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Sec. 3–5 Line Codes and Spectra 169
unipolar NRZ (f) = A2 T b
4
a sin pfT 2
b
b c1 +
pfT b
q
a e j2pkfT b
d
k=-q
But † q
a e jk2pfT b
= 1 q
(3–38)
T a d af - n b
b T b
Thus,
k=-q
n=-q
unipolar NRZ (f) = A2 T b
4
a sin pfT 2
b
b c1 + 1 pfT b T b
q
a
n=-q
daf - n T b
bd
(3–39a)
But because [sin (pfT b )(pfT b )] = 0 at f = nT b , for n Z 0, this reduces to
unipolar NRZ (f) = A2 T b
4
a sin pfT 2
b
b c1 + 1 d(f) d
pfT b T b
(3–39b)
If A is selected so that the normalized average power of the unipolar NRZ signal is unity, then ‡
A = 12. This PSD is plotted in Fig. 3–16a, where 1T b = R, the bit rate of the line code. The
disadvantage of unipolar NRZ is the waste of power due to the DC level and the fact that the
spectrum is not approaching zero near DC. Consequently, DC-coupled circuits are needed.
The advantages of unipolar signaling are that it is easy to generate using TTL and CMOS circuits
and requires the use of only one power supply.
Polar NRZ Signaling. For polar NRZ signaling, the possible levels for the a’s are
+A and -A V. For equally likely occurrences of +A and -A, and assuming that the data are
independent from bit to bit, we get
For k Z 0,
R(0) = a
2
i=1
(a n a n ) i P i = A 2 1 2 + (-A)2 1 2 = A2
Thus,
R(k) = a
4
i=1
(a n a n+k )P i = A 2 1 4 + (-A)(A) 1 4 + (A)(-A) 1 4 + (-A)2 1 4 = 0
† Equation (3–38) is known as the Poisson sum formula, as derived in Eq. (2–115).
‡ This can be demonstrated easily by using a line code with periodic data 10101010. Then, setting
P s = 1 T 0
s 2 1
(t) dt = A 2 T b equal to unity, we get A = 12.
T 0 L 0
2T b