# Finding the maximum phase and the maximum frequency deviation of an angle-modulated signal

I have an angle-modulated signal: $$s(t)=10\left[\cos\left(10^8\pi · t + 5\sin\left(2\pi\cdot10^3·t\right)\right)\right]$$

How would I go about finding the maximum phase and the maximum frequency deviation?

• well, you look at the phase of the signal and find the maximum deviation of that? Commented Feb 8, 2021 at 16:53
• @MarcusMüller Ok, I guess I should clarify as I'm having trouble finding the phase, as this is different from the general formula of a sin or cos function, how would I go about finding the phase? Commented Feb 8, 2021 at 18:05
• huh? this doesn't seem any different? Please edit your question to explain explicitly what you consider the "general formula", and how this formula doesn't fit that. Commented Feb 8, 2021 at 18:28

I think this is treating the $$10^8\pi\cdot t$$ as the carrier frequency and the $$5 sin(2\pi\cdot 10^3 t)$$ as the modulation. So the maximum phase difference of this signal to the carrier without modulation would be $$\pm5$$ radians.
To compute the change in frequency you need to find the compute $$\frac{d}{dt}$$ of the argument of the cos (divided by $$2\pi$$) and subtract the carrier frequency.
$$\frac{d}{dt} 10^8\pi\cdot t + 5 sin(2\pi\cdot 10^3 t) = 10^8\pi + 2\cdot 5\cdot 10^3\pi cos(2\pi\cdot 10^3 t)$$
Divide by $$2\pi$$ to get Hz instead of radians.
$$Frequency = 5\cdot 10^7 + 5\cdot 10^3 cos(2\pi \cdot 10^3t)$$
Subtract off the carrier of 50MHz and you have a maximum frequency excursion of $$\pm 5$$ k Hz