quiz1_solution

quiz1_solution - 1. P x ′ = 1 36 . 1 2 + 5 2 2 = 13 36 ....

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Department of Electrical Engineering EE 308: Communication Systems (Spring 2009) Course Instructor: Prof. Abhay Karandikar Solutions to Quiz 1 The envelope of the amplitude modulated signal should be positive for all t . Therefore, the given signal needs to be normalzed to get a signal x ( t ) which satiFes this criterion. The minimum value of x ( t ) should be greater than -1. The minimum value of x ( t ) is -6. Therefore, normalize x ( t ) to get x ( t ) = 1 6 [sin(2000 πt ) + 5 cos(4000 πt )]. Note: Since μ = 0 . 5, any normalizing factor greater than 3 will make the envelope positive.
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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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This note was uploaded on 02/05/2012 for the course EE EE308 taught by Professor B.k.dey during the Spring '09 term at IIT Bombay.

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