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Sec. 3–2 Pulse Amplitude Modulation 135
where f s = 1T s , vs = 2pf s , the duty cycle of s(t) is d = tT s , and W( f) = [w(t)] is the
spectrum of the original unsampled waveform.
PROOF.
Taking the Fourier transform of Eq. (3–1), we get
s(t) may be represented by the Fourier series
q
s(t) = a c n e jnv st
n =-q
where
sin npd
c n = d
npd
Since s(t) is periodic Eq. (2–109) may be used to get the spectrum:
Then Eq. (3–4) becomes
or
W s (f) = W(f) * a
q
a
n =-q
W s (f) = W(f) * S(f)
S(f) = [s(t)] =
c n d(f - nf s )b =
q
a c n d (f-nf s )
n =-q
q
a c n W(f) * d (f - nf s )
n =-q
(3–4)
(3–5a)
(3–5b)
(3–6)
q
W s (f) = a c n W(f - nf s )
(3–7)
n =-q
This equation becomes Eq. (3–3) upon substituting Eq. (3–5b).
The PAM waveform with natural sampling is relatively easy to generate, since it only
requires the use of an analog switch that is readily available in CMOS hardware (e.g., the
4016-quad bilateral switch). This hardware is shown in Fig. 3–2, where the associated waveforms
w(t), s(t), and w s (t) are as illustrated in Fig. 3–1.
Analog bilateral switch
w(t)
w s (t)=w(t)s(t)
s(t)
Clock
Figure 3–2
Generation of PAM with natural sampling (gating).