If we go back to basic theory, we have a carrier signal of the form :-
\$E_c\cos\phi_c\$
... and a sinusoidal modulation signal of the form ...
\$E_m\cos(\omega_mt)\$
and if we let the frequency deviation be proportional to the modulation amplitude, so
\$\Delta\omega\propto E_m\$
the instantaneous frequency is given by ->
\$\dot{\phi_c}=\omega_c+\Delta\omega.\cos(\omega_mt)\$
Integrating this to get the instantaneous phase ->
\$\phi_c=\omega_ct+\dfrac{\Delta\omega}{\omega_m}\sin(\omega_mt)\$
So the modulated output is ->
\$E_c\cos\Big[\omega_ct+\dfrac{\Delta\omega}{\omega_m}\sin(\omega_mt)\Big]\$
As you say, the modulation index is dependent upon \$\omega_m\$ so the relative amplitudes of the spectral components will vary with \$\omega_m\$, but the modulation index is also a measure of the peak phase deviation, so if you want the spectral amplitudes to be independent of \$\omega_m\$ you must have \$\omega_m\propto \Delta_\omega \propto E_m\$, ie phase modulation.
One technique of producing phase modulation is to use a frequency modulator with pre-emphasis of the modulating signal to get the amplitude proportional to the frequency.
I don't understand I in your formula.
Normally (traditionally) AM is: -
y(t) = [A + M cos(ωm t + φ)] . sin(ωc t)
where
- y(t) is the final modulated signal
- M is the amplitude of the modulating cosine (or sine to answer your
question)
- A is the amplitude of the carrier sine (or cosine to reinforce the
answer!!)
- φ is the phase displacement of the modulating sinewave but is
irrelevant all but mathematically
- ωm and ωc are the frequencies of modulation and carrier.
Maybe I just don't recognize your formula but the answer is, like Jim Dearden implies swap them up or use the same because carrier and modulator are not going to be the same frequency when dealing with AM.
Best Answer
Add a low-pass filter to either I or to Q. I helped a chip debug, where the sideband suppression was only 20dB, because of imbalanced parasitic capacitance on the I or on the Q paths.