I don't understand intuitively why the instantaneous frequency is obtained by calculating the time derivative of the phase

Question

I don't understand intuitively why the instantaneous frequency is obtained by calculating the time derivative of the phase

Answer 1

This question comes up frequently in the context of frequency modulation. You have a carrier frequency, say $\omega_0$, a frequency around which your antenna is "tuned", in other words, it can transmit efficiently. You make an oscillator that also oscillates at $\omega_0$, the oscillator produces a voltage dependent on time as $v_0(t)=A_0\cos(\omega_0t +\phi_0)$ where $\phi_0$ is an arbitrary fixed phase telling you the voltage at time $t=0$. We also assume that the amplitude $A_0$ of the oscillator is unchanging and given.

This signal $v_0(t)$, if transmitted, carries no information because you know everything about it: $A_0, \omega_0, \phi_0$.

Now instead of wasting time to transmit a signal without information you wish to add some information to it. One way of doing it is Armstrong's frequency modulation. An oscillator usually has an LC resonator in the feedback loop of an amplifier that compensates for the losses in the resonator. Most dielectrics in a capacitor are more or less sensitive to the the capacitor voltage so that a large enough bias can change its effective capacitance and thus the resonant frequency in the LC, and this change is to be much slower than the resonant frequency of the resonator.

Say, upon impressing a bias voltage $v_m(t)$ on the capacitor the resonant frequency is shifted by an amount $\omega_m = k_m v_m(t)$ where $k_m$ is a "material" quantity dependent on the dielectric. Of course, as you change the oscillator frequency you also change the instantaneous phase $\psi(t)$ of the oscillator. Without modulation the instantaneous phase is $\psi_0(t)=\omega_0t+\phi_0$ but when the resonant frequency is manipulated by $v_m(t)$ it becomes $\psi_1(t)=\omega_0t+\phi_0+\int_0^t dt' k_mv_m(t').$ Of course, $\frac{d\psi_0(t)}{dt}=\omega_0$ and $\frac{d\psi_1(t)}{dt}=\omega_0+k_mv_m(t)$ as it should be.

Now for any fixed bias voltage the resonant frequency is $\omega_0 + \omega_m = \omega_0 +k_m v_m$ and the transmitted signal from the antenna is then $v_1(t)=A_0\cos((\omega_0 +k_m v_m)t+\phi_0).$

Answer 2

If you have a sinusoidal plane wave of fixed frequency $\omega$ then the function describing the wave amplitude is $$A \cos(\vec k \cdot \vec x - \omega t + \phi)$$ where $A$ is the amplitude and $\vec k$ is the wavevector. The quantity $$\theta = \vec k \cdot \vec x - \omega t + \phi$$ is called the phase and the derivative of the phase $$\frac{\partial \theta}{\partial t} = \omega$$ is the frequency as expected.