ECE/Comp Sci 352 Digital System Fundamentals Quiz # 1 Solutions
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1 Last (Family) Name: KIME First (Given) Name: Student I: epartment of Electrical and omputer Engineering University of Wisconsin - Madison EE/omp Sci 352 igital System Fundamentals Quiz # Solutions October 4, 200 losed ook Examination 90 minutes ) No calculators, hand-held or laptop computers, cellular phones or pagers allowed. 2) If a box is provided, the contents of the box will be graded as the final answer. 3) Show your work for consideration of partial credit. 4) If you write anything on the back of a page and want it considered in the grading you must write See back of page (give page number) on the problem page. 5)The last page of the quiz is a sheet of identities and theorems for your use; it may be removed, but must be submitted with the exam. Problem Points Score Total 00
2 . a) (4 points) (F7.9) 6 = ( ) = (0) (2F.9) 6 = ( ) 4 b) (4 points) onvert the following radix 8 number to radix 7 representation. (607) 8 = (066) 7 6 x x = /7 = 55 rem 6 55/7 = 7 rem 6 7/7 = rem 0 /7 = 0 rem c) (4 points) Find the radix r such that: Hint: r is a power of 2. () r = 42 6 r2 + r + = 42 6 Note that the coefficient of 6 2 = 2 2, of 6 = 2 and of 6 0 =. So, the radix r since the coefficients are all is 6 x 2 = 32. More systematically, convert to binary = and examine for the number of bits that must be used to give. This number is 5 => base = 2 5 nswer: r = 32 2
3 2. a) (4 points) Place an above each of the given codes in the table below on the left that satisfies the properties of a Gray code for the decimal digits. (b) (4 points) Fill in the Excess 3 code for the decimal digits in the table below on the right. Excess NOT JENT UPLITE OEWOR c) (6 points) Perform subtraction corresponding to the decimal subtraction, showing LL details including all borrow values in the following table. Note that the correction when needed is 6, not + 6! orrow 0 0 operand Β οperand inary ifference or no correction ifference
4 3. a)(6 points) Find the algebraic expression in terms of,,, and for the sum of minterms (SOM) canonical form for the given function: f(,,, ) = ( + ) ( + ) ( + ) = ( + ) ( + ) = ( + + ) = + + = ( + ) + ( + ) + ( + ) = = f(,,, ) = b)(6 points) Find the algebraic expression for the product of maxterms (POM) canonical form for the given function: K(E, F, G, H) = Σ m (0,3,4,5,6,7,8,0,,2, 3,4) = Π M (,2,9,5) = (E + F + G + H)(E + F + G + H)(E + F + G + H)(E + F + G + H) K(E,F,G,H) = (E + F + G + H)(E + F + G + H)(E + F + G + H)(E + F + G + H) 4
5 4. a) ( 6 points) Find the minimum literal cost sum of products (SOP) expression for F(,,,) = Π Μ (,4,4,5) by using the Karnaugh map given below. F(,,,) = F 0 0 b) (4 points) Find the minimum literal cost sum of products (SOP) expression for: G(,,,) = Σ (3,6,7,9,,4,5) + Σ d (0,2,0,2,3) by using the Karnaugh map given below. G(,,,) = + + ( or c)(4 points) Find the minimum literal cost product of sums (POS) expression for: H(,,,) as given on the Karnaugh map given below. H ( or )+ + ( or ) G H(,,,) = (+)( + + OR ++)(++ OR ++) 5
6 5. (2 points) function F(W,, Y, Z) is given on a map below. F W W Z Y Z Y Z a) Find and list all prime implicants (PIs) of F (Hint: There are seven.). Prime Implicants: Z, Y Z, W, Y, W Y, W Z,W Y Z b) List all essential prime implicants of F and all less than prime implicants of F produced when the essential PIs are included in the solution. Stop at this point on this problem. Essential Prime Implicants: Y Z, W Less Than Prime Implicants: Z, W Y c) List an implicant of F that is NOT a prime implicant. Implicant (NOT Prime): W Z ny minterm or any 2-minterm implicant within a 4-minterm implicant is correct. b)(9 points) prime implicant table is given for function G(,,,). Find the subset of PIs that corresponds to a minimum literal SOP expression for F. On the right, fill in the type of each of the PIs as the solution is developed: Essential, Less Than, Secondary Essential, Redundant, Equivalent. e sure to follow the steps of the algorithm carefully. heck off the PIs to be used in the solution in the left column and check off the minterms covered in the bottom row. PIs in Solution PIs PI Type Secondary Essential 0 Less Than Essential Secondary Essential Less Than 0 0 Secondary Essential 0 0 Less Than Minterm heck Off 6
7 6. a) (6 points) Find a minimum literal cost two-level NN implementation for the function given on the following K-map. Your final answer may use both of the symbols for a NN gate as well as inverters. n inverter on the output is permitted if it gives a lower cost. ssume that the complements of the input variables are available. F nswer: F is cheaper. F = + F b)(6 points) Find a minimum gate multilevel NOR circuit for the given function. oth NORs and NOTs count as gates Your final answer may use both of the symbols for a NOR gate as well as inverters. ssume that the complements of the input variables are not available and must be generated within the circuit designed. G(,,,) = ( + + )( + + ) ( + + ) = (( ) + + )( + ( ) + ) ( + + ( )) nswer: 7
8 7. a) (5) Using oolean algebra only, simplify the following expression into a simplest sum-ofproducts (SOP) form. The simplest form contains 2 terms and 3 literal appearances. G(,,,,E) = E + E + + E = ( + ) E + + E = ( + ) E + + E = E + E + E + + E = E + E + + E = + + E = + E 8
9 7.b) (7) n engineer did a design in which the following equation was to be implemented. Function to be implemented: F(,,,,E) = E n implementation obtained for this equation is shown below. Implementation: E F F = ( ) ( + E) F = ( ) + + ( + E) = E ) Write a sum-of-products (SOP) expression for the circuit. F = E 2) Is the engineer s design correct? (ircle one) YES NO 3) If your answer is YES, show why it is correct. If your answer is NO, correct the logic diagram for the circuit. dd an inverter to the line, or complement, or move from the NN to the NOR. c) () TO E + TO E = 9
10 oolean lgebra Identities and Theorems. + 0 = 2. = Existence of and = 4. 0 = 0 Existence of and = 6. = Idempotence 7. + = 8. = 0 Existence of omplement 9. = Involution Y = Y +. Y = Y ommutative Laws 2. + (Y + Z) = ( + Y) + Z 3. (Y Z) = ( Y) Z ssociative Laws 4. (Y + Z) = Y + Z 5. + Y Z = ( + Y) ( + Z) istributive Laws 6. + Y = Y 7. Y = + Y emorgan s Laws Y = 9. ( + Y) = bsorption 20. Y + Y = Y 2. ( + Y) ( + Y) = Y Minimization Y = + Y 23. ( + Y) = Y Simplification 24. Y + Z + Y Z = Y + Z 25. ( + Y) ( + Z) (Y + Z) = ( + Y)( + Z) onsensus = 27. = 28. = = 30. Y = Y 3. Y = Y 32. Y = Y 33. ( Y) Ζ = (Y Ζ) = Y Ζ 0
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