ECE320_HomeWork_No_1

# ECE320_HomeWork_No_1 - 8 8’s Complement 7’s Complement...

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EE 320 - Homework No. 1 (Solution) R. Roosta 1. a) (1001001.001) 2 = 1x2 6 + 1x2 3 + 1x2 0 + 1x2 -2 + 1x2 -3 =(73.375) 10 b) (0.342) 6 = 3x6 -1 + 4x6 -2 + 2x6 -3 = (0.62037) 10 c) (12.0625) 10 = (1100.0001) 2 0.0625 x 2 = 0.125 0 0.125 x 2 = 0.25 0 0001 (fractional part) 1.25 x 2 = 0.5 0 0.5 x 2 = 1.0 1 d) (225.268) 10 = (011100001.010001001) 2 = (341.211) 8 e) (2AC5.D) 16 = (10101011000101.1101) 2 = (10949.8125) 10 2. Number 1’s Complement 2’s Complement 0111101101 1000010010 1000010011 111111 000000 000001 00000 11111 00000 10000 01111 10000 3. a) (765) 8 + [-(356) 8 ] = (765) 8 + 8’s Complement of (356)

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Unformatted text preview: 8 8’s Complement 7’s Complement 765 765 + 422 + 421 1 407 1 406 1 407 b)-(469) 10 + [-(237) 10 ] 10’s Complement 9’s Complement 531 530 + 763 + 762 1 294 1 292 10’s Comp 1 293 706 9’s Comp 706 c) (1101001) 2 + [-(1011011) 2 ] 2’s Complement 1’s Complement 1101001 1101001 0100101 0100100 1 0001110 1 0001101 1 0001110 4. Decimal 84-2-1 Parity Bit 0 0 0 0 1 1 0 1 1 1 2 0 1 1 0 1 3 0 1 0 1 1 4 0 1 0 0 5 1 0 1 1 6 1 0 1 0 1 7 1 0 0 1 1 8 1 0 0 0 9 1 1 1 1 1 5. F=(A+B).C...
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ECE320_HomeWork_No_1 - 8 8’s Complement 7’s Complement...

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