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 :
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 ?
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 ?
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?
