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.