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116 = H > B R G B D g Z F b g g h _ h e h ` d b y m g b \ _ j k b l _ l “ K \ B \ Z g J b e k d b”, L h f 53, K \I 1 1, F _ o Z g b a Z p b y, _ e _ d l j b n. X g r n p x g j c u x e x x x c ?. O m o n e 2 Corrections Two of the figures used in this lecture require corrections The first is the figure on board 2 which illustrates the symmetry for the imaginary part of the DFT The value indicated as a dashed line should have a height and polarity identical to X1 (0) since it represents the.

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N y(NT) = Z ak k=O (iii) If a > 1, this grows without bound as N (or t) increases Now we want the inverse system Recognize that we have actually solved this in part (b) of this problem gi = 1, g2 = a gi = 0, i # 0, 1 So the system appears as in Figure S411 y(t) _0_ x(t) Delay T. J N ̃` P b g z ē b j N ̃` P b g z ē b b c ⃄ L X ` P b g Ȃǂ̃ W O ϐ ` P b g ͂ ߁ANBA A A t g A A C X z b P ̊ϐ ` P b g A u h E F C ~ W J A I y A o G Ȃǂ̊ό ` P b g A W ` P b g z. , , ͂ , , F M, O p K, K, X e v , z b ` L X, C , X e V i ,stationery, ̃C X g f ށB N G ^ Y X N E F A ͒ z Ń_ E h ̃C X g f ޏW B V i lj \ ł B(005_0300).

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