Higher Engineering Mathematics

702 FOURIER SERIES

Imaginary axis

024 Real axis

ω = 2 rad/s

ω = 2 rad/s

Figure 74.10

(b) From equation (3), page 690,

cosθ=

1

2

(

ejθ +e−jθ

)

Hence, v=8 cos( 2 t− 1. 5 )

= 8

[

1

2

(

ej(^2 t−^1.^5 )+e−j(^2 t−^1.^5 )

)]

=4ej(2t−^1 .5)+4e−j(2t−^1 .5)

i.e. v=4e^2 te−j1.5+4e−j^2 tej1.5

This represents a phasor of length 4 and phase

angle−1.5 radians rotating anticlockwise (i.e. in

the positive direction) with an angular velocity

of 2 rad/s, and another phasor of length 4 and

phase angle+1.5 radians and rotating clockwise

(i.e. in the negative direction) with an angular

velocity of 2 rad/s. Figure 74.11 shows the two

phasors.

Imaginary axis

ω = 2 rad/s

ω = 2 rad/s

1.5 rad

1.5 rad Real axis

4

4

0

Figure 74.11

Problem 8. Determine – the pair of phasors

that can be used to represent the third harmonic

v=8 cos 3t−20 sin 3t

Using cost=

1

2

(

ejt+e−jt

)

and sint=

1

2 j

(

ejt−e−jt

)

from page 690

gives: v=8 cos 3t−20 sin 3t

= 8

[

1

2

(

ej^3 t+e−j^3 t

)

]

− 20

[

1

2 j

(

ej^3 t−e−j^3 t

)

]

=4ej^3 t+4e−j^3 t−

10

j

ej^3 t+

10

j

e−j^3 t

=4ej^3 t+4e−j^3 t−

10(j)

j(j)

ej^3 t+

10(j)

j(j)

e−j^3 t

=4ej^3 t+4e−j^3 t+ 10 jej^3 t− 10 je−j^3 t

sincej^2 =− 1

=( 4 +j 10 )ej^3 t+(4−j10) e−j^3 t

(4+j10)=

√

42 + 102 ∠tan−^1

(

10

4

)

= 10. 77 ∠ 1. 19

and (4−j10)

= 10. 77 ∠− 1. 19

Hence,v=10.77∠1.19+10.77∠−1.19

Thusvcomprises a phasor 10.77∠1.19 rotating anti-

clockwise with an angular velocity if 3 rad/s, and

a phasor 10.77∠−1.19 rotating clockwise with an

angular velocity of 3 rad/s.