### Aperiodic Signal Representation by Fourier Integral

An aperiodic function will never repeat, although technically speaking an aperiodic function can be considered similar to a periodic function with an infinite period. In order to show that an aperiodic signal can be expressed as a continuous sum (or integral) of infinite exponentials, a limiting process is applied. In order to represent an aperiodic signal f(t), such as the signal shown in Fig. 1.1a by infinite exponential signals, a new periodic signal $$f_{T_{0}}$$ must be formed by repeating the aperiodic signal f(t) every T0 seconds, shown in Fig. 1.1b. The period is made just long enough to not overlap between each repeating pulse. This periodic signal $$f_{T_{0}}$$ is represented by an exponential Fourier series. By letting $$T_{0} \rightarrow \infty$$, the pulses in the periodic signal repeat themselves after an infinite interval, therefore:

$$\lim_{T_{0}\rightarrow \infty } f_{T_{0}}(t) = f(t)$$

The Fourier series representing $$f_{T_{0}}$$ will thus also represent f(t) in the limit $$T_{0} \rightarrow \infty$$. The exponential Fourier series can be represented for $$f_{T_{0}}$$ as follows

$$f_{T_{0}}(t) = \sum_{n = - \infty }^{\infty }D_{n}e^{jn\omega _{0}t}$$     (1.1)

by which

$$D_{n} = \frac{1}{T_{0}}\int_{-\frac{T_{0}}{2}}^{\frac{T_{0}}{2}}f_{T_{0}}(t)e^{-jn\omega _{0}t}dt$$     (1.2a)

FIGURE 1.1. Construction of a periodic signal by periodic extension of f(t)

Figure 1.1 represents the construction of a periodic signal by periodic extension of f(t)

and

$$\omega _{0} = \frac{2\pi }{T_{0}}$$     (1.2b)

Figure 1.1a and 1.1b show that integrating $$f_{T_{0}}$$ over $$-\frac{T_{0}}{2},\frac{T_{0}}{2}$$ is exactly the same as if you were to integrate over $$\left (-\infty , \infty \right )$$. Simplifying the integration bounds, Eq 3.2a is now expressed by

$$D_{n} = \frac{1}{T_{0}}\int_{-\infty }^{\infty }f_{T_{0}}(t)e^{-jn\omega _{0}t}dt$$        (1.2c)

An interesting phenomenon is that the spectrum changes at T0 increases. To better understand this odd behavior, $$F(\omega )$$ is defined as a continuous function of $$\omega$$, as

$$F(\omega)= \int_{-\infty }^{\infty} f(t)e^{-j\omega t} dt$$     (1.3)

The last two equations above show that

$$D_{n} = \frac{1}{T_{0}}F(n\omega _{0})$$     (1.4)

What this shows is that the Fourier coefficients Dn are (1/T0 times) the samples of $$F(\omega )$$ spaced uniformly at intervals of $$\omega_{0}$$ rad/s, as shown in Fig. 1.2a. For simplicity, Dn and $$F(\omega )$$ are assumed to be real in Fig. 1.2.  Letting $$T_{0} \rightarrow \infty$$ by doubling T0 repeatedly, halves the fundamental frequency $$\omega_{0}$$; this operation is used so that there are twice as many components (or samples) in the spectrum. Consequently, by doubling T0 , the envelope $$(\frac{1}{T_{0}})F(\omega)$$ is halved, shown in Fig 1.2b. If T0 is doubled over and over again, the spectrum will become denser while its magnitude becomes smaller. Nothing that in the limits $$T_{0} \rightarrow \infty$$, $$\omega_{0} \rightarrow \infty$$, and $$D_{n} \rightarrow \infty$$, the relative shape of the envelope is kept the same. This means that the spectrum must be so dense that the spectral components are spaced at zero (or infinitesimal) level! Simultaneously, the amplitude of each component is also zero. This may seem peculiar at first glance; however, it will be shown that these are classic characteristics of a very familiar phenomenon.  By substituting Eq. 1.4 in Eq. 1.1, the following sum yields

$$f_{T_{0}}(t) = \sum_{n = - \infty }^{\infty}\frac{F(n\omega _{0})}{T_{0}}e^{jn\omega_{0}}$$     (1.5)

Here as $$T_{0} \rightarrow \infty$$, $$\omega_{0}$$ becomes extremely small $$(\omega_{0} \rightarrow 0)$$. Due to this limit, a more appropriate notation will replace $$\omega_{0}$$, $$\Delta \omega$$. With this new notion, Eq. 1.2b is now written as

$$\Delta \omega = \frac{2\pi }{T_{0}}$$

and Eq. 1.5 is now written as

$$f_{T_{0}}(t) = \sum_{n = - \infty}^{\infty} \left [ \frac{F(n\Delta \omega)\Delta \omega}{2\pi } \right ]e^{()jn\Delta \omega)t}$$     (1.6a)

Here, Eq. 1.6a shows that $$f_{T_{0}}(t))$$ may be expressed in terms of a sum of infinite exponentials with frequencies $$0, \pm \Delta \omega, \pm 2\Delta \omega, \pm 3\Delta \omega, ...$$, which is the Fourier series. In the limit as $$T_{0} \rightarrow \infty$$, $$\omega_{0} \rightarrow \infty$$, and $$D_{n} \rightarrow \infty$$, the amount of the component of frequency $$n\Delta\omega$$ is $$[F(n\Delta\omega)\Delta\omega]/2\pi$$. Thus,

$$f(t) = \lim_{T_{0}\rightarrow \infty}f_{T_{0}}(t) = \lim_{\Delta\omega \rightarrow 0} \frac{1}{2\pi}\sum_{n = - \infty}^{\infty} F(n\Delta\omega)e^{(nj\Delta\omega)t}\Delta\omega$$     (1.6b)

On the right side of Eq 1.6b, the sum is the area under the function $$F(\omega)e^{j\omega t}$$, as shown in Fig. 1.3. Thus,

$$f(t) = \frac{1}{2\pi} \int_{- \infty}^{\infty} F(\omega)e^{j\omega t}d\omega$$     (1.7)

FIGURE 1.2 Change in the Fourier spectrum when the period T0 in Fig 1.1 doubles.

FIGURE 1.3 The Fourier series becomes the Fourier integral in the limit as $$T_{0} \rightarrow \infty$$

The integral on the right side is known as the Fourier integral. This is the representation of an aperiodic signal f(t) by a Fourier integral, rather than a Fourier series. This Fourier integral is essentially a Fourier series (only in the limit) with fundamental frequency $$\Delta\omega \rightarrow 0$$, as shown in Eq. 1.6.

Assigning $$F(\omega)$$ as the direct  Fourier transform of f(t), and f(t) as the inverse Fourier transform of $$F(\omega)$$. Another way to convey this statement is by a Fourier transform pair as stated below

$$F(\omega )=\mathcal{F}[f(t)] \text{ and } f(t) = \mathcal{F}[F(\omega)]$$

or

$$f(t) \Leftrightarrow F(\omega)$$

To summarize,

$$F(t) = \int_{-\infty}^{\infty}f(\omega)e^{j\omega t}d\omega$$     (1.8a)

and

$$f(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}F(\omega)e^{j\omega t}d\omega$$     (1.8b)

The spectrum of $$F(\omega)$$ can also be plotted as a function of $$\omega$$. Because $$F(\omega)$$ is complex, both the amplitude and angle spectra are as follows:

$$F(\omega) = \left | F(\omega) \right |e^{j\theta_{0}(\omega)}$$

### Conjugate Symmetry Property

From Eq. 1.8a, if f(t) is a real function of t, then $$F(\omega)$$ and $$F(-\omega)$$ are known to be complex conjugates shown below.

$$F(-\omega) = F^{*}(\omega)$$     (1.9)

Therefore,

$$\left | F(-\omega) \right | = \left | F(\omega) \right |$$     (1.10a)

$$\theta_{f}(-\omega) = -\theta_{f}(\omega)$$     (1.10b)

Consequently, for real f(t), the amplitude spectrum $$\left | F(\omega) \right |$$ is an even function, and the phase spectrum $$\theta_{f}(\omega)$$ is an odd function of $$\omega$$. Only for real f(t), this property known as the conjugate symmetry property holds true.  This transform of $$F(\omega)$$ is the frequency domain specification of f(t).

Example

Find the Fourier transform of $$e^{-at}u(t)$$

By definition of Eq. 1.8a,

$$F(\omega)=\int_{-\infty }^{\infty }e^{-j\omega t}dt = \int_{0}^{\infty}e^{-(a+j\omega)t}dt = \frac{-1}{a+j\omega}e^{(a+j\omega)}t |^{\infty}_{0}$$

But $$\left | e^{-j\omega t} \right |= 1$$. Therefore, as $$t \rightarrow \infty, e^{-(a+j \omega)t} = e^{-at}e^{-j \omega t} = 0$$ if $$a>0$$. Therefore,

$$F(\omega) = \frac{1}{a+j \omega} \text{ } a>0$$

### Existence of the Fourier Transform

In the above example, it was shown that when < 0, the Fourier integral for $$e^{-at}u(t)$$ does not converge. Thus, the Fourier transform for $$e^{-at}u(t)$$ does not exist if < 0 (that is, exponentially growing). Observing from this example, not all signals are transformable. Any existence of the Fourier transform is assured for any f(t) that satisfies the Dirichlet conditions. The first of the conditions is as follows

$$\int_{-\infty}^{\infty}\left | f(t) \right | < \infty$$     (1.12)

In order to show this holds true, recall that $$\left | e^{-j\omega t} \right | = 1$$. Thus from Eq. 1.8a,

$$\left |F(\omega) \right | \leq \int_{-\infty}^{\infty} \left | f(t) \right |dt$$

By expressing $$a+j \omega$$ in the polar form as $$\sqrt{a^{2}+\omega^{2}}e^{j tan^{-1} (\frac{\omega}{a})}$$,

$$F(\omega) = \frac{1}{\sqrt{a^{2}+\omega^{2}}} e^{-j tan^{-1}(\frac{\omega}{a})}$$

Therefore

$$\left |F(\omega) \right | = \frac{1}{\sqrt{a^{2}+\omega^{2}}}$$

and

$$\theta_{f}(\omega) = -tan^{-1}(\frac{\omega}{a})$$

The amplitude spectrum of $$|F(\omega)|$$ and the phase spectrum $$\theta_{f}(\omega)$$ are shown in Figure 1.4b. Observe that $$|F(\omega)|$$ is an even function of $$\omega$$, and $$\theta_{f}(\omega)$$ is an odd function of $$\omega$$, as expected to be.

As long as condition 1.12 is satisfied, it shows that the existence of the Fourier transform is assured.

Linearity of the Fourier Transform

The Fourier transform can be considered linear if

$$f_{1}(t) \Leftrightarrow F_{1}(\omega)$$   and   $$f_{2}(t) \Leftrightarrow F_{2}(\omega)$$

then,

$$a_{1}f_{1}(t) + a_{2}f_{2}(t) \Leftrightarrow a_{1}F_{1}(\omega) + a_{2}F_{2}(\omega)$$      (1.13)

This result can be extended to any finite number of terms. This proof is trivial and follows exactly from Eq. 1.8a.

### Coming Up

As of now, you should have an understanding of what an aperiodic signal is and how it is represented by a Fourier Integral. By applying a limiting process, you should know how an aperiodic signal can be expressed as a continuous sum over everlasting exponentials, how the linearity of the Fourier Transform proof is satisfied, and how to find a Fourier transform using its spectra as well as the conjugate symmetry property. Next, an understanding of some useful functions, signal bandwidth, filtering (or interpolating), as well as the synthesis of a time-limited pulse signal.