2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Afternoon)

  • question_answer
    In an amplitude modulator circuit, the carrier wave is given by, \[C(t)=4sin(20000\pi t)\] while modulating signal is given by, \[m(t)=2sin(200\pi t)\]. The values of modulation index and lower side band frequency are: [JEE Main 12-4-2019 Afternoon]

    A) 0.5 and 9 kHz

    B) 0.5 and 10 kHz

    C) 0.3 and 9 kHz

    D) 0.4 and 10 kHz

    Correct Answer: A

    Solution :

    [a] Modulation index is given by \[m=\frac{{{\Alpha }_{m}}}{{{\Alpha }_{c}}}=\frac{2}{4}=0.5\] & carrier wave frequency is given by \[=2\pi {{f}_{c}}=2\times {{10}^{4}}\pi \] \[{{f}_{c}}=10\,kHz\] [b] modulating wave frequency \[({{f}_{m}})\] \[2\pi {{f}_{m}}=2000\pi \]\[\Rightarrow {{f}_{m}}=1\,kHz\] lower side band frequency \[\Rightarrow {{f}_{c}}-{{f}_{m}}\] \[\Rightarrow 10\,kHz-1\,kHz=9\,kHz\] Modulation index is given by \[m=\frac{{{\Alpha }_{m}}}{{{\Alpha }_{c}}}=\frac{2}{4}=0.5\] & carrier wave frequency is given by \[=2\pi {{f}_{c}}=2\times {{10}^{4}}\pi \] \[{{f}_{c}}=10\,kHz\] [b] modulating wave frequency \[({{f}_{m}})\] \[2\pi {{f}_{m}}=2000\pi \]\[\Rightarrow {{f}_{m}}=1\,kHz\] lower side band frequency \[\Rightarrow {{f}_{c}}-{{f}_{m}}\] \[\Rightarrow 10\,kHz-1\,kHz=9\,kHz\]          


You need to login to perform this action.
You will be redirected in 3 sec spinner