563489578934

462
Random Processes and Spectral Analysis Chap. 6
Furthermore, x and y are independent, because they are uncorrelated Gaussian random variables
[since the PSD of v(t) is symmetrical about f =;f c ]. Therefore, the joint PDF of x and y is
1 +y
f xy (x, y) =
2 )/(2s 2 )
(6–148)
2ps 2 e-(x2
The joint PDF for R and u is obtained by the two-dimensional transformation of x and y into
R and u.
f xy (x, y)
f Ru (R,u) =
ƒ J[(R, u)>(x, y)] ƒ
2
x = R cos u
y = R sin u
(x, y)
= f xy (x, y) 2 Ja
(R, u) b 22x = R cos u
y = R sin u
(6–149)
We will work with J[(x, y)/(R, u)] instead of J[(R, u)/(x, y)], because in this problem the partial
derivatives in the former are easier to evaluate than those in the latter. We have
0x
(x, y)
Ja
(R, u) b = Det ≥ 0R
0y
0x
0u
¥
0y
where x and y are related to R and u as shown in Fig. 6–13. Of course, R Ú 0, and u falls in the
interval (0, 2p), so that
(x, y)
u
Ja b = Det ccos - R sin u
(R, u) sin u R cos u d
0R
0u
= R[ cos 2 u + sin 2 u] = R
(6–150)
Substituting Eqs. (6–148) and (6–150) into Eq. (6–149), the joint PDF of R and u is
The PDF for the envelope is obtained by calculating the marginal PDF:
q
2p
R
f R (R) = f R (R, u) du =
L -q
L 0 2ps 2 e-R/(2s2) du, R Ú 0
or
R /2s 2
f R u (R, u) = 2ps 2 e-R2 , R Ú 0 and 0 … u … 2p
L
0, R and u otherwise
(6–151)
R /( 2s 2)
f R (R) = s 2 e-R2 , R Ú 0
(6–152)
L
0, R otherwise
This is called a Rayleigh PDF. Similarly, the PDF of u is obtained by integrating out R in
the joint PDF:
1
f u (u) = 2p , 0 … u … 2p
(6–153)
L
0, otherwise