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Sec. 4–19 Study-Aid Examples 303
The highest frequency in the down-converted signal (at the sampler input) is B = (f c + B T 2) - f 0 =
f c + B T 2 - f c + B T 2 = B T , and the lowest frequency (in the positive-frequency part of the downconverted
signal) is (f c - B T 2) - f 0 = f c - B T 2 - f c + B T 2 = 0. Using Eq. (2–168), we find that
the minimum sampling frequency is
(f s ) min = 2B T Method II
(4–131)
when the frequency of the LO is chosen to be f 0 = f c - B T 2. For this choice of LO frequency,
the bandpass filter becomes a low-pass filter with a cutoff frequency of B T . Note that Method
II gives a drastic reduction in the sampling frequency (an advantage) compared with Method I.
For example, if f c = 100 MHz and B T = 1 MHz, then the minimum sampling frequency is now
(f s ) min = 2 MHz, instead of the 201 MHz required in Method I. However, Method II requires
the use of a down converter (a disadvantage). Note also that (f s ) min of Method II, as specified
by Eq. (4–131), satisfies the (f s ) min given by the bandpass sampling theorem, as described by
Eq. (4–31).
Method II is one of the most efficient ways to obtain samples for a bandpass signal.
When the bandpass signal is reconstructed from the sample values with the use of Eq. (2–158)
and (2–160), the down-converted bandpass signal is obtained. To obtain the original bandpass
signal s(t), an up-converter is needed to convert the down-converted signal back to the original
bandpass region of the spectrum.
Method II can also be used to obtain samples of the quadrature (i.e., I and Q) components
of the complex envelope. From Fig. 4–32b, the IF signal at the input to the sampler is
v IF (t) = x(t) cos v IF t - y(t) sin v IF t
where f IF (t) = B T 2. Samples of x(t) can be obtained if v IF (t) is sampled at the times corresponding
to cos v IF t = ±1 (and sin v IF t = 0). This produces B T samples of x(t) per second. Likewise,
samples of y(t) are obtained at the times when sin v IF t = ±1 (and cos v IF t = 0). This produces B T
samples of y(t) per second. The composite sampling rate for the clock is f s = 2B T . Thus, the sampler
output contains the following sequence of I and Q values: x, -y, -x, y, x, -y, ... The sampling
clock can be synchronized to the IF phase by using carrier synchronization circuits. Method III
uses a similar approach.
Method III From Fig. 4–32c, Method III uses in-phase (I) and quadrature-phase (Q) product
detectors to produce the x(t) and y(t) quadrature components of s(t). (This was discussed in
Sec. 4–16 and illustrated in Fig. 4–31.) The highest frequencies in x(t) and y(t) are B = B T 2.
Thus, using Eq. (2–168), we find that the minimum sampling frequency for the clock of the I
and Q samplers is
(f s ) min = B T (each sampler) Method III
(4–132)
Because there are two samplers, the combined sampling rate is (f s ) min overall = 2B T . This also
satisfies the minimum sampling rate allowed for bandpass signals as described by Eq. (4–31).
Thus, Method III (like Method II) gives one of the most efficient ways to obtain samples of
bandpass signals. For the case of f c = 100 MHz and B T = 1 MHz, an overall sampling rate of
2 MHz is required for Method III, which is the same as that obtained by Method II. Because IQ
samples have been obtained, they may be processed by using DSP algorithms to perform