Higher Engineering Mathematics, Sixth Edition

640 Higher Engineering Mathematics

b 2 ≈

2

12

( 29. 43 )= 4. 91

and b 3 ≈

2

12

( 55 )= 9. 17

Substituting these values into the Fourier series:

f(x)=a 0 +

∑∞

n= 1

(ancosnx+bnsinnx)

gives: v= 17. 67 + 69 .66 cosθ− 6 .50cos2θ

− 8 .17 cos3θ+···− 46 .42 sinθ

+ 4 .91sin2θ+ 9 .17sin3θ+··· (4)

Note that in equation (4),(− 46 .42sinθ+ 69 .66cosθ)

comprises the fundamental,( 4 .91sin2θ− 6 .50cos2θ)

comprises the second harmonic and ( 9 .17sin3θ−

8 .17cos3θ)comprises the third harmonic.

It is shown in Chapter 17 that:

asinωt+bcosωt=Rsin(ωt+α)

where a=Rcosα, b=Rsinα, R=

√

a^2 +b^2 and

α=tan−^1

b

a

For the fundamental,R=

√

(− 46. 42 )^2 +( 69. 66 )^2

= 83. 71

If a=Rcosα,thencosα=

a

R

=

− 46. 42

83. 71

which is negative,

and if b=Rsinα,then sinα=

b

R

=

69. 66

83. 71

which is positive.

The only quadrant where cosαis negativeandsinαis

positive is the second quadrant.

Henceα=tan−^1

b

a

=tan−^1

69. 66

− 46. 42

= 123. 68 ◦or 2.16rad

Thus(− 46 .42sinθ+ 69 .66cosθ)

= 83 .71sin(θ+ 2. 16 )

By a similar method it may be shown that the second

harmonic

( 4 .91sin2θ− 6 .50cos2θ)= 8 .15sin( 2 θ− 0. 92 )

and the third harmonic

( 9 .17sin3θ− 8 .17cos3θ)= 12 .28sin( 3 θ− 0. 73 )

Hence equation (4) may be re-written as:

v= 17. 67 + 83 .71 sin(θ+ 2. 16 )

+ 8 .15sin( 2 θ− 0. 92 )

+ 12 .28sin( 3 θ− 0. 73 )volts

which is the form used in Chapter 25 with complex

waveforms.

Now try the following exercise

Exercise 234 Further problems on

numerical harmonic analysis

Determine the Fourier series to represent the peri-

odic functions given by the tables of values in

Problems 1 to 3, up to and including the third har-

monic and each coefficient correct to 2 decimal

places. Use 12 ordinates in each case.

1. Angleθ◦ 30 60 90 120 150180

Displacementy 40 43 38 30 23 17

Angleθ◦ 210240270300330360

Displacementy 11 9 10 13 21 32

⎡

⎣

y= 23. 92 + 7 .81cosθ+ 14 .61sinθ

+ 0 .17cos2θ+ 2 .31sin2θ

− 0 .33cos3θ+ 0 .50sin3θ

⎤

⎦

2. Angleθ◦ 0 30 60 90 120 150

Voltagev − 5. 0 − 1. 5 6.0 12.5 16.016.5

Angleθ◦ 180 210 240 270 300 330

Voltagev15.012.5 6.5 − 4. 0 − 7. 0 − 7. 5

⎡

⎣

v= 5. 00 − 10 .78cosθ+ 6 .83sinθ

− 1 .96cos2θ+ 0 .80sin2θ

+ 0 .58cos3θ− 1 .08sin3θ

⎤

⎦

3. Angleθ◦ 30 60 90 120 150 180

Currenti 0 − 1. 4 − 1. 8 − 1. 9 − 1. 8 − 1. 3

Angleθ◦ 210 240 270 300 330 360

Currenti 0 2.2 3.8 3.9 3.5 2.5

⎡

⎣

i= 0. 64 + 1 .58cosθ− 2 .73sinθ

− 0 .23cos2θ− 0 .42sin2θ

+ 0 .27cos3θ+ 0 .05sin3θ

⎤

⎦