146 Higher Engineering Mathematics
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
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
14.6 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 66 to 69). 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 70 on page 637. 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 100 V and frequency, f = ω/2π hertz
and a third harmonic component of maximum value
30 V and frequency = 3ω/2π(=3 f ), the fundamental
and third harmonics being initially in phase with each
other.
In Fig. 14.31, 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 30 V and for which three cycles are completed
in time T seconds. At zero time, 30 sin 3ωt is in phase
with 100 sinωt .
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 50 V and the third harmonic a value of 30 V. Adding gives a value of 80 V for
waveform v 1 at time T/12 seconds. Similarly, at time
T/4 seconds, the fundamental has a value of 100 V and
the third harmonic a value of −30 V. After addition,
the resultant waveform v 1 is 70 V 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
2
volts.
The peak value of the fundamental is 100 volts and the
peak value of the third harmonic is 30 V. However the
third harmonic has a phase displacement of
π
2
radian
leading (i.e. leading 30 sin 3ωt by
π
2
radian). Note that,
since the periodic time 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
4
cycle of the
third harmonic represents a time interval of (T/3) ÷ 4,
i.e. T/12 seconds.
Figure 14.32 shows graphs of 100 sin ωt and
30 sin
3ωt +
π
2
over the time for one cycle of the fundamental. When ordinates of the two graphs are added
at intervals, the resultant waveform v 2 is as shown.
If the negative half-cycle in Fig. 14.32 is reversed it
can be seen that the shape of the positive and negative
half-cycles are identical.
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
⎤
⎥
⎥
⎥
⎥
⎥
⎥
⎦
14.6 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 66 to 69). 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 70 on page 637. 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 100 V and frequency, f = ω/2π hertz
and a third harmonic component of maximum value
30 V and frequency = 3ω/2π(=3 f ), the fundamental
and third harmonics being initially in phase with each
other.
In Fig. 14.31, 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 30 V and for which three cycles are completed
in time T seconds. At zero time, 30 sin 3ωt is in phase
with 100 sinωt .
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 50 V and the third harmonic a value of 30 V. Adding gives a value of 80 V for
waveform v 1 at time T/12 seconds. Similarly, at time
T/4 seconds, the fundamental has a value of 100 V and
the third harmonic a value of −30 V. After addition,
the resultant waveform v 1 is 70 V 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
2
volts.
The peak value of the fundamental is 100 volts and the
peak value of the third harmonic is 30 V. However the
third harmonic has a phase displacement of
π
2
radian
leading (i.e. leading 30 sin 3ωt by
π
2
radian). Note that,
since the periodic time 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
4
cycle of the
third harmonic represents a time interval of (T/3) ÷ 4,
i.e. T/12 seconds.
Figure 14.32 shows graphs of 100 sin ωt and
30 sin
3ωt +
π
2
over the time for one cycle of the fundamental. When ordinates of the two graphs are added
at intervals, the resultant waveform v 2 is as shown.
If the negative half-cycle in Fig. 14.32 is reversed it
can be seen that the shape of the positive and negative
half-cycles are identical.