assign1b_soln

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Unformatted text preview: Solutions to Assign 1 questions 4 and 8    NOTE: handwritten numbers are incorrect.  Question 4.  a) 10000111 – 1011001 (135‐89 = 46)  0101 1001 2’s complement ‐> 1010 0111     1000 0111    + 1010 0111   1 0010 1110  (46)  ‐ note carry out so value is positive    b) 1011001 – 10000111 (89 – 135 = -46) 10000111 2’s complement -> 01111001    0101 1001    + 0111 1001      11010010 ‐ note no carry out so value is negative an din 2’s complement    11010010  2’s complement ‐>  ‐ 00101110 (‐46)    c) 0.1001 – 0.0101 (0.5625 – 0.3125 = 0.25)      0.0101 2’s complement ‐> 1.1011       0.1001    + 1.1011    10.0100 (0.25)‐ carry out so value is positive     d) .0101  ‐ 0.1001 (0.3125  ‐ 0.5625 = ‐0.25)    0.1001 2’s complement ‐> 1.0111       0.0101    + 1.0111  0 1.1100 – no carry out so value is negative    1.1100 2s complement ‐> ‐0.0100 (‐0.25)                      Question 8. 3 input NAND gate F = (ABC)’  ; cascade of two NAND gates G = ((AB)’C)’  F = (ABC)’ = A’ + B’ + C’  (deMorgans)  G = ((AB)’C)’ = ((A’ + B’)C)’ = (A’C + B’C)’ = (A’C)’(B’C)’ = (A + C’)(B+C’)       = AB + AC’ + BC’ + C’ = AB + AC’ + C’ = AB + C’ != F    using deMorgan’s, distribution and identity above    Truth table just to confirm    A  B  C  (ABC)’  (AB)’  ((AB)’C)’  0  0  0  1  1  1  0  0  1  1  1  0  0  1  0  1  1  1  0  1  1  1  1  0  1  0  0  1  1  1  1  0  1  1  1  0  1  1  0  1  0  1  1  1  1  0  0  1      ...
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## This note was uploaded on 12/21/2010 for the course EECE EECE 256 taught by Professor Sidney during the Spring '10 term at UBC.

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