Midterm 3 Solutions

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

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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 ..... .. .....
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## This document was uploaded on 01/24/2014.

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