TRIGONOMETRIC WAVEFORMS 163B
Figure 15.34
The currenticomprises three components—a 32 mA
d.c. component, a fundamental of amplitude 50 mA
and a second harmonic of amplitude 20 mA, lag-
ging by
π
2radian. The fundamental and secondharmonic are shown separately in Fig. 15.35. Adding
ordinates at intervals gives the complex waveform
50 sinωt+20 sin
(
2 ωt−π
2)
.
This waveform is then added to the 32 mA
d.c. component to produce the waveform i as
shown. The effect of the d.c. component is to
shift the whole wave 32 mA upward. The wave-
form approaches that expected from ahalf-wave
rectifier.
Problem 22. A complex waveformv com-
prises a fundamental voltage of 240 V rms and
frequency 50 Hz, together with a 20% third
harmonic which has a phase angle lagging
by 3π/4 rad at timet=0. (a) Write down an
expression to represent voltagev. (b) Use har-
monic synthesis to sketch the complex wave-
form representing voltagevover one cycle of
the fundamental component.(a) A fundamental voltage having an rms value of
240 V has a maximum value, or amplitude of√
2 (240) i.e. 339.4 V.
If the fundamental frequency is 50 Hz then
angular velocity,ω= 2 πf= 2 π(5 0)= 100 πrad/s.
Hence the fundamental voltage is represented
by 339.4 sin 100πtvolts. Since the fundamen-
tal frequency is 50 Hz, the time for one cycle
of the fundamental is given byT= 1 /f= 1 /50 s
or 20 ms.
The third harmonic has an amplitude equal to
20% of 339.4 V, i.e. 67.9 V. The frequency of the
third harmonic component is 3× 50 =150 Hz,
thus the angular velocity is 2π(150), i.e.
300 πrad/s. Hence the third harmonic voltage
is represented by 67.9 sin (300πt− 3 π/4) volts.
Thus
voltage,v=339.4 sin 100πt
+67.9 sin (300πt− 3 π/4) volts
(b) One cycle of the fundamental, 339.4 sin 100πt,
is shown sketched in Fig. 15.36, together with
three cycles of the third harmonic compon-
ent, 67.9 sin (300πt− 3 π/4) initially lagging
by 3π/4 rad. By adding ordinates at intervals,
the complex waveform representing voltage is
produced as shown.