Review of basic ideas from digital communication - Signal Processing and Optimization for Transceiver Systems - page 36

P ( j ω )

baseband
signal

(a)

P - P +

ω

−σ σ

0

cosine
modulation

(b)

P - P + P - P +

ω

−ω c ω c

0

sine
modulation

(c)

jP - jP +

ω c

ω

−ω c

0

− jP - − jP +

Figure 2.24 . Fourier transforms of (a) the baseband pulse signal p ( t ), (b) its cosine
modulated version, and (c) its sine modulated version.

Assume the real baseband signal p ( t ) has a bandlimited Fourier transform as
in Fig. 2.24(a). Since

cos ω c t = e jω c t + e −jω c t
2

sin ω c t = e jω c t

e −jω c t
2 j

−

and ,

the cosine modulated version has the Fourier transform

P ( j ( ω − ω c )) + P ( j ( ω + ω c ))
2

,

and the sine modulated version has the Fourier transform

P ( j ( ω − ω c )) − P ( j ( ω + ω c ))

= −

jP ( j ( ω − ω c )) + jP ( j ( ω + ω c ))

2

,

2 j

as indicated in Figs. 2.24(b) and 2.24(c). Since the cosine and sine modulated
versions carry different messages , we cannot eliminate any

of the side bands. Thus we need to retain both side bands, and the QAM signal
(2.40) is a DSB signal. The QAM signal therefore requires twice as much band-
width as a PAM/SSB signal. But, since the QAM signal carries two messages
s c ( n )and s s ( n ) at the same rate, the bandwidth used per real symbol is identical
for the two systems. One advantage of a QAM system is that it does not require
the Hilbert transformer. It is very easy to generate the QAM signal by using
Eq. (2.40), as shown schematically in Fig. 2.25.

{ s c ( n ) } and { s s ( n ) }