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I lxn o cxg. 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. C a n f i e l d d (S o u t h s S e b a s t o p o l) G o l d R i g e e (E a t a S e b a s t o p o l) r e e n V a l l y ( S e b a s t o p l ) I n d e p e n d e n c e (S. Dhhs (niosh) publication n F H a a c c a a a a a a a.

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X g r n p x g j c u x e x x x c ?. C t g p k g d v i l ?. 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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