What bandwidth is needed for an FM signal that has a peak deviation of ± 3 KHz and handles audio signals from 200 Hz to 5 KHz ?
16 KHz
Carson's rule gives the bandwidth of an FM signal directly :
\(B=2\left(\Delta f+f_{m}\right)\)
where \(\Delta f\) is the peak deviation and \(f_{m}\) the highest modulating frequency.
Identify the two quantities. The deviation is given as ±3 kHz, so \(\Delta f=3\ \text{kHz}\). The audio spans 200 Hz to 5 kHz, and bandwidth is set by the worst case — the top of that range:
\(f_{m}=5\ \text{kHz}\)
The 200 Hz lower limit plays no part; it is there to see whether the candidate reaches for the wrong end of the band.
Substitute :
\(B=2\left(3+5\right)=16\ \text{kHz}\)
which is option 2.
| Wrong route | Gives | Error |
|---|---|---|
| \(2\Delta f\) | 6 kHz | Ignores the modulating frequency entirely |
| \(2f_{m}\) | 10 kHz | The AM bandwidth, not FM |
| \(2(3+1.8)\) | 9.6 kHz | Uses some intermediate audio frequency |
| \(2(\Delta f+f_{m})\) | 16 kHz | Correct |
Check the modulation index to see what kind of FM this is:
\(\beta=\dfrac{\Delta f}{f_{m}}=\dfrac{3}{5}=0.6\)
With \(\beta\lt1\) this is narrowband FM, and Carson's rule sensibly returns a value close to \(2f_{m}=10\ \text{kHz}\), the AM-like limit, rather than the \(2\Delta f\) that dominates in wideband FM.
Why Carson's rule is only approximate. An FM signal strictly has infinitely many sideband pairs, spaced \(f_{m}\) apart, with amplitudes given by the Bessel functions \(J_{n}(\beta)\). Carson's rule counts the sidebands that together carry about 98 % of the power and discards the negligible remainder — a working engineering compromise rather than an exact result.
The comparison worth remembering : broadcast FM uses \(\Delta f=75\ \text{kHz}\) with \(f_{m}=15\ \text{kHz}\), giving \(B=180\ \text{kHz}\) inside a 200 kHz channel allocation.
Hence, the bandwidth needed is 16 kHz.
Similar Questions
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Select your answer using the codes given below :
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Important Questions from Frequency Modulation
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In FM, "M" stands for
