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Solved Examples · Example 23

Q.An FM signal has a resting frequency of 105 MHz and highest frequency of 105.03 MHz, when modulated by a signal of frequency of 5 kHz. Determine

(i) frequency deviation
(ii) carrier swing
(iii) modulation index
(iv) percent modulation and
(v) lowest frequency reached by the FM wave.
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[!TLDR]

The carrier deviates 30 kHz, so the swing is 60 kHz, the modulation index is 6, the percent modulation is 40 % of the 75 kHz maximum, and the lowest frequency reached is 104.97 MHz.

  1. Frequency deviation is the difference between the highest instantaneous frequency and the resting (carrier) frequency:

    Δf=fmax−fc=105.03−105=0.03 MHz=30 kHz\Delta f = f_{max}-f_c = 105.03-105 = 0.03\ \text{MHz} = 30\ \text{kHz}

  2. Carrier swing is twice the deviation: CS=2 Δf=60 kHzCS = 2\,\Delta f = 60\ \text{kHz}.
  3. Modulation index: mf=Δffm=30×1035×103=6m_f = \dfrac{\Delta f}{f_m} = \dfrac{30\times10^{3}}{5\times10^{3}} = 6. …

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