--- title: "Carson's rule; when FM bandwidth equals AM" subject: "Communication Systems" chapter: 3 key: "semester_6/communication_systems:carson-rule-fm-am-bandwidth" source: https://www.bctengineer.com/q/semester_6/communication_systems/carson-rule-fm-am-bandwidth site: BCT Engineering --- # Carson's rule; when FM bandwidth equals AM Communication Systems · Chapter 3: Angle Modulation - Asked in: 2072 Magh (CS I), 2+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/communication_systems/carson-rule-fm-am-bandwidth - Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3 ## Question Explain Carson's rule for determining the bandwidth of an FM wave. Under what condition the bandwidth of FM signal is same as that of AM? Explain. ## Short answer Carson's rule approximates 98% power bandwidth as B_T ≈ 2(Δf + f_m) = 2f_m(β+1). When β << 1 (narrowband FM), B_T ≈ 2f_m, same as AM, because only the carrier and one sideband pair are significant; the wave has constant amplitude with phase-reversed lower sideband. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What does Carson's rule state? A. About 98% of FM power lies in B = 2(Delta f + fm) B. B = 2 Delta f exactly always C. B = Delta f + fm / 2 D. 100% of power lies in 2 fm Answer: A. Equivalently B = 2 fm (beta + 1). (Key point: Carson's rule) 2. When is FM bandwidth equal to AM? A. When beta >> 1 B. When beta << 1 (NBFM) C. When Delta f = 75 kHz D. When fm = 0 Answer: B. Then B is about 2fm. (Key point: When equals AM) 3. Why does NBFM occupy only 2fm? A. J0 is about 0 and J1 about 1 B. All Jn are equal C. J0 is about 1, J1 about beta/2, and higher Jn about 0, leaving one sideband pair D. Carrier is suppressed Answer: C. Spectrum looks like AM. (Key point: Reason) 4. For large beta, what is B? A. About 2 fm B. About fm C. About beta D. About 2 Delta f Answer: D. Delta f dominates. 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