solution2 - one nonzero sample, g (0) = 1, so δ ∑ k g (...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: one nonzero sample, g (0) = 1, so δ ∑ k g ( kδ ) 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...
View Full Document

This note was uploaded on 10/07/2009 for the course ENSC 5210 taught by Professor Daniellee during the Spring '08 term at Simon Fraser.

Page1 / 3

solution2 - one nonzero sample, g (0) = 1, so δ ∑ k g (...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online