I have a small understanding issue here, so I would be very glad if someone could help me a bit!

We are given this FM modulated signal to volts;

Xfm (t) = cos[2*pi*(10^8)*t + 5*sin(pi*t(10^4)) + 3*sin(2*pi*t(10^4)) ] , on a R=50 Ohm.

Find phase and frequency deviation.

I have solved similar exercises, bet there was only one sin(...).

We know that Ufm(t) = Acos(ωt + βsinωt).

Which one here is the β? is it the max value?

For frequency deviation, I know that I have to take the derivative of phase in regard to time. But here is looks like there are two phases on the signal? Do I have to get the one with higher coefficient?

  • \$\begingroup\$ I'm not a radio guy, but is that even an FM signal? Shouldn't it be a central carrier 100MHz and sidebands. (FM sidebands are cool.. the modulation index contains Bessel functions.) to the OP, I'm sorry I don't know the definition of the frequency or phase deviation. The bandwidth of a real FM signal depends on the modulation index. \$\endgroup\$ Sep 15, 2014 at 22:45

1 Answer 1


We know that Ufm(t) = Acos(ωt + βsinωt)

It's better to distinguish the two omegas you have in the above equation as 1 and 2 (subscripts).

Back to the question....

To find the max frequency deviation it can be assumed that this will be when the two individual sine waves are both hitting maximum or minimum amplitude simulataneously. Unfortunately, the two sine terms are locked in phase with one double the frequency of the other so you need to differentiate the sum of the two sine terms and equate to zero. This should be fairly straightforward.

  • \$\begingroup\$ Cheers, first omega is for carrier second is for message, but I couldn't write the subscripts here. Now back to it. By differentiating those sines I am getting this: 10000pi[5cos(10000pit) +6cos(20000pit)] . Could I say that freq deviation is 10kpi/2pi = 5khz? But what about phase deviation ? \$\endgroup\$ Sep 15, 2014 at 21:00
  • \$\begingroup\$ I visualized it as the differential of [Sin (A) + Sin (2A)] - it's exactly the same but doing it algebraicly solves the problem then plug the numbers in - the numbers look OK though but, It's getting a little late tonight (long day etc..). \$\endgroup\$
    – Andy aka
    Sep 15, 2014 at 21:14
  • \$\begingroup\$ Oh I see, well derivation is correct. Anyway, when you have time, maybe you could help me to get the β term or somehow to find a way to get that phase deviation. \$\endgroup\$ Sep 15, 2014 at 21:46

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