Assignment07Fall2010

Assignment07Fall2010 - 525.412 Computer Architecture...

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Unformatted text preview: 525.412 Computer Architecture Assignment 7 Solutions 6.13 Repeat Exercise 6.8 but perform multiplications a. using Booth recoding. b. using bit-pair encoding. a. 101 101 110 100 b. 011 110 101 011 c. 111 100 011 101 d. 101 010 110 111 e. 011 100 100 010 f. 110 101 110 010 g. 110 001 011 011 h. 101 101 110 111 Solution a. In the solutions below the multiplier has been encoded using Booth encoding in which each bit has been replaced with a symbol based on the original bit as well as its neighbor to the right: Bit and its right-hand neighbor Booth encoding 00 01 1 10 1 11 Multiplying by 1 and adding is the same as adding in the twos complement of the multiplicand. a. 101 101 11 100 000 001 001 100 111 101 101 000 100 11 000 011 100 100- 19 - 12 = 228 b. 011 110 11 1 10 1 11 111 100 010 00 001 111 0 11 100 010 00 111 10 10 001 0 10 110 001 010 30 - 21 =- 630 c. 111 100 100 11 1 00 000 000 100 11 111 111 00 00 000 010 0 11 110 0 11 110 001 100- 4 29 =- 116 d. 101 010 11 00 1 0 000 010 110 1 101 010 0 101 10 0 011 000 110- 22 - 9 = 198 e. 011 100 100 1 10 11 111 001 000 00 001 110 0 10 010 0 10 010 111 000 28 - 30 =- 840 f. 110 101 10 1 10 0 000 010 110 1 111 010 1 0 010 11 0 010 011 010- 11 - 14 = 154 g. 110 001 10 1 10 1 00 000 001 111 11 111 000 1 00 001 111 11 000 1 11 001 101 011- 15 27 =- 405 h. 101 101 11 00 1 0 000 010 011 1 101 101 0 100 11 0 010 101 011- 19 - 9 = 171 1 b. In the solutions below the multiplier has been encoded using bit-pair encoding (Booth radix-4 encoding) in which each pair of bits has been replaced with a symbol based on the original pair of bits as well as its neighbor to the right: Bit and its Bit and its right-hand neighbor Bit-pair encoding right-hand neighbor Bit-pair encoding 000 00 100 10 001 01 101 1 010 01 110 1 011 10 111 00 Multiplying by 1 and adding is the same as adding in the twos complement of the multiplicand. a. 101 101 10 100 1 110 110 100 0 100 11 0 011 100 100- 19 - 12 = 228 b. 011 110 10 10 1 11 111 100 010 11 110 001 0 11 000 10 10 110 001 010 30 - 21 =- 630 c. 111 100 100 101 11 111 111 100 00 000 010 0 11 110 0 11 110 001 100- 4 29 =- 116 d. 101 010 11 00 1 0 010 110 1 101 010 0 101 10 0 011 000 110- 22 - 9 = 198 e.e....
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This note was uploaded on 11/25/2010 for the course ECE 525.412 taught by Professor Charlesb.cameron during the Spring '10 term at Johns Hopkins.

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Assignment07Fall2010 - 525.412 Computer Architecture...

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