Discussion of Discrete-time Fourier Transforms. Topics include comparison with analog transforms and discussion of Parseval's theorem.
The Fourier transform of the discrete-time signal
is defined to be
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. Real-valued signals have
conjugate-symmetric spectra:
.
A special property of the discrete-time Fourier transform is
that it is periodic with period one:
.
Derive this property from the definition of the DTFT.
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
.
When the signal is real-valued, we can further simplify our
plotting chores by showing the spectrum only over
;
the spectrum at negative frequencies can be derived from
positive-frequency spectral values.
When we obtain the discrete-time signal via sampling an analog
signal, the Nyquist frequency corresponds to the
discrete-time frequency
. To show this, note that a sinusoid having a
frequency equal to the Nyquist frequency
has a sampled waveform that equals
The exponential in the DTFT at frequency
equals
, meaning that discrete-time frequency equals analog
frequency multiplied by the sampling interval
and
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
narrows, the amplitudes of the signal's spectral
repetitions, which are governed by the Fourier series coefficients of
, become increasingly equal. Examination of the periodic pulse
signal reveals that as
decreases, the value of
,
the largest Fourier coefficient, decreases to zero:
.
Thus, to maintain a mathematically viable Sampling Theorem, the
amplitude
must increase as
, becoming infinitely large as the pulse duration
decreases. Practical systems use a small value of
, say
and use amplifiers to rescale the signal. Thus, the sampled
signal's spectrum becomes periodic with period
.
Thus, the Nyquist frequency
corresponds to the frequency
.
Let's compute the discrete-time Fourier transform of the
exponentially decaying sequence
,
where
is the unit-step sequence. Simply plugging the signal's
expression into the Fourier transform formula,
This sum is a special case of the geometric
series.
Thus, as long as
,
we have our Fourier transform.
Using Euler's relation, we can express the magnitude and phase
of this spectrum.
No matter what value of we
choose, the above formulae clearly demonstrate the periodic
nature of the spectra of discrete-time signals. [link] shows indeed that the spectrum
is a periodic function. We need only consider the spectrum
between
and
to unambiguously define it. When
,
we have a lowpass spectrum—the spectrum diminishes as
frequency increases from 0 to
—with increasing
leading to a greater low frequency
content; for
,
we have a highpass spectrum
([link]).
Spectrum of exponential signal
The spectrum of the exponential signal
() is shown over
the frequency range [-2, 2], clearly demonstrating the
periodicity of all discrete-time spectra. The angle has units
of degrees.
Spectra of exponential signals
The spectra of several exponential signals are shown. What is
the apparent relationship between the spectra for
and
?
Analogous to the analog pulse signal, let's find the spectrum
of the length- pulse sequence.
The Fourier transform of this sequence has the form of a
truncated geometric series.
For the so-called finite geometric series, we know that
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
.
which, after manipulation, yields the geometric sum formula.
The ratio of sine functions has the generic form of
,
which is known as the discrete-time sinc function
.
Thus, our transform can be concisely expressed as
. The discrete-time pulse's spectrum contains many
ripples, the number of which increase with
, the pulse's duration.
Spectrum of length-ten pulse
The spectrum of a length-ten pulse is shown. Can you explain
the rather complicated appearance of the phase?
The inverse discrete-time Fourier transform is easily derived
from the following relationship:
Therefore, we find that
The Fourier transform pairs in discrete-time are
The properties of the discrete-time 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.
To show this important property, we simply substitute the
Fourier transform expression into the frequency-domain
expression for power.
Using the orthogonality
relation, the integral equals
,
where
is the unit sample. Thus, the double sum collapses
into a single sum because nonzero values occur only when
,
giving Parseval's Theorem as a result. We term
the energy in the discrete-time signal
in spite of the fact that discrete-time signals don't consume
(or produce for that matter) energy. This terminology is a
carry-over from the analog world.
Suppose we obtained our discrete-time signal from values of
the product
,
where the duration of the component pulses in
is . How is
the discrete-time signal energy related to the total energy
contained in
?
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
. Thus, the energy in the sampled signal equals
the original signal's energy multiplied by
.