20111eeM214A_1_ps411-sol

20111eeM214A_1_ps411-sol - k is given by ℜ h k n = e an 2...

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UCLA Dept. of Electrical Engineering EE M214A, Winter 2011 Problem Set 4 Solution February 7, 2011 1. 7.13 Answer: 1
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2. 7.14 Answer: 3
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3. 7.18 Answer: 6
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4. Answer: Omitted. 5. (a) To avoid aliasing in the time domain, the number filters must be at least equal to the length of the window in samples. Thus N min =19. (b) The impulse response of the filter for channel k is given by: h k ( n ) = w k ( n ) e j 2 πnk N (1) h k ( n ) = e an 2 e j 2 πnk N (2) h k ( n ) = e an 2 + j 2 πnk N (3) (c) The spacing between adjacent filters is given by: Δ ω = 2 π N = 2 π 50 . (3) Thus, the spectral resolution in Hz is given by: Δ F = Δ ω 2 π F s = 2 π 50 × 2 π 10kHz = 200Hz . (3) (d) The real part of the impulse response of channel
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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)-1-0.5 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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