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<strong>The</strong> <strong>Sinc</strong> <strong>Dirichlet</strong> <strong>Function</strong><br />
February 20, 2013<br />
Diric (x) =<br />
sin nx<br />
2<br />
n sin x 2<br />
1 <strong>The</strong> L-point running-average filter<br />
y[n] = 1 ∑L−1<br />
x[n − k]<br />
L<br />
<strong>The</strong> output is the average of x[n] and the previous L − 1 samples (total L samples )<br />
k=0<br />
y[n] = 1 ( x[n] + x[n − 1] + x[n − 2] + ... + x[n − L + 2] + x[n − L + 1] )<br />
L<br />
This is an example of linear time-invariant system<br />
Consider the frequency response<br />
i.e. When input is x[n] = Ae j(nˆω+ϕ)<br />
y(ˆω) = 1 ∑L−1<br />
Ae j((n−k)ˆω+ϕ) =<br />
L<br />
k=0<br />
Apply the sum of geometric series<br />
H (ˆω) = 1 L<br />
(<br />
1 ∑L−1<br />
L<br />
∑L−1<br />
k=0<br />
k=0<br />
−j ˆωk<br />
e<br />
e −j ˆωk )<br />
Ae j(nˆω+ϕ)<br />
Thus<br />
Apply a trick<br />
N−1<br />
∑<br />
n=0<br />
H (ˆω) = 1 L<br />
= 1 L<br />
a n = 1 − aN<br />
1 − a<br />
∑L−1<br />
k=0<br />
−j ˆωL<br />
1 − e<br />
1 − e−j ˆω<br />
−j ˆωk<br />
e<br />
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<br />
− e<br />
ˆωN/2)<br />
1
H (ˆω) = 1 e ( −j ˆωL/2 e j ˆωL/2 −j<br />
− e<br />
ˆωL/2)<br />
L e −j ˆω/2 (e j ˆω/2 − e −j ˆω/2) )<br />
= 1 L<br />
(<br />
e<br />
j ˆωL/2 − e −j ˆωL/2) · 1<br />
2j<br />
(e j ˆω/2 − e −j ˆω/2) ) · 1<br />
2j<br />
= 1 sin ˆωL 2<br />
L sin ˆω 2<br />
·<br />
−j ˆω(L−1)/2<br />
e<br />
ˆωL/2 e−j<br />
e−j ˆω/2<br />
−j ˆω(L−1)/2<br />
H(ˆω) = Diric (ˆω) e<br />
Thus, the Diric function is the magnitude component of the L-point running average filter<br />
2 <strong>The</strong> <strong>Dirichlet</strong> <strong>Function</strong><br />
Diric (x) =<br />
sin nx<br />
2<br />
n sin x 2<br />
When x = 2kπ<br />
k ∈ Z<br />
=<br />
Diric (x) =<br />
sin nx<br />
2<br />
n x 2<br />
x<br />
2<br />
sin x 2<br />
sin nkπ<br />
n sin kπ<br />
=<br />
sinc nx<br />
2<br />
sinc x 2<br />
<strong>The</strong>refore<br />
lim<br />
x→0<br />
Diric (x) = lim<br />
x→0<br />
sinc nx<br />
2<br />
sinc x 2<br />
<strong>Sinc</strong>e the limit exists for both upper side and lower side<br />
Thus<br />
Also, it can be shown that<br />
=<br />
lim sincnx<br />
x→0 2<br />
lim sincx<br />
x→0 2<br />
= 1 1 = 1<br />
limDiric (x) = 1<br />
x→0<br />
max Diric (x) = 1<br />
−END−<br />
2