EE2301Dis1F09Sol

EE2301Dis1F09Sol - remainder of 2; 5 is the only...

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1 Department of Electrical and Computer Engineering University of Minnesota EE2301 Fall 2009 Introduction to Digital System Design L. L. Kinney Discussion I Solutions 1. (a) 10111 2 = 23 10 (b) 1101.101 2 = 13.625 10 (c) 1.0111 2 = 1.4375 10 2. (a) 356 10 = 101100100 2 (b) 0.00625 10 = 0.00000001100110011. .. 2 (c) 43.23 10 = 101011.0011101011100001010001. .. 2 (d) 0.51 10 = 0.1000001010001111010111 2 3. (a) 101001.11101 2 = 41.90625 10 = 51.72 8 = 29.E8 16 (b) 41.5 8 = 33.625 10 = 100001.101 2 = 21.A 16 (c) A5.CB2 16 = 165.7934569 10 = 10100101.110010110010 2 = 245.6262 8 4. (a) Since 6 + 3 produces a sum digit of 0, 9 must be a multiple of the base, in which case, 6 + 3 = 10, i.e., the sum digit is 0 with a carry of 1 to the next column. Then 1 + 5 + 4 is 11 in base 9. 1 + 4 + 1 is 6 in base 9. So base 9 works and there is no overflow because the carry from the most significant position is0. (b) Since 3 + 4 produces a sum digit of 2, 7 divided by the base must produce a
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Unformatted text preview: remainder of 2; 5 is the only possibility. Then, 1 + 4 + 2 equals 12 in base 5 and 1 + 1 + 4 equals 11 in base 5. Since the carry from the most significant position is 1, the sum overflows. (c) Since 3 + 4 produces a sum digit of 0, the base must divide 7 evenly; 7 is the only possibility. The result does not overflow. (d) Since 3 + 4 produces a sum digit of 1, the base must divide 7 with a remainder of 1; 6 is the only possibility. The result overflows. (e) The base can be 8 or any larger integer. The sum does not overflow. 5. (a) 0000000 1111111 0000000 (b) 1111111 0000000 0000001 (c) 00110011 11001100 11001101 (d) 1000000 0111111 1000000 6 (a) 11010 - 01101 01101* 01101* 01101 (b) 11010 - 10000 01010 01010 01010 (c) 10010 - 10011 11110 11111 11111* (d) 000100 - 110000 010011 010100 010100* * indicates overflow...
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This note was uploaded on 12/20/2009 for the course EE 2301 taught by Professor Larrykinney during the Fall '09 term at Minnesota.

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