hw02-f06-sol - EECS 20N Structure and Interpretation of...

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EECS 20N: Structure and Interpretation of Signals and Systems Department of Electrical Engineering and Computer Sciences U NIVERSITY OF C ALIFORNIA B ERKELEY Problem Set 2 SOLUTIONS HW 2.1 (a) The number of bits in the image is 1024 × 768 × 16 = 12 , 582 , 912 . It would take 12582912 / 28000 = 449 seconds to download over a 28 Kbps modem, 32.7 seconds over a 384 Kbps modem, and 1.25 seconds over a 10 Mbps LAN. (b) For 1 channel: 1 min × 60 s 1 min × 44 , 100 samples 1 s × 16 bits 1 sample × 1 byte 8 bits = 5 . 3 MB Thus, a one minute segment stereo audio content ccupies 10.6 MB (=10.09 MiB). (c) 4 . 7 × 10 9 bytes × 8 bits 1 byte × 1 sample 24 bits × 1 s 192 × 10 3 samples × 1 min 60 s = 136 min of 1 channel DVD audio or 68 min of 2 channel DVD audio 4 . 7 × 10 9 bytes × 8 bits 1 byte × 1 sample 16 bits × 1 s 44 , 100 samples × 1 min 60 s = 888 min of 1 channel CD audio or 444 min of 2 channel CD audio HW 2.2 (a) not one-to-one f (1) = f ( - 1) = e - 1 not onto x : e -| x | > 0 (b) not one-to-one Sinc (1) = Sinc (2) = 0 not onto t : Sinc ( t ) 1 (c) not one-to-one h ( Q ) = h ( R ) = 1 onto Every y B = { 1 , 2 , 3 } is in the image of h (d) one-to-one no element in B is mapped more than once onto Every element of B is in the image of r The inverse function is r - 1 , where graph ( r - 1 ) = { (1 , S ) , (2 , Q ) , (3 , R ) } (e) one-to-one and onto The inverse function is g - 1 , where g - 1 = y 1 3 1
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HW 2.3 (a) Let ( x, y ) A × ( B C ) x A (1) y B C (2) (2) y B (3) y C (4) (1) , (3) ( x, y ) ( A × B ) (5) (1) , (4) ( x, y ) ( A × C ) (6) (5) , (6) ( x, y ) ( A × B ) ( A × C ) A × ( B C ) ( A × B ) ( A × C ) . . . ? 1 Let ( x, y ) ( A × B ) ( A × C ) x A (7) y B
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