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 $$\begin{array}{}\text{(1)}& X(\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt\end{array}$$ The signal is said to be the inverse Fourier transform of $X(\omega )$. It can be shown that

$$\begin{array}{}\text{(2)}& x(t)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X(\omega )e^{j\omega t}d\omega \end{array}$$ It is helpful to keep in mind that the Fourier integral in $\text{(2)}$ is of the nature of a Fourier series with fundamental frequency $\mathrm{\Delta }\omega$ approaching zero [$\text{(???)}$]. 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 $$X(\omega )=\mathcal{F}[x(t)]\text{ and }x(t)={\mathcal{F}}^{-1}[X(\omega )]$$ or $$x(t)⟺X(\omega )$$

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 $$X(\omega )=|X(\omega )|e^{j\mathrm{\angle }X(\omega )}$$ in which $|X(\omega )|$ is the amplitude and $\mathrm{\angle }X(\omega )$ 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 $T_0\to \mathrm{\infty }$, the pulses in the periodic signal repeat after an infinite interval and, therefore, $$\underset{T_0\to \mathrm{\infty }}{lim}{x}_{T_0}(t)=x(t)$$ Thus, the Fourier series representing $x_{T_0}(t)$ will also represent $x(t)$ in the limit $T_0\to \mathrm{\infty }$. The exponential Fourier series for $x_{T_0}(t)$ is given by $$\begin{array}{}\text{(3)}& x_{T_0}(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{D}_n{e}^{jn{\omega }_{0}t}\end{array}$$ where ${\omega }_0=\frac{2\pi }{T_0}$ and $$\begin{array}{}\text{(4)}& D_n=\frac{1}{T_0}{\int }_{-T_0/2}^{T_0/2}{x}_{T_0}(t)e^{-jn{\omega }_{0}t}dt\end{array}$$

Observe that integrating $x_{T_0}(t)$ over $(-T_0/2,T_0/2)$ is the same as integrating $x(t)$ over $(-\mathrm{\infty },\mathrm{\infty })$. Therefore, $\text{(4)}$ can be expressed as $$\begin{array}{}\text{(5)}& D_n=\frac{1}{T_0}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-jn{\omega }_{0}t}dt\end{array}$$

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 $$\begin{array}{}\text{(6)}& X(\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{-j\omega t}dt\end{array}$$

A glance at Eqs. (7.3) and (7.4) shows that $$\begin{array}{}\text{(7)}& D_n=\frac{1}{T_0}X\left(n{\omega }_0\right)\end{array}$$

Substitution of $\text{(7)}$ in $\text{(3)}$ yields $$\begin{array}{}\text{(8)}& x_{T_0}(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\frac{X\left(n{\omega }_0\right)}{T_0}{e}^{jn{\omega }_{0}t}\end{array}$$

As $T_0\to \mathrm{\infty },{\omega }_0$ becomes infinitesimal $({\omega }_0\to 0$ ). Hence, we shall replace ${\omega }_0$ by a more appropriate notation, $\mathrm{\Delta }\omega$. In terms of this new notation, ${\omega }_0=\frac{2\pi }{T_0}$ becomes $$\mathrm{\Delta }\omega =\frac{2\pi }{T_0}$$ and $\text{(8)}$ becomes $$x_{T_0}(t)=\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}\left[\frac{X(n\mathrm{\Delta }\omega )\mathrm{\Delta }\omega }{2\pi }\right]e^{(jn\mathrm{\Delta }\omega )t}$$

This equation shows that $x_{T_0}(t)$ can be expressed as a sum of everlasting exponentials of frequencies $0,\pm \mathrm{\Delta }\omega ,\pm 2\mathrm{\Delta }\omega ,\pm 3\mathrm{\Delta }\omega ,\dots$ (the Fourier series). The amount of the component of frequency $n\mathrm{\Delta }\omega$ is $[X(n\mathrm{\Delta }\omega )\mathrm{\Delta }\omega ]/2\pi$. In the limit as $T_0\to \mathrm{\infty },\mathrm{\Delta }\omega \to 0$ and $x_{T_0}(t)\to x(t)$. Therefore, $$x(t)=\underset{T_0\to \mathrm{\infty }}{lim}{x}_{T_0}(t)=\underset{\mathrm{\Delta }\omega \to 0}{lim}\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}X(n\mathrm{\Delta }\omega )e^{(jn\mathrm{\Delta }\omega )t}\mathrm{\Delta }\omega$$ 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, $$x(t)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}X(\omega )e^{j\omega t}d\omega$$

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 $\mathrm{\Delta }\omega \to 0$, as seen from $$x(t)=\underset{T_0\to \mathrm{\infty }}{lim}{x}_{T_0}(t)=\underset{\mathrm{\Delta }\omega \to 0}{lim}\frac{1}{2\pi }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}X(n\mathrm{\Delta }\omega )e^{(jn\mathrm{\Delta }\omega )t}\mathrm{\Delta }\omega .$$ The amount of the exponential $e^{jn\mathrm{\Delta }\omega t}$ is $X(n\mathrm{\Delta }\omega )\mathrm{\Delta }\omega /2\pi$. Thus, the function $X(\omega )$ given by $\text{(6)}$ acts as a spectral function.

The conjugate symmetry property of the Fourier transform

According to $\text{(1)}$, $$X(-\omega )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}x(t)e^{j\omega t}dt$$

Taking the conjugates of both sides yields $$x^{\ast }(t)⟺X^{\ast }(-\omega )$$

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

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

Thus, for real $x(t)$, the amplitude spectrum $|X(\omega )|$ is an even function, and the phase spectrum $\mathrm{\angle }X(\omega )$ 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.