GE Power Management PQM Power Quality Meter A-7
APPENDIX A A.1 PQM APPLICATION NOTES
A
A.1.3 PHASORS IMPLEMENTATION
APPLICATION NOTE PQMAN03: PQM PHASORS IMPLEMENTATION
The purpose of the function Calc_Phasors within the PQM firmware is to take a digitally sampled periodic sig-
nal and generate the equivalent phasor representation of the signal. In the conventional sense, a phasor
depicts a purely sinusoidal signal which is what we’re interested in here; we wish to calculate the phasor for a
given signal at the fundamental power system frequency. The following Discrete Fourier Series equations cal-
culate the phasor in rectangular co-ordinates for an arbitrary digitally sampled signal. The justification for the
equations is beyond the
scope
of this document but can be found in some form in any text on signal analysis.
where: Re(
g
) = Real component of phasor
Im(
g
) = Imaginary component of phasor
g
= Set of N digital samples = {
g
0
,
g
1
,...,
g
N
-1
}
g
n
= nth sample from
g
N
= Number of samples
f
0
= Fundamental frequency in Hertz
ω
0
= 2
π
f
0
= Angular frequency in radians
T
=
1 /
(f
0
N)
= Time between samples
The PQM Trace Memory feature is employed to calculate the phasors. The Trace Memory feature samples 16
times per cycle for two cycles for all current and voltage inputs. Substituting
N
= 16 (samples/cycle) into the
equations yields the following for the real and imaginary components of the phasor:
The number of multiples in the above equation can be reduced by using the symmetry inherent in the sine and
cosine functions which is illustrated as follows:
Let
k
1
= cos(
π
/8),
k
2
= cos(
π
/4),
k
3
= cos(3
π
/8); the equations for the real and imaginary components are
reduced to:
Re
g
()
2
n
---
g
n
ω
0
nT
()
cos
⋅
n
0=
N
1–
∑
=
; Im
g
()
2
n
---
g
n
ω
0
nT
()
sin
⋅
n
0=
N
1–
∑
=
Re
g
()
1
8
---
g
0
0cos
g
1
π
8
---
cos
g
2
2
π
8
------
cos
…
g
31
31
π
8
----------
cos
++ ++
=
Im
g
()
1
8
---
g
0
0sin
g
1
π
8
---
sin
g
2
2
π
8
------
sin
…
g
31
31
π
8
----------
sin
++ ++
=
φ
cos
πφ
–
()
cos
–
πφ
+
()
cos
–
2
πφ
–
()
cos
===
φ
sin
πφ
–
()
sin
πφ
+
()
sin
–
2
πφ
–
()
sin
–== =
φ
cos
π
2
---
φ
–
sin
=
Re
g
()
1
8
---
k
1
g
1
g
7
–
g
9
–
g
15
g
17
g
23
–
g
25
–
g
31
++ +
()
k
2
g
2
g
6
–
g
10
–
g
14
g
18
g
22
–
g
26
–
g
30
++ +
()
k
3
g
3
g
5
–
g
11
–
g
13
g
19
g
21
–
g
27
–
g
29
++ +
()
g
0
g
8
–
g
16
g
24
–+
()
++
+
(
)
=
Im
g
()
1
8
---
k
1
g
3
g
5
g
11
–
g
13
–
g
19
g
21
g
27
–
g
29
–+++
()
k
2
g
2
g
6
g
10
–
g
14
–
g
18
g
22
g
26
–
g
30
–+++
()
k
3
g
1
g
7
g
9
–
g
15
–
g
17
g
23
g
25
–
g
31
–+++
()
g
4
g
12
–
g
20
g
28
–+
()
++
+
(
)
=
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