When we obtain the discrete-time signal via sampling an analog
signal, the Nyquist frequency corresponds to the discrete-time
frequency
12
1
2
. To show this, note that a sinusoid at the Nyquist frequency
12
T
s
1
2
T
s
has a sampled waveform that equals
cos2π12
T
s
n
T
s
=cosπn=-1n
2
1
2
T
s
n
T
s
n
1
n
(1)
The exponential in the DTFT at frequency
12
1
2
equals
ⅇ-ⅈ2πn2=ⅇ-ⅈπn=-1n
2
n
2
n
1
n
, meaning that the correspondence between analog and
discrete-time frequency is established:
f
D
=
f
A
T
s
f
D
f
A
T
s
(2)
where
f
D
f
D
and
f
A
f
A
represent discrete-time and analog frequency
variables, respectively. The aliasing figure provides
another way of deriving this result. As the duration of each
pulse in the periodic sampling signal
p
T
s
t
p
T
s
t
narrows, the amplitudes of the signal's spectral
repetitions, which are governed by the Fourier series coefficients
of
p
T
s
t
p
T
s
t
, become increasingly equal.
Thus, the sampled signal's spectrum becomes periodic with period
1
T
s
1
T
s
. Thus, the Nyquist frequency
12
T
s
1
2
T
s
corresponds to the frequency
12
1
2
.
The inverse discrete-time Fourier transform is easily derived
from the following relationship:
∫-1212ⅇ-ⅈ2πfmⅇ+ⅈπfndf=1ifm=n0ifm≠n
1
2
1
2
f
2
f
m
f
n
1
m
n
0
m
n
(3)
Therefore, we find that
∫-1212Sⅇⅈ2πfⅇ+ⅈ2πfndf=∫-1212∑msmⅇ-ⅈ2πfmⅇ+ⅈ2πfndf=∑msm∫-1212ⅇ-ⅈ2πfm−ndf=sn
f
1
2
1
2
S
2
f
2
f
n
f
1
2
1
2
m
m
s
m
2
f
m
2
f
n
m
m
s
m
f
1
2
1
2
2
f
m
n
s
n
(4)
The Fourier transform pairs in discrete-time are
Sⅇⅈ2πf=∑nsnⅇ-ⅈ2πfn
S
2
f
n
n
s
n
2
f
n
(5)
sn=∫-1212Sⅇⅈ2πfⅇ+ⅈ2πfndf
s
n
f
1
2
1
2
S
2
f
2
f
n
(6)
"My introduction to signal processing course at Rice University."