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Unformatted text preview: Polytechnic University, Dept. Electrical and Computer Engineering EE3414 Multimedia Communication System I Spring 2006, Yao Wang __________________________________________________________________________________________________ Homework 11 (Digital Communication) (Solution) Written Assignment : 1. Consider the two digital modulation schemes shown in the figure below. For each scheme, determine the minimum distance between two symbols d min , the average energy per symbol E av , and the average energy per transmitted bit E b . Which scheme is more efficient in terms of transmission energy usage? Solution: For 4ary PAM, the distance between any two adjacent symbols (e.g. “A” and “3A”) is 2A, therefore d min =2A. The energy for symbols “A” and “A” are both A 2 ; and the energy for symbols “3A” and “3A” are both 9A 2 . Therefore the average energy per symbol is E av =(A 2 +9A 2 )/2=5A 2 . Because each symbol carries 2 bits, the average energy per bit is E b = E av /2=5A 2 /2. For 4QAM, the distance between two closest symbols (say “00” and “01”) is 2A, therefore d min =2A. The length from =2A....
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This note was uploaded on 01/22/2012 for the course EE 3414 taught by Professor Wang during the Spring '06 term at NYU Poly.
 Spring '06
 Wang

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