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Unformatted text preview: ECE 5670 : Digital Communications ECE department, Cornell University, Spring 2011 Homework 5 Solutions Instructor: Salman Avestimehr Office 325 Rhodes Hall 1. The channel has bandwidth W = 4 kHz (i.e. bandwidth is between [ 2 kHz, 2 kHz ] ) and we are using raised cosine pulse shape with β = 0 . 5 . Therefore the symbol rate should satisfy 1 T (1 + β ) = W (1) Therefore 1 T = W 1 + β = 4000 1 . 5 = 2667 symbols/sec (2) (a) Since 1 bit/symbol is transmitted in BPSK, the bit rate is 2667 bits/sec. (b) We send 4 bits/symbol in a 16PAM. Therefore, the bit rate is 4 × 2667 = 10668 bits/sec. 2. The bandwidth of the channel is W = 4 kHz . Hence, the rate of transmission should be less or equal to 4000 symbols/sec. If an MPAM constellation is employed, then the required symbol rate is 1 T = 9600 log M symbols/sec (3) If a signal pulse with raised cosine spectrum with rolloff factor β is used for shaping, then we should have 1 T (1 + β ) ≤ 4000 (4) Therefore we should have 9600 log M (1 + β ) ≤ 4000 (5) Therefore log M ≥ 9600(1 + β ) 4000 (6) And since we want to have β ≥ . 5 we need to set log M such that log M ≥ 9600(1 + β ) 4000 ( β ≥ . 5) ≥ 9600 × 1 . 5 4000 = 3 . 6 (7) So we set M = 16 and β = 0 . 66 so that (5) is satisfied with equality. Therefore the symbol rate is 1 T = 9600 log M = 2400 symbols/sec (8) Now in the error probability of MPAM is P M = 2( M 1) M Q s 3 p average ( M 2 1) σ 2 ! (9) Therefore since M = 16 and σ 2 = N 2 W = 4 × 10 7 we should have 10 6 = 30 16 Q r 3 p average 1 . 02 × 10 4 !...
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This note was uploaded on 10/02/2011 for the course ECE 5670 taught by Professor Scaglione during the Spring '11 term at Cornell University (Engineering School).
 Spring '11
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