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Sec. 3–5 Line Codes and Spectra 173
Example 3–10 PLOT THE PSD FOR A BIPOLAR RZ LINE CODE
Evaluate and plot the PSD for a bipolar RZ line code with R = 1. For the solution, use Eq. (3–45).
See Example3_09.m for the resulting plot. Compare this result with Fig. 3–16d.
The clock signal can easily be extracted from the bipolar waveform by converting the
bipolar format to a unipolar RZ format by the use of full-wave rectification. The resulting
(unipolar) RZ signal has a periodic component at the clock frequency. (See Fig. 3–16c.)
Bipolar signals are not transparent. That is, a string of zeros will cause a loss in the clocking
signal. This difficulty can be prevented by using high-density bipolar n (HDBn) signaling, in
which a string of more than n consecutive zeros is replaced by a “filling” sequence that contains
some pulses. † The calculation of the PSD for HDBn codes is difficult, because the R(k)’s
have to be individually evaluated for large values of k [Davis and Barber, 1973].
Bipolar signals also have single-error detection capabilities built in, since a single error
will cause a violation of the bipolar line code rule. Any violations can easily be detected by
receiver logic.
Two disadvantages of bipolar signals are that the receiver has to distinguish between
three levels (+A, -A, and 0), instead of just two levels in the other signaling formats previously
discussed. Also, the bipolar signal requires approximately 3 dB more signal power than
3
a polar signal for the same probability of bit error. [It has
2
the error of unipolar signaling as
described by Eq. (7–28).]
Manchester NRZ Signaling.
The Manchester signal uses the pulse shape
f(t) =∑a t + T b>4
T b >2
b -∑a t - T b>4
b
T b >2
(3–46a)
and the resulting pulse spectrum is
or
F(f) = T b
2 c sin (pfT b>2
de jwTb>4 - T b
pfT b >2
2 c sin (pfT b>2
pfT b >2
de -jwT b>4
F(f) = jT b c sin (pfT b>2)
d sin a wT b
pfT b >2 4 b
(3–46b)
† For example, for HDB3, the filling sequences used to replace n + 1 = 4 zeros are the alternating sequences
000V and 100V, where the 1 bit is encoded according to the bipolar rule and the V is a 1 pulse of such polarity as to
violate the bipolar rule. The alternating filling sequences are designed so that consecutive V pulses alternate in sign.
Consequently, there will be a 0 DC value in the line code, and the PSD will have a null at f = 0. To decode the HDB3
code, the bipolar decoder has to detect the bipolar violations and count the number of zeros preceding each violation
so that the substituted 1’s can be deleted.