Review of basic ideas from digital communication - Signal Processing and Optimization for Transceiver Systems - page 62

it simple. Since the frequency bands of the real signal x bp ( t ) do not in general
exhibit symmetry around ω c , the baseband equivalent X lp ( jω ) has no Hermitian
symmetry around ω =0 . Thus the baseband equivalent x lp ( t ) is not necessarily
real, even though the bandpass signal x bp ( t )isreal.
We will see that it is extremely useful to have the definition of a baseband
equivalent signal. Among other things, it allows us to perform e cient digital
signal processing in the complex domain, both at the transmitter and at the
receiver. Assume now that a real bandpass signal x bp ( t ) is passed through a real
bandpass channel h bp ( t ) as shown in Fig. 2.44. Then the output is the bandpass
signal y bp ( t )=( h bp ∗

x bp )( t ) with Fourier transform Y bp ( jω )= H bp ( jω ) X bp ( jω ) .
So we can define the baseband equivalent y lp ( t ) of the output signal exactly as
we did in Fig. 2.45. The baseband equivalents are then related as

Y lp ( jω )= H lp ( jω ) X lp ( jω ) . (2 . 94)

Thus, the communication of the modulated signal x bp ( t ) over a bandpass chan-
nel h bp ( t ) is mathematically equivalent to the communication of the lowpass
baseband signal x lp ( t ) over the lowpass baseband channel h lp ( t ) . Notice again
that the baseband channel h lp ( t ) is in general complex , even though the physical
channel h bp ( t ) and the signal x lp ( t ) may be real.

2.A.1 Baseband channel model for QAM

In QAM transmission the real bandpass signal transmitted has the form

x bp ( t )=
k

s c ( k ) p ( t − kT )cos ω c t − s s ( k ) p ( t − kT )sin ω c t

k

s c ( k )+ js s ( k ) p ( t

=Re
k

kT ) e jω c t , −

where s c ( k ) ,s s ( k ), and p ( t ) are real. Here p ( t ) is the baseband waveform and
s c ( k )+ js s ( k ) represents the complex QAM symbol. The complex baseband
equivalent is therefore
x lp ( t )=
k

s c ( k )+ js s ( k ) p ( t

− kT ) . (2 . 95)

Assume this is transmitted on a real bandpass channel h bp ( t ) with lowpass equiv-
alent h lp ( t ):

h bp ( t )=Re h lp ( t ) e jω c t . (2 . 96)

Then the channel output (ignoring noise) is:

y bp ( t )= h bp ∗ x bp ( t )

=Re ( x lp ∗ h lp )( t ) e jω c t