Methods of adding alternating waveforms 267
Hence, the sinusoidal expression for the resultant i 1 + i 2
is given by:
i R = i 1 + i 2 = 26.5 sin(ωt + 0.33) A
Now try the following exercise
Exercise 107 Further problems on plotting
periodic functions
1. Plot the graph of y = 2 sin A from A = 0 ◦
to A = 360 ◦ . On the same axes plot
y = 4 cos A. By adding ordinates at intervals
plot y = 2 sin A + 4 cos A and obtain a sinusoidal expression for the waveform.
[4.5 sin(A + 63.5 ◦ )]
2. Two alternating voltages are given by
v 1 = 10 sin ωt volts and v 2 = 14 sin(ωt + π/3)
volts. By plotting v 1 and v 2 on the same axes
over one cycle obtain a sinusoidal expression
for (a) v 1 + v 2 (b) v 1 − v 2 .
(a) 20.9 sin(ωt + 0.63) volts
(b) 12.5 sin(ωt − 1.36) volts
3. Express 12 sin ωt + 5 cos ωt in the form
A sin(ωt ± α) by drawing and measurement.
[13 sin(ωt + 0.395)]
25.3 Determining resultant phasors
by drawing
The resultant of two periodic functions may be found
from their relative positions when the time is zero.
For example, if y 1 = 4 sinωt and y 2 = 3 sin(ωt − π/3)
then each may be represented as phasors as shown in
Fig. 25.5, y 1 being 4 units long and drawn horizontally
and y 2 being 3 units long, lagging y 1 by π/3 radians or
60 ◦ . To determine the resultant of y 1 + y 2 , y 1 is drawn
horizontally as shown in Fig. 25.6 and y 2 is joined to the
end of y 1 at 60 ◦ to the horizontal. The resultant is given
by y R . This is the same as the diagonal of a parallelogram
that is shown completed in Fig. 25.7.
Resultant y R , in Figs. 25.6 and 25.7, may be determined
by drawing the phasors and their directions to scale and
measuring using a ruler and protractor.
608 or /3 rads
y 1 5 4
y 2 5 3
Figure 25.5
y 1 5 4
y 2 5 3
0
608
y R
Figure 25.6
y 1 5 4
y 2 5 3
y R
Figure 25.7
In this example, y R is measured as 6 units long and angle
φ is measured as 25 ◦ .
25
◦
= 25 ×
π
180
radians = 0.44 rad
Hence, summarising, by drawing: y R = y 1 + y 2 =
4 sinωt + 3 sin(ωt − π/3) = 6 sin(ωt − 0.44)
If the resultant phasor y R = y 1 − y 2 is required, then y 2
is still 3 units long but is drawn in the opposite direction,
as shown in Fig. 25.8.
608
608
y 1 5 4
2y 2 5 3
y R
y 2
Figure 25.8
Hence, the sinusoidal expression for the resultant i 1 + i 2
is given by:
i R = i 1 + i 2 = 26.5 sin(ωt + 0.33) A
Now try the following exercise
Exercise 107 Further problems on plotting
periodic functions
1. Plot the graph of y = 2 sin A from A = 0 ◦
to A = 360 ◦ . On the same axes plot
y = 4 cos A. By adding ordinates at intervals
plot y = 2 sin A + 4 cos A and obtain a sinusoidal expression for the waveform.
[4.5 sin(A + 63.5 ◦ )]
2. Two alternating voltages are given by
v 1 = 10 sin ωt volts and v 2 = 14 sin(ωt + π/3)
volts. By plotting v 1 and v 2 on the same axes
over one cycle obtain a sinusoidal expression
for (a) v 1 + v 2 (b) v 1 − v 2 .
(a) 20.9 sin(ωt + 0.63) volts
(b) 12.5 sin(ωt − 1.36) volts
3. Express 12 sin ωt + 5 cos ωt in the form
A sin(ωt ± α) by drawing and measurement.
[13 sin(ωt + 0.395)]
25.3 Determining resultant phasors
by drawing
The resultant of two periodic functions may be found
from their relative positions when the time is zero.
For example, if y 1 = 4 sinωt and y 2 = 3 sin(ωt − π/3)
then each may be represented as phasors as shown in
Fig. 25.5, y 1 being 4 units long and drawn horizontally
and y 2 being 3 units long, lagging y 1 by π/3 radians or
60 ◦ . To determine the resultant of y 1 + y 2 , y 1 is drawn
horizontally as shown in Fig. 25.6 and y 2 is joined to the
end of y 1 at 60 ◦ to the horizontal. The resultant is given
by y R . This is the same as the diagonal of a parallelogram
that is shown completed in Fig. 25.7.
Resultant y R , in Figs. 25.6 and 25.7, may be determined
by drawing the phasors and their directions to scale and
measuring using a ruler and protractor.
608 or /3 rads
y 1 5 4
y 2 5 3
Figure 25.5
y 1 5 4
y 2 5 3
0
608
y R
Figure 25.6
y 1 5 4
y 2 5 3
y R
Figure 25.7
In this example, y R is measured as 6 units long and angle
φ is measured as 25 ◦ .
25
◦
= 25 ×
π
180
radians = 0.44 rad
Hence, summarising, by drawing: y R = y 1 + y 2 =
4 sinωt + 3 sin(ωt − π/3) = 6 sin(ωt − 0.44)
If the resultant phasor y R = y 1 − y 2 is required, then y 2
is still 3 units long but is drawn in the opposite direction,
as shown in Fig. 25.8.
608
608
y 1 5 4
2y 2 5 3
y R
y 2
Figure 25.8