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

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

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

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 $e = 5 (1 + 0. 6Cos1 000 π t) Cos5 \times 10^{6} π t$ . Frequencies of sideband is

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$2.505 \times 10^{6} H z, 2.495 k H z$

$2.505 \times 10^{6} H z, 2.495 \times 10^{6} H z$

$2.505 \times 10^{6} H z, 2.4995 \times 10^{6} H z$

$2.5005 \times 10^{6} H z, 2.4995 \times 10^{6} H z$

answer is A.

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Detailed Solution

$G i v e n e q u a t i o n e = 5 (1 + 0. 6Cos1 000 π t) Cos5 \times 10^{6} π t \Rightarrow e = (5 + 3 C o s 1000 π t) . C o s 5 \times 10^{6} π t S \tan d a r d e q u a t i o n o f m o d u l a t i n g w a v e E = (E_{c} + E_{m} C o s w_{m} t) C o s w_{c} t C o m p a r i n g w e g e t, E_{c} = 5, E_{m} = 3 w_{m} = 1000 π \Rightarrow 2 π n_{m} = 1000 π \Rightarrow n_{m} = 500 H z S i m i l a r l y, w_{c} = 5 \times 10^{6} π \Rightarrow 2 π n_{c} = 5 \times 10^{6} π \Rightarrow n_{c} = 25 \times 10^{5} H z U p p e r s i d e b a n d = f_{c} + f_{m} = n_{c} + n_{m} = 2500000 + 500 = 2.5005 \times 10^{6} H z L o w e r s i d e b a n d = f_{c} - f_{m} = n_{c} - n_{m} = 2500000 - 500 = 2.4995 \times 10^{6} H z$