Measurement Algorithms for Digital Protection - Digital Signal Processing in Power System Protection and Control - page 160

:

DX
X 1

¼ 1
p

y S ð n Þ y C ð n N 1 = 2 Þ y C ð n Þ y S ð n N 1 = 2 Þ
y S ð n Þ y C ð n N 1 = 4 Þ y C ð n Þ y S ð n N 1 = 4 Þ

ð 8 : 158 Þ

Described method can and may be used to measure both frequency and fre-
quency deviations. The latter can use simplified equations giving reasonable
accuracy for frequency deviations that are met in practice.

Example 8.17 Provide examples of simplified and accurate algorithms of fre-
quency measurement with use of orthogonal components. Assume sampling at
1000 Hz. For measured frequencies 48 and 45 Hz determine relative measurement
error of the simplified method (related to the nominal value of 50 Hz). Draw the
course of this error in the frequency range from 45 to 55 Hz.

Solution Measurement of frequency can be done according to Eq. 8.156 .

Assuming possibly shortest delay (k ¼ 1) one obtains:

;

X ¼ 2pf
f S

0 : 5 y S ð n Þ y C ð n 2 Þ y C ð n Þ y S ð n 2 Þ
y S ð n Þ y C ð n 1 Þ y C ð n Þ y S ð n 1 Þ

¼ arccos

whereas the orthogonal signal components for the adopted sampling frequency are
calculated from:

y C ð n Þ¼ X

19

x ð n k Þ cos ½ 0 : 1p ð 9 : 5 k Þ;

k ¼ 0

y S ð n Þ¼ X

19

x ð n k Þ sin ½ 0 : 1p ð 9 : 5 k Þ:

k ¼ 0

Instead of function arccos one can apply the simplified version, under the
condition that the delay is equivalent to a quarter of signal cycle (k ¼ N 1 = 4 ¼ 5),
which gives:

DX
X 1

¼ Df
f 1

¼ 1

p y S ð n Þ y C ð n 10 Þ y C ð n Þ y S ð n 10 Þ

y S ð n Þ y C ð n 5 Þ y C ð n Þ y S ð n 5 Þ

and finally the frequency deviation can be obtained from:

:

Df ¼ 2f 1
p

pf md
2f 1

cos

The error of measurement resulting from the adopted simplification is equal:

D ¼ f 1 þ Df f md ¼ 50 þ Df f md

where f 1 is a nominal power frequency (50 Hz), Df is a frequency deviation
calculated with the simplified method, f md is an accurate value of the measured