Higher Engineering Mathematics, Sixth Edition

656 Higher Engineering Mathematics

2 rad/s

2 rad/s

Real axis

Imaginary axis

024

Figure 71.10

negative direction) with an angular velocity of

2 rad/s. Both phasors have zero phase angle.

Figure 71.10 shows the two phasors.

(b) From equation (3), page 644,

cosθ=

1

2

(

ejθ+e−jθ

)

Hence,v=8cos( 2 t− 1. 5 )

= 8

[

1

2

(

ej(^2 t−^1.^5 )+e−j(^2 t−^1.^5 )

)]

=4ej(^2 t−^1.^5 )+4e−j(^2 t−^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 71.11 shows the two phasors.

Real axis

4

4

Imaginary axis

0

 5 2 rad/s

 5 2 rad/s

1.5 rad

1.5 rad

Figure 71.11

Problem 8. Determine – the pair of phasors that

can be used to represent the third harmonic

v=8cos3t−20sin3t.

Using cost=

1

2

(

ejt+e−jt

)

and sint=

1

2 j

(

ejt−e−jt

)

from page 644

gives: v=8cos3t−20sin3t

= 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 −j 10 )e−j^3 t

( 4 +j 10 )=

√

42 + 102 ∠tan−^1

(

10

4

)

= 10. 77 ∠ 1. 19

and( 4 −j 10 )

= 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 rotatingclockwise with anangular

velocity of 3rad/s.