Classical transceiver techniques - Signal Processing and Optimization for Transceiver Systems - page 170

y ( t )

s ( n )

h ( t ) r ( n )

D/C

C/D
T T

t

T 2 T

Figure 5.2 . Example of a channel h ( t ) without any receiver filter g ( t ) . In this case the
transfer function from s ( n )to r ( n ) is zero, and no information about y ( t ) is retained
after sampling.

Proof. The Fourier transform at the output of G ( jω ) in Fig. 5.1(b) is
S ( e jωT ) H ( jω ) G ( jω )= S ( e jωT ) |H ( jω ) | 2 .

The Fourier transform of the sampled version r ( n )istherefore

H j ( ω +2 πk )
T

∞

R ( e jω )= S ( e jω )
T

2 .

(5 . 4)

k = −∞

If the summation above is nonzero for all ω, define
A 2 ( e jωT )= 1
T

H j ω + 2 πk
T

2 − 1 .

∞

(5 . 5 a )

k = −∞

We can then reconstruct s ( n )simplybypassing r ( n ) through the inverse
filter A 2 ( e jω ) (Fig. 5.3). From s ( n ) we can then obtain y ( t )byusingits
very definition as shown in the figure.
More generally we have to allow for the possibility that the sum in (5.4)
can be zero for some ω. So let us define

A 2 ( e jωT )

⎧
⎨

1
T

H j ω + 2 πk
T

2 − 1

∞

when H ( jω ) =0

= (5 . 5 b )

⎩

k = −∞

0 otherwise.

When H ( jω ) = 0 for some ω the sum in (5.4) is positive and the inverse

indicated in (5.5b) exists for that ω. With A 2 ( e jωT ) as above, (5.4) implies

R ( e jωT )= S ( e jωT )
A 2 ( e jωT )

when H ( jω ) =0 .