tut4_solutions_09_03_09

# tut4_solutions_09_03_09 - Department of Electrical...

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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.

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tut4_solutions_09_03_09 - Department of Electrical...

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