Measurement Algorithms for Digital Protection - Digital Signal Processing in Power System Protection and Control - page 118

e N e N ¼ X

N 1

e n ! min : ð 8 : 19 Þ

n ¼ 0

It can easily be derived and the parameters of signal model are then calculated
from:

1 h N x N ¼ P N x N ;

X ¼ h N h N

ð 8 : 20 Þ

where X is a approximate vector of sought parameters determined with minimum
least square error and P N is a LSE transformation matrix.
It should be noted that the matrix P N in ( 8.20 ) for given numbers of K and N is
fully determined before the calculations start and thus it can be prepared well in
advance and stored in look-up tables in the protection memory. Application of the
estimation procedure in on-line mode requires execution of ( 8.20 ) at consecutive
time instants, with new sets of signal samples captured within fixed-data window.

Example 8.2 For the signal model ( 8.15 ) containing the fundamental frequency
component only determine the model parameters and calculate the signal mag-
nitude assuming availability of N = 2, 6, 12 signal samples. Assume sampling
frequency f S = 600 Hz, unity level of the signal and its initial phase equal to p = 6.
Draw the course of measured signal magnitude for the case of pure fundamental
frequency signal and for the case of signal distorted by the second harmonic
component (10% of the fundamental).

Solution One should note that the minimum window length for estimation of
signal parameters is here equal to two sampling periods (two parameters of the
model are sought). Extending-data window length makes the problem overdeter-
mined which should result in improved accuracy, especially when the signal dif-
fers from assumed model (distortions, noise). The window length N = N 1 = 12
covers full cycle of the signal, thus the expected results should be close to the ones
obtained when full-cycle Fourier filters are applied.
For the assumed sampling rate the relative fundamental angular frequency is
X 1 ¼ 2pf 1 = f S ¼ 2p50 = 600 ¼ p = 6 : The signal model can be expressed in the form:

x ð n Þ¼ X 1C cos ð nX 1 Þþ X 1S sin ð nX 1 Þþ e ð n Þ
¼ X 1C cos ð np = 6 Þþ X 1S sin ð np = 6 Þþ e ð n Þ:

For the case when there are no distortions the signal samples are equal:

x ð n Þ¼ X 1m cos ð nX 1 þ u Þ¼ cos ð np = 6 þ p = 4 Þ
¼ 0 : 707 cos ð np = 6 Þ 0 : 707 sin ð np = 6 Þ:

For the window lengths N = 2, 6, 12 the matrix h N is equal: