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

10.1 1 Noncoherent Detection 583
Because the matched filter is used, A p
= E p
and a; = NEp/2. Moreover, for ASK there are,
on the average, only R b /2 nonzero pulses per second. Thus, E b = E p
/2. Hence,
and
( Ap 2
) =
2E p
= 4 E b
a n N N
(10. 139a)
Observe that the optimum threshold is not constant but depends on Eb/N. This is a serious
drawback in a fading channel. For a strong signal, Eb/N » 1,
( 10. 139b)
and
P(E lm = 0) = r X) Pr (rlm = 0) dr
]A p/2
= (X) !_ e -r z 1 2 0}; dr
]A p/2 a;
= e -A /80};
Also,
P(E lm = 1) =
} _ 00
/2
r p Pr (rlm = l) dr
(10.140)
Evaluation of this integral is somewhat cumbersome. 4 For a strong signal (that is, for Eb/ N »
1), the Ricean PDF can be approximated by the Gaussian PDF [Eq. (9.86c)], and
I A p/2
2 2
P(E lm = 1) --- ! e-(r-Ap) /Zan dr
a n ,/iir -oo
(10.141)
As a result,
Pb = Pm (0)(E l m = 0) + Pm (l)P(E lm = 1)