Bird J. Electrical Circuit Theory and Technology
Подождите немного. Документ загружается.
638 Electrical Circuit Theory and Technology
From Figures 36.2 to 36.6, it is seen that a waveform can change its
shape considerably as a result of changes in both phase and magnitude of
the harmonics.
Example 7
Consider the complex current expression given by
i
a
= 10sin!t Y 4sin2!t amperes
Current i
a
consists of a fundamental component, 10 sin ωt, and a second
harmonic component, 4sin2ωt, the components being initially in phase
with each other. Current i
a
contains 40% second harmonic. The funda-
mental and second harmonic are shown plotted separately in Figure 36.8.
By adding ordinates at intervals, the complex waveform representing i
a
is
produced as shown. It is noted that if all the values in the negative half-
cycle were reversed then this half-cycle would appear as a mirror image of
the positive half-cycle about a vertical line drawn through time, t D T/2.
Figure 36.8
Example 8
Consider the complex current expression given by
i
b
= 10sin!t Y 4sin2!t Y 3sin4!t amperes
The waveforms representing (10 sin ωt C 4sin2ωt) and the fourth
harmonic component, 3 sin 4ωt, are each shown separately in Figure 36.9,
the former waveform having been produced in Figure 36.8. By adding
Complex Waveforms 639
Figure 36.9
ordinates at intervals, the complex waveform for i
b
is produced as shown
in Figure 36.9. If the half-cycle between times T/2andT is reversed then
it is seen to be a mirror image of the half-cycle lying between 0 and T/2
about a vertical line drawn through the time, t D T/2.
Example 9
Consider the complex current expressions given by
i
c
= 10sin!t Y 4sin
2!t Y
p
2
amperes
The fundamental component, 10 sin ωt, and the second harmonic compo-
nent, having an amplitude of 4 A and a phase displacement of /2 radian
leading (i.e., leading 4sin2ωt by /2 radian or T/8 seconds), are shown
plotted separately in Figure 36.10. By adding ordinates at intervals, the
complex waveform for i
c
is produced as shown. The positive and negative
half-cycles of the resultant waveform i
c
are seen to be quite dissimilar.
Example 10
Consider the complex current expression given by
i
d
= 10sin!t Y 4sin.2!t Y p/amperes
The fundamental, 10sin ωt, and the second harmonic component which
leads 4 sin 2ωt by rad are shown separately in Figure 36.11. By adding
ordinates at intervals, the resultant waveform i
d
is produced as shown. If
the negative half-cycle is reversed, it is seen to be a mirror image of the
positive half-cycle about a line drawn vertically through time, t D T/2.
640 Electrical Circuit Theory and Technology
Figure 36.10
Figure 36.11
General conclusions on examples 7 to 10
Whenever even harmonics are added to a fundamental component:
(a) if the harmonics are initially in phase or if there is a phase-shift of
rad, the negative half-cycle, when reversed, is a mirror image of the
positive half-cycle about a vertical line drawn through time, t D T/2;
Complex Waveforms 641
(b) if the harmonics are initially out of phase with each other (i.e., other
than rad), the positive and negative half-cycles are dissimilar.
These are features of waveforms containing the fundamental and even
harmonics.
Example 11
Consider the complex voltage expression given by
v
g
= 50sin!t Y 25sin2!t Y 15sin3!t volts
The fundamental and the second and third harmonics are each shown
separately in Figure 36.12. By adding ordinates at intervals, the resultant
waveform
v
g
is produced as shown. If the negative half-cycle is reversed,
it appears as a mirror image of the positive half-cycle about a vertical
line drawn through time D T/2.
Figure 36.12
Example 12
Consider the complex voltage expression given by
v
h
= 50sin!t Y 25sin.2!t − p/ Y 15sin
3!t Y
p
2
volts
The fundamental, the second harmonic lagging by radian and the third
harmonic leading by /2 radian are initially plotted separately, as shown
642 Electrical Circuit Theory and Technology
in Figure 36.13. Adding ordinates at intervals gives the resultant wave-
form
v
h
as shown. The positive and negative half-cycles are seen to be
quite dissimilar.
Figure 36.13
General conclusions on examples 11 and 12
Whenever a waveform contains both odd and even harmonics:
(a) if the harmonics are initially in phase with each other, the negative
cycle, when reversed, is a mirror image of the positive half-cycle
about a vertical line drawn through time, t D T/2;
(b) if the harmonics are initially out of phase with each other, the posi-
tive and negative half-cycles are dissimilar.
Example 13
Consider the complex current expression given by
i
= 32 Y 50sin!t Y 20sin
2!t −
p
2
mA
The current i comprises three components — a 32 mA d.c. component, a
fundamental of amplitude 50 mA and a second harmonic of amplitude
20 mA, lagging by /2 radian. The fundamental and second harmonic
are shown separately in Figure 36.14. Adding ordinates at intervals gives
the complex waveform 50sinωt C 20sin2ωt /2).
Complex Waveforms 643
Figure 36.14
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 seen to be
to shift the whole wave 32 mA upward. The waveform approaches that
expected from a half-wave rectifier (see Section 36.8).
Problem 1. A complex waveform v comprises a fundamental
voltage of 240 V rms and frequency 50 Hz, together with a 20%
third harmonic which has a phase angle lagging by 3/4 rad at
time D 0. (a) Write down an expression to represent voltage
v.(b)
Use harmonic synthesis to sketch the complex waveform repre-
senting voltage over one cycle of the fundamental component.
(a) A fundamental voltage having an rms value of 240 V has a maximum
value, or amplitude of (
p
2)(240), i.e., 339.4 V.
If the fundamental frequency is 50 Hz then angular velocity,
ω D 2f D 250 D 100 rad/s. Hence the fundamental voltage
is represented by 339.4sin100t volts. Since the fundamental
frequency is 50 Hz, the time for one cycle of the fundamental is
given by T D 1/f D 1/50sor20ms.
644 Electrical Circuit Theory and Technology
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 ð 50 D 150 Hz, thus the angular velocity is 2 (150), i.e.,
300 rad/s. Hence the third harmonic voltage is represented by
67.9sin300t 3/4 volts. Thus
voltage,
v = 339.4sin100pt Y 67.9sin
300pt −
3p
4
volts
(b) One cycle of the fundamental, 339.4sin100t, is shown sketched
in Figure 36.15, together with three cycles of the third harmonic
component, 67.9sin300t 3/4 initially lagging by 3/4 rad.
By adding ordinates at intervals, the complex waveform representing
voltage is produced as shown. If the negative half-cycle is reversed,
it is seen to be identical to the positive half-cycle, which is a feature
of waveforms containing the fundamental and odd harmonics.
Figure 36.15
Problem 2. For each of the periodic complex waveforms shown
in Figure 36.16, suggest whether odd or even harmonics (or both)
are likely to be present.
(a) If in Figure 36.16(a) the negative half-cycle is reversed, it is seen to
be identical to the positive half-cycle. This feature indicates that the
complex current waveform is composed of a fundamental and odd
harmonics only (see examples 1 to 6).
(b) In Figure 36.16(b) the negative half-cycle is quite dissimilar to the
positive half-cycle.
This indicates that the complex voltage waveform comprises either
(i) a fundamental and even harmonics, initially out of phase with
each other (see example 9), or
Figure 36.16
Complex Waveforms 645
(ii) a fundamental and odd and even harmonics, one or more of
the harmonics being initially out of phase (see example 12).
(c) If in Figure 36.16(c) the negative half-cycle is reversed, it is seen
to be a mirror image of the positive half-cycle about a vertical line
drawn through time T/2. This feature indicates that the complex
e.m.f. waveform comprises either:
(i) a fundamental and even harmonics initially in phase with each
other (see examples 7 and 8), or
(ii) a fundamental and odd and even harmonics, each initially in
phase with each other (see example 11).
Further problems on harmonic synthesis may be found in Section 36.9,
problems 1 to 6, page 671
36.4 Rms value, mean
value and the form factor
of a complex wave
Rms value
Let the instantaneous value of a complex current, i, be given by
i D I
1m
sinωt C
1
C I
2m
sin2ωt C
2
CÐÐÐCI
nm
sinnωt C
n
amperes
The effective or rms value of this current is given by
I D
p
mean value of i
2
i
2
D [I
1m
sinωt C
1
C I
2m
sin2ωt C
2
CÐÐÐCI
nm
sinnωt C
n
]
2
i.e., i
2
D I
2
1m
sin
2
ωt C
1
C I
2
2m
sin
2
2ωt C
2
CÐÐÐCI
2
nm
sin
2
nωt C
n
C 2I
1m
I
2m
sinωt C
1
sin2ωt C
2
CÐÐÐ 36.3
Without writing down all terms involved when squaring current i, it can
be seen that two types of term result, these being:
(i) terms such as I
2
1m
sin
2
ωt C
1
, I
2
2m
sin
2
2ωt C
2
, and so on, and
(ii) terms such as 2I
1m
I
2m
sinωt C
1
sin2ωt C
2
, i.e., products of
different harmonics.
The mean value of i
2
is the sum of the mean values of each term in
equation (36.3).
Taking an example of the first type, say I
2
1m
sin
2
ωt C
1
, the mean
value over one cycle of the fundamental is determined using integral
calculus:
646 Electrical Circuit Theory and Technology
Mean value of I
2
1m
sin
2
ωt C
1
D
1
2
2
0
I
2
1m
sin
2
ωt C
1
dωt
(since the mean value of y D fx between x D a and x D b is given by
1
b a
b
a
ydx
D
I
2
1m
2
2
0
1 cos 2ωt C
1
2
dωt,
(since cos2x D 1 2sin
2
x, from which sin
2
x D 1 cos2x/2,
D
I
2
1m
4
ωt
sin2ωt C
1
2
2
0
D
I
2
1m
4
2
sin22 C
1
2
0
sin20 C
1
2
D
I
2
1m
4
2
sin22 C
1
2
C
sin2
1
2
D
I
2
1m
4
2 D
I
2
1m
2
Hence it follows that the mean value of I
2
nm
sin
2
nωt C
n
is given by
I
2
nm
/2.
Taking an example of the second type, say,
2I
1m
I
2m
sinωt C
1
sin2ωt C
2
,
the mean value over one cycle of the fundamental is also determined
using integration:
Mean value of 2I
1m
I
2m
sinωt C
1
sin2ωt C
2
D
1
2
2
0
2I
1m
I
2m
sinωt C
1
sin2ωt C
2
dωt
D
I
1m
I
2m
2
0
1
2
f
cosωt C
2
1
cos3ωt C
2
C
1
g
dωt
(since sin A sin B D
1
2
[cosA B cosA C B], and taking
A D 2ωt C
2
and B D ωt C
1
D
I
1m
I
2m
2
sinωt C
2
1
sin3ωt C
2
C
1
3
2
0
D
I
1m
I
2m
2
sin2 C
2
1
sin6 C
2
C
1
3
sin
2
1
sin
2
C
1
3
D
I
1m
I
2m
2
[0] D 0 36.4
Complex Waveforms 647
Hence it follows that all such products of different harmonics will have
a mean value of zero. Thus
mean value of i
2
D
I
2
1m
2
C
I
2
2m
2
CÐÐÐC
I
2
nm
2
Hence the rms value of current,
I D
I
2
1m
2
C
I
2
2m
2
CÐÐÐC
I
2
nm
2
i.e.,
I =
I
2
1m
Y I
2
2m
Y ···Y I
2
nm
2
36.5
For a sine wave, rms value D 1/
p
2 maximum value, i.e., maximum
value D
p
2 rms value. Hence, for example, I
1m
D
p
2I
1
, where I
1
is the
rms value of the fundamental component, and I
1m
2
D
p
2I
1
2
D 2I
2
1
.
Thus, from equation (36.5), rms current
I D
2I
2
1
C 2I
2
2
CÐÐÐC2I
2
n
2
i.e.,
I =
p
.I
2
1
Y I
2
2
Y ···Y I
2
n
/
36.6
where I
1
,I
2
,...,I
n
are the rms values of the respective harmonics.
By similar reasoning, for a complex voltage waveform represented by
v D V
1m
sinωt C
1
C V
2m
sin2ωt C
2
CÐÐÐCV
nm
sinnωt C
n
volts
the rms value of voltage, V, is given by
V =
V
2
1m
Y V
2
2m
Y ···Y V
2
nm
2
36.7
or
V =
.V
2
1
Y V
2
2
Y ···Y V
2
n
/
36.8
where V
1
,V
2
,...,V
n
are the rms values of the respective harmonics.
From equations (36.5) to (36.8) it is seen that the rms value of a
complex wave is unaffected by the relative phase angles of the harmonic
components. For a d.c. current or voltage, the instantaneous value, the