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Unformatted text preview: k is given by: ℜ ( h k ( n )) = e an 2 cos p 2 πnk 50 P . (3) ±or the example channels k = 0 , 5, the real part oF the impulse response reduces to: ℜ ( h ( n )) = e an 2 (4) ℜ ( h 5 ( n )) = e an 2 cos p 2 πn 10 P (5) These Functions are illustrated in ±igure 1. 7 0.2 0.4 0.6 0.8 1 h (n)10.5 0.5 1 5 Figure 1: Real Parts of the Filters h k ( n ) for Channels k =0 and 5. 6. a) 8 b) 3: female; 4: male. c) 3: 210 Hz 4: 140 Hz 9...
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This note was uploaded on 04/19/2011 for the course EE 214A taught by Professor Alwan during the Winter '11 term at UCLA.
 Winter '11
 ALWAN
 Aliasing

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