Environmental Engineering Reference
In-Depth Information
harmonics in the current
i
because of non-linear loads and/or the pulse width modulation built
in the inverter, which cause harmonic voltage drops on the output impedance
Z
o
. If there
are no corresponding harmonic voltage components provided by the reference
v
r
, then the
harmonic voltage components appear in the output voltage, which degrades the voltage quality
and causes a high THD. Another source of high THD of the output voltage is from the voltage
reference
v
r
if it contains significant harmonic components.
In order to obtain low THD for
v
o
, there are three options:
1. to keep
v
r
clean and maintain a small output impedance
Z
o
over the frequency range of the
major harmonic current components;
2. to bypass the harmonic current components so that the current
i
is clean; and
3. to make sure that the reference voltage
v
r
is able to provide the right amount of harmonic
voltages to compensate the harmonic voltage dropped on the output impedance.
The first option will be discussed in detail in the next subsection, with control strategies
presented in Chapter 7. The second option will be discussed in Chapter 8 while the third option
will be discussed in Chapter 21.
It is believed that the repetitive control strategies discussed in (Chen
et al.
2008; Costa-
Castello
et al.
2004; Escobar
et al.
2008; Garcia-Cerrada
et al.
2007; Hornik and Zhong,
2010b,2011; Tzou
et al.
1999; Weiss
et al.
2004; Zhou and Wang, 2003; Zhou
et al.
2009)
and those in Chapters 3-6 should lead to small output impedances but this fact has not been
well documented. Note that the harmonics injection discussed in (Borup
et al.
2001) belongs
to the third option as well.
2.1.3 Role of Inverter Output Impedance
Assume that the output current is
√
2
h
=
1
I
h
sin(
h
i
=
ω
t
+
φ
h
)
,
where
ω
is the system freque
nc
y. Then the amplitude of the
h
-th harmonic voltage dropped
on the output impedance is
√
2
I
h
|
|
Z
o
(
jh
ω
)
. Moreover, assume that the voltage reference
v
r
is clean and sinusoidal and is described as
√
2
E
sin(
v
r
=
ω
+
δ
.
t
)
Then the fundamental component of the output voltage is
1
v
1
=
√
2
E
sin(
−
√
2
I
1
|
ω
t
+
δ
)
Z
o
(
j
ω
)
|
sin(
ω
t
+
φ
1
+
θ
)
√
2
V
1
sin(
=
ω
t
+
β
)
,
1
See the trigonometric identities at http://en.wikipedia.org/wiki/List of trigonometric identities.
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