Angle modulation; single tone FM time and frequency domain

Angle modulation

Angle modulation is modulation in which the angle (phase) of the carrier is varied according to the message while its amplitude stays constant:

$$s(t) = A_c\cos\theta_i(t)$$

It has two forms: phase modulation (PM), $\theta_i = \omega_ct + k_pm(t)$, and frequency modulation (FM), $\theta_i = \omega_ct + 2\pi k_f\int m(\tau)d\tau$.

Time-domain expression of single-tone FM

Let $m(t) = A_m\cos2\pi f_mt$. Instantaneous frequency:

$$f_i(t) = f_c + k_fA_m\cos2\pi f_mt = f_c + \Delta f\cos2\pi f_mt$$

Instantaneous phase:

$$\theta_i(t) = 2\pi\int_0^tf_i(\tau)d\tau = 2\pi f_ct + \frac{\Delta f}{f_m}\sin2\pi f_mt$$

So

$$s(t) = A_c\cos(2\pi f_ct + \beta\sin2\pi f_mt), \qquad \beta = \frac{\Delta f}{f_m} = \frac{k_fA_m}{f_m}$$

Frequency-domain expression

Write $s(t) = \text{Re}[A_ce^{j\beta\sin\omega_mt}e^{j\omega_ct}]$. The periodic term has Fourier series

$$e^{j\beta\sin\omega_mt} = \sum_{n=-\infty}^{\infty}J_n(\beta)e^{jn\omega_mt}, \quad J_n(\beta) = \frac{1}{2\pi}\int_{-\pi}^{\pi}e^{j(\beta\sin x - nx)}dx$$

Therefore

$$s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[2\pi(f_c + nf_m)t]$$

Taking the Fourier transform:

$$S(f) = \frac{A_c}{2}\sum_{n=-\infty}^{\infty}J_n(\beta)\left[\delta(f - f_c - nf_m) + \delta(f + f_c + nf_m)\right]$$

The spectrum has a carrier of amplitude $A_cJ_0(\beta)$ and sideband pairs at $f_c \pm nf_m$ with amplitudes $A_c|J_n(\beta)|$. Total power is $A_c^2/2$, and practical bandwidth is $2(\beta + 1)f_m$.