tailieunhanh - Proakis J. (2002) Communication Systems Engineering - Solutions Manual (299s) Episode 9

Tham khảo tài liệu 'proakis j. (2002) communication systems engineering - solutions manual (299s) episode 9', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | or since N 2V x2 x2 DoM x x 1 4pb 4v - 1 xN 1 4Pb N2 - 1 4 SNR E X2 E X2 3N2 Dtotal xmax 1 4Pb N2 - w If we let X X then EX E X2 X2. Hence SNR 3N2X 3 4X2 S R 1 4pb N2 - 1 1 4pb 4v - 1 Problem 1 log 1 Id-X g x iOe 1 ri Differentiating the previous using natural logarithms we obtain g x 1 l xmax ln 1 l 1 g xmax sgn2 x Since for the 1-law compander ymax g xmax 1 we obtain D - .y x í 00 fX Xi 2 dx 3 X 4V .X g x 2 xmax ln 1 l 2 A 2 x 2 9 lxl X 3 Xv P2 X 1 . w Jfx x dx _ xmax ln 1 l -J I .2p i Y o. fa n will 3 X 4Vg2-- 1 E X 2lE X J _ xmax ln 1 l -1 .2pi Y21m9 Zp II will 3 X N2 2---- Í1 1 E X 2lE X J where N2 4V and X X xmax. 2 SQNR E X 2 D E X2 i23 N2 xmax ln 1 i 2 i2E JX2 2iE JX 1 3i2N 2E X 2 ln 1 i 2 i2E X2 2gE XT 1 3 Since SQNRunif 3 N2E X2 we have u2 SQNRMiaw SQNRunif ln 1 i 2 i2E X2 2gE XT 1 SQNRunif G g Xc 158 where we identify G g X _P2_ ln 1 g 2 p2E X2 2ụE JV 1 3 The truncated Gaussian distribution has a PDF given by fy y K ỵ 2nơx -xL e where the constant K is such that K f r 1 e dx 1 K J 4ơX 2n x 1 1 0 4 10001 Hence E X I K G x ỵ 2nơx J 4 X x I xX--Ị e 2ơX dx 2K 4ự2nơ2 Ị-4 X E- I xe Xơx dx 0 K 2 2 22 2e x Xơ X 4ơx 0 7 1 e 2 W2n In the next figure we plot 10log10 SQNRunif and 10log10 SQNR law vs. 10log10 E X2 when the latter varies from 100 to 100 db. As it is observed the -law compressor is insensitive to the dynamic range of the input signal for E X2 1. Problem The optimal compressor has the form g x ymax 2 ỉl fx v 3 dv JZ fx v 3 dv where ymax g xmax g 1 . Z fx v 3 dv - f1 r p MÌ Ấ fb . i J fx v 3 dv J v 1 3 dv J v 1 3 dv x3 dx 3 2 159 If x 0 then í fx v 3 dv J tt fx . r 1 1 3 4 y v 1 3 dv y z3 dz - z3 3 4 4 x 1 3 If x 0 then I fx v 3 dv J tt i v 1 3 dv i v 1 3 dv 1 0 1 L x Hence g 1 g 1 _x4 x 1 3 1 1 1 x 3 The next figure depicts g x for g 1 1. Since the resulting distortion is see Equation 1 3 D m 1 AV 12 X 4V i fx x 3 dx J infty 3 4 1 x 0 0 x 1 1 3 Y 12 X 4V V2J we have x 1 0 SQNR E x2 32 X 4VE X2 32