Finite Impulse Response Filters - Digital Signal Processing in Power System Protection and Control - page 74

(a) (b)

1

w 0 k )

0

k

-1

0

N 21 − N −1

1

w 1 k )

0 k

-1

0

N 41 − N 21 − N −1

Fig. 6.6 Normalized impulse (a) and frequency responses (b) of half cycle Walsh filters w 1 ; w 2

¼ p

2 N 1 = 2 1

X 1 ¼þ p arg W 1 ð jX 1 Þ ½

N 1 :

ð 6 : 25c Þ

2

These two cases of pairs of full and a half cycle Walsh orthogonal functions are
very important for practical applications in power system protection devices.
Identical filter gains and full orthogonality allow to get very simple algorithms of
protection criterion values. The normalized frequency responses of the filters are
shown in Fig. 6.6 .

Example 6.3 Calculate relative gains of the first and second order Walsh filters
(with respect to the gains for the values for the fundamental frequency 50 Hz) for
the third and fifth harmonic components. Assume sampling frequency 1000 Hz.
Solution The number of samples in one cycle of fundamental frequency equals
N 1 ¼ f S = f 1 ¼ 20. Thus, from Eqs. 6.20 , 6.23 and 6.24a , b , c one gets:

X 1 ¼ 0 : 1p ; X 3 ¼ 0 : 3p ; X 5 ¼ 0 : 5p ;

2
sin ð X 1 = 2 Þ ¼ 12 : 785 ;

j

W 1 ð jX 1 Þ

j ¼ W 2 ð jX 1 Þ j j ¼

2 sin 2 ð 0 : 25N 1 X 3 Þ
W 1 ð jX 1 Þ

j W 1 ð jX 3 Þ j 2 : 2

12 : 785 ¼ 0 : 17 ; j ¼ j sin ð 0 : 5X 3 Þ ¼ j W 1 ð jX 1 Þ j

j W 1 ð jX 5 Þ j

j ¼ 0 : 11 ;

j W 1 ð jX 1 Þ

j W 2 ð jX 3 Þ j

j ¼ 2 sin ð 0 : 25N 1 X 3 Þ 1 cos ð 0 : 25N 1 X 3 Þ

½ 5 : 28

12 : 785 ¼ 0 : 41 ;

¼

j W 2 ð jX 1 Þ j W 1 ð jX 1 Þ j sin ð 0 : 5X 3 Þ

j W 2 ð jX 5 Þ j

j ¼ 0 : 266 : j

W 2 ð jX 1 Þ