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Sec. 3–13 Study-Aid Examples 227
(d) For the unipolar NRZ line code, there are N = 5 pulses in T 0 = 0.3472 ms, or
D = 5>0.3472 ms = 14,400 baud
R = D because the unipolar NRZ line code is binary (i.e., L = ; or / = 1). Thus, R = 14,400
bitss. The null bandwidth is
B null = R>/ =D = 14,400 Hz
SA3–4 Bandwidth of RS-232 Signals The RS-232 serial port on a personal computer is
transmitting data at a rate of 38,400 bitss using a polar NRZ line code. Assume that binary 1’s
and 0’s are equally likely to occur. Compute and plot the PSD for this RS-232 signal. Use a dB
scale with the PSD being normalized so that 0 dB occurs at the peak of the PSD plot. Discuss the
bandwidth requirements for this signal.
Solution. Referring to Eq. (3–41), set A 2 T b equal to 1 so that 0 dB occurs at the peak. Then the
PSD, in dB units, is
dB (f) = 10 log ca sin pfT 2
b
b d
pfT b
where T b = 1R and R = 38,400 bitss.
This result is plotted in Fig. 3–48 using a dB scale. The plot reveals that the spectrum is broad for
this case of digital signaling with rectangular pulse shapes. Although the null bandwidth is 38,000 Hz
(B null = R), it gives a false sense that the spectrum is relatively narrow, because the first sidelobe
peak (at f = 57,600 Hz = 1.5R) is down by only 13.5 dB from the main lobe, and the second sidelobe
2 /
0
Power Spectral Density
–5
–10
–15
(f) dB
–20
–25
–30
–35
–40
–45
–50
–2
Figure 3–48
–1.5 –1 –0.5 0 0.5 1 1.5 2
f(Hz)
* 10 5
PSD of an RS-232 signal with a data rate of 38,400 bitss. (See SA3_4.m.)