trigonometry

160 GEOMETRY AND TRIGONOMETRY
current at t = 8 ms (d) the time when the current
is first a maximum (e) the time when the
current first reaches 3A. Sketch one cycle of
the waveform showing relevant points.
⎡
(a) 5 A, 20 ms, 50 Hz,
24 ◦ 45 ′ lagging
(b) −2.093 A
⎢
⎣
(c) 4.363 A
(d) 6.375 ms
(e) 3.423 ms
⎤
⎥
⎦
15.6 Harmonic synthesis with complex
waveforms
A waveform that is not sinusoidal is called a complex
wave. Harmonic analysis is the process of resolving
a complex periodic waveform into a series of
sinusoidal components of ascending order of frequency.
Many of the waveforms met in practice
can be represented by the following mathematical
expression.
v = V 1m sin(ωt + α 1 ) + V 2m sin(2ωt + α 2 )
+···+V nm sin(nωt + α n )
and the magnitude of their harmonic components
together with their phase may be calculated using
Fourier series (see Chapters 69 to 72). Numerical
methods are used to analyse waveforms for
which simple mathematical expressions cannot be
obtained. A numerical method of harmonic analysis
is explained in the Chapter 73 on page 683. In a laboratory,
waveform analysis may be performed using a
waveform analyser which produces a direct readout
of the component waves present in a complex wave.
By adding the instantaneous values of the fundamental
and progressive harmonics of a complex
wave for given instants in time, the shape of a
complex waveform can be gradually built up. This
graphical procedure is known as harmonic synthesis
(synthesis meaning ‘the putting together of parts
or elements so as to make up a complex whole’).
Some examples of harmonic synthesis are considered
in the following worked problems.
Problem 17. Use harmonic synthesis to construct
the complex voltage given by:
v 1 = 100 sin ωt + 30 sin 3ωt volts.
The waveform is made up of a fundamental wave
of maximum value 100V and frequency, f = ω/2π
hertz and a third harmonic component of maximum
value 30V and frequency = 3ω/2π(=3f ), the fundamental
and third harmonics being initially in phase
with each other.
In Figure 15.31, the fundamental waveform is
shown by the broken line plotted over one cycle, the
periodic time T being 2π/ω seconds. On the same
axis is plotted 30 sin 3ωt, shown by the dotted line,
having a maximum value of 30V and for which three
cycles are completed in time T seconds.At zero time,
30 sin 3ωt is in phase with 100 sin ωt.
The fundamental and third harmonic are combined
by adding ordinates at intervals to produce
the waveform for v 1 , as shown. For example, at time
T/12 seconds, the fundamental has a value of 50V
and the third harmonic a value of 30V.Adding gives a
value of 80V for waveform v 1 at time T/12 seconds.
Similarly, at time T/4 seconds, the fundamental has
a value of 100V and the third harmonic a value of
−30V. After addition, the resultant waveform v 1 is
70V at T/4. The procedure is continued between
t = 0 and t = T to produce the complex waveform for
v 1 . The negative half-cycle of waveform v 1 is seen
to be identical in shape to the positive half-cycle.
If further odd harmonics of the appropriate amplitude
and phase were added to v 1 a good approximation
to a square wave would result.
Problem 18. Construct the complex voltage
given by:
(
v 2 = 100 sin ωt + 30 sin 3ωt + π )
volts.
2
The peak value of the fundamental is 100 volts
and the peak value of the third harmonic is 30V.
However the third harmonic has a phase displacement
of π radian leading (i.e. leading 30 sin 3ωt
2
by π radian). Note that, since the periodic time
2
of the fundamental is T seconds, the periodic time
of the third harmonic is T/3 seconds, and a phase
displacement of π 2 radian or 1 cycle of the third harmonic
represents a time interval of (T/3) ÷ 4, i.e.
4
T/12 seconds.
Figure ( 15.32 shows graphs of 100 sin ωt and
30 sin 3ωt + π )
over the time for one cycle of the
2
fundamental. When ordinates of the two graphs are
added at intervals, the resultant waveform v 2 is as