A high frequency signal is frequency modulated by n number of modulating signals. The ideal number of sidebands in the modulated signal will be :
Infinite
The key fact about FM: it is a non-linear process. Unlike amplitude modulation, where each tone simply produces one pair of sidebands, frequency modulation generates an unlimited set of them. Expanding a single-tone FM wave in Bessel functions,
\(s(t)=A_c\sum_{n=-\infty}^{\infty}J_n(\beta)\cos\left(\omega_c+n\omega_m\right)t\)
The sum runs from −∞ to +∞, so even one modulating tone theoretically produces an infinite number of sideband pairs at \(\omega_c\pm n\omega_m\). Adding more modulating signals cannot reduce that — it adds cross-products as well. The ideal answer is therefore option 4.
Why the count is infinite while the bandwidth is not. The amplitudes \(J_n(\beta)\) fall away rapidly once n exceeds the modulation index β, so in practice only a finite set matters. The usual criterion keeps every sideband whose amplitude exceeds 1 % of the unmodulated carrier, which leads to Carson's rule:
\(BW\approx2(\Delta f+f_m)=2f_m(\beta+1)\)
So "infinite sidebands, finite bandwidth" is not a contradiction — the tail is simply negligible.
Contrast with AM, which is where the distractors come from.
| Scheme | Sidebands from n tones | Bandwidth |
|---|---|---|
| AM / DSB | 2n — one pair per tone | 2fm,max |
| SSB | n | fm,max |
| FM | infinite | Carson's rule |
Options 1 and 2 are the SSB and AM answers respectively; option 3 has no standard basis. The whole point of the question is that FM does not follow the linear one-pair-per-tone rule.
A striking consequence of the Bessel behaviour. \(J_0(\beta)\) passes through zero at β = 2.405, so at that modulation index the carrier itself disappears and all the transmitted power sits in the sidebands. This carrier-null condition is the standard laboratory method for calibrating frequency deviation on a spectrum analyser.
Hence, the ideal number of sidebands is infinite.
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 :
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 :
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 ?
Which of the following techniques are used to generate frequency modulated signal :
1. Armstrong
2. Foster-Sealy Discriminator
3. Balanced Modulator
4. Reactance ModulatorWhich one of the following is true ?
Important Questions from Frequency Modulation
Which of the following statements is true for FM?
What is the modulation index in a frequency modulated signal with a modulating frequency of 500 Hz and frequency deviation of 10 kHz?
Which of the following rules states that the bandwidth required to transmit an angle modulated wave is twice the sum of the peak frequency deviation and highest modulating signal frequency?
If f mis modulating frequency and m fis modulation index, then by the Carson's rule, the bandwidth of an FM signal at the input of a conventional discriminator will be:
In FM, "M" stands for
