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Problem 4.5 from Wozencraft and Jacobs
One of the two signals s0 = −1, s1 = +1 is transmitted over the channel shown in Fig. P4.5a. The
two noise random variables n1 and n2 are statistically independent of the transmitted signals and
of each other. Their density functions are
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pn1 (α) = pn2 (α) = e−|α| .
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a. Prove that the optimum decision regions for equally likely messages are as shown in Fig. P4.5b.
Hint. Use geometric reasoning and the fact that |ρ1 − 1| + |ρ2 − 1| = a + b, as shown in Fig. P4.5d.
b. A receiver decides that s1 was transmitted if and only if (r1 + r2 ) > 0. Is this receiver
optimum for equally likely messages? What is its probability of error?
c. Prove that the optimum decision regions are modified as indicated in Fig. P4.5c when
P [s1 ] > 12 .
d. The channel may be discarded without affecting P [²]min if P [s1 ] ≥ q. Evaluate q.
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Problem 4.10 from Wozencraft and Jacobs
One of two equally likely messages is to be transmitted over an additive white Gaussian noise
channel with Sn (f ) = 0.05 by means of binary pulse position modulation. Specifically,
s0 (t) = p(t),
s1 (t) = p(t − 2),
in which the pulse p(t) is shown in Fig. P4.10.
a. What mathematical operations are performed by the optimum receiver?
b. What is the resulting probability of error?
c. Indicate two methods of implementing the receiver, each of which uses a single linear filter
followed by a sampler and comparison device. Method I requires that two samples from the filter
output be fed into the comparison device. Method II requires that just one sample be used.
For each method sketch the impulse response of the appropriate filter and its responses to p(t).
Which of these methods is most easily extended to M -ary pulse position modulation, where si (t) =
p(t − 2i), i = 0, 1, . . . , M − 1?
d. Suggest another pair of waveforms that require the same energy as the binary pulse-position
waveforms and yield the same error probability; yield a lower error probability.
e. Calculate the minimum attainable probability of error if
s0 (t) = p(t) and s1 (t) = p(t − 1).
Repeat for
s0 (t) = p(t) and s1 (t) = −p(t − 1).
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