FM transmission bandwidth by Carson rule
BlinkNBuild practice problem · GATE standard, authored and verified in-house
An FM signal has a peak frequency deviation of $\Delta f = 75\ \text{kHz}$ and the modulating signal has a maximum frequency of $f_m = 15\ \text{kHz}$.
The transmission bandwidth by Carson's rule, in kHz, is ________.
Show the step-by-step derivation
Answer
180 kHz kHz
Step-by-step derivation
- Compute the modulation index: $$\beta = \frac{\Delta f}{f_m} = \frac{75}{15} = 5.$$ Since $\beta \gg 1$, this is wideband FM - the commercial broadcast standard.
- Carson's rule estimates the bandwidth containing about $98\%$ of the signal power: $$B = 2\,(\Delta f + f_m).$$
- Substitute: $$B = 2\,(75 + 15) = 2 \times 90 = \mathbf{180\ kHz}.$$
- Equivalent form. Writing it as $B = 2f_m(\beta + 1) = 2(15)(6) = 180$ kHz gives the same answer and makes the two limits obvious: as $\beta \to 0$ it collapses to $2f_m$, the AM bandwidth, and for large $\beta$ it approaches $2\Delta f$.
- Reality check. FM broadcast channels are spaced $200$ kHz apart precisely to accommodate this $180$ kHz occupancy plus a guard band. ✓
The trap this question is built around
Answering $2\Delta f = 150$ kHz. That is the large-$\beta$ asymptote and it always underestimates. Carson's rule keeps the $+f_m$ term precisely because it matters at the modulation indices actually used.