Measurement Algorithms for Digital Protection - Digital Signal Processing in Power System Protection and Control - page 154

The same result can be achieved by increasing the number of counted pulses by
increasing the number of half periods. Having k half periods measured frequency is:

f m ¼ kf S

2M k ;

ð 8 : 139 Þ

where M k is a number of samples in k half-periods.
Measurement error is given by the equation:

d k ¼ 1
M k

1
kM 0 : 5

¼ ð 8 : 140 Þ

and is k-time less than in case of one-half period.
Evident disadvantage is k-time is longer measurement time.

Example 8.15 Applying the frequency measurement method by counting of
impulses one should assure the error of measurement of the frequencies close to
50 Hz not higher than 1%. What should be the sampling frequency, so that the
required error should not exceed during impulse counting within single half-per-
iod? How many half-periods one should consider when the sampling frequency is
1000 Hz?

Solution The required sampling frequency results directly from the error

Eq. 8.138 :

d 0 : 5 ¼ 2f m
f S

2f 1
f S

¼ 100
f S

¼ 0 : 01 ;

therefore: f S = 10 kHz.
On the other hand, for the assumed sampling at 1000 Hz, where the number of
samples per cycle equals 20 and for a half-cycle is 10, the frequency measurement
error by counting over k half-cycles amounts to ( 8.140 ):

1
10k ¼ 0 : 01 ;

d ¼

thus k ¼ 10 : Therefore, in order to get the error not higher than 1% one should
perform impulse counting over ten half-cycles, i.e. over 100 ms.
The third method of decreasing the error relies on reaching greater precision of
zero crossing between sampling instants. The situation is presented in Fig. 8.8 .
Assuming straight line interpolation and the number of samples of the same sign
equal to M 0 : 5 one gets more precise value of half period of the signal:
T m
2 ¼ M 0 : 5 T S þ t a2 t b2 :

ð 8 : 141 Þ

The equation can be rewritten in the form:

T m
2T S

u k þ 1
u k þ 1 u k

u m þ 1
u m þ 1 u m

¼ M 0 : 5 þ ; ð 8 : 142 Þ

p p þ 1