B. P. Lathi, Zhi Ding - Modern Digital and Analog Communication Systems-Oxford University Press (2009)

5.4 Demodulation of FM Signals 233
Because the operation is on the slope of IH(f)I, this method is also called slope detection.
Since, however, the slope of IH (f) I is linear over only a small band, there is considerable
distortion in the output. This fault can be partially corrected by a balanced discriminator
formed by two slope detectors. Another balanced demodulator, the ratio detector, also widely
used in the past, offers better protection against carrier amplitude variations than does the
discriminator. For many years ratio detectors were standard in almost all FM receivers. 1 0
Zero-crossing detectors are also used because of advances in digital integrated circuits.
The first step is to use the amplitude limiter of Fig. 5.9a to generate the rectangular pulse
output of Fig. 5.9c. The resulting rectangular pulse train of varying width can then be applied
to trigger a digital counter. These are the frequency counters designed to measure the instantaneous
frequency from the number of zero crossings. The rate of zero crossings is equal to
the instantaneous frequency of the input signal.
FM Demodulation via PLL
Consider a PLL that is in lock with input signal sin [w e t + 0; (t)] and output error signal e 0 (t).
When the input signal is an FM signal,
f-00
t
J(
0i (t) = kt m(a) da + -
2
(5.27)
then,
0 0 (t) = kt f m(a) dct + O.5n - 0e (t) 00
With PLL in lock we can assume a small frequency error 0 e (t) R:: 0. Thus, the loop filter output
signal is
1 . 1 d [
ft ] k t
e 0 (t) = -0 0 (t) = - - kt m(a) da + O.5n - 0e (t) =::: -m(t)
C C dt -(X) C
(5.28)
Thus, the PLL acts as an FM demodulator. If the incoming signal is a PM wave, then
e 0 (t) = kpin(t)/c. In this case we need to integrate e 0 (t) to obtain the desired signal m(t) .
To more precisely analyze PLL behavior as an FM demodulator, we consider the case of
a small error (linear model of the PLL) with H (s) = 1. For this case, feedback analysis of the
small-error PLL in Chapter 4 becomes
AK H(s)
AK
0 0 (s) = ---- 0;(s) = -- 0;(s)
s + AK H(s) s + AK
If En (s) and M (s) are Laplace transforms of e 0 (t) and m(t), respectively, then from Eqs. (5.27)
and (5.28) we have
kt M(s)
0i(S) = -­
S
and
s0 0 (s) = cE0 (s)