0 x1 3 8 10 x2 3 8 110 x3 0 0 1 8 0 111 x4

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: minimum average bit rate is 3 3 R(Q(Xn )) = 8 (1) + 8 (2) + 1 (3) + 1 (3) = 15 8 8 8 = 1:875 Problem 4 a) Since s (t) = s (t), the dimension of the signal space is 3. We will use the Gram-Schmidt procedure on fsi (t)gji to construct an orthonormal basis for the set. 3 =1 4 3 The energy of s1 (t) is Es1 = ZT 0 sin2 2T t dt = ZT 0 3 1 cos 4T t dt = 1 T 2 2 So the rst basis signal is r s1 (t) = 2 s (t) 1 (t) = p T1 Es1 s1 (t) and s2 (t) are orthogonal (s1 (t) s2 (t) = 0), hence r s2 (t) = 2 s (t) 2 (t) = p T2 Es2 The component of the projection of s3 (t) on 2(t) is 0, but for the projection of s3 (t) on s3 (t) Z T =2 = 1 (t) r 0 p 2T = So, a signal orthogonal to 1 (t) and 2(t) is d3 (t) = s3 (t) Its energy is Ed3 = = = = p p 2T (t) 1 p Then the third basis signal is 2T 1=2 2 (t), 2 2t 2T T sin T dt = T 2T (t)] s (t) s3(t) 1 3 p (s3 (t) s3 (t)) 2 2T (s3 (t) pp T 2 2T 2T + 2T 2 2 T 2T 2 2 T 3 (t) = 2 1 r s3 (t) 2T (t)] 1 2T ( (t) 1 (t)) + 1 2 p 2T (t) 1 1 (t)) ! The decorrelator looks like this -k 6t) ( - -k 6t) ( -k 6t) ( ..... .... .. .. .. .. Y (t) 1 .... .... .. .. .. .. 2 ..... .. .....
View Full Document

This document was uploaded on 01/24/2014.

Ask a homework question - tutors are online