tut2_solutions_31_01_09

# tut2_solutions_31_01_09 - Department of Electrical...

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

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

View Full Document
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

## 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.

### 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
Ask a homework question - tutors are online