Higher Engineering Mathematics, Sixth Edition

612 Higher Engineering Mathematics

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx) (1)

where for the range−πtoπ:

and

a 0 =

1

2 π

∫π

−π

f(x)dx

an=

1

π

∫π

−π

f(x)cosnxdx

(n= 1 , 2 , 3 ,...)

bn=

1

π

∫π

−π

f(x)sinnxdx

(n= 1 , 2 , 3 ,...)

(ii) a 0 ,anandbnare called theFourier coefficients

of the series and if these can be determined, the

series of equation (1) is called theFourier series

corresponding tof(x).

(iii) An alternative way of writing the series is by

using theacosx+bsinx=csin(x+α)relation-

ship introduced in Chapter 17, i.e.

f(x)=a 0 +c 1 sin(x+α 1 )+c 2 sin( 2 x+α 2 )

+···+cnsin(nx+αn),

wherea 0 is a constant,

c 1 =

√

(a^21 +b 12 ),...cn=

√

(a^2 n+b^2 n)

are the amplitudes of the various components,

and phase angle

αn=tan−^1

an

bn

(iv) For the series of equation (1): the term

(a 1 cosx+b 1 sinx)orc 1 sin(x+α 1 )is called the

first harmonicor thefundamental,theterm

(a 2 cos2x+b 2 sin2x)orc 2 sin( 2 x+α 2 )is called

thesecond harmonic, and so on.

For an exact representation of a complex wave, an infi-

nite number of terms are, in general, required. In many

practical cases, however, it is sufficient to take the first

few terms only (see Problem 2).

The sum of a Fourier series at a point ofdiscontinuity

isgivenbythearithmeticmeanofthetwolimitingvalues

off(x)asxapproaches the point of discontinuityfrom

the two sides. For example, for the waveform shown in

x

f(x)

0 3 /2

8

23

2  2 /2 /2 

Figure 66.2

Fig. 66.2, the sum of the Fourier series at the points of

discontinuity (i.e. at

π

2

,π,...is given by:

8 +(− 3 )

2

=

5

2

or 2

1

2

66.4 Worked problemson Fourier

series of periodicfunctions of

period 2π

Problem 1. Obtain a Fourier series for the

periodic functionf(x)defined as:

f(x)=

{

−k, when −π<x< 0

+k, when 0<x<π

The function is periodic outside of this range with

period 2π.

The square wave function defined is shown in Fig. 66.3.

Sincef(x)is given by two different expressions in the

two halves of the range the integration is performed in

two parts, one from−πto 0 and the other from 0 toπ.

x

f(x)

0 2 

k

2 k

22  2 

Figure 66.3