Finite Impulse Response Filter Design - Digital Signal Processing: Fundamentals and Applications - page 236

U H ¼ 2pf H T ¼ 2p 2; 400=8; 000 ¼ 0:6p radians

Since 2M þ 1 ¼ 5 in this case, using the equation in Table 7.1 yields

<
:

p U H þ U L
p

n ¼ 0

hðnÞ¼

sinðU H n Þ
np

þ sinðU L n Þ
np

n s 0 2 n 2

When n ¼ 0, we have

hð0Þ¼ p U H þ U L
p

¼ p 0:6p þ 0:5p
p

¼ 0:9

The other computed filter coefficients for the previous expression are listed below:
hð1Þ¼ sin½0:5p 1
1 p

sin½0:6p 1
1 p

¼ 0:01558

hð2Þ¼ sin½0:5p 2
2 p

sin½0:6p 2
2 p

¼ 0:09355

Using symmetry leads to

hð1Þ¼hð1Þ¼0:01558

hð2Þ¼hð2Þ¼0:09355

Applying the Hamming window function in Equation (7.18) , we have

0 p
2

w ham ð0Þ¼0:54 þ 0:46 cos ¼ 1:0

1 p
2

w ham ð1Þ¼0:54 þ 0:46 cos ¼ 0:54

2 p
2

w ham ð2Þ¼0:54 þ 0:46 cos ¼ 0:08

Using the symmetry of the window function gives

w ham ð1Þ¼w ham ð1Þ¼0:54

w ham ð2Þ¼w ham ð2Þ¼0:08

The windowed impulse response is calculated as

h w ð0Þ¼hð0Þw ham ð0Þ¼0:9 1 ¼ 0:9