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Find the quotient and remainder when 1111101 is divided by 1101 in modulo 2 arithmetic… Now we bring down the next bit of the dividend. We see that 00101 is not divisible by 1101. So we place a zero in the quotient. 2.8 Error Detection and Correction 精品文档 Find the quotient and remainder when 1111101 is divided by 1101 in modulo 2 arithmetic… 1010 is divisible by 1101 in modulo 2. We perform the modulo 2 subtraction. 2.8 Error Detection and Correction 精品文档 Find the quotient and remainder when 1111101 is divided by 1101 in modulo 2 arithmetic… We find the quotient is 1011, and the remainder is 0010. This procedure is very useful to us in calculating CRC syndromes. 2.8 Error Detection and Correction Note: The divisor in this example corresponds to a modulo 2 polynomial: X 3 + X 2 + 1. 精品文档 Suppose we want to transmit the information string: 1111101. The receiver and sender decide to use the (arbitrary) polynomial pattern, 1101. The information string is shifted left by one position less than the number of positions in the divisor. The remainder is found through modulo 2 division (at right) and added to the information string: 1111101000 + 111 = 1111101111. 2.8 Error Detection and Correction 精品文档 If no bits are lost or corrupted, dividing the received information string by the agreed upon pattern will give a remainder of zero. We see this is so in the calculation at the right. Real applications use longer polynomials to cover larger information strings. Some of the standard poly-nomials are listed in the text. 2.8 Error Detection and Correction 精品文档 Data transmission errors are easy to fix once an error is detected. Just ask the sender to transmit the data again. In computer memory and data storage, however, this cannot be done. Too often the only copy of something important is in memory or on disk. Thus, to provide data integrity over the long term, error correcting codes are required. 2.8 Error Detection and Correction 精品文档 Hamming codes and Reed-Soloman c
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