Frequency-Domain Analysis - Introduction to Digital Signal Processing and Filter Design - page 120

There is some ambiguity in the published literature regarding the definition of
what is called the Nyquist frequency . Most of the topics define half the sampling
frequency as the Nyquist frequency and 2 f b as the Nyquist rate, which is the
minimum sampling rate required to avoid aliasing. Because of this definition for
the Nyquist rate, some authors erroneously define f b as the Nyquist frequency.
In our example, when the signal is sampled at 8 kHz, we have 4 kHz as the
Nyquist frequency (or the folding frequency) and 6.8 kHz as the Nyquist rate.
If we sample the analog signal at 20 kHz, the Nyquist frequency is 10 kHz, but
the Nyquist rate is still 6.8 kHz. We will define half the sampling frequency as
the Nyquist frequency throughout this topic. Some authors define the Nyquist
frequency as the bandwidth of the corresponding analog signal, whereas some
authors define 2 f b as the bandwidth.

3.2.1 Sampling of Bandpass Signals

Suppose that we have an analog signal that is a bandpass signal (i.e., it has a
Fourier transform that is zero outside the frequency range ω 1 ≤ ω ≤ ω 2 ); the

bandwidth of this signal is B ω 1 , and the maximum frequency of this

signal is ω 2 . So it is bandlimited, and according to Shannon's sampling theorem,
one might consider a sampling frequency greater than 2 ω 2 ; however, it is not
necessary to choose a sampling frequency ω s

= ω 2 −

≥ 2 ω 2 in order to ensure that we
can reconstruct this signal from its sampled values. It has been shown [3] that
when ω 2 is a multiple of B , we can recover the analog bandpass signal from its
samples obtained with only a sampling frequency ω s

≥ 2 B . For example, when
the bandpass signal has a Fourier transform between ω 1 = 4500 and ω 2 = 5000,
we don't have to choose ω s > 10 , 000. We can choose ω s > 1000, since ω 2 =
10 B in this example.

Example 3.1

e − 0 . 2 t u(t) that has the Fourier Consider a continuous-time signal x a (t) =

transform X(jω) = 1 /(j ω + 0 . 2 ) . The magnitude | X(jω) | = | 1 /(j ω + 0 . 2 ) | =

1 /(ω 2

0 . 04 ) , and when we choose a frequency of 200 π , we see that the
magnitude is approximately 0 . 4 ( 10 − 3 ) . Although the function x a (t)

+

e − 0 . 2 t u(t)
is not bandlimited, we can assume that it is almost bandlimited with bandwidth
of 200 π and choose a sampling frequency of 400 π rad/s or 200 Hz. So the sam-
pling period T

=

1
200

= 400 π rad/s. To verify
that (3.11) and (3.16) both give the same result, let us evaluate the DTFT at
ω

= = 0 . 005 second and ω s = 2 π/T

= 0 . 5 rad/s. According to (3.11), the DTFT of x(nT ) is

∞ ∞

e − 0 . 2 (nT ) e − jωnT e − 0 . 001 n e − jωn( 0 . 005 ) =

n = 0 n = 0

1

X(e jωT )

= =

(3.20)

e − 0 . 001 e − j( 0 . 005 ω)

1 −