Finite Impulse Response Filter Design Equations by the Frequency Sampling Design Method - Digital Signal Processing: Fundamentals and Applications - page 818

N 1
n¼ 0 ðW N e j U Þ n ¼

1 e jð U 2 p k=NÞN
1 e jð U 2 p k=NÞ

(E.6)

Using the Euler formula, Equation (E.6) becomes

N 1

n¼ 0 ðW N e j U Þ n ¼ e jNð U 2 p k=NÞ= 2

ðe jNð U 2 p k=NÞ= 2 e jNð U 2 p k=NÞ= 2 Þ= 2 j

e jð U 2 p k=NÞ= 2 ðe jð U 2 p k=NÞ= 2 e jð U 2 p k=NÞ= 2 Þ= 2 j

¼ e jNð U 2 p k=NÞ= 2 sin ½Nð U 2 p k=NÞ= 2
e jð U 2 p k=NÞ= 2 sin ½ð U 2 p k=NÞ= 2

(E.7)

Substituting Equation (E.7) into Equation (E.5) leads to

N e jðN 1 Þ U = 2 N 1

k ¼ 0 HðkÞe jðN 1 Þk p =N sin ½Nð U 2 p k=NÞ= 2

1

Hðe j U Þ¼ (E.8)

sin ½ð U 2 p k=NÞ= 2 Þ

2 p m
N

Let U ¼ U m ¼ , and substitute it into Equation (E.8) to get

N e jðN 1 Þ 2 p m=ð 2 NÞ N 1

1

k ¼ 0 HðkÞe jðN 1 Þk p =N sin ½Nð 2 p m=N 2 p k=NÞ= 2

Hðe j U m Þ¼ (E.9)

sin ½ð 2 p m=N 2 p k=NÞ= 2 Þ

Clearly, when m s k , the last term of the summation in Equation (E.9) becomes

sin ½Nð 2 p m=N 2 p k=NÞ= 2
sin ½ð 2 p m=N 2 p k=NÞ= 2 Þ ¼

sin ð p ðm kÞÞ
sin ð p ðm kÞ=NÞ ¼

0
sin ð p ðm kÞ=NÞ ¼ 0

When m ¼ k , using L'Hospital's rule we have

sin ½Nð 2 p m=N 2 p k=NÞ= 2
sin ½ð 2 p m=N 2 p k=NÞ= 2 Þ ¼

sin ðN p ðm kÞ=NÞ
sin ð p ðm kÞ=NÞ

sin ðNxÞ
sin ðxÞ

¼ lim
x / 0

¼ N

Then Equation (E.9) is simplified to

1

Hðe j U k Þ¼

N e jðN 1 Þ p k=N HðkÞe jðN 1 Þk p =N N ¼ HðkÞ

that is,

Hðe j U k Þ¼HðkÞ; 0 k N 1 (E.10)

2 p k
N

where U k ¼ , corresponding to the k th DFT frequency component. The fact is that if we specify

the desired frequency response, Hð U k Þ ,0 k N 1, at the equally spaced sampling frequency
determined by U k ¼

2 p k
N

, they are actually the DFT coefficients; that is, HðkÞ ,0 k N 1, via
Equation (E.10) . Furthermore, the inverse of the DFT calculated using (E.10) will give the desired
impulse response, hðnÞ ,0 n N 1.