Verified Solution • 2025–26 Session moderate

QJEEM19Phca7d: JEE Main 2019 Physics - Class 12 JEEM19Phca7d Solution: A message signal of frequency 100 MHz and peak voltage 100 V is …

A message signal of frequency 100 MHz and peak voltage 100 V is used to execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V. The modulation index and difference between the two side band frequencies are :

Multiple Choice Options

  • A 0.25; 1 × 108 Hz
  • B 4; 2 × 108 Hz
  • C 4; 1 × 108 Hz
  • D 0.25; 2 × 108 Hz

Step-by-Step Verified Solution

fm = 100 MHz = 108 Hz, (Vm)0 = 100 V

fc = 300 GHz, (Vc)0 = 400 V

∴\therefore UBF – LBF = 2fm = 2 × 108 Hz

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