Correct Answer - Option 3 : Both (1) and (2)
The frequency deviation in a phase-modulated carrier is proportional to the amplitude and frequency of the modulating signal both.
Derivation:
A general expression for a phase-modulated wave is:
xPM (t) = A cos [2πfct + kpm(t)]
The instantaneous angle is given as:
ϕi(t) = 2πfct + kp m(t)
The instantaneous frequency (in Hz) will be obtained as:
\({f_i}\left( t \right) = \frac{1}{{2\pi }}\frac{{d{\phi _i}\left( t \right)}}{{dt}}\)
\({f_i}\left( t \right) = \frac{1}{{2\pi }}\frac{d}{{dt}}\left[ {2\pi {f_c}t + {k_p}m\left( t \right)} \right]\)
\( = \frac{1}{{2\pi }} \cdot 2\pi {f_c} + \frac{{{k_p}}}{{2\pi }} \cdot \frac{{d\;m\left( t \right)}}{{dt}}\)
\({f_i}\left( t \right) = {f_c} + \frac{{{k_p}}}{{2\pi }}\frac{{dm\left( t \right)}}{{dt}}\)
Frequency deviation is given by:
\(Δ f= \frac{{{k_p}}}{{2\pi }}\frac{{dm\left( t \right)}}{{dt}}\)
For a single tone modulation:
m(t) = Am sin ωmt
The frequency deviation becomes:
\(Δ f= \frac{{{k_p}}}{{2\pi }}\frac{{d(A_msin\omega _mt)}}{{dt}}\)
\(Δ f= \frac{{{k_pA_m\omega _mcos\omega _mt}}}{{2\pi }} \)
Δf ∝ Am fm