Unformatted text preview: 1. P x ′ = 1 36 . 1 2 + 5 2 2 = 13 36 . The e±ciency of modulation is given by η = μ 2 .P x ′ 1 + μ 2 .P x ′ 2. P c = 1 2 A 2 c = 50 W . Therefore, A c = 10. 3. The amplitude modulated signal is 10(1 + . 5 6 [sin(2000 πt ) + 5 cos(4000 πt )]) cos(1600000 πt ) = 10 cos(1600000 πt ) + 5 12 sin(1602000 πt )5 12 sin(1598000 πt ) + 25 12 sin(1604000 πt ) + 25 12 cos(1596000 πt ) 4. The envelope of the modulated signal is: 10(1 + . 5 6 [sin(2000 πt ) + 5 cos(4000 πt )]) 1...
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 Spring '09
 B.K.Dey
 Electrical Engineering, Carrier wave, minimum value, Electrical Engineering EE, Prof. Abhay Karandikar

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