solution2 - one nonzero sample g(0 = 1 so δ k g(kδ = 0 =...

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one nonzero sample, g (0) = 1, so δ k g ( ) 6 = 0 = R g ( t ) dt . Problem 5.2 (a) Since u ( t ) has a nice orthogonal waveform expansion, to find the coefficient associated with a particular waveform, just take the projection of the function onto the normalized waveform. All the other components will be zero since the waveforms are orthogonal. (This is just like determining a coefficient of a particular frequency of a Fourier Series expansion.) u k = 1 A k Z -∞ u ( t ) θ * k ( t ) dt To see that this indeed extracts out the value of u k , substitute in u ( t ) = j =1 u j θ j ( t ) above to get 1 A k Z -∞ X j =1 u j θ j ( t ) θ * k ( t ) dt = 1 A k X j =1 u j Z -∞ θ j ( t ) θ * k ( t ) dt = 1 A k u k A k = u k (b) Z -∞ | u ( t ) | 2 dt = Z -∞ u ( t ) u ( t ) * dt = Z -∞ X k =1 u k θ k ( t ) X j =1 u * j θ * j ( t ) (3) = X k =1 u k X j =1 u * j Z -∞ θ k ( t ) θ * j ( t ) dt (4) = X k =1 | u ( t ) | 2 A k (5) (c) Z -∞ u ( t ) v * ( t ) = Z -∞ X k =1 u k θ k ( t ) X j =1 v * j θ * j ( t ) dt (6) = X k =1 X j =1 u k v * j Z -∞ θ k ( t ) θ * j ( t ) dt (7) = X k =1 u k v * k A k (8) Problem 5.3 (a) Recall that ˆ u A ( f ) = R A - A u ( t ) e - 2 πift dt , then almost everywhere u A ( t ) = u ( t ) for | t | < A and u A ( t ) = 0 for | t | > A . Furthermore, lim A →∞ u A ( t ) = u ( t ) a.e. (almost everywhere) and lim A →∞ ˆ u A ( f
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