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Sec. 6–7 Bandpass Processes 461
Thus, Eq. (6–142) reduces to property 15. Taking the Fourier transform of Eq. (6–135a), we
obtain properties 16 and 17.
As demonstrated by property 6, the detected mean values x(t) and y(t) are zero
when v(t) is independent of u 0 . However, from Eqs. (6–139) and (6–140), we realize that a
less restrictive condition is required. That is, x(t) or y(t) will be zero if v(t) is orthogonal to
cos(v c t u 0 ) or sin(v c t u 0 ), respectively; otherwise, they will be nonzero. For example,
suppose that v(t) is
v(t) = 5 cos (v c t + u c )
(6–146)
where u c is a random variable uniformly distributed over (0, 2p). If the reference 2 cos
(v c t u 0 ) of Fig. 6–11 is phase coherent with cos (v c t u c ) (i.e., u 0 u c ), the output of
the upper LPF will have a mean value of 5. On the other hand, if the random variables u 0
and u c are independent, then v(t) will be orthogonal to cos (v c t u 0 ), and xq of the output
of the upper LPF of Fig. 6–11 will be zero. The output DC value (i.e., the time average) will
be 5 cos (u 0 - u c ) in either case.
No properties have been given pertaining to the autocorrelation or the PSD of R(t) and
u(t) as related to the autocorrelation and PSD of v(t). In general, this is a difficult problem,
because R(t) and u(t) are nonlinear functions of v(t). The topic is discussed in more detail
after Example 6–13.
As indicated previously, x(t), y(t), and g(t) are Gaussian processes when v(t) is
Gaussian. However, as we will demonstrate, R(t) and u(t) are not Gaussian processes when
v(t) is Gaussian.
Example 6–13 PDF FOR THE ENVELOPE AND PHASE FUNCTIONS OF A GAUSSIAN
BANDPASS PROCESS
Assume that v(t) is a wide-sense stationary Gaussian process with finite PSD that is symmetrical
about f =;f c . We want to find the one-dimensional PDF for the envelope process R(t). Of course,
this is identical to the process that appears at the output of an envelope detector when the input is
a Gaussian process, such as Gaussian noise. Similarly, the PDF for the phase u(t) (the output of a
phase detector) will also be obtained.
The problem is solved by evaluating the two-dimensional random-variable transformation
of x = x(t) and y = y(t) into R = R(t) and u = u(t), which is illustrated in Fig. 6–13. Because v(t)
is Gaussian, we know that x and y are jointly Gaussian. For v(t) having a finite PSD that is
symmetrical about f =;f c , the mean values for x and y are both zero and the variances of both are
s 2 = s x 2 = s y 2 = R v (0)
(6–147)
x
y
Nonlinear transformation
x=R cos ¨
y=R sin ¨
R
¨
Figure 6–13
Nonlinear (polar) transformation of two Gaussian random variables.