Sources:

B. P. Lathi & Roger Green. (2018). Chapter 7: Continuous-Time Signal Analysis. Signal Processing and Linear Systems (3nd ed., pp. 699-708). Oxford University Press.

The Fourier transform

For a signal \(x(t)\), its Fourier transform is defined by \[ \begin{equation} \label{eq7_9} \color{red} {X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j \omega t} d t} \end{equation} \] The signal is said to be the inverse Fourier transform of \(X(\omega)\). It can be shown that

\[ \begin{equation} \label{eq7_10} \color{blue} {x(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) e^{j \omega t} d \omega} \end{equation} \] It is helpful to keep in mind that the Fourier integral in \(\eqref{eq7_10}\) is of the nature of a Fourier series with fundamental frequency \(\Delta \omega\) approaching zero [\(\eqref{eq7_7}\)]. Therefore, most of the discussion and properties of Fourier series apply to the Fourier transform as well. The transform \(X(\omega)\) is the frequency-domain specification of \(x(t)\).

The same information is conveyed by the statement that \(x(t)\) and \(X(\omega)\) are a Fourier transform pair. Symbolically, this statement is expressed as \[ X(\omega)=\mathcal{F}[x(t)] \quad \text { and } \quad x(t)=\mathcal{F}^{-1}[X(\omega)] \] or \[ x(t) \Longleftrightarrow X(\omega) \]

We can plot the spectrum \(X(\omega)\) as a function of \(\omega\). Since \(X(\omega)\) is complex, we have both amplitude and angle (or phase) spectra \[ X(\omega)=|X(\omega)| e^{j \angle X(\omega)} \] in which \(|X(\omega)|\) is the amplitude and \(\angle X(\omega)\) is the angle (or phase) of \(X(\omega)\). # Derivation of the Fourier transform

Applying a limiting process, we now show that an aperiodic signal can be expressed as a continuous sum (integral) of everlasting exponentials. To represent an aperiodic signal \(x(t)\) such as the one depicted in Fig. 7.1a by everlasting exponentials, let us construct a new periodic signal \(x_{T_0}(t)\) formed by repeating the signal \(x(t)\) at intervals of \(T_0\) seconds, as illustrated in Fig. 7.1b.

Figure 7.1

The period \(T_0\) is made long enough to avoid overlap between the repeating pulses. The periodic signal \(x_{T_0}(t)\) can be represented by an exponential Fourier series. If we let \(T_0 \rightarrow \infty\), the pulses in the periodic signal repeat after an infinite interval and, therefore, \[ \lim _{T_0 \rightarrow \infty} x_{T_0}(t)=x(t) \] Thus, the Fourier series representing \(x_{T_0}(t)\) will also represent \(x(t)\) in the limit \(T_0 \rightarrow \infty\). The exponential Fourier series for \(x_{T_0}(t)\) is given by \[ \begin{equation} \label{eq7_1} x_{T_0}(t)=\sum_{n=-\infty}^{\infty} D_n e^{j n \omega_0 t} \end{equation} \] where \(\omega_0=\frac{2 \pi}{T_0}\) and \[ \begin{equation} \label{eq7_2} D_n=\frac{1}{T_0} \int_{-T_0 / 2}^{T_0 / 2} x_{T_0}(t) e^{-j n \omega_0 t} d t \end{equation} \]

Observe that integrating \(x_{T_0}(t)\) over \(\left(-T_0 / 2, T_0 / 2\right)\) is the same as integrating \(x(t)\) over \((-\infty, \infty)\). Therefore, \(\eqref{eq7_2}\) can be expressed as \[ \begin{equation} \label{eq7_3} D_n=\frac{1}{T_0} \int_{-\infty}^{\infty} x(t) e^{-j n \omega_0 t} d t \end{equation} \]

It is interesting to see how the nature of the spectrum changes as \(T_0\) increases. To understand this fascinating behavior, let us define \(X(\omega)\), a continuous function of \(\omega\), as \[ \begin{equation} \label{eq7_4} X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j \omega t} d t \end{equation} \]

A glance at Eqs. (7.3) and (7.4) shows that \[ \begin{equation} \label{eq7_5} D_n=\frac{1}{T_0} X\left(n \omega_0\right) \end{equation} \]

Substitution of \(\eqref{eq7_5}\) in \(\eqref{eq7_1}\) yields \[ \begin{equation} \label{eq7_6} x_{T_0}(t)=\sum_{n=-\infty}^{\infty} \frac{X\left(n \omega_0\right)}{T_0} e^{j n \omega_0 t} \end{equation} \]

As \(T_0 \rightarrow \infty, \omega_0\) becomes infinitesimal \(\left(\omega_0 \rightarrow 0\right.\) ). Hence, we shall replace \(\omega_0\) by a more appropriate notation, \(\Delta \omega\). In terms of this new notation, \(\omega_0=\frac{2 \pi}{T_0}\) becomes \[ \Delta \omega=\frac{2 \pi}{T_0} \] and \(\eqref{eq7_6}\) becomes \[ x_{T_0}(t)=\sum_{n=-\infty}^{\infty}\left[\frac{X(n \Delta \omega) \Delta \omega}{2 \pi}\right] e^{(j n \Delta \omega) t} \]

This equation shows that \(x_{T_0}(t)\) can be expressed as a sum of everlasting exponentials of frequencies \(0, \pm \Delta \omega, \pm 2 \Delta \omega, \pm 3 \Delta \omega, \ldots\) (the Fourier series). The amount of the component of frequency \(n \Delta \omega\) is \([X(n \Delta \omega) \Delta \omega] / 2 \pi\). In the limit as \(T_0 \rightarrow \infty, \Delta \omega \rightarrow 0\) and \(x_{T_0}(t) \rightarrow x(t)\). Therefore, \[ x(t)=\lim _{T_0 \rightarrow \infty} x_{T_0}(t)=\lim _{\Delta \omega \rightarrow 0} \frac{1}{2 \pi} \sum_{n=-\infty}^{\infty} X(n \Delta \omega) e^{(j n \Delta \omega) t} \Delta \omega \] the sumon the right-hand side of which can be viewed as the area under the function \(X(\omega) e^{j \omega t}\), as illustrated in Fig. 7.3.

Figure 7.3

Therefore, \[ x(t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} X(\omega) e^{j \omega t} d \omega \]

The integral on the right-hand side is called the Fourier integral. We have now succeeded in representing an aperiodic signal \(x(t)\) by a Fourier integral (rather than a Fourier series). This integral is basically a Fourier series (in the limit) with fundamental frequency \(\Delta \omega \rightarrow 0\), as seen from \[ x(t)=\lim _{T_0 \rightarrow \infty} x_{T_0}(t)=\lim _{\Delta \omega \rightarrow 0} \frac{1}{2 \pi} \sum_{n=-\infty}^{\infty} X(n \Delta \omega) e^{(j n \Delta \omega) t} \Delta \omega . \] The amount of the exponential \(e^{j n \Delta \omega t}\) is \(X(n \Delta \omega) \Delta \omega / 2 \pi\). Thus, the function \(X(\omega)\) given by \(\eqref{eq7_4}\) acts as a spectral function.

The conjugate symmetry property of the Fourier transform

According to \(\eqref{eq7_9}\), \[ X(-\omega)=\int_{-\infty}^{\infty} x(t) e^{j \omega t} d t \]

Taking the conjugates of both sides yields \[ x^*(t) \Longleftrightarrow X^*(-\omega) \]

This property is known as the conjugation property. Now, if \(x(t)\) is a real function of \(t\), then \(x(t)=x^*(t)\), and from the conjugation property, we find that \[ X(-\omega)=X^*(\omega) \]

This is the conjugate symmetry property of the Fourier transform, applicable to real \(x(t)\). Therefore, for real \(x(t)\), \[ |X(-\omega)|=|X(\omega)| \quad \text { and } \quad \angle X(-\omega)=-\angle X(\omega) \]

Thus, for real \(x(t)\), the amplitude spectrum \(|X(\omega)|\) is an even function, and the phase spectrum \(\angle X(\omega)\) is an odd function of \(\omega\). These results were derived earlier for the Fourier spectrum of a periodic signal [Eq. (6.22)] and should come as no surprise.