A message signal of frequency 10 kHz and peak voltage 10 V is used to modulate a carrier wave of frequency 1 MHz and peak voltage 20 V. Determine the modulation index and the sideband frequencies.

A. = 0.5, f_USB = 1010 kHz, f_LSB = 990 kHz — The modulation index is the ratio of signal amplitude to carrier amplitude, and sidebands are at carrier frequency plus or minus signal frequency.

B. = 0.5, f_USB = 1001 kHz, f_LSB = 999 kHz

C. = 2.0, f_USB = 1010 kHz, f_LSB = 990 kHz

D. = 0.25, f_USB = 1010 kHz, f_LSB = 990 kHz

JEE Main · PhysicsMedium

$A message signal of frequency f_{m} = 10 kHz and peak voltage A_{m} = 10 V is used to modulate a carrier wave of frequency f_{c} = 1 MHz and peak voltage A_{c} = 20 V . Determine the modulation index μ and the sideband frequencies f_{U S B} and f_{L S B} .$

A. $μ = 0.5, f_{U S B} = 1010 kHz, f_{L S B} = 990 kHz$
B. $μ = 0.5, f_{U S B} = 1001 kHz, f_{L S B} = 999 kHz$
C. $μ = 2.0, f_{U S B} = 1010 kHz, f_{L S B} = 990 kHz$
D. $μ = 0.25, f_{U S B} = 1010 kHz, f_{L S B} = 990 kHz$

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Correct answer: A — $μ = 0.5, f_{U S B} = 1010 kHz, f_{L S B} = 990 kHz$

Solution

$The modulation index is given by μ = A_{m} / A_{c} = 10 V /20 V = 0.5 .$
$The sideband frequencies are f_{c} + f_{m} and f_{c} - f_{m} .$
$f_{U S B} = 1000 kHz + 10 kHz = 1010 kHz .$
$f_{L S B} = 1000 kHz - 10 kHz = 990 kHz .$
$Hence the answer is (A).$

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A message signal of frequency fm​=10 kHz and peak voltage Am​=10 V is used to modulate a carrier wave of frequency fc​=1 MHz and peak voltage Ac​=20 V. Determine the modulation index μ and the sideband frequencies fUSB​ and fLSB​.

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$A message signal of frequency f_{m} = 10 kHz and peak voltage A_{m} = 10 V is used to modulate a carrier wave of frequency f_{c} = 1 MHz and peak voltage A_{c} = 20 V . Determine the modulation index μ and the sideband frequencies f_{U S B} and f_{L S B} .$