A) 10.62 MHz
B) 10.62 kHz
C) 5.31 MHz
D) 5.31 kHz
Correct Answer: D
Solution :
Cut−off frequency is given by \[{{f}_{c}}=\frac{1}{2\pi RC}\] \[=\frac{1}{2\times 3.14\times {{10}^{5}}\times 250\times {{10}^{-12}}}\] \[=6.37\text{ }KHZ.\] Modulation factor\[=\frac{\sqrt{1-{{m}^{2}}}}{m}\] \[(n=0.6)\] \[=\frac{4}{3}\] So, maximum frequency that can be detected should be less than modulation factor\[\times \text{ }{{f}_{c}}\] i.e., Less than\[6.37\times \frac{4}{3}=8.5\,KH\] \[\therefore \]Option (4) is correct.
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