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# solution3 - Question 1 Vector space V is an Abelian group...

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Question 1: Vector space V is an Abelian group under +. Let θ be the identity element. Then, for an arbitrary element vV , vv += . An arbitrary has its inverse, denoted as v , () +− = . First we prove the following proposition: If wv v , then w = . Proof: ( ) ( ) wv v v v w w θθ ⇒+ + = + −⇒+ = ⇒= . Now, 0* 1* (0 1)* v + = + = = Hence, from the above proposition v = .

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Problem 7.3 a) To show that the waveforms ψ n ( t ), n =1 ,..., 3 are orthogonal we have to prove that Z −∞ ψ m ( t ) ψ n ( t ) dt =0 ,m 6 = n Clearly, c 12 = Z −∞ ψ 1 ( t ) ψ 2 ( t ) dt = Z 4 0 ψ 1 ( t ) ψ 2 ( t ) dt = Z 2 0 ψ 1 ( t ) ψ 2 ( t ) dt + Z 4 2 ψ 1 ( t ) ψ 2 ( t ) dt = 1 4 Z 2 0 dt 1 4 Z 4 2 dt = 1 4 × 2 1 4 × (4 2) Similarly, c 13 = Z −∞ ψ 1 ( t ) ψ 3 ( t ) dt = Z 4 0 ψ 1 ( t ) ψ 3 ( t ) dt = 1 4 Z 1 0 dt 1 4 Z 2 1 dt 1 4 Z 3 2 dt + 1 4 Z 4 3 dt and c 23 = Z −∞ ψ 2 ( t ) ψ 3 ( t ) dt = Z 4 0 ψ 2 ( t ) ψ 3 ( t ) dt = 1 4 Z 1 0 dt 1 4 Z 2 1 dt + 1 4 Z 3 2 dt 1 4 Z 4 3 dt Thus, the signals ψ n ( t ) are orthogonal. b) We Frst determine the weighting coeﬃcients x n = Z −∞ x ( t ) ψ n ( t ) dt, n , 2 , 3 x 1 = Z 4 0 x ( t ) ψ 1 ( t ) dt = 1 2 Z 1 0 dt + 1 2 Z 2 1 dt 1 2 Z 3 2 dt + 1 2 Z 4 3 dt x 2 = Z 4 0 x ( t ) ψ 2 ( t ) dt = 1 2 Z 4 0 x ( t ) dt x 3 = Z 4 0 x ( t ) ψ 3 ( t ) dt = 1 2 Z 1 0 dt 1 2 Z 2 1 dt + 1 2 Z 3 2 dt + 1 2 Z 4 3 dt As it is observed, x ( t ) is orthogonal to the signal waveforms ψ n ( t ), n , 2 , 3 and thus it can not represented as a linear combination of these functions.
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solution3 - Question 1 Vector space V is an Abelian group...

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