Problem 9.3. For the same received signal power, compare the post-detection SNRs of DSB-SC with coherent detection and envelope detection with ka = 0.2 and 0.4. Assume the average message power is P = 1. Solution From Eq. (9.23), the post-detection SNR of DSB-SC with received power 2 2 PAc DSB is WN PASNR c DSB DSB post 0 2 2 = From Eq. (9.30), the post-detection SNR of AM with received power ( )PkA acAM 22 12 + is WN PkASNR ac AM AM post 0 22 2 = So, by equating the transmit powers for DSB-Sc and AM, we obtain ( )2 2 2 2 2 2 1 2 2 2 2 1 DSB AM c c a AM DSB c c a A P A k P A A P k P = + \u21d2 = + Substituting this result into the expression for the post-detection SNR of AM, 2 2 AM DSB post post2 0 SNR SNR 2 1 DSB c a a A P k P N W k P \u239b \u239e= = \u2206\u239c \u239f+\u239d \u23a0 Where the factor \u2206 is Pk Pk a a 2 2 1+=\u2206 With ka= 0.2 and P= 1, the AM SNR is a factor ( ) 04. 04.1 2. 2 ==\u2206 less. With ka = 0.4 and P = 1, the AM SNR is a factor ( ) 14.0 16.1 16. 16.1 4. 2 \u2248=+=\u2206 less. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.