Finite Impulse Response Filters - Digital Signal Processing in Power System Protection and Control - page 78

(a) (b)

A ( j Ω )

1

ak c () A ( j Ω )

0

1

k

0

0.8

- 1

0

N 0 1 − 21

0

N −

0.6

a C

1

ak

s () 0.4

a S

0 k

0.2

- 1 0

21
0

N − 0

N 0 1 −

0 2 4 6 8

Ω Ω

0

Fig. 6.9 Characteristics of double-cycle sine/cosine filters: a impulse responses b frequency

spectra

window to full cycle (by the same natural frequency) one gets excellent band-pass
filter of second harmonic.

Example 6.4 Calculate the level of suppression of the frequencies f a ¼ 125 and
f b ¼ 175 Hz by the full cycle sine, cosine filters (in respect to filter gains for the
fundamental frequency f 1 ¼ 50 Hz). Compare results for two sampling frequencies
f s ¼ 1000 and 600 Hz.
Solution Equations 6.30 , 6.31 and 6.33a , b , c , d , e are used for the calculations

(1) Sampling frequency 1000 Hz

Number of samples per cycle is N 1 ¼ f S = f 1 ¼ 20 and:

X a ¼ 2pf a = f S ¼ 2p 125 = 1000 ¼ 0 : 25p ; X b ¼ 0 : 35p ; X 1 ¼ 0 : 1p :

From ( 6.30 ) and ( 6.31 ) one obtains:

j ¼ 0 : 5 sin ½ 0 : 5N 1 ð X X 1 Þ
sin ½ 0 : 5 ð X X 1 Þ

þ 0 : 5 sin ½ 0 : 5N 1 ð X þ X 1 Þ
sin ½ 0 : 5 ð X þ X 1 Þ

j A c ð jX Þ ;

j ¼ 0 : 5 sin ½ 0 : 5N 1 ð X X 1 Þ
sin ½ 0 : 5 ð X X 1 Þ

0 : 5 sin ½ 0 : 5N 1 ð X þ X 1 Þ
sin ½ 0 : 5 ð X þ X 1 Þ

j A s ð jX Þ

and from ( 6.33a , b , c , d , e ): j A c ð jX 1 Þ j ¼ A s ð jX 1 Þ j j ¼ N 1 = 2 ¼ 10 :

Substituting the calculated values yields:

j

A c ð jX a Þ

j

sin ½ 5 ð 0 : 25p 0 : 1p Þ
sin ½ 0 : 5 ð 0 : 25p 0 : 1p Þ þ 0 : 5

sin ½ 5 ð 0 : 25p þ 0 : 1p Þ
sin ½ 0 : 5 ð 0 : 25p þ 0 : 1p Þ j ¼ 0 : 10 : 5 j A c ð jX 1 Þ

¼ 0 : 05 ð 4 : 27 þ 1 : 91 Þ¼ 0 : 309 ;