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Baseband Pulse and Digital Signaling Chap. 3
(a) What is the null bandwidth of the PCM signal if a polar line code is used?
(b) What is the average SNR of the recovered analog signal at the receiving end?
Solution.
(a) M = 64 quantizing steps generate 6-bit PCM words because M = 2 n . Using Eq. (3–15b),
we find what the null bandwidth is
B null = nf s = 6(7,000) = 42 kHz
Note: If sin xx pulse shapes were used, the bandwidth would be
B null = 1 2 nf s = 21 kHz
(b) Using Eq. (3–16b) with M = 64 and P e = 10 -4 yields
a S N b = M 2
1 + 4(M 2 - 1)P e
=
4,096
1 + 1.64
= 1,552 = 31.9 dB
(Note: The 1 in the denominator represents quantization noise, and the 1.64 represents
noise in the recovered analog signal caused by bit errors at the receiver. In this example,
both noise effects contribute almost equally. For the case of M = 64, if the BER was less
than 10 -5 , the quantizing noise would dominate, or if the BER was larger than 10 -3 , noise
resulting from receiver bit errors would dominate.)
SA3–3 Properties of NRZ Line Codes A unipolar NRZ line code is converted to a multilevel
signal for transmission over a channel as illustrated in Fig. 3–13. The number of possible
values in the multilevel signal is 32, and the signal consists of rectangular pulses that have a pulse
width of 0.3472 ms. For the multilevel signal,
(a) What is the baud rate?
(b) What is the equivalent bit rate?
(c) What is the null bandwidth?
(d) Repeat (a) to (c) for the unipolar NRZ line code.
Solution.
(a) Using Eq. (3–28) where N = 1 pulse occurs in T 0 = 0.3452 ms, we get
D = N>T 0 = 1>0.3472 ms = 2,880 baud
(b) Because L = 32 = 2 , = 5. Using Eq. (3–34),
R =/D = 5(2880) = 14,400 bits/s
(c) Using Eq. (3–54), we find that the null bandwidth is
B null = R>/ =D = 2,880 Hz