The Sinc Dirichlet Function

The Sinc Dirichlet Function
February 20, 2013
Diric (x) =
sin nx
2
n sin x 2
1 The L-point running-average filter
y[n] = 1 ∑L−1
x[n − k]
L
The output is the average of x[n] and the previous L − 1 samples (total L samples )
k=0
y[n] = 1 ( x[n] + x[n − 1] + x[n − 2] + ... + x[n − L + 2] + x[n − L + 1] )
L
This is an example of linear time-invariant system
Consider the frequency response
i.e. When input is x[n] = Ae j(nˆω+ϕ)
y(ˆω) = 1 ∑L−1
Ae j((n−k)ˆω+ϕ) =
L
k=0
Apply the sum of geometric series
H (ˆω) = 1 L
(
1 ∑L−1
L
∑L−1
k=0
k=0
−j ˆωk
e
e −j ˆωk )
Ae j(nˆω+ϕ)
Thus
Apply a trick
N−1
∑
n=0
H (ˆω) = 1 L
= 1 L
a n = 1 − aN
1 − a
∑L−1
k=0
−j ˆωL
1 − e
1 − e−j ˆω
−j ˆωk
e
1 − e j ˆωN = e ( j ˆωN/2 e −j ˆωN/2 − e j ˆωN/2) 1 − e −j ˆωN = e ( −j ˆωN/2 e j ˆωN/2 −j
− e
ˆωN/2)
1

H (ˆω) = 1 e ( −j ˆωL/2 e j ˆωL/2 −j
− e
ˆωL/2)
L e −j ˆω/2 (e j ˆω/2 − e −j ˆω/2) )
= 1 L
(
e
j ˆωL/2 − e −j ˆωL/2) · 1
2j
(e j ˆω/2 − e −j ˆω/2) ) · 1
2j
= 1 sin ˆωL 2
L sin ˆω 2
·
−j ˆω(L−1)/2
e
ˆωL/2 e−j
e−j ˆω/2
−j ˆω(L−1)/2
H(ˆω) = Diric (ˆω) e
Thus, the Diric function is the magnitude component of the L-point running average filter
2 The Dirichlet Function
Diric (x) =
sin nx
2
n sin x 2
When x = 2kπ
k ∈ Z
=
Diric (x) =
sin nx
2
n x 2
x
2
sin x 2
sin nkπ
n sin kπ
=
sinc nx
2
sinc x 2
Therefore
lim
x→0
Diric (x) = lim
x→0
sinc nx
2
sinc x 2
Since the limit exists for both upper side and lower side
Thus
Also, it can be shown that
=
lim sincnx
x→0 2
lim sincx
x→0 2
= 1 1 = 1
limDiric (x) = 1
x→0
max Diric (x) = 1
−END−
2