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Unformatted text preview: Department of Electrical Engineering EE 308: Communication Systems (Spring 2009) Course Instructor: Prof. Abhay Karandikar Solutions to Problem Set 2 1. Since X c ( f ) = A c 2 [ M ( f f c ) + M ( f + f c )] , (1) we have V ( f ) = A c 2 [ M ( f f c ) + M ( f + f c )] H ( f ) , (2) The output of the product modulator is V o ( f ) = 1 2 [ V ( f f c ) + V ( f + f c )] A c 4 M ( f ) [ H ( f f c ) + H ( f + f c )] , + A c 4 [ M ( f 2 f c ) H ( f f c ) + M ( f + 2 f c ) H ( f + f c )] (3) which, after lowpass filtering results in A c 4 M ( f ) [ H ( f f c ) + H ( f + f c )] (4) From (4), it is obvious that m ( t ) can be perfectly recovered if H ( f f c ) + H ( f + f c ) = 4 A c (5) 2. x c ( t ) can be expressed as A c ( t ) cos( ω c t + θ c ( t )), where A c ( t ) = radicalbig 1 + 0 . 04 cos 2 ω m t θ c ( t ) = tan 1 (0 . 2 cos ω m t ) (6) As we can see, both the amplitude and phase are functions of the message cos ω m t ....
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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.
 Spring '09
 B.K.Dey
 Electrical Engineering

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