563489578934

Sec. 3–5 Line Codes and Spectra 171
R polar (k) = e A2 , k = 0
0, k Z 0 f
(3–40)
Then, substituting Eqs. (3–40) and (3–37a) into Eq. (3–36a), we obtain the PSD for the polar
NRZ signal:
polar NRZ (f) = A 2 T b a sin pfT 2
b
b
pfT b
(3–41)
If A is selected so that the normalized average power of the polar NRZ signal is unity, then
A = 1, and the resulting PSD is shown in Fig. 3–16b, where the bit rate is R = 1T b .
The polar signal has the disadvantage of having a large PSD near DC. On the other
hand, polar signals are relatively easy to generate, although positive and negative power
supplies are required, unless special-purpose integrated circuits are used that generate dual
supply voltages from a single supply. The probability of bit error performance is superior to
that of other signaling methods. (See Fig. 7–14.)
Unipolar RZ. The autocorrelation for unipolar data was calculated previously and is
given by Eq. (3–37a). For RZ signaling, the pulse duration is T b 2 instead of T b , as used in
NRZ signaling. That is, for RZ,
F(f) = T b
(3–42)
2 a sin (pfT b>2)
b
pfT b >2
Then, referring to Eqs. (3–37b) and (3–39a), we get the PSD for the unipolar RZ line code:
unipolar RZ (f) = A2 T b
(3–43)
16 a sin (pfT 2
b>2
b c1 + 1 q
pfT b >2 T a daf - n bd
b n=-q T b
If A is selected so that the normalized average power of the unipolar RZ signal is unity, then A = 2.
The PSD for this unity-power case is shown in Fig. 3–16c, where R = 1T b .
As expected, the first null bandwidth is twice that for unipolar or polar signaling,
since the pulse width is half as wide. There is a discrete (impulse) term at f = R.
Consequently, this periodic component can be used for recovery of the clock signal. One
disadvantage of this scheme is that it requires 3 dB more signal power than polar signaling
for the same probability of bit error. (See Chapter 7.) Moreover, the spectrum is not negligible
for frequencies near DC.
Bipolar RZ Signaling. The PSD for a bipolar signal can also be obtained using
Eq. (3–36a). The permitted values for a n are + A, -A, and 0, where binary 1’s are represented
by alternating + A and -A values and a binary 0 is represented by a n = 0. For k = 0, the products
a n a n are A 2 and 0, where each of these products occurs with a probability of Thus,
R(0) = A2
2
1
2 .