The Fourier transform of the discretetime signal
sn
s n
is defined to be
Sej2πf=∑
n
=−∞∞sne−(j2πfn)
S
2
f
n
s
n
2
f
n
(1)
Frequency here has no units. As should be expected, this
definition is linear, with the transform of a sum of signals
equaling the sum of their transforms. Realvalued signals have
conjugatesymmetric spectra:
Se−(j2πf)=Sej2πf*
S
2
f
S
j
2
f
.
A special property of the discretetime Fourier transform is
that it is periodic with period one:
Sej2π(f+1)=Sej2πf
S
2
f
1
S
2
f
.
Derive this property from the definition of the DTFT.
Sej2π(f+1)=∑
n
=−∞∞sne−(j2π(f+1)n)=∑
n
=−∞∞e−(j2πn)sne−(j2πfn)=∑
n
=−∞∞sne−(j2πfn)=Sej2πf
S
2
f
1
n
s
n
2
f
1
n
n
2
n
s
n
2
f
n
n
s
n
2
f
n
S
2
f
(2)
Because of this periodicity, we need only plot the spectrum over
one period to understand completely the spectrum's structure;
typically, we plot the spectrum over the frequency range
−12
12
1
2
1
2
.
When the signal is realvalued, we can further simplify our
plotting chores by showing the spectrum only over
0
12
0
1
2
;
the spectrum at negative frequencies can be derived from
positivefrequency spectral values.
When we obtain the discretetime signal via sampling an analog
signal, the Nyquist frequency corresponds to the
discretetime frequency
12
1
2
. To show this, note that a sinusoid having a
frequency equal to the Nyquist frequency
12Ts
1
2
Ts
has a sampled waveform that equals
cos2π×12TsnTs=cosπn=−1n
2
1
2
T
s
n
T
s
n
1
n
The exponential in the DTFT at frequency
12
1
2
equals
e−j2πn2=e−(jπn)=−1n
2
n
2
n
1
n
, meaning that discretetime frequency equals analog
frequency multiplied by the sampling interval
f
D
=
f
A
Ts
f
D
f
A
Ts
(3)
f
D
f
D
and
f
A
f
A
represent discretetime 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
Ts
t
p
Ts
t
narrows, the amplitudes of the signal's spectral
repetitions, which are governed by the
Fourier series coefficients of
p
Ts
t
p
Ts
t
, become increasingly equal. Examination of the
periodic pulse
signal reveals that as
ΔΔ decreases, the value of
c
0
c
0
,
the largest Fourier coefficient, decreases to zero:

c
0
=AΔTs
c
0
A
Δ
Ts
.
Thus, to maintain a mathematically viable Sampling Theorem, the
amplitude
AA must increase as
1Δ
1
Δ
, becoming infinitely large as the pulse duration
decreases. Practical systems use a small value of
ΔΔ, say
0.1·
Ts
0.1
·
Ts
and use amplifiers to rescale the signal. Thus, the sampled
signal's spectrum becomes periodic with period
1Ts
1
Ts
.
Thus, the Nyquist frequency
12Ts
1
2
Ts
corresponds to the frequency
12
1
2
.
Let's compute the discretetime Fourier transform of the
exponentially decaying sequence
sn=anun
s
n
a
n
u
n
,
where
un
u
n
is the unitstep sequence. Simply plugging the signal's
expression into the Fourier transform formula,
Sej2πf=∑
n
=−∞∞anune−(j2πfn)=∑
n
=0∞ae−(j2πf)n
S
2
f
n
a
n
u
n
2
f
n
n
0
a
2
f
n
(4)
This sum is a special case of the geometric
series.
∑
n
=0∞αn=11−α ,
α<1
n
0
α
n
α
α
1
1
1
α
(5)
Thus, as long as
a<1
a
1
,
we have our Fourier transform.
Sej2πf=11−ae−(j2πf)
S
2
f
1
1
a
2
f
(6)
Using Euler's relation, we can express the magnitude and phase
of this spectrum.
Sej2πf=11−acos2πf2+a2sin22πf
S
2
f
1
1
a
2
f
2
a
2
2
f
2
(7)
∠Sej2πf=−tan1asin2πf1−acos2πf
S
2
f
a
2
f
1
a
2
f
(8)No matter what value of aa we
choose, the above formulae clearly demonstrate the periodic
nature of the spectra of discretetime signals. Figure 1 shows indeed that the spectrum
is a periodic function. We need only consider the spectrum
between
−12
1
2
and
12
1
2
to unambiguously define it. When
a>0
a
0
,
we have a lowpass spectrum—the spectrum diminishes as
frequency increases from 0 to
12
1
2
—with increasing
aa leading to a greater low frequency
content; for
a<0
a
0
,
we have a highpass spectrum
(Figure 2).
Analogous to the analog pulse signal, let's find the spectrum
of the lengthNN pulse sequence.
sn={1 if 0≤n≤N−10 otherwise
s
n
1
0
n
N
1
0
(9)
The Fourier transform of this sequence has the form of a
truncated geometric series.
Sej2πf=∑
n
=0N−1e−(j2πfn)
S
2
f
n
0
N
1
2
f
n
(10)
For the socalled finite geometric series, we know that
∑
n
=
n
0
N+
n
0
−1αn=α
n
0
1−αN1−α
n
n
0
N
n
0
1
α
n
α
n
0
1
α
N
1
α
(11)
for
all values of α.
Derive this formula for the finite geometric series sum.
The "trick" is to consider the difference between the
series' sum and the sum of the series multiplied by
αα.
α∑n=
n
0
N+
n
0
−1αn−∑n=
n
0
N+
n
0
−1αn=αN+
n
0
−α
n
0
α
n
n
0
N
n
0
1
α
n
n
n
0
N
n
0
1
α
n
α
N
n
0
α
n
0
which, after manipulation, yields the geometric sum formula.
Applying this result yields (Figure 3.)
Sej2πf=1−e−(j2πfN)1−e−(j2πf)=e−(jπf(N−1))sinπfNsinπf
S
2
f
1
2
f
N
1
2
f
f
N
1
f
N
f
(12)
The ratio of sine functions has the generic form of
sinNxsinx
N
x
x
,
which is known as the
discretetime sinc function
dsincx
dsinc
x
.
Thus, our transform can be concisely expressed as
Sej2πf=e−(jπf(N−1))dsincπf
S
2
f
f
N
1
dsinc
f
. The discretetime pulse's spectrum contains many
ripples, the number of which increase with
NN, the pulse's duration.
The inverse discretetime Fourier transform is easily derived
from the following relationship:
∫−1212e−(j2πfm)ej2πfnd
f
={1 if m=n0 if m≠n=δm−n
1
2
1
2
f
2
f
m
2
f
n
1
m
n
0
m
n
δ
m
n
(13)
Therefore, we find that
∫−1212Sej2πfej2πfnd
f
=∫−1212∑
m
msme−(j2πfm)ej2πfnd
f
=∑
m
msm∫−1212e(−(j2πf))(m−n)d
f
=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
(14)
The Fourier transform pairs in discretetime are
Sej2πf=∑n=−∞∞sne−(j2πfn)
sn=∫−1212Sej2πfej2πfndf
S
2
f
n
s
n
2
f
n
s
n
f
1
2
1
2
S
2
f
2
f
n
(15)
The properties of the discretetime Fourier transform mirror
those of the analog Fourier transform. The
DTFT properties table
shows similarities and differences. One important common
property is Parseval's Theorem.
∑
n
=−∞∞sn2=∫−1212Sej2πf2d
f
n
s
n
2
f
1
2
1
2
S
2
f
2
(16)
To show this important property, we simply substitute the
Fourier transform expression into the frequencydomain
expression for power.
∫−1212Sej2πf2d
f
=∫−1212∑
n
nsne−(j2πfn)∑
m
msn*ej2πfmd
f
=∑
n,m
n,msnsn*∫−1212ej2πf(m−n)d
f
f
1
2
1
2
S
2
f
2
f
1
2
1
2
n
n
s
n
2
f
n
m
m
s
n
2
f
m
,
n
m
,
n
m
s
n
s
n
f
1
2
1
2
2
f
m
n
(17)
Using the
orthogonality
relation, the integral equals
δm−n
δ
m
n
,
where
δn
δ
n
is the
unit sample. Thus, the double sum collapses
into a single sum because nonzero values occur only when
n=mnm,
giving Parseval's Theorem as a result. We term
∑
n
ns2n
n
n
s
n
2
the energy in the discretetime signal
sn
s
n
in spite of the fact that discretetime signals don't consume
(or produce for that matter) energy. This terminology is a
carryover from the analog world.
Suppose we obtained our discretetime signal from values of
the product
st
p
T
s
t
s
t
p
T
s
t
,
where the duration of the component pulses in
p
T
s
t
p
T
s
t
is ΔΔ. How is
the discretetime signal energy related to the total energy
contained in
st
s
t
?
Assume the signal is bandlimited and that the sampling rate
was chosen appropriate to the Sampling Theorem's conditions.
If the sampling frequency exceeds the Nyquist frequency, the
spectrum of the samples equals the analog spectrum, but over
the normalized analog frequency
fT
f
T
. Thus, the energy in the sampled signal equals
the original signal's energy multiplied by
TT.
"Electrical Engineering Digital Processing Systems in Braille."