[email protected] wrote:
> Hi,
>
> It's recently come to my attention
> that the FM index of modulation is
> also the peak phase deviation in radians.
> This was in regards to a question about the
> linear region of phase/frequency detector.
>
> My question is (and excuse me if
> the answer is simple):
>
> How can you show that this is
> indeed the case? A mathematical
> proof would be interesting.
There's nothing to prove; it's a matter of definition. The modulation
index (beta) is defined for a single sinusoidal modulating signal.
Instantaneous frequency == w_i
Carrier frequency == w_c
Modulating frequency == w_m
k is a constant that depends on the amplitude w_m and the fixed gain of
the modulator.
Then by the definition of FM, w_i = w_c + k*cos(w_m) and it is clear
that k is the maximum frequency deviation. Phase deviation is the time
integral of frequency deviation. Because the choice of phase reference
is arbitrary, the constant of integration may be set to zero.
theta(t) = integral(w_i*dt). The integral of the modulated term
k*cos(w_m) is (k/w_m)*sin(w_m); the term k/w_m) is called beta, and its
value is as clearly the peak phase deviation as k is the peak frequency
deviation.
Be warned that FM is not a linear process and superposition doesn't
apply. The relative amplitudes of sidebands produced by a single
modulating frequency depends on its amplitude, and the analysis of even
two modulating frequencies is very complex. In general, the sidebands
are not even symmetric. So "modulation index" is a useful idealization
that doesn't exactly apply in practice.
Jerry
--
Engineering is the art of making what you want from things you can get.
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