The approximate rule for transmission of an FM signal generated by a single-tone modulating signal of frequency fm, modulation index β and maximum frequency deviation Δf, is defined as :
\(2\Delta f\left(1+\dfrac{1}{\beta}\right)\)
This is Carson's rule, and it is easiest to recognise in its plain form first :
\(B=2\left(\Delta f+f_{m}\right)\)
— twice the sum of the peak deviation and the highest modulating frequency. Every version in the options is this same statement rewritten using the modulation index
\(\beta=\dfrac{\Delta f}{f_{m}}\qquad\Rightarrow\qquad f_{m}=\dfrac{\Delta f}{\beta}\)
Substitute and factorise :
\(B=2\left(\Delta f+\dfrac{\Delta f}{\beta}\right)=2\Delta f\left(1+\dfrac{1}{\beta}\right)\)
which is option 1. Factorising the other way gives the equally valid \(B=2f_{m}(1+\beta)\) — note that options 2 and 4 are exactly these two correct forms with \(\Delta f\) and \(f_{m}\) swapped, which is the trap.
| Form | Expression | Correct? |
|---|---|---|
| Plain | \(2(\Delta f+f_{m})\) | ✓ |
| In terms of Δf | \(2\Delta f\left(1+\tfrac{1}{\beta}\right)\) | ✓ option 1 |
| In terms of fm | \(2f_{m}(1+\beta)\) | ✓ |
| Option 2 | \(2\Delta f(1+\beta)\) | ✗ mixes the two |
Why an approximation is needed at all. Strictly, an FM signal has infinitely many sidebands, spaced fm apart, with amplitudes given by the Bessel functions \(J_{n}(\beta)\). Carson's rule is the working compromise: it counts only the sidebands carrying about 98 % of the total power and discards the rest, which is why it is called approximate.
The two limits are worth checking. For narrowband FM, \(\beta\ll1\), the rule gives \(B\approx2f_{m}\) — the same as AM, since only the first sideband pair survives. For wideband FM, \(\beta\gg1\), it gives \(B\approx2\Delta f\) — the bandwidth is set by the deviation alone.
A worked example : commercial FM broadcasting uses \(\Delta f=75\ \text{kHz}\) and \(f_{m}=15\ \text{kHz}\), so \(\beta=5\) and \(B=2(75+15)=180\ \text{kHz}\) — comfortably inside the 200 kHz channel allocation.
Hence, the required bandwidth is 2Δf(1 + 1/β).
Similar Questions
In FM
Consider the following :
ST1 : F.M. signal produces more side bands than A.M.
ST2 : The carrier in a F.M. signal can never be dropped to zero amplitude.Which of the following is valid ?
Which of the following is true ?
The FM transmitters have the following blocks as per the following correct sequence :
(A) Crystal Oscillator
(B) Antenna
(C) Frequency multiplier
(D) Phase modulate / Audio source
(E) Power amplifierChoose the most appropriate answer from the options given below :
Consider an FM signal
\(s(t) = 10\sin\left(4\pi \times 10^6 t + 9\cos\left(2\pi \times 10^3 t\right)\right)\)
the frequency deviation and bandwidth of FM wave are
In case of wideband FM, the modulation index value is :
A fm kc/s modulating frequency provides Mf = 2 (significant Bessel functions i.e. n = 4) in F.M. wave. What bandwidth is required for passing this wave keeping ∆f constant, if frequency of modulating signal is doubled what will be its effect on Mf ?
A high frequency signal is frequency modulated by n number of modulating signals. The ideal number of sidebands in the modulated signal will be :
Assertion (A) : The FM radio broadcast of analog signals provides higher fidelity.
Reason (R) : FM uses significantly larger channel bandwidth for signal transmission.
Select your answer using the codes given below :
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 ?
Important Questions from Frequency Modulation
Which of the following is NOT the advantage of frequency modulation ?
Which of the following statements is true for FM?
The modulation technique in which frequency of the carrier wave is changed with respect to the modulating wave is called:
A phase locked loop can be used to demodulate
What is the modulation index in a frequency modulated signal with a modulating frequency of 500 Hz and frequency deviation of 10 kHz?
