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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 GramSchmidt 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.
 Spring '09

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