }\vert f(x)\vert \,dx <\infty$ , the Fourier transform of f is defined, for each real number ω, by

$\displaystyle{ \mathcal{F}f(\omega ):=\int _{ -\infty }^{\infty }f(x)\,e^{-i\omega x}\,dx. }$

(5.1)

The idea behind this definition is that, for each value of ω, the value of $\,\mathcal{F}f(\omega )\,$ captures the component of f that has the frequency ω∕(2π) (and period 2π∕ω ).

Example 5.1.

The Fourier transform of a Gaussian. Let $f(x) = e^{-Ax^{2} }$ , for some positive constant A > 0. Then we have

$\displaystyle{ \mathcal{F}f(\omega ) = \sqrt{ \frac{\pi } {A}}\,e^{- \frac{\omega ^{2}} {4A} }. }$

(5.2)

To prove this, we first need the following fact.

Lemma 5.2.

For A ≠ 0, we have $\int _{-\infty }^{\infty }e^{-Ax^{2} }\,dx = \sqrt{ \frac{\pi }{A}}$ .

Proof.

Squaring the integral, we get

$\displaystyle\begin{array}{rcl} \left (\int _{-\infty }^{\infty }e^{-Ax^{2} }\,dx\right )^{2}& =& \left (\int _{ -\infty }^{\infty }e^{-Ax^{2} }\,dx\right )\,\left (\int _{-\infty }^{\infty }e^{-Ax^{2} }\,dx\right ) {}\\ & =& \left (\int _{-\infty }^{\infty }e^{-Ax^{2} }\,dx\right )\,\left (\int _{-\infty }^{\infty }e^{-Ay^{2} }\,dy\right ) {}\\ & =& \int _{-\infty }^{\infty }\int _{ -\infty }^{\infty }e^{-A(x^{2}+y^{2}) }\,dx\,dy {}\\ \mathrm{(polar\ coordinates)}& =& \int _{\theta =0}^{2\pi }\int _{ r=0}^{\infty }e^{-Ar^{2} }\,r\,dr\,d\theta {}\\ & =& \int _{\theta =0}^{2\pi }\left (\lim _{ b\rightarrow \infty }\frac{1 - e^{-Ab^{2} }} {2A} \right )\,d\theta {}\\ & =& \int _{\theta =0}^{2\pi } \frac{1} {2A}\,d\theta {}\\ & =& \frac{\pi } {A}. {}\\ \end{array}$

Taking square roots proves the lemma.

Now to compute the Fourier transform for this example. For each ω,

$\displaystyle\begin{array}{rcl} \mathcal{F}f(\omega )& =& \int _{-\infty }^{\infty }e^{-Ax^{2} }\,e^{-i\omega x}\,dx {}\\ & =& \int _{-\infty }^{\infty }e^{-A(x^{2}+i\omega x/A) }\,dx {}\\ (\mathrm{complete\ the\ square})& =& \int _{-\infty }^{\infty }e^{-A(x^{2}+i\omega x/A+(i\omega /2A)^{2} }\,e^{A(i\omega /2A)^{2} }\,dx {}\\ & =& e^{-\omega ^{2}/4A }\,\int _{-\infty }^{\infty }e^{-A(x+i\omega /2A)^{2} }\,dx {}\\ & =& e^{-\omega ^{2}/4A }\,\int _{-\infty }^{\infty }e^{-Au^{2} }\,du\ \ \mathrm{with}\ u = x + i\omega /2A {}\\ & =& \sqrt{ \frac{\pi } {A}}\,e^{-\omega ^{2}/4A }\ \ \mathrm{by\ the\ lemma.} {}\\ \end{array}$

This establishes the result we were after. □

Observe that if we take $$A = 1/2$$

, then $\,f(x) = e^{-x^{2}/2 }\,$ and $\mathcal{F}f(\omega ) = \sqrt{2\pi }e^{-\omega ^{2}/2 }$ , a constant multiple of f itself. In the language of linear algebra, this function f is an eigenvector, or eigenfunction, of the Fourier transform.

Additional examples are considered in the exercises. Let us look at some basic properties of the Fourier transform.

5.2 Properties and applications

Additivity. Because the integral of a sum of functions is equal to the sum of the integrals of the functions separately, it follows that

$\displaystyle{ \mathcal{F}(f + g)(\omega ) = \mathcal{F}f(\omega ) + \mathcal{F}g(\omega )\, }$

(5.3)

for all integrable functions f and g and every real number ω.

Constant multiples. Because the integral of c ⋅ f is equal to c times the integral of f, we see that

$\displaystyle{ \mathcal{F}(cf)(\omega ) = c \cdot \mathcal{F}f(\omega )\, }$

(5.4)

for all integrable functions f, all (complex) numbers c, and every real number ω.

The two properties (5.3) and (5.4) taken together prove the following.

Proposition 5.3.

The Fourier transform acts as a linear transformation on the space of all absolutely integrable functions. That is, for two such functions f and g and any constants α and β, we get that

$\displaystyle{ \mathcal{F}(\alpha f +\beta g) =\alpha \mathcal{F}(f) +\beta \mathcal{F}(g). }$

(5.5)

Shifting/translation. For an integrable function f and fixed real number α, let $g(x) = f(x-\alpha )$ . (So the graph of g is the graph of f shifted or translated to the right by α units.) Then

$\displaystyle{ \mathcal{F}g(\omega ) = e^{-i\omega \alpha }\mathcal{F}f(\omega ). }$

(5.6)

Proof.

For each ω, we get

$\displaystyle\begin{array}{rcl} \mathcal{F}g(\omega )& =& \int _{-\infty }^{\infty }f(x-\alpha )\,e^{-i\omega x}\,dx {}\\ (\mathrm{let\ }u = x-\alpha )\ & =& \int _{-\infty }^{\infty }f(u)\,e^{-i\omega (u+\alpha )}\,du {}\\ & =& e^{-i\omega \alpha }\,\int _{ -\infty }^{\infty }f(u)\,e^{-i\omega u}\,du {}\\ & =& e^{-i\omega \alpha }\,\mathcal{F}f(\omega )\ \ \mathrm{as\ claimed.} {}\\ \end{array}$

Since the graph of g is a simple translation of the graph of f, the magnitude of the component of g at any given frequency is the same as that of f. However, the components occur at different places in the two signals, so there is a phase shift or delay in the Fourier transform. Also, the fixed translation α encompasses more cycles of a wave at a higher frequency than at a lower one. (For instance, an interval of width α = 2π contains two cycles of the wave y = sin(2x), but only one cycle of the wave y = sin(x).) Therefore, the larger the value of α is relative to the wavelength 2π∕ω the larger will be the phase delay in the transform. That is, the phase delay in the transform is proportional to ω, which explains the factor of e ^{−i ω α} in the transform of the shifted function g.

Shifting/modulation. For a given function f and a fixed real number ω ₀, let $h(x) = e^{i\omega _{0}x}\ f(x)$ . (So h “modulates” f by multiplying f by a periodic function of a fixed frequency ω ₀∕2π.) Then

$\displaystyle{ \mathcal{F}h(\omega ) = \mathcal{F}f(\omega -\omega _{0}). }$

(5.7)

Proof.

For each ω, we get

$\displaystyle\begin{array}{rcl} \mathcal{F}h(\omega )& =& \int _{-\infty }^{\infty }e^{i\omega _{0}x}\,f(x)\,e^{-i\omega x}\,dx {}\\ & =& \int _{-\infty }^{\infty }f(x)\,e^{-i(\omega -\omega _{0})x}\,dx {}\\ & =& \mathcal{F}f(\omega -\omega _{0})\ \ (\mathrm{by\ definition!}). {}\\ \end{array}$

These two shifting properties show that a translation of f results in a modulation of $\,\mathcal{F}f\,$ while a modulation of f produces a translation of $\mathcal{F}f$ .

Scaling. For a given function f and a fixed real number a ≠ 0, let ϕ(x) = f(ax). Then

$\displaystyle{ \mathcal{F}\phi (\omega ) = \frac{1} {\vert a\vert }\mathcal{F}f(\omega /a). }$

(5.8)

Proof.

Assume that a > 0 for now. For each ω, we get

$\displaystyle\begin{array}{rcl} \mathcal{F}\phi (\omega )& =& \int _{-\infty }^{\infty }f(ax)\,e^{-i\omega x}\,dx {}\\ (\mathrm{let}\ u = ax)\ & =& \frac{1} {a}\int _{-\infty }^{\infty }f(u)\,e^{-i\omega u/a}\,du {}\\ & =& \frac{1} {a}\int _{-\infty }^{\infty }f(u)\,e^{-i(\omega /a)u}\,du {}\\ & =& \frac{1} {a}\mathcal{F}f(\omega /a)\ \ (\mathrm{by\ definition!}). {}\\ \end{array}$

A similar argument applies when a < 0. (Why do we get a factor of 1∕ | a | in this case?)

Even and odd functions. A function f, defined on the real line, is even if $\,f(-x) = f(x)\,$ for every x. Similarly, a function g is odd if $\,g(-x) = -g(x)\,$ for every x. For example, the cosine function is even while the sine function is odd.

Using Euler’s formula (4.4), $e^{\,i\theta } =\cos (\theta ) + i\sin (\theta )$ , we may write the Fourier transform of a suitable real-valued function f as

$\displaystyle{\mathcal{F}f(\omega ) =\int _{ -\infty }^{\infty }f(x)\cos (\omega x)\,dx - i \cdot \int _{ -\infty }^{\infty }f(x)\sin (\omega x)\,dx\,.}$

Now, if f is even, then, for fixed ω, the function $\,x\mapsto f(x)\sin (\omega x)\,$ is odd, whence $\int _{-\infty }^{\infty }f(x)\sin (\omega x)\,dx = 0$ . Thus, an even function has a real-valued Fourier transform.

Similarly, if f is odd, then $\,x\mapsto f(x)\cos (\omega x)\,$ is also odd for each fixed ω. Thus, $\int _{-\infty }^{\infty }f(x)\cos (\omega x)\,dx = 0$ . It follows that an odd function has a purely imaginary Fourier transform.

Transform of the complex conjugate. For a complex-number-valued function f defined on the real line $\mathbb{R}$ , the complex conjugate of f is the function $\,\overline{f}\,$ defined by

$\displaystyle{ \overline{f}(x) = \overline{f(x)}\ \ \mathrm{for\ every\ real\ number}\ x. }$

(5.9)

To uncover the relationship between the Fourier transform of $\,\overline{f}\,$ and that of f, let ω be an arbitrary real number. Then we have

$\displaystyle\begin{array}{rcl} \mathcal{F}\overline{f}(\omega )& =& \int _{-\infty }^{\infty }\overline{f(x)}\,e^{-i\omega x}\,dx {}\\ & =& \int _{-\infty }^{\infty }\overline{f(x)}\,e^{\,i(-\omega )x}\,dx {}\\ & =& \int _{-\infty }^{\infty }\overline{f(x)}\,\overline{e^{-i(-\omega )x}}\,dx {}\\ & =& \overline{\int _{-\infty }^{\infty }f(x)\,e^{-i(-\omega )x}\,dx} {}\\ & =& \overline{\mathcal{F}f(-\omega )} {}\\ & =& \overline{\mathcal{F}f}(-\omega ). {}\\ \end{array}$

This proves the following proposition.

Proposition 5.4.

For an integrable function f defined on the real line, and for every real number ω,

$\displaystyle{ \mathcal{F}\overline{f}(\omega ) = \overline{\mathcal{F}f}(-\omega ). }$

(5.10)

Example 5.5.

Let $$$\,f(x) = \left \{\begin{array}{cc} 1&\mbox{ if $ - 1 \leq x \leq 1$,}\\ 0 & \mbox{ if $\vert x\vert> 1$.} \end{array} \right.$$ ” src=”/wp-content/uploads/2016/10/A183987_2_En_5_Chapter_IEq24.gif”></SPAN> Then the Fourier transform of <SPAN class=EmphasisTypeItalic>f</SPAN> is <SPAN id=IEq25 class=InlineEquation><IMG alt=$ 2 sin(ω)∕ω. Now let ϕ(x) = f(ax), where a > 0. That is, $$$\,\phi (x) = \left \{\begin{array}{cc} 1&\mbox{ if $-1/a \leq x \leq 1/a$,}\\ 0 & \mbox{ if $\vert x\vert> 1/a$.} \end{array} \right.$$ ” src=”/wp-content/uploads/2016/10/A183987_2_En_5_Chapter_IEq26.gif”></SPAN> </DIV><br /> <DIV class=Para>So we already know from earlier work that <SPAN id=IEq27 class=InlineEquation><IMG alt=$ , which is the same as $(1/a)\mathcal{F}f(\omega /a)$ . This agrees with the scaling result (5.8).

Example 5.6.

Let $$$\,f(x) = \left \{\begin{array}{cc} 1&\mbox{ if $ - 1 \leq x \leq 1$,}\\ 0 & \mbox{ if $\vert x\vert> 1$} \end{array} \right.$$ ” src=”/wp-content/uploads/2016/10/A183987_2_En_5_Chapter_IEq29.gif”></SPAN> as in the previous example. So, again, the Fourier transform of <SPAN class=EmphasisTypeItalic>f</SPAN> is <SPAN id=IEq30 class=InlineEquation><IMG alt=$ . Now let

$$$\displaystyle{g(x) = f(x-2) = \left \{\begin{array}{cc} 1& \mbox{ if $1 \leq x \leq 3$,}\\ 0 &\mbox{ if $x <1$ or $x> 3$.} \end{array} \right.}$$ ” src=”/wp-content/uploads/2016/10/A183987_2_En_5_Chapter_Equb.gif”></DIV></DIV></DIV>By the shifting/translation result (<SPAN class=InternalRef><A href=$ 5.6), we get

$\displaystyle{\mathcal{F}g(\omega ) = e^{-2i\omega }\mathcal{F}f(\omega ) = 2e^{-2i\omega }\frac{\sin (\omega )} {\omega }.}$

Example 5.7.

As an application of the shifting/modulation property (5.7), observe that

$\displaystyle{\cos (\omega _{0}x) = (1/2)(e^{i\omega _{0}x} + e^{-i\omega _{0}x}).}$

Thus, for any suitable f, take h(x) = f(x)cos(ω ₀ x). That is,

$\displaystyle{h(x) = (1/2)e^{i\omega _{0}x}f(x) + (1/2)e^{i(-\omega _{0})x}f(x).}$

It follows that

$\displaystyle{\mathcal{F}h(\omega ) = (1/2)\mathcal{F}f(\omega -\omega _{0}) + (1/2)\mathcal{F}f(\omega +\omega _{0}).}$

For a specific example, let $$$\,f(x) = \left \{\begin{array}{cc} 1&\mbox{ if $ - 1 \leq x \leq 1$,}\\ 0 & \mbox{ if $\vert x\vert> 1$} \end{array} \right.\,$$ ” src=”/wp-content/uploads/2016/10/A183987_2_En_5_Chapter_IEq31.gif”></SPAN> as in the previous examples. So, again, the Fourier transform of <SPAN class=EmphasisTypeItalic>f</SPAN> is <SPAN id=IEq32 class=InlineEquation><IMG alt=$ . With h(x) = f(x)cos(ω ₀ x), we get

$\displaystyle\begin{array}{rcl} \mathcal{F}h(\omega )& =& (1/2)\mathcal{F}f(\omega -\omega _{0}) + (1/2)\mathcal{F}f(\omega +\omega _{0}) {}\\ & =& (1/2)\left \{2\frac{\sin (\omega -\omega _{0})} {(\omega -\omega _{0})} + 2\frac{\sin (\omega +\omega _{0})} {(\omega +\omega _{0})}\right \} {}\\ & =& \frac{\sin (\omega -\omega _{0})} {(\omega -\omega _{0})} + \frac{\sin (\omega +\omega _{0})} {(\omega +\omega _{0})} {}\\ \end{array}$