Calculate the DTFS
ck
c
k
using:
ck=1N∑n=0N−1fne−(i2πNkn)
c
k
1
N
n
0
N
1
f
n
2
N
k
n
(2)
Just like continuous time Fourier series, we can take the summation
over any interval, so we have
c
k
=1N∑n=−
N
1
N
1
e−(i2πNkn)
c
k
1
N
n
N
1
N
1
2
N
k
n
(3)
Let
m=n+
N
1
m
n
N
1
(so we can get a geometric series starting at 0)
c
k
=1N∑m=02
N
1
e−(i2πN(m−
N
1
)k)=1Nei2πNk∑m=02
N
1
e−(i2πNmk)
c
k
1
N
m
0
2
N
1
2
N
m
N
1
k
1
N
2
N
k
m
0
2
N
1
2
N
m
k
(4)
Now, using the "partial summation formula"
∑n=0Man=1−aM+11−a
n
0
M
a
n
1
a
M
1
1
a
(5)
c
k
=1Nei2πN
N
1
k∑m=02
N
1
e−(i2πNk)m=1Nei2πN
N
1
k1−e−(i2πN(2
N
1
+1))1−e−(ik2πN)
c
k
1
N
2
N
N
1
k
m
0
2
N
1
2
N
k
m
1
N
2
N
N
1
k
1
2
N
2
N
1
1
1
k
2
N
(6)
Manipulate to make this look like a sinc function (distribute):
c
k
=1Ne−(ik2π2N)(eik2πN(
N
1
+12)−e−(ik2πN(
N
1
+12)))e−(ik2π2N)(eik2πN12−e−(ik2πN12))=1Nsin2πk(
N
1
+12)NsinπkN=
digital sinc
c
k
1
N
k
2
2
N
k
2
N
N
1
1
2
k
2
N
N
1
1
2
k
2
2
N
k
2
N
1
2
k
2
N
1
2
1
N
2
k
N
1
1
2
N
k
N
digital sinc
(7)