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
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 :
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 :
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?
