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Sec. 7–8 Output Signal-to-Noise Ratios for Analog Systems 545
Evaluating the derivative and solving the resulting equation for m ' (t), we obtain
m ' (t) = a
KD f
2p + KD v
bm(t)
Substituting this expression for m ' (t) into Eq. (7–133), we get
' A c A 0
e (t) =
2
t
1
cos cv if t + a
b D
1 + (K>2p)D f m(l) dl d
v L
-q
(7–134)
'
This demonstrates that the modulation index of e (t) is exactly 1[1 + (K/2p) D v ] of the modulation
index of v in (t). The threshold extension provided by the FMFB receiver is on the order
of 5 dB, whereas that of a PLL receiver is on the order of 3 dB (when both are compared with
the threshold of an FM discriminator). Although this is not a fantastic improvement, it can be
quite significant for systems that operate near the threshold, such as satellite communication
systems. A system that uses a threshold extension receiver instead of a conventional receiver
may be much less expensive than the system that requires a double-sized antenna to provide
the 3-dB signal gain. Other threshold extension techniques have been described and analyzed
in the literature [Klapper and Frankle, 1972].
FM Systems with Deemphasis
The noise performance of an FM system can be improved by preemphasizing the higher
frequencies of the modulation signal at the transmitter input and deemphasizing the output of
the receiver (as first illustrated in Fig. 5–16). This improvement occurs because the PSD of
the noise at the output of the FM detector has a parabolic shape, as shown in Fig. 7–23b.
Referring to Fig. 7–21, we incorporate deemphasis into the receiver by including a
deemphasis response in the LPF characteristic. Assume that the transfer function of the
LPF is
H(f) =
1
1 + j(f>f 1 )
(7–135)
over the message bandwidth of B Hz. For standard FM broadcasting, a 75-µs deemphasis
filter is used so that f 1 = 1/[(2p) (75 × 10 -6 )] = 2.1 kHz. Using Eq. (7–125), we see that the
noise power out of the receiver with deemphasis is
or
[n ' B
0(t)] 2 = |H(f)| 2 n0 (f) df = a K B
1
b
2N 0 c
L A c L 1 + (f>f 1 ) 2 df2 df
-B
[n ' 0(t)] 2 = 2a K 2
3
b N 0 f
A 1
c B
c f 1
-B
- tan -1 a B f 1
bd
(7–136)
In a typical application, B>f and so tan -1 1 1,
(B/f 1 ) ≈ p/2, which is negligible when compared
with B/ f 1 . Thus, Eq. (7–136) becomes