I've been reading about FM and I keep coming to an expression, where the instantaneous phase

$$\Theta=2\pi f_it+\phi\text, \tag1$$ where $$f_i=f_c+k_fm(t)\text,\tag2$$ ($f_c$ being the base carrier frequency and $k_f$ being some proportionality constant) is written as some sort of integral expression, like here:

$$\Theta_i\{t\}=2\pi f_ct+2\pi k_f\int_0^t m\{t\}\,dt$$

No clear explanation is given, apart from:

The instantaneous phase of the modulated wave at any instant can be obtained by substituting 12.6 into 12.3 and integrating to get [the integral expression]

(12.3 and 12.6 being my equations (1) and (2) respectively).

Why do we need to integrate the modulator signal?

Why is the following not an accurate expression for the phase: $\Theta =2\pi f_ct+2\pi k_fm(t)t+\phi$, (simply plugging equation (2) into equation (1))?


You need to integrate the modulating signal because frequency is the time derivative of phase. Therefore, the typical relationship from introductory calculus holds:

$$ \phi_i(t) = \int_{-\infty}^{t} \frac{d\phi_i(\tau)}{d\tau} d\tau = \int_{-\infty}^{t} 2\pi f_i(\tau) d\tau $$

For causal signals, the lower limit on the integral changes to zero:

$$ \phi_i(t) = \int_{0}^{t} 2\pi f_i(\tau) d\tau $$

$$ \phi_i(t) = \int_{0}^{t} 2\pi\left(f_c + k_fm(\tau)\right) d\tau $$

$$ \phi_i(t) = 2\pi f_ct +\int_{0}^{t} 2\pi k_f m(\tau) d\tau $$

which is the answer you were trying to get to.

  • 1
    $\begingroup$ Thanks for the answer! Where does the 2π term inside the integral in your last equation come from? $\endgroup$ – Ivan Zabrodin Dec 31 '19 at 16:20
  • $\begingroup$ I adjusted the answer to make it more clear. Phase is naturally measured in radians, but frequency when expressed with variable $f$ is typically in units of Hertz (cycles per second). There are $2\pi$ radians per cycle, so you get a scaling factor of $2\pi$ in the integrand. $\endgroup$ – Jason R Dec 31 '19 at 21:41
  • $\begingroup$ Another question - why does plugging equation 2 into equation 1 not give you an accurate expression for the modulated signal? Is it because the frequency is only the coefficient of t when the frequency is a constant? $\endgroup$ – Ivan Zabrodin Jan 4 '20 at 15:39

Frequency by definition is the derivative of phase with respect to time (a change in phase divided by the change in time is frequency). You see this with the radian expression for frequency given by $2\pi f$: A frequency of 1 Hz is 1 cycle per second which is $2\pi$ radians per second. So similarly phase versus time is the integral of frequency versus time.

When you FM modulate, you are converting your units of magnitude (for example could be volts) versus time into units of instantaneous frequency versus time. So for example a sine wave that is oscillating over 1Vpp that is FM modulated using a modulation index that provides 10Hz/V would then be a signal of some arbitrary carrier frequency that is oscillating back and forth +/- 5Hz. So in your equation $m(t)$ which is your modulation signal is directly proportional to frequency and together with the translation constants shown has magnitude units of frequency.

Now if we are interested in the instantaneous phase of that signal versus time, which your equation is showing, then we would need to integrate the frequency versus time signal to get phase versus time.


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