sol-exam2.2011

# sol-exam2.2011 - 1 EE 7615 Solutions to Exam II Nov 2011...

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Unformatted text preview: 1 EE 7615 Solutions to Exam II Nov. 2011 Problem 1 1) By inspection we can see that the first three signals are orthogonal and normalized. Also all the other signals can be written as a linear combination of these three signals. Thus our orthonormal basis is given by S = { s 1 ( t ) , s 2 ( t ) , s 3 ( t ) } The vector representation are given by (see figure below) s 1 = (1 , , 0) , s 2 = (0 , 1 , 0) , s 3 = (0 , , 1) , s 4 = (0 , , 0) s 5 = (1 , 1 , 0) , s 6 = (1 , , 1) , s 7 = (0 , 1 , 1) , s 8 = (1 , 1 , 1) 1 (1 ) s t g (1 ) t 2 Compute || || Choose ( ) r t ˆ M (1 ) s t g 2 (1 ) s t g 2 || || 1,2,...,8 i i u i G g G r s smallest 3 (1 ) s t u i = || r − s i || 2 = ( r 1 − s i 1 ) 2 + ( r 2 − s i 2 ) 2 + ( r 3 − s i 3 ) 2 ˆ V = i iff u i ≤ u j ∀ j negationslash = i 2) The signal set corresponds to the vertices of a hypercube (see figure). Thus P ( E ) = 1 − (1 − p ) 3 and P ( C ) = (1 − p ) 3 where p = Q parenleftBig d √ 2 N parenrightBig = Q parenleftBig 1 √ 2...
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sol-exam2.2011 - 1 EE 7615 Solutions to Exam II Nov 2011...

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