563489578934

Sec. 6–2 Power Spectral Density 431
With this result, Eq. (6–58) becomes
and, with T = 2(N + 1 2 ) T b,
x (f) = ƒ F(f) ƒ
2
lim a 1 N
T: q T a 1b
n = -N
or
x (f) = ƒ F(f) ƒ 2 lim
N:qc
2N + 1
d
(2N + 1)T b
x (f) =
1 T b
ƒ F(f) ƒ
2
(polar signaling)
(6–60)
For the rectangular pulse shape shown in Fig. 6–5b,
F(f) = T b a sin pfT b
b
(6–61)
pfT b
Thus, the PSD for a polar signal with a rectangular pulse shape is
x (f) = T b a sin pfT 2
b
b
(6–62)
pfT b
This PSD is plotted in Fig. 6–5c. † The null bandwidth is B = 1T b = R, where R is the bit rate.
Note that Eq. (6–62) satisfies the properties for a PSD function listed previously.
Method 2 will now be used to evaluate the PSD of the polar signal. This involves calculating
the autocorrelation function and then evaluating the Fourier transform of R x (t) to obtain the
PSD. When we use Eq. (6–56), the autocorrelation is
R x (t, t + t) = x(t)x(t + t)
By using Eq. (6–59), this equation reduces to
R x (t, t + t) =
q
a f(t - nT b )f(t + t - nT b )
(6–63)
n = -q
Obviously, x(t) is not a wide-sense stationary process, since the autocorrelation function depends
on absolute time t. To reduce Eq. (6–63), a particular type of pulse shape needs to be designated.
Assume, once again, the rectangular pulse shape
The pulse product then becomes
= a
q
n = -q
a n f(t - nT b )
= a n
a m
a n a m f(t - nT b )f(t + t - mT b )
f(t) = e 1, |t| … T b> 2
0, t elsewhere
f(t - nT b )f(t + t - nT b ) = e 1, if |t - nT b| … T b > 2 and |t + t - nT b | … T b > 2
0, otherwise
q
a
m =-q
a m f(t + t - mT b )
† The PSD of this polar signal is purely continuous because the positive and negative pulses are assumed
to be of equal amplitude.