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

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516 PERFORMANCE ANALYSIS OF DIGITAL COMMUNICATION SYSTEMS

The optimum threshold a 0 is obtained by substituting Eqs. (10.17a, b) and (10.22a) into

Eq. (10.19a) and recognizing (via variable substitution) that

1_: P (f)Q (-f) df = 1_: P (-f) Q (j) df = E pq (10.26)

This gives

(10.27)

In deriving the optimum binary receiver, we assumed a certain receiver structure (the threshold

detection receiver in Fig. 10.4). It is not clear yet whether there exists another structure that

may have better performance than that in Fig. 10.4. It will be shown later (in Sec. 10.6) that for

a Gaussian noise, the receiver derived here is the definite optimum. Equation (10.25b) gives P b

for the optimum receiver when the channel noise is white Gaussian. For the case of nonwhite

noise, P b is obtained by substituting f3 max from Eq. (10.21a) into Eq. (10.25a).

Equivalent Optimum Binary Receivers

For the optimum receiver in Fig. 10.4a,

H(f) = P(-f)e - j 2 nfTb _ Q(-f)e - j 2 nfTb

This filter can be realized as a parallel combination of two filters matched to p(t) and q(t),

respectively, as shown in Fig. 10.5a. Yet another equivalent form is shown in Fig. 10.5b.

Because the threshold is (E p

- E q

)/2, we subtract E p

/2 and E q

/2, respectively, from the two

matched filter outputs. This is equivalent to shifting the threshold to 0. In the case of E p = E q

,

we need not subtract E p

/2 and E q

/2 from the two outputs, and the receiver simplifies to that

shown in Fig. 10.5c.

10.2.2 Performance Analysis of General Binary Systems

In this section, we analyze the performance of several typical binary digital communication

systems by applying the techniques derived in the last section for general binary receivers.

Polar Signaling

For the case of polar signaling, q(t) = -p(t). Hence,

and

Substituting these results into Eq. (10.25b) yields

E pq

= -1 00 p 2 (t) dt = -E p

-oo

(10.28)

(10.29)

Also from Eq. (10.22b ),

h(t) = 2p(Tb - t) (10.30)

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