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-ⅈ2πNm-
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-ⅈ2πNkm=1Nⅇⅈ2πN
N
1
k1-ⅇ-ⅈ2πN2
N
1
+11-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-ⅈk2π2Nⅇⅈk2πN
N
1
+12-ⅇ-ⅈk2πN
N
1
+12ⅇ-ⅈk2π2Nⅇⅈk2πN12-ⅇ-ⅈk2πN12=1Nsin2πk
N
1
+12NsinπkN=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)
note:
It's periodic!
Figure 3,
Figure 4, and
Figure 5show our
above function and coefficients for various values of
N
1
N
1
.
"My introduction to signal processing course at Rice University."