A) \[{{f}_{c}}=200\,Hz;\,{{f}_{w}}=50\,Hz;\,\,\mu =\frac{1}{2}\]
B) \[{{f}_{c}}=150\,Hz;\,{{f}_{w}}=50\,Hz;\,\,\mu =\frac{2}{3}\]
C) \[{{f}_{c}}=150\,Hz;\,{{f}_{w}}=30\,Hz;\,\,\mu =\frac{1}{3}\]
D) \[{{f}_{c}}=200\,Hz;\,{{f}_{w}}=30\,Hz;\,\,\mu =\frac{1}{2}\]
Correct Answer: B
Solution :
[b] Comparing the given equation with standard modulated signal wave equation, \[m={{A}_{c}}\sin {{\omega }_{c}}t+\frac{\mu {{A}_{c}}}{2}\cos ({{\omega }_{c}}-{{\omega }_{s}})t\] \[-\frac{\mu {{A}_{c}}}{2}\cos ({{\omega }_{c}}+{{\omega }_{s}})t\] \[\mu \frac{{{A}_{c}}}{2}=10\Rightarrow \mu =\frac{2}{3}\] (modulation index) \[{{A}_{c}}=30\] \[{{\omega }_{c}}-{{\omega }_{s}}=200\pi \] \[{{\omega }_{c}}+{{\omega }_{s}}=400\pi \] \[\Rightarrow \,\,\,\,\,\,\,\,{{f}_{c}}=150,\,\,{{f}_{s}}=50\,Hz.\]
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