Problem 2 HOME WORK 4: DUE MARCH 16 WEDProblem 1. Using Karnaugh maps find a minimal sum of products expression and a minimal product of sumsexpression for each of the following logic functions(a) F(W.X.Y.Z) = (m,.m,,m,,m,,m,, mm3. m)(b) FW.X.Y.Z) = 1(M.M3, M.M, M.M,. M,.M.M.M(c) F(W.X.Y.Z) = (m,,m,.m,.m, ,m)+ d(m.m.m,.m.m;)(d) F(WXY) = TICM.M.M.M.M,)Note: d(m.) means F(W.X.Y.Z) for the row of the truth table corresponding to the i'th minterm is "don'tcare"Problem 2. Given a sum of products or product of sums expression for F. it is possible to fill in the Is in a K-map forF directly and simplify the expression using K-map simplification. Using this approach, simplify the followingexpressions for F(a) F= (W + Z (W +Y+ZXX +Y +Z) (Wakerly Problem 4.58. part (d))(b) ABÇD+ACD+BCD+ ABD+ ABC (Wakerly Problem 4.58. part (e))Problem 3: Perform the indicated subtraction with the following unsigned binary numbers by taking the 2'scomplement of the subtrahend.(a) 11010 - 100001(b) 11110 - 1110(c) 101001- 101Hint: You necd to make the no of bits in the two numbers equal by padding Os to the left of the subtrahendbefore taking its 2's complement.Problem 4: Assuming that the numbers below are signed 2's complement numbers with the leftmost bit representinghe sign bit, indicate the presence of overflow in any of the following calculations(a) 11010 +011II(b) 11110 + 10010(c) 111II1O + 0000010(d) 101001 + 000010

2)a) GivenF=(W+Z')(W'+Y'+Z')(X'+Y'+Z')

Answered: a) F=(W+ZXW +Y+ZXX+Y+Z) (Wake )…

Engineering

Electrical Engineering

a) F=(W+ZXW +Y+ZXX+Y+Z) (Wake ) ABCD+ACD+ BCD+ ABD+ ABC (Wak

Electric Motor Control

10th Edition

ISBN:9781133702818

Author:Herman

Publisher:Herman

Chapter22: Sequence Control

Section: Chapter Questions

Problem 6SQ: Draw a symbol for a solid-state logic element AND.

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Question

Problem 2

HOME WORK 4: DUE MARCH 16 WED
Problem 1. Using Karnaugh maps find a minimal sum of products expression and a minimal product of sums
expression for each of the following logic functions
(a) F(W.X.Y.Z) = (m,.m,,m,,m,,m,, mm3. m)
(b) FW.X.Y.Z) = 1(M.M3, M.M, M.M,. M,.M.M.M
(c) F(W.X.Y.Z) = (m,,m,.m,.m, ,m)+ d(m.m.m,.m.m;)
(d) F(WXY) = TICM.M.M.M.M,)
Note: d(m.) means F(W.X.Y.Z) for the row of the truth table corresponding to the i'th minterm is "don't
care"
Problem 2. Given a sum of products or product of sums expression for F. it is possible to fill in the Is in a K-map for
F directly and simplify the expression using K-map simplification. Using this approach, simplify the following
expressions for F
(a) F= (W + Z (W +Y+ZXX +Y +Z) (Wakerly Problem 4.58. part (d))
(b) ABÇD+ACD+BCD+ ABD+ ABC (Wakerly Problem 4.58. part (e))
Problem 3: Perform the indicated subtraction with the following unsigned binary numbers by taking the 2's
complement of the subtrahend.
(a) 11010 - 100001
(b) 11110 - 1110
(c) 101001- 101
Hint: You necd to make the no of bits in the two numbers equal by padding Os to the left of the subtrahend
before taking its 2's complement.
Problem 4: Assuming that the numbers below are signed 2's complement numbers with the leftmost bit representing
he sign bit, indicate the presence of overflow in any of the following calculations
(a) 11010 +011II
(b) 11110 + 10010
(c) 111II1O + 0000010
(d) 101001 + 000010

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