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Sec. 6–7 Bandpass Processes 453
certain PSD characteristics. In this case, it can be shown that Re{(-j)g(t) e jvct } gives the
same PSD as Re{g(t) e jvct } when v(t) is a WSS process [Papoulis, 1984, pp. 314–322].
Consequently, g(t) is not unique, and it can be chosen to satisfy some additional desired condition.
Yet, properties 1 through 14 will still be satisfied when the conditions of Eq. (6–129)
are satisfied.
In some applications, conditions 1 through 3 of Eq. (6–129) are not satisfied. This will
be the case, for example, when the x(t) and y(t) quadrature components do not have the same
power, as in an unbalanced quadrature modulation problem. Another example is when x(t) or
y(t) has a DC value. In these cases, the bandpass random process model described by
Eq. (6–126) would be nonstationary. Consequently, one is faced with the following question:
Can another bandpass model be found that models a WSS v(t), but yet does not require
conditions 1 through 3 of Eq. (6–129)? The answer is yes. Let the model of Eq. (6–126)
be generalized to include a phase constant u c that is a random variable. Then we have the
following theorem.
THEOREM.
If x(t) and y(t) are jointly WSS processes, the real bandpass process
v(t) = Re{g(t)e j(v ct+u c ) } = x(t) cos (v c t + u c ) - y(t) sin (v c t + u c )
(6–130)
will be WSS when u c is an independent random variable uniformly distributed over (0, 2p).
This modification of the bandpass model should not worry us, because we can argue
that it is actually a better model for physically obtainable bandpass processes. That is, the
constant u c is often called the random start-up phase, since it depends on the “initial conditions”
of the physical process. Any noise source or signal source starts up with a randomphase
angle when it is turned on, unless it is synchronized by injecting some external signal.
Proof. Using Eq. (6–130) to model our bandpass process, we now demonstrate that
this v(t) is wide-sense stationary when g(t) is wide-sense stationary, even though the conditions
of Eq. (6–129) may be violated. To show that Eq. (6–130) is WSS, the first requirement
is that v(t) be a constant:
v(t) = Re{g(t)e j(v ct+u c ) } = ReEg(t)e jvc t e
ju cF
But e ju c
= 0, so we have v(t) = 0, which is a constant. The second requirement is that R v (t, t + t)
be only a function of t:
R v (t, t + t) = v(t)v(t + t)
= Re{g(t)e j(v ct+u c ) } Re{g(t + t)e j(v ct+v c t+u c ) }
Using the identity Re(c 1 ) Re(c 2 ) = 1 2 Re(c 1c 2 ) + 1 2 Re(c*c 1 2) and recalling that u c is an independent
random variable, we obtain
R v (t, t + t) = 1 2 Re {g(t)g(t + t)ej(2v ct)+v c t) e j2u c }
+ 1 2 Re{g*(t)g(t + t) e jv ct }