Q&A - Ask Doubts and Get Answers
Q

Help me please, - Communication Systems - JEE Main

A modulated signal Cm(t) has the form Cm(t)=30 sin 300πt+10 (cos 200πt −cos 400πt).  The carrier frequencyf_{c{}'}, the modulating frequency (message frequency)f_{w{}'}, and the modulation index µ are respectively given by :

  • Option 1)

    f_{c} = 200 Hz \, ; f_{w} = 50Hz\, ; \mu = \frac{1}{2}

  • Option 2)

    f_{c} = 150 Hz \, ; f_{w} = 50Hz\, ; \mu = \frac{2}{3}

  • Option 3)

    f_{c} = 150 Hz \, ; f_{w} = 30Hz\, ; \mu = \frac{1}{3}

  • Option 4)

    f_{c} = 200 Hz \, ; f_{w} = 30Hz\, ; \mu = \frac{1}{2}

 
Answers (1)
125 Views

As we learnt in

Modulated Wave -

e= E_{e}\sin \omega _{c}t

+\frac{m_{0}E_{c}}{2}\cdot \cos \left ( \omega _{c}-\omega _{m} \right )t

-\frac{m_{0}E_{e}\cdot}{2} \cos \left ( \omega _{c}+\omega _{m} \right )t

- wherein

Resultant is summation of three sinusoidal wave.

 

 Comparing the given equation with standard modulated signal wave equation 

m=A_{c}sin\ \omega_{c}+\frac{\mu A_{c}}{2}

cos(\omega_{c}-\omega_{s})t-\frac{\mu A_{c}}{2}cos(\omega_{c}+\omega_s)t

\frac{\mu A_{c}}{2}=10\ \; \Rightarrow\ \; \mu=\frac{2}{3}(modulated\ index)

Ac = 30

\omega_{c}-\omega_{s}=200\pi

\omega_{c}+\omega_{s}=400\pi

fc = 150,    fs = 50 Hz

Correct option is 2.

 


Option 1)

f_{c} = 200 Hz \, ; f_{w} = 50Hz\, ; \mu = \frac{1}{2}

This is an incorrect option.

Option 2)

f_{c} = 150 Hz \, ; f_{w} = 50Hz\, ; \mu = \frac{2}{3}

This is the correct option.

Option 3)

f_{c} = 150 Hz \, ; f_{w} = 30Hz\, ; \mu = \frac{1}{3}

This is an incorrect option.

Option 4)

f_{c} = 200 Hz \, ; f_{w} = 30Hz\, ; \mu = \frac{1}{2}

This is an incorrect option.

Exams
Articles
Questions