tut2_solutions_31_01_09

tut2_solutions_31_01_09 - Department of Electrical...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 3

tut2_solutions_31_01_09 - Department of Electrical...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online