Carson's rule; when FM bandwidth equals AM

Carson's rule

In theory an FM wave has infinitely many sidebands at $f_c \pm n f_m$, with amplitudes $A_c J_n(\beta)$. In practice, sidebands beyond about $n = \beta + 1$ are very small. Carson's rule says that about 98% of the total power of an FM wave lies within

$$B_T \approx 2(\Delta f + f_m) = 2 f_m(\beta + 1)$$

where $\Delta f$ is the peak frequency deviation and $f_m$ the highest modulating frequency.

• For large $\beta$ (WBFM): $B_T \approx 2\Delta f$ (set by the deviation).
• For small $\beta$ (NBFM): $B_T \approx 2 f_m$.

Example: commercial FM with $\Delta f = 75$ kHz and $f_m = 15$ kHz gives $B_T = 2(75 + 15) = 180$ kHz (200 kHz channels are allotted).

When FM bandwidth equals AM bandwidth

When $\beta \ll 1$ (narrowband FM, $\beta < 0.3$), $\Delta f \ll f_m$, so

$$B_T = 2(\Delta f + f_m) \approx 2 f_m$$

which is the same as the bandwidth of AM ($2f_m$).

Reason: for small $\beta$, $J_0(\beta) \approx 1$, $J_1(\beta) \approx \beta/2$, and $J_n(\beta) \approx 0$ for $n \ge 2$. So the NBFM wave is

$$s(t) \approx A_c\cos 2\pi f_c t + \frac{\beta A_c}{2}\left[\cos 2\pi(f_c + f_m)t - \cos 2\pi(f_c - f_m)t\right]$$

It has only the carrier and one pair of sidebands, exactly like AM, so it occupies $2f_m$. The only difference is that the lower sideband is phase-reversed, so the resultant changes the phase (not the amplitude) of the carrier.