EE Practice Problems for Exam : Solutions, Fall 28. ircle T (true) or F (false) for each of these oolean equations. (a). T FO + = (b). T FO + = ( + )( + ) (c). TO F = (d). TO F () = () (e). TO F + + = 2. Evaluate the following: 2(a). onvert to hex:. 2 =? Integer Part: 2 = H 2 = 9H Fractional Part: 2 = H Verification of Fractional Part: 2 = 2 + 2 3 + 2 4 = /2 + /8 + /6 = 8/6 + 2/6 + /6 = /6 = 6 = 6 Verification is for your own interest; you are not expected to show this on an exam problem.. 2 =9.H 2(b). onvert to decimal:. 2 =? Integer: 2 = 2 + 2 2 + 2 4 + 2 5 = + 4 + 6 + 32 = 53 Fraction:. 2 = 2 2 = /4 =.25. 2 = 53.25
EE Practice Problems for Exam : Solutions, Fall 28 2 2(c). onvert to binary: 98.25 =? Using sum-of-powers method, we obtain for the integer and fractional parts: Integer: 2 6 = 64 < 98 98 64 = 34 2 5 = 32 < 34 34 32 = 2 2 = 2 2 2 = Integer: 98 = 2 6 + 2 5 + 2 = 2 Fractional Part: 2 2 = /4 =.25.25.25 = Fraction:.25 = 2 2 =. 2 Note: It is sufficient to write that.25 = 2 2 =. 2 for the fractional part. 98.25 =. 2 2(d). onvert to decimal: 2.4 6 =? Either convert directly from hex: 2.4 6 = 2 6 + 6 + 4 6 = 32 + + 4/6 = 32 + + /4 = 42.25 or convert to binary first and then to decimal: 2.4 6 =. 2 = 2 5 + 2 3 + 2 + 2 2 = 32 + 8 + 2 + /4 = 42.25 2.4 6 = 42.25
EE Practice Problems for Exam : Solutions, Fall 28 3 3. Write the oolean expression equivalent to the following logic circuit. Do not simplify! $% D $% = ( + ) + D 4(a). Draw the logic circuit realization of the following oolean expression as stated. Do not simplify! ou may draw inverters explicitly or use inversion bubbles, as you choose. = f(,, ) = ( + )( + ) "# 4(b). Write the complete truth table for the oolean expression of 4(a).
EE Practice Problems for Exam : Solutions, Fall 28 4 4(c). onvert the oolean equation of 4(a) to its DeMorgan equivalent. There are two (very similar) ways to apply DeMorgan s theorem to 4(a). ( + )( + ) = ( + )( + ) = ( + ) + ( + ) = ( + ) + ( + ) = ( + ) + = ( + ) + OR ( + )( + ) = ( + )( + ) = ( + ) + ( + ) = ( ) + = ( ) + stopping here gives circuit 4.d.(alt) = ( + ) + = ( + ) + = ( + ) + OR = ( ) + for SOP 4(d). Draw the logic circuit for the DeMorgan equivalent oolean equation you found in 4(c). ou may use inverters or inversion bubbles, as you choose. "# OR (alt)
EE Practice Problems for Exam : Solutions, Fall 28 5 5. Simplify the following oolean expression as far as possible, using the postulates and theorems of oolean algebra. DO NOT use a Karnaugh map except possibly to check your work. ou do not have to justify each step by stating the theorem or postulate used, but you must show each step in your simplification. f(w, x, y) = wxy + wx + wy + wxy wxy + wx + wy + wxy = wxy + wx + wy + wyx associative (b) = wxy + wx + wy + wyx identity (b), = = wxy + wx + wy( + x) distributive (b), ( + ) = + = wxy + wx + wy Nullity Thm (a), + = = wxy + wx + wy identity (b) = w(xy + x) + wy distributive (b) = w(x + y) + wy Reduction Thm (a), + = + = w(x + y + y) distributive (b) = w(x + ) complement (a), + = = w Nullity Thm (a) = w identity (b) wxy + wx + wy + wxy = w Simplified! f(w, x, y, z) = w Note: ou may approach this problem in a different order, for example, by factoring out w initially. This is fine, as long as you include all steps and obtain the same final answer. 6. Simplify the following expression using the postulates and theorems of oolean algebra. Eliminate all group complements. Justify each step by stating or referrring to the oolean theorem or postulate you use. Don t skip any steps! Do NOT use a Karnaugh map to simplify the expressions. ( )( + )( + ) Hint: Remember DeMorgan s theorem! ( )( + )( + ) = ( ) ombining Thm (b), ( + )( + ) = = ( + + ) DeMorgan s Theorem, = + = + + distributive (b) = + + complement (b), = = + identity (a), + = = ( + ) distributive (b) ( )( + )( + ) = ( + ) Simplified!
EE Practice Problems for Exam : Solutions, Fall 28 6 7. Given = f(w, x, y, z) = M(,, 3, 5, 3), 7(a). Write the complete truth table for = f(w, x, y, z). w x y z 7(b). Write = f(w, x, y, z) in POS canonical form. (Do not use Σ or Π notation in your final answer.) = (w + x + y + z)(w + x + y + z)(w + x + y + z)(w + x + y + z)(w + x + y + z) 7(c). Write = f(w, x, y, z) in shorthand SOP form. (Use Σ or Π notation in your final answer.) = m(2, 4, 6, 7, 8, 9,,, 2, 4, 5) 7(d). Write = f(w, x, y, z) in SOP canonical form. (Do not use Σ or Π notation in your final answer.) =w x yz + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z + w x y z
EE Practice Problems for Exam : Solutions, Fall 28 7 8. For = f(w, x, y, z) = M(,, 3, 5, 3) as given in Problem 7, 8(a). Fully label and complete the Karnaugh map below with as given above. Then derive a minimized POS expression for = f(w, x, y, z). wx yz POS Groupings: red horizontal pair=w + x + y; magenta vertical pair=x + y + z blue horizontal pair=w + x + z POS: = (w + x + y)(w + x + z)(x + y + z) 8(b). Fully label and complete the Karnaugh map below with as given above. Then derive a minimized SOP expression for. wx yz SOP Groupings: dark blue horizontal quad=wx; red vertical quad=yz light blue square quad=xy; green wraparound square quad=xz SOP: = wx + yz + xy + xz