hw4_Solutions_correctedb

hw4_Solutions_correctedb - Homework #4 Solution Problem 1...

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Unformatted text preview: Homework #4 Solution Problem 1 p⊕q p'q pq' Therefore p⊕q p'q pq': Problem 2 p’q: pq’: a) f a⊕b c⨁d = (a’b + b’a)(c’d + d’c) = a’bc’d + a’bcd’ + ab’c’d + ab’cd’ b) f a⊕b ⊕ c⨁d a'b b'a ⊕ c'd d'c = (a’b + b’a)’(c’d + d’c) + (a’b + b’a)( c’d + d’c)’ = (a’b)’(b’a)’(c’d + d’c) + (a’b + b’a)(c’d)’(d’c)’ = (a + b’)(b + a’)(c’d + d’c) + ( a’b + b’a)(c + d’)(d + c’) = (ab + aa’ + bb’ + b’a’)(c’d + d’c) + (a’b + b’a)(cd + cc’ + dd’ + c’d’) = (ab + a’b’)(c’d + d’c) + (a’b + b’a)(cd + c’d’) = abc’d + abcd’ + a’b’c’d + a’ b’cd’ + a’bcd + a’bc’d’ + ab’cd + ab’c’d’ ’ Problem 3 a) Given the gate f = a’c + b’c’ + a’b To implement the NOT operation: To implement OR: a 0 1 f = a’(1) + (0)’(1)’ + a’(0) = a’ To implement AND: Problem 3 part b To implement XNOR we can use a combination of each of these gates: Problem 4 Is a⨁b c⨁d a'b'd' a'cd a'b'c'd' a'cd' a'b'c a'bd b'cd' ab'c'd ? Simplify: a⨁b c⨁d a'b'd' a'cd = (a’b’ + b’a)(c’d + d’c) + a’b’d’ + a’cd = a’bc’d + a’bcd’ + ab’c’d + ab’cd’ + a’b’d’ + a’cd expand a’bc’d + a’bcd’ + ab’c’d + ab’cd’ + a’b’(c + c’)d’ + a’(b + b’)cd = = a’bc’d + a’bcd’ + ab’c’d + ab’cd’ + a’b’cd’ + a’b’c ’d’ + a’bcd + a’b’cd = a’bc’d + a’bcd’ + ab’c’d + ab’cd’ + a’b’cd’ + a’b’c d’ + a’b’cd’ + a’b’c’d’ + a’bcd + a’b’cd simplify= a’b’c’d’ + a’cd’ + a’b’c + a’bd + ab’c’d + b’cd’ RHS = LHS Therefore the two expressions are equal. Problem 5 f(a,b,c,d) = Σm(0,2,6,7,8,9,14,15) Karnaugh Map: Therefore: f = bc + ab’c’ + a’b’d’ Problem 6 a) Function EQU NOR XOR NAND Minimum SOP: C1 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 C2 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 X1 X2 Z 0 0 1 0 1 0 1 0 0 1 1 1 0 0 1 0 1 0 1 0 0 1 1 0 0 0 0 0 1 1 1 0 1 1 1 0 0 0 1 0 1 1 1 0 1 1 1 0 b) Karnaugh Map: f C1 ' X1 ' X 2 ' C1 C2 X 1 ' C1 X 1 ' X 2 C1 ' C2 ' X 1 X 2 C1 X 1 X 2 ' To build the circuit we use the following topology: f C1 ' X1 ' X 2 ' C1 C2 X 1 ' C1 X 1 ' X 2 C1 ' C2 ' X 1 X 2 C1 X 1 X 2 ' Problem 7 a) f(a,b,c,d) = Σm(0,2,3,5,7,8,10,13) f = b’ d’ + a’cd + bc’d b’d’ iis essential due to m(0,8,10) bc’d is essential due to m13 Problem 7 continued b) f(a,b,c,d) = ΠM(2,3,6,11,13,14,15) f = ab’d’ + a’bd + c’d’ + b’c’ ab’d’ is essential due to m(10) a’bd is essential due to m(7) b’c’ is essential due to m(9) c’d’ is essential due to m(12) Problem 8 a) f(a,b,c,d) = ΠM(0,1,5,8,10,13) x ΠD(2,6,7,11) Problem 8 continued b) f(a,b,c,d) = Σm(0,2,8,14,15) + Σd(3,4,10,11,13) Problem 9 a) f(a,b,c,d) = Σm(0,3,4,5,6,7,8,12,13,14,16,21,23,24 ,29,31) Problem 10 a) F(a,b,c,d) = a’c’d’ + b’cd’ + bcd’ + a’bcd = a’bc’d’ + a’b’c’d’ + ab’cd’ + a’b’cd’ + a bcd’ + a’bcd’ + a’bcd = m0 + m2 + m4 + m6 + m7 + m10 + m 14 Therefore the maxterm expression is simply the missing numbers: F = ΠM(1,3,5,8,9,11,12,13,15) In SOP: F = a’b’c’d + a’b’cd + a’bc’d + ab’c’d’ + ab’c’d + ab’cd + a bc’d’ + abc’d + abcd In POS: F = (a + b + c + d’)(a + b + c’ + d’)(a + b’ + c + d’)(a’ + b + c + d)(a’ + b + c + d’)(a’ + b + c’ + d’)(a’ + b’ + c + d)(a’ + b’ + c + d’)(a’ + b’ + c’ + d’) b) When we calculate F’ the maxterms become minterms Therefore F’ = Σm(1,3,5,8,9,11,12,13,15) F’ = b’d + c’d + ac’ + ad c) Minimum POS: F’’ = F = (b + d’)(c + d’)(a’+ c)(a’+ d’) ...
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This note was uploaded on 02/26/2011 for the course ECE 15A taught by Professor M during the Winter '08 term at UCSB.

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