HW01_sol

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Unformatted text preview: (255.55)8 = (011 101 101 100 . 111 110)2 d) 29/8 = 3 r 5 3/8 = 0 r 3 109/8 = 13 r 5 13/8 = 1 r 5 1/8 = 0 r 1 0.30 x 8 = 2.4 (109.30)10 ≈ (155.23)8 0.4 x 8 = 3.2 (155.23)8 = (001 101 101 . 010 011)2 0.2 x 8 = 1.6 Part II (2’s Complement Representations & BCD): 1) Find the 1’s Complement and 2’s Complement using the appropriate method in the table below: 1’s Complement 2’s Complement Bus/Register Reverse all Reverse bits Method Width (# of 2n – 1 – x 2n – x 1’s Comp. +1 bits left of 1st ‘1’ bits n) Example (1001)2 +(0001)2 0110 24 ‐1‐6 = (9)10 0110 24 ‐6 = (10)10 4 (110)2 =(1001)2 1010 1001 =(1010)2 =(1010)2 (10010111)2 8 28‐1‐104 = 01101000 01101000 28‐104 = (152)10 (151)10 +(00000001)2 (01101000)2 10011000 10010111 =(10011000)2 =(10011000)2 =(10010111)2 4 24‐1‐1 = (14)10 0001 24‐1 = (15)10 (1110)2 +(0001)2 0001 (000001)2 1110 = (1110)2 = (1111)2 =(1111)2 1111 4 (1111)2 24‐0 = (16)10 24‐1‐0 = (15)10 +(0001)2 0000 0000 (0000)2* = (10000)2 ‐> 0000 = (1111)2 1111 =(10000)2 ‐> (0000)2 in 4 bits* (0000)2 in 4 bits* 8 (11110111)2 00001000 28‐1‐8 = (248)10 28‐1‐8 = (247)10 00001000 (1000)2** +(00000001)2 11111000 = (11110111)2 11110111 = (11111000)2 =(11111000)2 *Zero is special in 2’s complement. If you use the formula you get 2n which requires n+1 bits. We only count th...
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