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Optical Modulators and Modulation Schemes 143

ρ m ( f) ρ m ( f)

f f
–3B –2B –B 0 B 2B 3B –3B –2B –B 0 B 2B 3B
(a) (b)

Figure 4.3 Spectrum of a polar signal: (a) NRZ, and (b) RZ with 50% duty cycle. B = bit rate.

When x = 1 (NRZ), the sinc function has nulls at f = l∕T , l ≠ 0, which coincide with the locations of the
b
delta functions. Therefore, the PSD of a unipolar NRZ signal is
2
A T [ ]
NRZ 0 b 2 (f)
(f)= sinc (T f) 1 + . (4.20)
m b
4 T
b
Fig. 4.4 shows the PSD of unipolar NRZ and RZ signals. The PSD has continuous and discrete components
corresponding to the first and second terms on the right-hand side of Eq. (4.19). For a RZ signal, the discrete
components are located at f = l∕T . However, the PSD of a unipolar NRZ signal has only a d.c. component
b
(f = 0). The origin of the discrete components can be understood as follows. The unipolar NRZ signal can
be imagined as a polar signal with constant bias. The PSD of this constant bias is the discrete component at
f = 0. In the case of unipolar RZ, it can be imagined as a polar signal added to a periodic pulse train. Since
the Fourier series expansion of the periodic pulse train leads to frequency components at the clock frequency
1∕T , and its harmonics, the PSD has discrete components at these frequencies.
b

ρ ( f) ρ ( f)
m
m

f f
–3B –2B –B 0 B 2B 3B –3B –2B –B 0 B 2B 3B
(a) (b)

Figure 4.4 Spectrum of a unipolar signal: (a) NRZ, and (b) RZ with 50% duty cycle. B = bit rate. The arrows indicate
delta functions.