Hw2Soln

# Hw2Soln - ECE 5670 Digital Communications ECE department...

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ECE 5670 : Digital Communications ECE department, Cornell University, Spring 2011 Homework 2 Solutions Instructor: Salman Avestimehr Ofﬁce 325 Rhodes Hall 1. We have doubled our transmit energy constraint – so the transmit voltage range has in- creased by a factor of 2 . Let us consider an increase in data rate of r from the existing rate of R bits per unit time. So, the distance between the nearest voltage levels is now 2 2 E 2 R + r (1) as compared to the previous distance of 2 E 2 R . (2) The new error probability is Q 2 2 E σ 2 R + r ! 1 2 e - 4 E 2 σ 2 2 R + r (3) as compared to the previous level of p e def = Q 2 2 E σ 2 2( R + r ) ! (4) 1 2 e - 2 E 2 σ 2 2 2 R . (5) So, the new error probability (cf. Equation (3)) can be written as p 2 1 - 2 r e . (6) We then conclude that the set ( r,d ) is one that satisﬁes: 0 r 0 . 5 (7) 0 d 2 1 - 2 r . (8) 2. The overall unreliability level for sequential communication is given by: P [ E ] = 1 - (1 - p e ) n (9)

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where p e = Q SNR 2 R - 1 ! (10) Setting P [ E ] equal to 10 - 5 and solving for SNR, we get: 10 - 5 = 1 - 1 - Q SNR 2 R - 1 !! n (11) 1 - Q SNR 2 R - 1 ! = (1 - 10 - 5 ) 1 /n (12) SNR 2 R - 1 = Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) (13) SNR = [(2 R - 1) Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) ] 2 (14) If you tried plotting 10 log 10 SNR , you would realize that some numbers are so large that MATLAB cannot handle it (especially values like 2 8192 ). Thus we need to simplify by hand. So when we take the 10 log 10 SNR , and approximate 2 R - 1 by 2 R , we get the following: 10 log 10 SNR = 10 log 10 [2 R Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) ] 2 (15) = 20 log 10 [2 R Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) ] (16) = 20 log 10 2 R + 20 log 10 Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) (17) = 20 R log 10 2 + 20 log 10 Q - 1 ( 1 - (1 - 10 - 5 ) 1 /n ) . (18)
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Hw2Soln - ECE 5670 Digital Communications ECE department...

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