Fourier transformations may be used to convert functions of time into functions of frequency. This procedure is used in analyzing data from a frequency response test. We consider periodic functions first, then non­periodic functions.

A periodic function that satisfies certain conditions! may be expressed as a Fourier series. The trigonometric form of the Fourier series is

A 00

f(t) = — Г2 + X cos comt + Bm sin comt (2.5.1)

Z m= 1

where com = Ъпк/Т is the angular frequency (radians/time) and T the period

of/(0-

The virtue of this sort of representation of the function is that certain orthogonality relations exist that facilitate the evaluation of the coefficients Ak and Bk. These are

0	if	k Ф m
772	if	k = m
0	if	k Ф m
T/2	if	k = m
0	if	k Ф m
0	if	k = m

(2.5.2) (2.5.3) (2.5.4)

t See Churchill (3).

tThese conditions, called the Dirichlet conditions, are that f(t) must (1) be single valued, (2) be finite, (3) have a finite number of maxima and minima, and (4) have at most a finite num­ber of finite discontinuities. It is clear that Dirichlet’s conditions are satisfied by signals of prac­tical importance.

The integration shown here has limits — T/2 to T/2 but any section of the function of duration T is suitable.

Cofficient Ak is evaluated by multiplying both sides of Eq. (2.5.1) by cos[(2kn/T)t] and integrating from — T/2 to T/2:

Therefore,

2 rT/2

= t /(0

і J — T/2

Example 2.5.1. Consider the square wave shown in Fig. 2.17. The coefficients Ak are given by

Ak = 0

The coefficients Bk are given by

Bk = —[1 — cos kn — cos kn + 1] kn

. 2n 1.6n 1 . 10я

sin—t + — sin—t + — sin—-t +
T З T 5 T

Note that only sin(o;tt) terms appear in the Fourier series representation of the square wave of Fig. 2.17. Any odd function [i. e.,/(t) = —/( —t)] has only sin(wkt) terms in its Fourier series. Even functions [i. e.,/(t) = /( —t)] have only cos(o>kt) terms. ■

Fig. 2.17. A square wave.

It is also possible to write the Fourier series in a more compact exponential form:

oo

f(t)= £ C*expO'(ott) (2.5.7)

к = — oo

Where

Q — 2 (Л к — jBk) C-k = j(Ak + jBk)

Example 2.5.2. Compute the complex Fourier coefficients for the square wave considered previously:

1 f° 1 ГТІ2

Ck = — exp (~ja)kt)dt+ — exp (-jcokt)dt

‘ J-T/2 t J0

2 _ 2[exp(jcokT/2) + exp(—jcokT/2)]

2

Use the Euler formulas to give

C =_L_

k jcokT

Since cok = 2nk/T,

Ck = —r[l — cos kn] jnk

0 if к is even

Ck = 2

—- if к is odd

(jnk

We can compare this with the value of Ck predicted from the trigonometric form by Eqs. (2.5.9) and (2.5.10).

Ck = j(Ak — jBk) = ^(0 — 0) = 0 if к is even

= i(A — jBk) = £[0 — j(4/knj] if к is odd

Thus, the value of Ck for odd values of к is

Ck = 2/jkn

which agrees with the previous results. For negative k, we get C-k = i(A + jBk) = £(0 + 0) = 0 if к is even

= i(A + }Bk) = £[0 + j(4/kn)] if к is odd

The value obtained for — к is then

C-k = -2/jkn

which also agrees with the previous results.

We have seen that there is a definite set of Fourier coefficients associated with a given periodic function and that there is likewise a definite periodic function associated with a given set of Fourier coefficients. This unique relationship between a periodic signal and its Fourier coefficients means that the signal may be completely characterized by its Fourier coefficients (harmonics). Since the Fourier coefficients are related to a specific frequency a>k, harmonic analysis is also called frequency-domain analysis.

The set of Fourier coefficient amplitudes, taken as a function of frequency, is called the amplitude spectrum of the periodic signal. We have seen pre­viously that the Fourier coefficients of a square wave have zero values except at certain frequencies. The amplitude spectrum of the square wave is shown in Fig. 2.18 for the complex and the trigonometric forms. Note that

the complex spectrum extends from — oo to +oo while the trigonometric form extends from 0 to oo.

Modified square waves and their amplitude spectra are shown in Fig. 2.19. Note that the amplitude of the harmonics becomes smaller, and the number of harmonics in a given frequency range becomes greater, as T

increases. In general, as the period increases, the harmonic amplitude diminishes and the number of frequencies increases in the same ratio:

C(co)T = constant (2.5.11)

N/T = constant (2.5.12)

where C(a>) is the Fourier coefficient at frequency to, N the number of harmonics in a given frequency range, and T the fundamental period. The fundamental period is the shortest time T for which f(t) = f(t + T). The signal is also periodic with period 2T, ЗT,…, but an analysis using these longer periods will not change the harmonic amplitude or frequency spacing of signal harmonics.

A periodic function becomes nonperiodic if the fundamental period is allowed to become infinite. We have observed that the harmonic amplitude decreases and the number of harmonics increases as the period increases. Thus the Fourier coefficients approach zero as T -> oo. However, we noted
that the product of the Fourier coefficient and the fundamental period is a constant. Therefore, for the complex form, let us consider the function CkT as T -> oo and denote it by F{jto):

ЛТ/2

F{ja>} = lim CkT = lim f(t)exp{-j(okt)dt

T -* oo T-*ooJ—j/2

(2.5.13)

This is the definition of the Fourier transform F(jw). The inverse transform is obtained by a similar limiting process:

/(0 = lim £ (CkT) ехр(М0(1/Л

T~* 00 к = — oo

We observe that the increment in со is given by

сok+1 — cok = 2 n/T = Acu

We can substitute this into the equation for/(t) and use the fact that limj.,,^ is the same as 1ітДш_0 to obtain

J oo 1 Л00

/(t) = — lim £ (CkT) exp(M0 Act) = — F(jco)eia>t doo (2.5.14)

2,71 A(0~* <X) QQ 271 щ) —————— QQ

We saw earlier that the spectrum of a Fourier series has nonzero values only at discrete frequencies. The spacing of harmonics decreases to zero as the fundamental period goes to infinity. Thus the Fourier transform of a nonperiodic signal has a continuous spectrum. We also saw that the product CkT is a constant. Since CkT becomes F(jto) in the limiting process, we see that the amplitude of the Fourier transform is identical with the envelope of the curve through the amplitude points of CkT for the Fourier series.