JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    A signal of frequency 20 kHz and peak voltage of 5 volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 volts. Choose the correct statement. [JEE Online 08-04-2017]

    A)  Modulation index = 5, side frequency bands are at 1400 kHz and 1000 kHz

    B)  Modulation index = 0.8, side frequency bands are at 1180 kHz and 1220 kHz

    C)  Modulation index = 0.2, side frequency bands are at 1200 kHz and 1180 kHz

    D)  Modulation index = 5, side frequency bands are at 21/2 kHz and 18.8 kHz

    Correct Answer: C

    Solution :

    Modulation \[idex=m=\frac{{{V}_{m}}}{{{V}_{0}}}\] \[=\frac{1}{5}=0.2\] Frequency \[=12\times {{10}^{3}}kHz\] F = 12.00 kHz \[{{\text{F}}_{\text{1}}}\] = 1200 - 20 = 1180 kHz \[{{\text{F}}_{2}}\] = 1200 + 20 = 1220 kHz


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