# FM INDEX OF MODULATION QUESTION

Started by October 11, 2005
```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

How can you show that this is
indeed the case?  A mathematical
proof would be interesting.

Thanks!

Slick

```
```radio913@aol.com 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
>
>      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.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```Jerry Avins 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.

So if your modulator is a Voltage Controlled Oscillator,
then "k"= (Kvco) X (peak modulating amplitude in Volts),

where Kvco is in units of MHz/volt, or (radians/sec)/volt.

>
> 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.
>

Ahh, i knew it was something simple.

Thanks, Jerry!

Slick

```
```radio913@aol.com wrote:
> Jerry Avins wrote:

...

>>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.
>>
>
>
>      Ahh, i knew it was something simple.
>
>      Thanks, Jerry!

You're welcome.

Right after I got my BEE (before, actually, but I hadn't realized it) I
discovered that if I remembered definitions, I didn't really need formulas.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```