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

490 RANDOM PROCESSES AND SPECTRAL ANALYSIS
Substitution of this value of K into Eqs. (9.59a) and (9.59b) yields
2 SrJSn (f)/Sm lf)
IH
p
(f) lopt =
IHc lf) I f [JSmlf)Sn lf)/IH c(f) I] df
G 2 2 fcxJJSmlf)Sn (f)/IHclf) I] df
IHd (f) lopt =
Sr lH c lf) I JSn lf)/Sm (f)
(9.60a)
(9.60b)
The output SNR under optimum conditions is given by Eq. (9.56) with its denominator replaced
by the right-hand side of Eq. (9.57). Finally, substituting IH p (f)H d (f) I = G/ IHc(f) l leads to
( S0 ) Sr f oo Sm(f) df
No opt - (!
00
[JSm (f">Sn (f)/IH c (f) l] df )
2
(9.60c)
Equations (9.60a) and (9.60b) give the magnitudes of the optimum filters H p (f) and H d (f).
The phase functions must be chosen to satisfy the condition of distortionless transmission
[Eq. (9.53b)].
Observe that the preemphasis filter in Eq. (9.59a) boosts frequency components where
the signal is weak and suppresses frequency components where the signal is strong. The
deemphasis filter in Eq. (9.59b) does exactly the opposite. Thus, the signal is unchanged but
the noise is reduced.
Example 9. 12 Consider the case with a = 1400;r,
The channel noise is white with PSD
ll l :s 4000
ll l ::: 4000
N
Sn (f) = 2
(9.61a)
(9.61b)
The channel is assumed to be ideal [H elf) = 1 and G
(0-4000 Hz).
l] over the band of interest
Without preemphasis-deemphasis, we have
So =
14000
Sm (f)df
-4000
{ 4000 c
= 2 Jo (2;r f ) 2 + a
= 10- 4 c
Also, because G = 1, the transmitted power Sr = S0 ,
S0 = Sr = 10- 4 c
2 df a= 1400;r