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JEE Main & Advanced JEE Main Paper (Held On 10 April 2016)

  • question_answer
    A modulated signal \[{{C}_{m}}\,\,(t)\]has the form \[{{C}_{m}}\,\,(t)=30\,\sin \,300\pi t+10\,(\cos \,200\pi t-\cos 400\pi t)\].The carrier frequency fc, the modulating frequency (message frequency) \[{{f}_{\omega }}\], and the modulation index \[\mu \]are respectively given by :   JEE Main Online Paper (Held On 10 April 2016)

    A) \[{{f}_{c}}=200\,Hz\,;\,{{f}_{\omega }}=30Hz\,;\,\mu =\frac{1}{2}\]                

    B) \[{{f}_{c}}=150\,Hz\,;\,{{f}_{\omega }}=50Hz\,;\,\mu =\frac{2}{3}\]

    C) \[{{f}_{c}}=200\,Hz\,;\,{{f}_{\omega }}=50Hz\,;\,\mu =\frac{1}{2}\]                

    D) \[{{f}_{c}}=150\,Hz\,;\,{{f}_{\omega }}=30Hz\,;\,\mu =\frac{1}{3}\]

    Correct Answer: B

    Solution :

                 \[{{\ell }_{m}}(t)=30\,\sin \,(300\,\pi t)+10\,\cos \,(400\,\pi t)\]              \[f=\frac{\omega }{2\pi }=\frac{300\pi }{2\pi }=150\]              \[=\frac{20\pi }{2\pi }=100\]                        \[\frac{400}{2\pi }=200\]              \[{{f}_{c}}=150\,Hz\]              \[{{f}_{m}}=50\,Hz\]                       \[\mu \frac{2}{3}\]                


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