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))?


3 Answers 3


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
    Dec 31, 2019 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, 2019 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
    Jan 4, 2020 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.


Our plan is to come up with a mathematical representation of a frequency modulated oscillations of carrier waves that incorporate given sound waves.

Sound waves are changes in the air pressure and can be represented by some function of time m(t) (m stands for modulation, as we want this signal to modulate the frequency of sinusoidal carrier signal).

Practically, m(t) is the changes in the current of some circuit that contains a microphone, that converts the air pressure changes into synchronous changes in the electric current.

Let's discuss the concept of angular frequency of the carrier signal in more mathematical terms.
Frequency f, if constant, is the number of periods of oscillations per unit of time (second).
It's often expressed in radians per second ω=2πf and unmodulated oscillations of an electric current in the carrier LC circuit that produces base oscillations would be
I(t) = A·cos(ω·t)

This sinusoidal function of time I(t) that represents oscillations of electric current can be viewed as an X-coordinate of a point on a circle of radius A (amplitude of oscillations) with a center at the origin of coordinates, that rotates along a circle counterclockwise with constant angular speed ω radians per second, assuming that at t=0 it is located on the X-axis at point (A,0).

So, angular frequency in more mechanical terms is angular speed.

From now on we will consider the above presented rotation of a point as mathematical representation of sinusoidal oscillations.

Another important characteristic of an oscillation is its phase φ(t).
The definition of a phase is an angle a rotating point has rotated to during the time of rotation from its start up to a current moment t, which can be expressed as a time-dependent function φ(t).

From this definition immediately follows the analogy between kinematics terms distance, speed and concepts phase, angular frequency used in radio electronics.

This analogy is complete in a sense that the relationship between a phase φ(t) and angular frequency ω(t) is similar to that between a distance S(t) covered by a moving object and its instantaneous speed V(t).

The instantaneous speed at moment in time t in, generally, a non-uniform movement, as a function of time, is a derivative of a distance covered by a moving object from the start of movement up to a position at time t, as a function of time:
V(t) = dS/dt = S'(t)

Similarly, angular frequency (sometimes called instantaneous angular frequency or simply instantaneous frequency) is the first derivative of a phase (angle of rotation) as a function of time:
ω(t) = dφ/dt = φ'(t)

Knowing speed V(t) of a moving object at every moment of time from start to t, we can restore the distance S(t) covered by this object as a function of time
S(t) = [0,t]V(τ)·dτ

Similarly, we can restore the phase φ(t) (that is, an angle a point has rotated from the start of its rotation to a moment in time t), knowing the instantaneous angular frequency ω(t) at each moment of time.
φ(t) = [0,t]ω(τ)·dτ

Consider the main equation of oscillations of an electric current in the LC circuit of a carrier without any sound modulation
I(t) = A·cos(ω0·t)

The argument to a function cos() is a product of a constant angular frequency (speed) ω0 by time, which is an angular distance of rotation or, using terminology introduced above, a phase of the rotation at time t
φ(t) = ω0·t

Therefore, our representation of carrier signal can be expressed in a more general form, suitable even for non-uniform rotation:
I(t) = A·cos(φ(t))

In a non-uniform rotation with variable instantaneous angular frequency ω(t) we can always derive this frequency from the phase:
ω(t) = φ'(t)

If we want to combine the carrier signal I(t) with some frequency modulating signal m(t) in such a way that the resulting variable instantaneous frequency ωmod(t) of a modulated signal reflected the modulation, we need to satisfy the following equation:
ωmod(t) = ω0 + m(t)
ω0 is the carrier own unmodulated constant frequency determined by it main LC circuit,
m(t) is a modulating add-on to reflect the sound waves to be transmitted.

We can even vary the degree by which the modulating affects the output signal by adding a modulating index λ as a factor to a modulator m(t):
ωmod(t) = ω0 + λ·m(t)

Knowing the target instantaneous frequency ωmod(t) and the above expression of a phase in terms of this frequency
φ(t) = [0,t]ω(τ)·dτ
we can express the modulated phase as
φmod(t) = [0,t]ωmod(τ)·dτ

This modulated phase will be an argument to a modulated signal of a carrier
Imod(t) = A·cos[φmod(t)] =
= A·cos
[[0,t]ωmod(τ)·dτ] =
= A·cos
[[0,t]0+λ·m(τ))·dτ] =
= A·cos

The role of modulation index λ in this formula is to define how significantly base carrier frequency should change with a change in sound waves frequency.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.