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 ?
Mf would reduce to \(\dfrac{1}{2}\) of original value
The one formula this question rests on. The modulation index of an FM wave is
\(M_f=\dfrac{\Delta f}{f_m}\)
where ∆f is the peak frequency deviation (set by the amplitude of the modulating signal) and fm is the modulating frequency.
Step 1 — hold ∆f fixed and double fm. The question states explicitly that ∆f is kept constant, so Mf and fm are inversely proportional:
\(M_f'=\dfrac{\Delta f}{2f_m}=\dfrac{1}{2}\cdot\dfrac{\Delta f}{f_m}=\dfrac{M_f}{2}\)
With Mf = 2 originally, the new index is \(M_f'=1\) — exactly half.
Step 2 — the bandwidth part of the question. With n = 4 significant Bessel terms, the bandwidth is
\(BW=2n f_m=8f_m\ \text{kc/s}\)
Carson's rule gives the same order of magnitude:
\(BW=2(\Delta f+f_m)=2f_m(M_f+1)=6f_m\ \text{kc/s}\)
the Bessel count being the more conservative of the two.
Step 3 — the physical reason Mf falls. Deviation measures how far the carrier swings; the modulating frequency measures how fast it swings. Doubling the rate while keeping the swing the same means each excursion covers the same ground in half the time, so relative to the new modulating frequency the swing looks half as large. That ratio is precisely the modulation index.
Contrast with AM — the classic point of confusion. In amplitude modulation the modulation index depends only on the amplitude ratio, not on fm at all, and the bandwidth is always 2fm. In FM the index depends on both amplitude (through ∆f) and frequency, which is why FM bandwidth is not simply proportional to the message bandwidth and why an FM system must be specified by its deviation ratio.
Eliminating the distractors. Option 1 would require Mf to be proportional to fm — the relation is inverse. Option 2 describes what happens to ∆f, not to Mf. Option 4 would need fm to be quadrupled, not doubled.
Hence, doubling fm at constant ∆f makes Mf fall to half of its original value.
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 ?
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(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 :
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\(s(t) = 10\sin\left(4\pi \times 10^6 t + 9\cos\left(2\pi \times 10^3 t\right)\right)\)
the frequency deviation and bandwidth of FM wave are
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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?
