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Sec. 7–3 Coherent Detection of Bandpass Binary Signals 509
only one of the filters, since the order of linear operations may be exchanged without affecting
the results. The mark (binary 1) and space (binary 0) signals are
s 1 (t) = A cos (v 1 t + u c ), 0 6 t … T (binary 1)
s 2 (t) = A cos (v 2 t + u c ), 0 6 t … T (binary 0)
(7–39a)
(7–39b)
where the frequency shift is 2 ∆F = f 1 - f 2 , assuming that f 1 7 f 2 . This FSK signal plus
white Gaussian noise is present at the receiver input. The PSD for s 1 (t) and s 2 (t) is shown
in Fig. 7–8b.
First, evaluate the performance of a receiver that uses an LPF H(f) with a DC gain of 1.
Assume that the equivalent bandwidth of the filter is 2/T B 6 ∆F. The LPF, when combined
with the frequency translation produced by the product detectors, acts as dual bandpass
filters—one centered at f = f 1 and the other at f = f 2 , where each has an equivalent bandwidth
of B p = 2B. Thus, the input noise that affects the output consists of two narrowband components
n 1 (t) and n 2 (t), where the spectrum of n 1 (t) is centered at f 1 and the spectrum of n 2 (t) is
centered at f 2 , as shown in Fig. 7–8. Furthermore, n(t) = n 1 (t) + n 2 (t), where, using
Eq. (6–130),
n 1 (t) = x 1 (t) cos (v 1 t + u c ) - y 1 (t) sin (v 1 t + u c )
(7–40a)
and
n 2 (t) = x 2 (t) cos (v 2 t + u c ) - y 2 (t) sin (v 2 t + u c )
(7–40b)
The frequency shift is 2∆F 7 2B, so that the mark and space signals may be separated
by the filtering action. The input signal and noise that pass through the upper channel
in Fig. 7–8a are described by
r 1 (t) = e s 1(t), binary 1
(7–41)
0, binary 0 f + n 1(t)
and the signal and noise that pass through the lower channel are described by
0, binary 1
r 2 (t) = e (7–42)
s 2 (t), binary 0 f + n 2(t)
where r(t) = r 1 (t) + r 2 (t). The noise power of n 1 (t) and n 2 (t) is n 2 1 (t) = n 2 2 (t) =
= (N 0 >2) (4B) = 2N 0 B. Thus, the baseband analog output is
+A, 0 6 t … T, binary 1
r 0 (t) = e (7–43)
-A, 0 6 t … T, binary 0 f + n 0(t)
where s 01 =+A, s 02 =-A, and n 0 (t) = x 1 (t) - x 2 (t). The optimum threshold setting is V T = 0.
Furthermore, the bandpass noise processes n 1 (t) and n 2 (t) are independent, since they have
spectra in nonoverlapping frequency bands (see Fig. 7–8) and they are white. (See Prob. 7–29
for the verification of this statement.) Consequently, the resulting baseband noise processes
x 1 (t) and x 2 (t) are independent, and the output noise power is
n 0 2 (t) = s 0 2 = x 1 2 (t) + x 2 2 (t) = n 1 2 (t) + n 2 2 (t) = 4N 0 B
(7–44)