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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    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 : [JEE Main 10-4-2019 Morning]

    A) \[4;1\times {{10}^{8}}Hz\]             

    B) \[0.25;1\times {{10}^{8}}Hz\]

    C) \[4;2\times {{10}^{8}}Hz\]             

    D) \[0.25;\,2\times {{10}^{8}}Hz\]

    Correct Answer: D

    Solution :

    \[{{f}_{m}}=100MHz={{10}^{8}}Hz,\]\[{{({{V}_{m}})}_{0}}=100V\]           \[{{f}_{c}}=300GHz\]                \[,{{({{V}_{c}})}_{0}}=400V\] Modulaton Index \[=\frac{{{({{V}_{m}})}_{0}}}{{{({{V}_{c}})}_{0}}}=\frac{100}{400}=\frac{1}{4}=0.25\] Upper band frequency (UBF) \[={{f}_{c}}+{{f}_{m}}\] Lower band frequency (LBF) \[={{f}_{c}}-{{f}_{m}}\] \[\therefore UBF-LBF=2{{f}_{m}}=2\times {{10}^{8}}Hz\]


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