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Q.

The expression from an AM voltage is e=51+0.6cos1000πtcos5x106πt . Frequency of sidebands is

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a

2.5005 × 106 Hz,  2.4995 ×106 Hz

b

2.505 × 106 Hz,  2.495 ×106 Hz

c

2.505 × 106 kHz,  2.495 ×106 kHz

d

2.505 MHz,  2.495 kHz

answer is A.

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Detailed Solution

fc=5×1062=2.5×106 Hz fm=10002=500 Hz=0.0005×106 Hz LSB=fc-fm= 2.5×106 Hz-0.0005×106 Hz=2.4995×106 Hz USB=fc+fm= 2.5×106 Hz+0.0005×106 Hz=2.5005×106 Hz
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