B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

118 ANALYSIS AND TRANSMISSION OF SIGNALS
The transfer function of an ideal differentiator is H (f) = j2rrJ . If the noise at the
demodulator output is n 0 (t), then from Eq. (3.91),
The output PSD S n0 (f) is parabolic, as shown in Fig. 3.38c. The output noise power N 0 is
the area under the output PSD. Therefore,
! B iB
N o = K(2rrf) 2 dJ = 2K (2rrf) 2 8rr
dJ = ---
2 B 3 K
-B O 3
3.8.4 PSD of Modulated Signals
Following the argument in deriving Eqs. (3.70) and (3.71) for energy signals, we can derive
similar results for power signals by taking the time averages. We can show that for a power
signal g(t), if
<p(t) = g(t) cos 2rrJot
then the PSD S rp (f) of the modulated signal <p(t) is given by
(3.92)
The detailed derivation is provided in Sec. 7.8. Thus, modulation shifts the PSD of g(t) by
±Jo. The power of <p(t) is half the power of g (t), that is,
Jo ?:. B
(3.93)
3.9 NUMERICAL COMPUTATION OF FOURIER
TRANSFORM: THE OFT
To compute G(f), the Fourier transform of g(t), numerically, we have to use the samples of
g(t). Moreover, we can determine G(f) only at some finite number of frequencies. Thus, we
can compute only samples of G (f). For this reason, we shall now find the relationships between
samples of g(t) and samples of G(f).
In numerical computations, the data must be finite. This means that the number of samples
of g(t) and G(f) must be finite. In other words, we must deal with time-limited signals. If the
signal is not time-limited, then we need to truncate it to make its duration finite. The same is
true of G(f). To begin, let us consider a signal g(t) of duration r seconds, starting at t = 0,
as shown in Fig. 3.39a. However, for reasons that will become clear as we go along, we shall
consider the duration of g(t) to be To, where To ?:. r, which makes g(t) = 0 in the interval
r < t ::S To, as shown in Fig. 3.39a. Clearly, this makes no difference in the computation of
G(f). Let us take samples of g(t) at uniform intervals of Ts seconds. There are a total of No
samples, where
To
No = ­ Ts
(3.94)