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Sec. 7–11 Study-Aid Examples 559
(d) As discussed in the solution for SA7–2, if an inexpensive RC LPF is used with B = 2R,
then an inexpensive bit synchronizer (with imprecise sampling times) is adequate for use
in the receiver. As found in (a), this gives a BER of 9.9 × 10 -3 .
If better performance (i.e., a lower BER) is desired, the bandwidth of the LPF can
be reduced to B = R, and a precise (more expensive) bit synchronizer is required. This
lowers the BER to 4.9 × 10 -4 , as found in part (b).
If even better performance is desired, an MF (more expensive than an LPF) can be
used. This is implemented by using an integrate and dump filter as previously shown in
Fig. 6–17. This MF requires a bit synchronizer to provide the clocking signals that reset
the integrator and dump the sampler. This MF receiver gives the optimum (lowest BER)
performance. In this case, the BER is reduced to 1.6 × 10 -6 , as found in part (c). However,
it is realized that in some applications, such as some fiber-optic systems, the noise is negligible.
In such a case, the simple LPF of part (a) with B = 2/T = 2R is adequate because
the BER will be approach zero.
SA7–4 Output S/N Improvement for FM with Preemphasis In FM signaling systems,
preemphasis of the modulation at the transmitter input and deemphasis at the receiver output is
often used to improve the output SNR. For 75 µs emphasis, the 3-dB corner frequency of the
receiver deemphasis LPF is f 1 = 1/(2p75µs) = 2.12 kHz. The audio bandwidth is B = 15 kHz.
Derive a formula for the improvement in dB, I dB , for the SNR of a FM system with preemphasis–
deemphasis when compared with the SNR for a FM system without preemphasis–deemphasis.
Compute I dB for f 1 = 2.12 kHz and B = 15 kHz.
Solution Refer to Eqs. (7–121)–(7–127) for the development of the (S/N) out equation for FM
with no deemphasis and to Eqs. (7–135)–(7–138) for the development of the (S/N) out equation
for FM with deemphasis. Then,
I = (S>N) with deemphasis
(S>N) no deemphasis
=
a s 0 2
n 0
2 b with deephasis
a s 0 2
n 0
2 b no deemphasis
(7–152)
From Sec. 5–6 and Fig. 5–15, it is seen that the inclusion of two filters—a deemphasis filter at the
receiver, H d ( f ), and a preemphasis filter at the transmitter, H p ( f )—has no overall effect on the frequency
response (and power) of the output audio signal, s 0 (t), because H p ( f ) H d ( f ) = 1 over the audio
bandwidth, B, where B 6 f 2 . Thus, 1s 2 0
2 with deemphasis = 1s 2 0 2 no deemphasis, and Eq. (7–152) reduces to
I = (n 0 2 ) no deemphasis
(n 0 2 ) with deemphasis
(7–153)
The output noise is different for the receiver with deemphasis (compared with a receiver without
deemphasis) because the deemphasis filter attenuates the noise at the higher audio frequencies.
(The transmitter preemphasis filter has no effect on the received noise.) Using Eq. (7–126) for
An and Eq. (7–136) for An 2 02 B no deemphasis
0 B with deemphasis, we see that Eq. (7–153) becomes
I =
2
3 a K 2
b N 0 B 3
A c
2a K 2
b N
A 0 f 3 1 c B
c f 1
- tan -1 a B f 1
bd