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Random Processes and Spectral Analysis Chap. 6
Working with the inequalities, we get a unity product only when
1n - 1 2 2T b … t … 1n + 1 2 2T b
and
1n - 1 2 2T b - t … t … 1n + 1 2 2T b - t
Assume that t 0; then we will have a unity product when
1n - 1 2 2T b … t … 1n + 1 2 2T b - t
provided that t T b . Thus, for 0 t T b ,
q
R x (t, t + t) = a e 1, 1n - 1 2 2T b … t … 1n + 1 2 2T b - t
f (6–64)
n = - q 0, otherwise
We know that the Wiener–Khintchine theorem is valid for nonstationary processes if we let
R x (t) = R x (t, t t). Using Eq. (6–64), we get
R x (t) = 8R x (t, t + t)9 = lim
T: q
or
where T>2 = 1N - 1 2 2 T b. This reduces to
or
R x (t) =
1
T L
R x (t) = lim
T: q
T/2
-T/2 n = - q
1
T
q
a
a
N
n = -N
e 1, 1n - 1 2 2T b … t … 1n + 1 2 2T b - t
f dt
0, elsewhere
(n+1/2)T b - t
a 1 dtb
L (n-1/2) T b
lim c 1
(2N + 1) e 1T b - t2, 0 … t … T b
fd
N: q (2N + 1) T b 0, t 7 T b
T b - t
, 0 … t … T b
R x (t) = T b (6–65)
L
0, t > T b
Similar results can be obtained for t 6 0. However, we know that R x ( - t) = R x (t), so Eq. (6–65)
can be generalized for all values of t. Thus,
T b - |t|
, |t| … T
R x (t) = T b
b (6–66)
L
0, otherwise
which shows that R x (t) has a triangular shape. Evaluating the Fourier transform of Eq. (6–66), we
obtain the PSD for the polar signal with a rectangular bit shape:
x (f) = T b a sin pfT 2
b
b
(6–67)
pfT b
This result, obtained by the use of method 2, is identical to Eq. (6–62), obtained by using method 1.