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Performance of Communication Systems Corrupted by Noise Chap. 7
Polar Signaling
As shown in Fig. 7–4c, the baseband polar signaling waveform is
s 1 (t) = +A, 0 6 t … T (binary 1)
(7–25a)
s 2 (t) = -A, 0 6 t … T (binary 0)
(7–25b)
The polar signal is an antipodal signal, since s 1 (t) =-s 2 (t).
The performance of an LPF receiver system is obtained by using Eq. (7–17). Assuming
that the equivalent bandwidth of the LPF is B 2/T, we realize that the output signal samples
are s 01 (t 0 ) ≈ A and s 02 (t 0 ) ≈ -A at the sampling time t = t 0 . In addition, s 2 0 = (N 0 B). The
optimum threshold setting is now V T = 0. Thus, the BER for polar signaling is
P e =
A
Q¢ 2
(7–26a)
C N 0 B ≤ (low-pass filter)
where B is the equivalent bandwidth of the LPF.
The performance of the matched-filter receiver is obtained, once again, by using
Eq. (7–20), where t 0 = T. (The integrate-and-dump matched filter for polar signaling was
given in Fig. 6–17.) Since the energy of the difference signal is E d = (2A) 2 T, the BER is
2A 2 T
P e = Q¢ ≤ = Q¢ 2a E b
b ≤
C N 0 C N 0
(matched filter)
(7–26b)
where the average energy per bit is E b = A 2 T. The optimum threshold setting is V T = 0.
A plot of the BER for unipolar and polar baseband signaling is shown in Fig. 7–5. It is
apparent that polar signaling has a 3-dB advantage over unipolar signaling, since unipolar
signaling requires a E b /N 0 that is 3 dB larger than that for polar signaling for the same P e .
Bipolar Signaling
For bipolar NRZ signaling, binary 1’s are represented by alternating positive and negative
values, and the binary 0’s are represented by a zero level. Thus,
s 1 (t) = ;A, 0 6 t … T (binary 1)
(7–27a)
s 2 (t) = 0, 0 6 t … T (binary 0)
(7–27b)
where A 7 0. This is similar to unipolar signaling, except two thresholds, +V T and -V T , are
needed as shown in Fig. 7–6. Figure 7–6b illustrates the error probabilities for the case of
additive Gaussian noise. The BER is
P e = P(error| + A sent)P(+A sent) + P(error| - A sent)P(-A sent)
+ P(error|s 2 sent)P(s 2 sent)
and, by using Fig. 7–6b, we find that
P e L c2Qa A - V T
s 0
bd 1 4 + c2Qa V T
s 0
bd 1 2