Single tone FM in Bessel functions; average power

Single-tone FM in terms of Bessel functions

For $m(t) = A_m\cos\omega_mt$, the instantaneous frequency is $f_i = f_c + k_fA_m\cos\omega_mt$, so the phase is $\omega_ct + \beta\sin\omega_mt$ with $\beta = k_fA_m/f_m = \Delta f/f_m$:

$$s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt) = \text{Re}\left[A_ce^{j\beta\sin\omega_mt}e^{j\omega_ct}\right]$$

$e^{j\beta\sin\omega_mt}$ is periodic with period $1/f_m$; its Fourier series is

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

(substituting $x = \omega_mt$). Therefore

$$s(t) = A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos[(\omega_c + n\omega_m)t]$$

i.e. a carrier $A_cJ_0(\beta)$ at $f_c$ and sidebands $A_cJ_n(\beta)$ at $f_c + nf_m$.

Average power with components from −n to +n

Each component is a cosine of amplitude $A_cJ_k(\beta)$; the components are at different frequencies, so their powers add (1 Ω load):

$$P_n = \sum_{k=-n}^{n}\frac{A_c^2J_k^2(\beta)}{2} = \frac{A_c^2}{2}\left[J_0^2(\beta) + 2\sum_{k=1}^{n}J_k^2(\beta)\right]$$

using $J_{-k}^2 = J_k^2$. When all components are included, $\sum_{k=-\infty}^{\infty}J_k^2(\beta) = 1$, so

$$P = \frac{A_c^2}{2}$$

the same as the unmodulated carrier. The fraction of power within $\pm n$ sidebands is $J_0^2 + 2\sum_{1}^{n}J_k^2$. Example: for $\beta = 1$, $n = 2$ gives $0.7652^2 + 2(0.4401^2 + 0.1149^2) = 0.9992$, i.e. 99.9% of the power.