tut4_solutions_09_03_09

tut4_solutions_09_03_09 - Department of Electrical...

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

View Full Document Right Arrow Icon
Department of Electrical Engineering EE 308: Communication Systems (Spring 2009) Course Instructor: Prof. Abhay Karandikar Solutions to Problem Set 4 1. Solution: The spectrum of the PAM signal is given by S ( f ) = f s H ( f ) s k = -∞ M ( f - kf s ) (1) where M ( f ) = Δ p f 800 P H ( f ) = 10 - 4 sinc(10 - 4 f ) exp( - jπf 10 - 4 ) (2) and f s = 1 kHz. A sketch of | S ( f ) | is available in Fig. 1. 2. Solution: (a) Two message signals can be transmitted over two sidebands. Then twelve message signals can be transmitted over six upper sidebands and six lower sidebands. Thus, the required bandwidth is 12 × 10 = 120 kHz. (b) The bandwidth required for TDM, PAM is given by N × B , where N is the number of message signals to be multiplexed and B is the bandwidth of each. Using this relation, the required bandwidth is 12 × 10 = 120 kHz. In general, the bandwidth requirement of FDM, SSB is equal to that of TDM, PAM. 3.
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the 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 / 4

tut4_solutions_09_03_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