Classical Signal Theory

Abstract

In this chapter, we outline the Signal Theory as it is usually presented in textbooks, where each class of signals is separately developed. “Separately” means that, for each class, definitions and development are presented independently. For convenience, we call this approach the Classical Signal Theory in contrast to the approach of this book, the Unified Signal Theory. The classes of signals we will consider are:

• 1. Aperiodic continuous-time signals,

• 2. Periodic continuous-time signals,

• 3. Periodic discrete-time signals,

• 4. Periodic discrete-time signals.

These classes are introduced to give readers the classical background, before dealing with the Unified Theory.

Appendix: Fourier Transform of the Signum Signal sgn(t)

The Fourier transform definition (2.55b) yields

$$\int_{-\infty}^{+\infty}\mathop{\mathrm{sgn}}\nolimits (t)\mathrm{e}^{-\mathrm{i}2\pi ft}\, \mathrm{d} t =\frac{2}{\mathrm{i}} \int_0^{\infty} \sin2\pi ft\, \mathrm{d} t .$$

These integrals do not exist. However, sgn (t) can be expressed as the inverse Fourier transform of the function 1/(iπf), namely

$$\mathop{\mathrm{sgn}}\nolimits (t) = \int_{-\infty}^{+\infty }\frac{1}{\mathrm{i}\pi f}\mathrm{e}^{\mathrm{i}2\pi ft}\, \mathrm{d} f\,\buildrel \varDelta \over= \, x(t) $$

(2.113a)

provided that the integral is interpreted as a Cauchy principal value, i.e.,

$$x(t)= \int_{-\infty}^{+\infty}\frac{1}{\mathrm{i}\pi f}\mathrm{e}^{\mathrm{i}2\pi ft}\, \mathrm{d} f =\lim_{F\rightarrow\infty} \int_{-F}^F\frac{1}{\mathrm{i}\pi f}\mathrm{e}^{\mathrm{i}2\pi ft}\, \mathrm{d} f . $$

(2.113b)

Using Euler’s formula, we get

$$x(t)=\int_{-\infty}^{+\infty}\frac{1}{\mathrm{i}\pi f}\cos(2\pi ft)\,\mathrm{d} f +\int_{-\infty}^{+\infty}\frac{1}{\pi f}\sin(2\pi ft)\,\mathrm{d} f$$

where the integrand (1/i2πf)cos (2πf) in an odd function of f, and therefore the integral is zero. Then

$$x(t)=\int_{-\infty}^{+\infty}\frac{\sin(2\pi ft)}{\pi f}\,\mathrm {d} f .$$

Now, for t=0 we find x(0)=0. For t≠0, letting

$$2ft \to u,\qquad\mathrm{d} f \to\frac{\mathrm{d} u}{2t},$$

we obtain

$$x(t)=\begin{cases}\int_{-\infty}^{+\infty}\frac{\sin(\pi u)}{\pi u}\,\mathrm{d} u&\hbox{for}\ t>0;\cr \int_{+\infty}^{-\infty}\frac{\sin(\pi u)}{\pi u}\,\mathrm{d} u =-\int_{-\infty}^{+\infty}\frac{\sin(\pi u)}{\pi u}\,\mathrm{d} u&\hbox{for}\ t<0.\end{cases}$$

It remains to evaluate the integral

$$I=\int_{-\infty}^{+\infty}\frac{\sin(\pi u)}{\pi u}\,\mathrm{d} u=\int_{-\infty}^{+\infty}\mathop{\mathrm{sinc}}\nolimits (u)\,\mathrm{d} u .$$

To this end, we use the rule (2.61) giving for a Fourier pair s(t),S(f)

$$\mathop{\mathrm{area}}\nolimits (S)=\int_{-\infty}^{+\infty}S(f)\,\mathrm{d} f=s(0)$$

with s(t)=rect(t), S(f)=sinc(f) (see (2.64a, 2.64b)). Hence, we obtain

$$\int_{-\infty}^{+\infty}\mathop{\mathrm{sinc}}\nolimits (f)\,\mathrm{d} f=s(0)=\mathop{\mathrm{rect}}\nolimits (0)=1.$$

Combination of the above results gives x(t)=sgn(t).