# Representation of Aperiodic Signals - the Continuous-Time Fourier Transform

This is a continuation from the previous tutorial - examples of discrete-time filters described by difference equations.

In the last few tutorials, we developed a representation of periodic signals as linear combinations of complex exponentials. We also saw how this representation can be used in describing the effect of LTI systems on signals.

In this tutorial, we extend these concepts to apply to signals that are not periodic. As we will see, a rather large class of signals, including all signals with finite energy, can also be represented through a linear combination of complex exponentials.

Whereas for periodic signals the complex exponential building blocks are harmonically related, for aperiodic signals they are infinitesimally close in frequency, and the representation in terms of a linear combination takes the form of an integral rather than a sum.

The resulting spectrum of coefficients in this representation is called the Fourier transform, and the synthesis integral itself, which uses these coefficients to represent the signal as a linear combination of complex exponentials, is called the inverse Fourier transform.

The development of this representation for aperiodic signals in continuous time is one of Fourier's most important contributions, and our development of the Fourier transform follows very closely the approach he used in his original work.

In particular, Fourier reasoned that an aperiodic signal can be viewed as a periodic signal with an infinite period. More precisely, in the Fourier series representation of a periodic signal, as the period increases the fundamental frequency decreases and the harmonically related components become closer in frequency. As the period becomes infinite, the frequency components form a continuum and the Fourier series sum becomes an integral.

In the next tutorial we develop the Fourier series representation for continuous-time periodic signals, and in the tutorials that follow we build on this foundation as we explore many of the important properties of the continuous-time Fourier transform that form the foundation of frequency-domain methods for continuous-time signals and systems. In later tutorials, we parallel this development for discrete-time signals.

## 1. Development of the Fourier Transform Representation of an Aperiodic Signal

To gain some insight into the nature of the Fourier transform representation, we begin by revisiting the Fourier series representation for the continuous-time periodic square wave examined in Example 3.5 [refer to the Fourier series representation of continuous-time periodic signals tutorial].

Specifically, over one period,

$x(t)=\begin{cases}1,\qquad|t|\lt{T_1}\\0,\qquad{T_1}\lt|t|\lt{T/2}\end{cases}$

and periodically repeats with period $$T$$, as shown in Figure 4.1.

As determined in Example 3.5 [refer to the Fourier series representation of continuous-time periodic signals tutorial], the Fourier series coefficients $$a_k$$ for this square wave are

$\tag{4.1}[\text{eq. }(3.44)]\qquad\qquad\qquad{a_k}=\frac{2\sin(k\omega_0T_1)}{k\omega_0T}$

where $$\omega_0=2\pi/T$$. In Figure 3.7 [refer to the Fourier series representation of continuous-time periodic signals tutorial], bar graphs of these coefficients were shown for a fixed value of $$T_1$$ and several different values of $$T$$.

An alternative way of interpreting eq. (4.1) is as samples of an envelope function,
specifically,

$\tag{4.2}Ta_k=\left.\frac{2\sin\omega{T_1}}{\omega}\right|_{\omega=k\omega_0}$

That is, with $$\omega$$ thought of as a continuous variable, the function $$(2\sin\omega{T_1})/\omega$$ represents the envelope of $$Ta_k$$, and the coefficients $$a_k$$ are simply equally spaced samples of this envelope. Also, for fixed $$T_1$$, the envelope of $$Ta_k$$ is independent of $$T$$.

In Figure 4.2, we again show the Fourier series coefficients for the periodic square wave, but this time as samples of the envelope of $$Ta_k$$, as specified in eq. (4.2).

From the figure, we see that as $$T$$ increases, or equivalently, as the fundamental frequency $$\omega_0=2\pi/T$$ decreases, the envelope is sampled with a closer and closer spacing.

As $$T$$ becomes arbitrarily large, the original periodic square wave approaches a rectangular pulse (i.e., all that remains in the time domain is an aperiodic signal corresponding to one period of the square wave).

Also, the Fourier series coefficients, multiplied by $$T$$, become more and more closely spaced samples of the envelope, so that in some sense (which we will specify shortly) the set of Fourier series coefficients approaches the envelope function as $$T\rightarrow\infty$$.

This example illustrates the basic idea behind Fourier's development of a representation for aperiodic signals.

Specifically, we think of an aperiodic signal as the limit of a periodic signal as the period becomes arbitrarily large, and we examine the limiting behavior of the Fourier series representation for this signal.

In particular, consider a signal $$x(t)$$ that is of finite duration. That is, for some number $$T_1$$, $$x(t)=0$$ if $$|t|\gt{T_1}$$, as illustrated in Figure 4.3(a).

From this aperiodic signal, we can construct a periodic signal $$\tilde{x}(t)$$ for which $$x(t)$$ is one period, as indicated in Figure 4.3(b).

As we choose the period $$T$$ to be larger, $$\tilde{x}(t)$$ is identical to $$x(t)$$ over a longer interval, and as $$T\rightarrow\infty$$, $$\tilde{x}(t)$$ is equal to $$x(t)$$ for any finite value of $$t$$.

Let us now examine the effect of this on the Fourier series representation of $$\tilde{x}(t)$$. Rewriting eqs. (3.38) and (3.39) [refer to the Fourier series representation of continuous-time periodic signals tutorial] here for convenience, with the integral in eq. (3.39) carried out over the interval $$-T/2\le{t}\le{T/2}$$, we have

$\tag{4.3}\tilde{x}(t)=\sum_{k=-\infty}^{+\infty}a_ke^{jk\omega_0t}$

$\tag{4.4}a_k=\frac{1}{T}\int_{-T/2}^{T/2}\tilde{x}(t)e^{-jk\omega_0t}\text{d}t$

where $$\omega_0=2\pi/T$$. Since $$\tilde{x}(t)=x(t)$$ for $$|t|\lt{T/2}$$, and also, since $$x(t)=0$$ outside this interval, eq. (4.4) can be rewritten as

$a_k=\frac{1}{T}\int_{-T/2}^{T/2}x(t)e^{-jk\omega_0t}\text{d}t=\frac{1}{T}\int_{-\infty}^{+\infty}x(t)e^{-jk\omega_0t}\text{d}t$

Therefore, defining the envelope $$X(j\omega)$$ of $$Ta_k$$ as

$\tag{4.5}X(j\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega{t}}\text{d}t$

We have, for the coefficients $$a_k$$,

$\tag{4.6}a_k=\frac{1}{T}X(jk\omega_0)$

Combining eqs. (4.6) and (4.3), we can express $$\tilde{x}(t)$$ in terms of $$X(j\omega)$$ as

$\tilde{x}(t)=\sum_{k=-\infty}^{+\infty}\frac{1}{T}X(jk\omega_0)e^{jk\omega_0t}$

or equivalently, since $$2\pi/T=\omega_0$$,

$\tag{4.7}\tilde{x}(t)=\frac{1}{2\pi}\sum_{k=-\infty}^{+\infty}X(jk\omega_0)e^{jk\omega_0t}\omega_0$

As $$T\rightarrow\infty$$, $$\tilde{x}(t)$$ approaches $$x(t)$$, and consequently, in the limit eq. (4.7) becomes a representation of $$x(t)$$.

Furthermore, $$\omega_0\rightarrow0$$ as $$T\rightarrow\infty$$, and the right-hand side of eq. (4.7) passes to an integral. This can be seen by considering the graphical interpretation of the equation, illustrated in Figure 4.4.

Each term in the summation on the right-hand side of eq. (4.7) is the area of a rectangle of height $$X(jk\omega_0)e^{jk\omega_0t}$$ and width $$\omega_0$$. (Here, $$t$$ is regarded as fixed.)

As $$\omega_0\rightarrow0$$, the summation converges to the integral of $$X(j\omega)e^{j\omega{t}}$$. Therefore, using the fact that $$\tilde{x}(t)\rightarrow{x}(t)$$ as $$T\rightarrow\infty$$, we see that eqs. (4.7) and (4.5) respectively become

$\tag{4.8}x(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}X(j\omega)e^{j\omega{t}}\text{d}\omega$

and

$\tag{4.9}X(j\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega{t}}\text{d}t$

Equations (4.8) and (4.9) are referred to as the Fourier transform pair, with the function $$X(j\omega)$$ referred to as the Fourier Transform or Fourier integral of $$x(t)$$ and eq. (4.8) as the inverse Fourier transform equation.

The synthesis equation (4.8) plays a role for aperiodic signals similar to that of eq. (3.38) [refer to the Fourier series representation of continuous-time periodic signals tutorial] for periodic signals, since both represent a signal as a linear combination of complex exponentials.

For periodic signals, these complex exponentials have amplitudes $$\{a_k\}$$. as given by eq. (3.39) [refer to the Fourier series representation of continuous-time periodic signals tutorial], and occur at a discrete set of harmonically related frequencies $$k\omega_0$$, $$k=0,\pm1,\pm2,\ldots.$$

For aperiodic signals, the complex exponentials occur at a continuum of frequencies and, according to the synthesis equation (4.8), have "amplitude" $$X(j\omega)(\text{d}\omega/2\pi)$$.

In analogy with the terminology used for the Fourier series coefficients of a periodic signal, the transform $$X(j\omega)$$ of an aperiodic signal $$x(t)$$ is commonly referred to as the spectrum of $$x(t)$$, as it provides us with the information needed for describing $$x(t)$$ as a linear combination (specifically, an integral) of sinusoidal signals at different frequencies.

Based on the above development, or equivalently on a comparison of eq. (4.9) and eq. (3.39) [refer to the Fourier series representation of continuous-time periodic signals tutorial], we also note that the Fourier coefficients $$a_k$$ of a periodic signal $$\tilde{x}(t)$$ can be expressed in terms of equally spaced samples of the Fourier transform of one period of $$\tilde{x}(t)$$.

Specifically, suppose that $$\tilde{x}(t)$$ is a periodic signal with period $$T$$ and Fourier coefficients $$a_k$$. Let $$x(t)$$ be a finite-duration signal that is equal to $$\tilde{x}(t)$$ over exactly one period—say, for $$s\le{t}\le{s+T}$$ for some value of $$s$$—and that is zero otherwise.

Then, since eq. (3.39) [refer to the Fourier series representation of continuous-time periodic signals tutorial] allows us to compute the Fourier coefficients of $$\tilde{x}(t)$$ by integrating over any period, we can write

$a_k=\frac{1}{T}\int_{s}^{s+T}\tilde{x}(t)e^{-jk\omega_0t}\text{d}t=\frac{1}{T}\int_{s}^{s+T}x(t)e^{-jk\omega_0t}\text{d}t$

Since $$x(t)$$ is zero outside the range $$s\le{t}\le{s+T}$$ we can equivalently write

$a_k=\frac{1}{T}\int_{-\infty}^{+\infty}x(t)e^{-jk\omega_0t}\text{d}t$

Comparing with eq. (4.9) we conclude that

$\tag{4.10}a_k=\left.\frac{1}{T}X(j\omega)\right|_{\omega=k\omega_0}$

where $$X(j\omega)$$ is the Fourier transform of $$x(t)$$.

Equation 4.10 states that the Fourier coefficients of $$\tilde{x}(t)$$ are proportional to samples of the Fourier transform of one period of $$\tilde{x}(t)$$.

## 2. Convergence of Fourier Transforms

Although the argument we used in deriving the Fourier transform pair assumed that $$x(t)$$ was of arbitrary but finite duration, eqs. (4.8) and (4.9) remain valid for an extremely broad class of signals of infinite duration.

In fact, our derivation of the Fourier transform suggests that a set of conditions like those required for the convergence of Fourier series should also apply here [refer to the convergence of Fourier series part in the Fourier series representation of continuous-time periodic signals tutorial], and indeed, that can be shown to be the case.

Specifically, consider $$X(j\omega)$$ evaluated according to eq. (4.9), and let $$\hat{x}(t)$$ denote the signal obtained by using $$X(j\omega)$$ in the right-hand side of eq. (4.8). That is,

$\hat{x}(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}X(j\omega)e^{j\omega{t}}\text{d}\omega$

What we would like to know is when eq. (4.8) is valid [i.e., when is $$\hat{x}(t)$$ a valid representation of the original signal $$x(t)$$?].

If $$x(t)$$ has finite energy, i.e., if it is square integrable, so that

$\tag{4.11}\int_{-\infty}^{+\infty}|x(t)|^2\text{d}t\lt\infty$

then we are guaranteed that $$X(j\omega)$$ is finite [i.e., eq. (4.9) converges] and that, with $$e(t)$$ denoting the error between $$\hat{x}(t)$$ and $$x(t)$$ [i.e., $$e(t)=\hat{x}(t)-x(t)$$],

$\tag{4.12}\int_{-\infty}^{+\infty}|e(t)|^2\text{d}t=0$

Equations (4.11) and (4.12) are the aperiodic counterparts of eqs. (3.51) and (3.54) for periodic signals [refer to the Fourier series representation of continuous-time periodic signals tutorial]. Thus, in a manner similar to that for periodic signals, if $$x(t)$$ has finite energy, then, although $$x(t)$$ and its Fourier representation $$\hat{x}(t)$$ may differ significantly at individual values of $$t$$, there is no energy in their difference.

Just as with periodic signals, there is an alternative set of conditions which are sufficient to ensure that $$\hat{x}(t)$$ is equal to $$x(t)$$ for any $$t$$ except at a discontinuity, where it is equal to the average of the values on either side of the discontinuity.

These conditions, again referred to as the Dirichlet conditions, require that:

1. $$x(t)$$ be absolutely integrable; that is, $\tag{4.13}\int_{-\infty}^{+\infty}|x(t)|\text{d}t\lt\infty$
2. $$x(t)$$ have a finite number of maxima and minima within any finite interval.
3. $$x(t)$$ have a finite number of discontinuities within any finite interval. Furthermore, each of these discontinuities must be finite.

Therefore, absolutely integrable signals that are continuous or that have a finite number of discontinuities have Fourier transforms.

Although the two alternative sets of conditions that we have given are sufficient to guarantee that a signal has a Fourier transform, we will see in the next tutorial that periodic signals, which are neither absolutely integrable nor square integrable over an infinite interval, can be considered to have Fourier transforms if impulse functions are permitted in the transform.

This has the advantage that the Fourier series and Fourier transform can be incorporated into a common framework, which we will find to be very convenient in subsequent tutorials. Before examining the point further in the next tutorial, however, let us consider several examples of the Fourier transform.

## 3.  Examples of Continuous-Time Fourier Transforms

Example 4.1

Consider the signal

$x(t)=e^{-at}u(t)\qquad{a\gt0}$

From eq. (4.9),

$X(j\omega)=\int_0^{\infty}e^{-at}e^{-j\omega{t}}\text{d}t=\left.-\frac{1}{a+j\omega}e^{-(a+j\omega)t}\right|_0^\infty$

That is,

$X(j\omega)=\frac{1}{a+j\omega}\qquad{a\gt0}$

Since this Fourier transform is complex valued, to plot it as a function of $$\omega$$, we express $$X(j\omega)$$ in terms of its magnitude and phase:

$|X(j\omega)|=\frac{1}{\sqrt{a^2+\omega^2}},\qquad\measuredangle{X(j\omega)}=-\tan^{-1}\left(\frac{\omega}{a}\right)$

Each of these components is sketched in Figure 4.5.

Note that if $$a$$ is complex rather than real, then $$x(t)$$ is absolutely integrable as long as $$\mathcal{Re}\{a\}\gt0$$, and in this case the preceding calculation yields the same form for $$X(j\omega)$$. That is,

$X(j\omega)=\frac{1}{a+j\omega},\qquad\mathcal{Re}\{a\}\gt0$

Example 4.2

Let

$x(t)=e^{-a|t|},\qquad{a\gt0}$

This signal is sketched in Figure 4.6. The Fourier transform of this signal is

\begin{align}X(j\omega)&=\int_{-\infty}^{+\infty}e^{-a|t|}e^{-j\omega{t}}\text{d}t=\int_{-\infty}^0e^{at}e^{-j\omega{t}}\text{d}t+\int_{0}^{\infty}e^{-at}e^{-j\omega{t}}\text{d}t\\&=\frac{1}{a-j\omega}+\frac{1}{a+j\omega}\\&=\frac{2a}{a^2+\omega^2}\end{align}

In this case $$X(j\omega)$$ is real, and it is illustrated in Figure 4.7.

Example 4.3

Now let us determine the Fourier transform of the unit impulse

$\tag{4.14}x(t)=\delta(t)$

Substituting into eq. (4.9) yields

$\tag{4.15}X(j\omega)=\int_{-\infty}^{+\infty}\delta(t)e^{-j\omega{t}}\text{d}t=1$

That is, the unit impulse has a Fourier transform consisting of equal contributions at all frequencies.

Example 4.4

Consider the rectangular pulse signal

$\tag{4.16}x(t)=\begin{cases}1,\qquad|t|\lt{T_1}\\0,\qquad|t|\gt{T_1}\end{cases}$

as shown in Figure 4.8(a). Applying eq. (4.9), we find that the Fourier transform of this signal is

$\tag{4.17}X(j\omega)=\int_{-T_1}^{T_1}e^{-j\omega{t}}\text{d}t=2\frac{\sin\omega{T_1}}{\omega}$

as sketched in Figure 4.8(b).

As we discussed at the beginning of this tutorial, the signal given by eq. (4.16) can be thought of as the limiting form of a periodic square wave as the period becomes arbitrarily large.

Therefore, we might expect that the convergence of the synthesis equation for this signal would behave in a manner similar to that observed in Example 3.5 [refer to the Fourier series representation of continuous-time periodic signals tutorial] for the square wave. This is, in fact, the case.

Specifically, consider the inverse Fourier transform for the rectangular pulse signal:

$\hat{x}(t)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}2\frac{\sin\omega{T_1}}{\omega}e^{j\omega{t}}\text{d}\omega$

Then, since $$x(t)$$ is square integrable,

$\int_{-\infty}^{+\infty}|x(t)-\hat{x}(t)|^2\text{d}t=0$

Furthermore, because $$x(t)$$ satisfies the Dirichlet conditions, $$\hat{x}(t)=x(t)$$, except at the points of discontinuity, $$t=\pm{T_1}$$, where $$\hat{x}(t)$$ converges to $$1/2$$, which is the average of the values of $$x(t)$$ on both sides of the discontinuity.

In addition, the convergence of $$\hat{x}(t)$$ to $$x(t)$$ exhibits the Gibbs phenomenon, much as was illustrated for the periodic square wave in Figure 3.9 [refer to the Fourier series representation of continuous-time periodic signals tutorial].

Specifically, in analogy with the finite Fourier series approximation, eq. (3.47) [refer to the Fourier series representation of continuous-time periodic signals tutorial], consider the following integral over a finite-length interval of frequencies:

$\frac{1}{2\pi}\int_{-W}^W2\frac{\sin\omega{T_1}}{\omega}e^{j\omega{t}}\text{d}\omega$

As $$W\rightarrow\infty$$, this signal converges to $$x(t)$$ everywhere, except at the discontinuities. Moreover, the signal exhibits ripples near the discontinuities. The peak amplitude of these ripples does not decrease as $$W$$ increases, although the ripples do become compressed toward the discontinuity, and the energy in the ripples converges to zero.

Example 4.5

Consider the signal $$x(t)$$ whose Fourier transform is

$\tag{4.18}X(j\omega)=\begin{cases}1,\qquad|\omega|\lt{W}\\0,\qquad|\omega|\gt{W}\end{cases}$

This transform is illustrated in Figure 4.9(a). Using the synthesis equation (4.8), we can then determine

$\tag{4.19}x(t)=\frac{1}{2\pi}\int_{-W}^We^{j\omega{t}}\text{d}\omega=\frac{\sin{Wt}}{\pi{t}}$

which is depicted in Figure 4.9(b).

Comparing Figures 4.8 and 4.9 or, equivalently, eqs. (4.16) and (4.17) with eqs. (4.18) and (4.19), we see an interesting relationship.

In each case, the Fourier transform pair consists of a function of the form $$(\sin{a\theta})/b\theta$$ and a rectangular pulse. However, in Example 4.4, it is the signal $$x(t)$$ that is a pulse, while in Example 4.5, it is the transform $$X(j\omega)$$.

The special relationship that is apparent here is a direct consequence of the duality property for Fourier transforms, which we discuss in detail in a later tutorial.

Functions of the form given in eqs. (4.17) and (4.19) arise frequently in Fourier analysis and in the study of LTI systems and are referred to as sinc functions. A commonly used precise form for the sinc function is

$\tag{4.20}\text{sinc}(\theta)=\frac{\sin\pi\theta}{\pi\theta}$

The sinc function is plotted in Figure 4.10. Both of the signals in eqs. (4.17) and (4.19) can be expressed in terms of the sinc function:

$\frac{2\sin\omega{T_1}}{\omega}=2T_1\text{sinc}\left(\frac{\omega{T_1}}{\pi}\right)$

$\frac{\sin{Wt}}{\pi{t}}=\frac{W}{\pi}\text{sinc}\left(\frac{Wt}{\pi}\right)$

Finally, we can gain insight into one other property of the Fourier transform by examining Figure 4.9, which we have redrawn as Figure 4.11 for several different values of $$W$$.

From this figure, we see that as $$W$$ increases, $$X(j\omega)$$ becomes broader, while the main peak of $$x(t)$$ at $$t=0$$ becomes higher and the width of the first lobe of this signal (i.e., the part of the signal for $$|t|\lt\pi/W$$) becomes narrower.

In fact, in the limit as $$W\rightarrow\infty$$, $$X(j\omega)=1$$ for all $$\omega$$, and consequently, from Example 4.3, we see that $$x(t)$$ in eq. (4.19) converges to an impulse as $$W\rightarrow\infty$$.

The behavior depicted in Figure 4.11 is an example of the inverse relationship that exists between the time and frequency domains, and we can see a similar effect in Figure 4.8, where an increase in $$T_1$$ broadens $$x(t)$$ but makes $$X(j\omega)$$ narrower.

In a later tutorial, we provide an explanation of this behavior in the context of the scaling property of the Fourier transform.

The next tutorial discusses about Fourier transform for continuous-time periodic signals.