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Q.

The expression for an AM voltage is  e=51+0.6Cos1000πtCos5×106πt .  Frequencies of sideband 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 Hz , 2.4995×106 Hz

d

2.505×106 Hz,  2.495  kHz

answer is A.

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Detailed Solution

Given equation e=51+0.6Cos1000πtCos5×106πt ⇒e=(5+3Cos1000πt).Cos5×106πt  Standard equation of modulating wave  E=(Ec+EmCoswmt)Coswct    Comparing  we get,   Ec=5 , Em = 3  wm = 1000π ⇒2πnm = 1000π ⇒nm=500 Hz   Similarly,  wc = 5×106π ⇒2πnc = 5×106π ⇒nc = 25×105 Hz  Upper side band  = fc+fm=nc+nm = 2500000+500 =2.5005×106Hz Lower side band = fc−fm=nc−nm =2500000-500  =2.4995×106Hz
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The expression for an AM voltage is  e=51+0.6Cos1000πtCos5×106πt .  Frequencies of sideband is