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504 Performance of Communication Systems Corrupted by Noise Chap. 7 For bipolar NRZ signaling, the energy in the different signal is E d = A 2 T = 2E b where E b is the average energy per bit. Thus, for a matched-filter receiver, the BER is P e = 3 2 Q¢ E b ≤ CN 0 (matched filter) (7–28b) For bipolar RZ signaling, E d = A 2 T/4 = 2E b , so that the resulting BER formula is identical to 3 Eq. (7–28b). These results show that the BER for bipolar signaling is just 2 that for unipolar signaling as described by Eq. (7–24b). Example 7–2 BER FOR POLAR, UNIPOLAR, AND BIPOLAR NRZ SIGNALING WITH MATCHED FILTERING Evaluate and plot the bit error rate (BER) for the cases of polar NRZ, unipolar NRZ, and bipolar NRZ signaling in the presence of additive white Gaussian noise (AWGN). In all cases assume that matched filtering is used at the receiver. See Example7_02.m for the solution. Compare these results with those shown in Fig. 7–5. Example 7–3 BER FOR POLAR, UNIPOLAR, AND BIPOLAR NRZ SIGNALING WITH LOW-PASS FILTERING Repeat Example 7–2, except assume that low-pass filtering (LPF) is used in all of the receivers where B = 2/T. See Example7_03.m for the solution. 7–3 COHERENT DETECTION OF BANDPASS BINARY SIGNALS On–Off Keying From Fig. 5–1c, an OOK signal is represented by s 1 (t) = A cos (v c t + u c ), 0 6 t … T (binary 1) (7–29a) or s 2 (t) = 0, 0 6 t … T (binary 0) (7–29b) For coherent detection, a product detector is used as illustrated in Fig. 7–7. Actually, in RF applications, a mixer would convert the incoming OOK signal to an IF, so that stable highgain amplification could be conveniently achieved and then a product detector would translate the signal to baseband. Figure 7–7, equivalently, represents these operations by converting the incoming signal and noise directly to baseband. Assume that the OOK signal plus white (over the equivalent bandpass) Gaussian noise is present at the receiver input. As developed in Chapter 6, this bandpass noise may be represented by n(t) = x(t) cos (v c t + u n ) - y(t) sin (v c t + u n )
- Page 1006: ` Sec. 6-11 Study-Aid Examples 479
- Page 1010: Problems 481 SA6-4 PSD for a Bandpa
- Page 1014: Problems 483 The power of n 1 (t) i
- Page 1018: Problems 485 1 2 x (f) = e N 0, ƒ
- Page 1022: Problems 487 6-42 A bandpass WSS ra
- Page 1026: Problems 489 6-54 A narrowband-sign
- Page 1030: Problems 491 6-60 Let be a wideband
- Page 1034: Sec. 7-1 Error Probabilities for Bi
- Page 1038: Sec. 7-1 Error Probabilities for Bi
- Page 1042: Sec. 7-1 Error Probabilities for Bi
- Page 1046: Sec. 7-2 Performance of Baseband Bi
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- Page 1064: 508 Upper channel Receiver r(t)=s(t
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- Page 1084: 518 DPSK signal plus noise in Bandp
- Page 1088: 520 QPSK signal plus noise (data ra
- Page 1092: TABLE 7-1 COMPARISON OF DIGITALSIGN
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504<br />
Performance of Communication Systems Corrupted by Noise Chap. 7<br />
For bipolar NRZ signaling, the energy in the different signal is E d = A 2 T = 2E b where E b is<br />
the average energy per bit. Thus, for a matched-filter receiver, the BER is<br />
P e = 3 2 Q¢ E b<br />
≤<br />
CN 0<br />
(matched filter)<br />
(7–28b)<br />
For bipolar RZ signaling, E d = A 2 T/4 = 2E b , so that the resulting BER formula is identical to<br />
3<br />
Eq. (7–28b). These results show that the BER for bipolar signaling is just 2 that for unipolar<br />
signaling as described by Eq. (7–24b).<br />
Example 7–2 BER FOR POLAR, UNIPOLAR, AND BIPOLAR<br />
NRZ SIGNALING WITH MATCHED FILTERING<br />
Evaluate and plot the bit error rate (BER) for the cases of polar NRZ, unipolar NRZ, and bipolar<br />
NRZ signaling in the presence of additive white Gaussian noise (AWGN). In all cases assume<br />
that matched filtering is used at the receiver. See Example7_02.m for the solution. Compare<br />
these results with those shown in Fig. 7–5.<br />
Example 7–3 BER FOR POLAR, UNIPOLAR, AND BIPOLAR NRZ<br />
SIGNALING WITH LOW-PASS FILTERING<br />
Repeat Example 7–2, except assume that low-pass filtering (LPF) is used in all of the receivers<br />
where B = 2/T. See Example7_03.m for the solution.<br />
7–3 COHERENT DETECTION OF BANDPASS BINARY SIGNALS<br />
On–Off Keying<br />
From Fig. 5–1c, an OOK signal is represented by<br />
s 1 (t) = A cos (v c t + u c ), 0 6 t … T (binary 1) (7–29a)<br />
or<br />
s 2 (t) = 0, 0 6 t … T (binary 0) (7–29b)<br />
For coherent detection, a product detector is used as illustrated in Fig. 7–7. Actually, in RF<br />
applications, a mixer would convert the incoming OOK signal to an IF, so that stable highgain<br />
amplification could be conveniently achieved and then a product detector would translate<br />
the signal to baseband. Figure 7–7, equivalently, represents these operations by converting the<br />
incoming signal and noise directly to baseband.<br />
Assume that the OOK signal plus white (over the equivalent bandpass) Gaussian noise<br />
is present at the receiver input. As developed in Chapter 6, this bandpass noise may be<br />
represented by<br />
n(t) = x(t) cos (v c t + u n ) - y(t) sin (v c t + u n )
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