hwsol1 - 4 8 145 2 144 11=B 3(1pt for each part 2.7(a 2.7(c...

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Solution to ECSE 2610 Homework Set 1 For problem in items 2 to 4 , process should be shown to get full credits. 1. (1.6) (1 pt) Input Output ------------------------------- 0 0 0 0 1 1 1 0 1 1 1 1 (2 pts) The same function is performed by a single 2-input OR gate. (2 pts) 2. (1 pt for each part) 2.1(a) 110 2 =6 16 , 1011 2 =B 16 , so 1101011 2 =6B 16 2.1(f) F 16 =1111 2 , 3 16 =0011 2 , A 16 =1010 2 , 5 16 =0101 2 , so F3A5 16 =1111001110100101 2 2.3(c) A 16 =1010 2 , B 16 =1011 2 , C 16 =1100 2 , D 16 =1101 2 , so ABCD 16 =1010101111001101 2 1 010 101 111 001 101 2 1 2 5 7 1 5 8 so ABCD 16 =125715 8 2.5(a) 1101011 2 =1x2 6 +1x2 5 +1x2 3 +1x2 1 +1x2 0 =107 10 2.5(f) F3A5 16 =15x16 3 +3x16 2 +10x16 1 +5x16 0 =62373 10 2.6(a) 125 10 =1111101 2 6 2 3 1 1 5 7 3 1 2 )125 2 ) 62 2 )31 2 ) 15 2 ) 7 2 ) 3 12 6 2 2 14 6 2 5 0 1 1 1 1 1 4 1 0 1 1 2.6(f) 23851 10 =5D2B 16 1490 93 5 16 ) 23851 16 ) 1490 16 ) 93 16 144 80 78 50 13=D 64
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Unformatted text preview: 4 8 145 2 144 11=B 3. (1pt for each part) 2.7(a) 2.7(c) 2.7(d) C 1100010 C 111111110 C 11000000 110101 11011101 1110010 + 11001 + 1100011 + 1101101 1001110 101000000 11011111 2.10(a) 2.10(b) C 0000 C 10100 1372 4F1A5 + 4631 + B 8 D 5 59A3 5AA7A 4. (2.11) Use -49 as example. 49 10 =00110001 2 , so the signed-magnitude of -49 is 10110001. The complement bits of 00110001 is 11001110, so the two’s complement of -49 is 11001110+1 = 11101111. The answer: Decimal +18 +115 +79 -49 -3 -100 Signed-magnitude 00010010 01110011 01001111 10110001 10000011 11100100 Two’s complement 00010010 01110011 01001111 11001111 11111101 10011100 (deduct 0.5 pt for every number wrong)...
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This homework help was uploaded on 04/22/2008 for the course ECSE 2610 taught by Professor Ji during the Spring '08 term at Rensselaer Polytechnic Institute.

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hwsol1 - 4 8 145 2 144 11=B 3(1pt for each part 2.7(a 2.7(c...

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