CNT 3004 - Notes - 6-29

# CNT 3004 - Notes - 6-29 - CNT 3004 6/29/2010 Min hamming...

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CNT 3004 – 6/29/2010 Min hamming distance: d m = distance minimum s = # of errors Detection : d m = 5+1 Correction : Hamming codes – correct 1 error Design hamming codes : usually talking about correction errors (1) suppose we have m-bits |x-bits [ how do we get x-bits?] 2 r ≥ m + r + 1 If m =5 what is the value of r …r =3 ? 2 3 = 8…the min will be the number of bits your looking for so r=4 cause 2 4 is 16 > or = to 10 [try and get as close as possible to the minimum] Code Based on Parity: d d d d x x x 6 5 4 3 2 1 0 Power of 2, 0, 1, 2 The three parity bits (xxx) will have the position of the bit in error [computations in this] EXAM: Max grade was an 87 Extra credit will be given: 51(GIFT). an image is 1024 x 768 pixels with 3 bytes/pixel. Assume the image is uncompressed. There are 30 images per second to be broadcasted through a communication channel at 100MHz bandwidth (A) what would be the min. (s/n) in db required to accomplish this transmission the image is 1024 x 768 pixels 3 bytes/pixel 30 images/sec we are trying to get it through a (_________) chanel of 100 Mhz The same type of problem was done in class # of bits in a still image is 768 * 1024 * 3 bytes * 8 = 18,874,368 in bits

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## This note was uploaded on 10/05/2010 for the course CNT CNT 3004 taught by Professor K during the Spring '10 term at University of Central Florida.

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CNT 3004 - Notes - 6-29 - CNT 3004 6/29/2010 Min hamming...

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