exam1_practice_solutions - Exam 1 Study Question Solutions...

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WSF Page 1 of 12 Exam 1 Study Question Solutions August 6, 2007 1. A3.12 16 = 10•16 1 + 3•16 0 + 1•16 -1 + 2•16 -2 = 160 + 3 + 0.0625 + 0.0078125 = 163.0703125 10 2. 10110.101 2 = 2 4 + 2 2 + 2 1 + 2 -1 + 2 -2 = 16 + 4 + 2 + 0.5 + 0.125 = 22.625 10 3. 762.125 10 = 762.125 - 512 1•2 9 250.125 - 0 0•2 8 250.125 - 128 1•2 7 122.125 - 64 1•2 6 58.125 - 32 1•2 5 26.125 - 16 1•2 4 1011111010.001 2 10.125 - 8 1•2 3 2.125 - 0 0•2 2 2.125 - 2 1•2 1 0.125 - 0 0•2 0 0.125 - 0 0•2 -1 0.125 - 0 0•2 -2 0.125 - 0 .125 1•2 -3 0
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WSF Page 2 of 12 4. First convert 1362 7 to decimal, then to hex. 1•7 3 + 3•7 2 + 6•7 1 + 2•7 0 = 343 + 147 + 42 + 2 = 534 10 534 - 512 2•16 2 22 -16 1•16 1 216 16 6 -6 6•16 0 0 5. 1234 10 = 0001 0010 0011 0100 BCD 6. 0011 1001 0101 0111 BCD 3957 10 3 9 5 7 7. A.) Not valid. Two bits change from 3 4. B.) Valid. Notice that it must also have only one bit change when going from 7 back to 0. C.) Not valid. The code “011” appears twice. 8. A.) AND D.) OR G.) NOT (inverter) J.) AND B.) XOR E.) XNOR (equivalence) H.) PASS (we didn’t really talk about this gate) C.) NAND F.) NOR I.) XOR 9. OR (+) AND (•) XOR ( ) XNOR or equivalence ( ) 10. We can read the unminimized equation directly from the truth table. F = I 2 ’I 1 ’I 0 ’ + I 2 ’I 1 I 0 ’ + I 2 ’I 1 I 0 + I 2 I 1 ’I 0 ’ + I 2 I 1 I 0 This is the minterm expansion. Or we could write it as Σ m(0, 2, 3, 4, 7). 11. a.) Neither d.) Maxterm b.) Maxterm e.) Neither c.) Minterm f.) Minterm 12. a.) xy + x’yz’ + yz = y(x + x’z’ + z) = y[(x+x’)(x+z’) + z] Or use Theorem 11D = y(x + z’ + z) b.) (xy’ + z)(x + y’)z = (xy’x + xy’y’ + zx + zy’)z Multiply out = (xy’ + xy’ + xz + y’z)z Eliminate literals = (xy’ + xz + y’z)z Eliminate duplicate terms = xy’z + xz + y’z Distribute = xz + y’z Eliminate redundant term (Theorem 10)
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WSF Page 3 of 12 c.) xy’ + z + (x’+y)z’ = xy’ + [z + (x’ + y)](z + z’) = xy’ + z + x’ + y or use Theorem 11D = x’ + xy’ + y + z = x’ + y’ + y + z Theorem 11D = x’ + 1 + z Theorem 5 = 1 d.) a’d(b’ + c) + a’d’(b + c’) + (b’ + c)(b + c’) = a’b’d + a’cd + a’bd’ + a’c’d’ + b’b + b’c’ + bc + cc’ = a’b’d + a’cd + a’bd’ + a’c’d’ + b’c’ + bc Hmmm… This is looking ugly. Let’s try a K-map. cd ab 00 01 11 10 00 1 1 1 01 1 1 1 11 1 1 10 1 1 The remaining terms can be covered in several ways. = bc + b’c’ + a’b’d + a’c’d’ = bc + b’c’ + a’cd + a’c’d’ Four minimum solutions. = bc + b’c’ + a’b’d + a’bd’ All are valid simplifications. = bc + b’c’ + a’cd + a’bd’ e.) w’x’ + x’y’ + yz + w’z’ Expand the w’x’ term and see what we get… = w’x’yz + w’x’yz’ + w’x’y’z + w’x’y’z’ + x’y’ + yz + w’z’ = w’x’yz + w’x’yz’ + w’x’y’z + w’x’y’z’ + x’y’ + yz + w’z’ Eliminate redundant terms using Theorem 10 = x’y’ + yz + w’z’ We could have also used a 4-variable K-map. f.) A’BCD + A’BC’D + B’EF + CDE’G + A’DEF + A’B’EF Use Theorem 9 Use Theorem 10 = A’BD + B’EF + A’DEF + CDE’G Use Theorem 17 (Consensus) = A’BD + B’EF + CDE’G
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WSF Page 4 of 12 g.) [(a’ + d’ + b’c)(b + d + ac’)]’ + b’c’d’ + a’c’d = ad(b + c’) + b’d’(a’ + c) + b’c’d’ + a’c’d DeMorgan’s Law = abd + ac’d + a’b’d’ + b’cd’ + b’c’d’ + a’c’d Distribute = abd + c’d + a’b’d’ + b’d’ Theorem 9 = abd + c’d + b’d’ Theorem 10 13. (A’ + B + C’)(A + B’ + C) = AA’ + A’B’ + A’C + AB + BB’ + BC + AC’ + B’C’ + CC’
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