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
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 idea behind this question
An FM signal in theory has infinitely many sidebands, but almost all of its power lies within a bandwidth of about $2(\Delta f + f_m)$. That is Carson's rule. The peak deviation $\Delta f$ is how far the carrier swings; $f_m$ is the highest frequency in the message.
Try a variation
With $\Delta f = 50$ kHz and $f_m = 10$ kHz, what is the Carson bandwidth?
Show the answer
Answer: $120$ kHz
$2(50 + 10) = 120$ kHz.
Other mistakes to avoid
- Using the modulation index $\beta$ in place of the deviation $\Delta f$ in the formula.
- Answering $2f_m$, the AM bandwidth, which ignores the frequency swing altogether.