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ECE 124 Assignment #2: Algebric simplification and applications - - PowerPoint PPT Presentation
ECE 124 Assignment #2: Algebric simplification and applications - - PowerPoint PPT Presentation
ECE 124 Assignment #2: Algebric simplification and applications 05/15/18 Presented by: Mahmoud KhalafAlla Supervised by: Prof. Cathy Gebotys ECE 124 Assignment #2 Outline Brief Introduction to simplification theorems Assignment
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Brief Introduction to Simplification theoreoms
§ Consensus theorem
XY + X’Z + YZ = XY + X’Z XY + X’Z + YZ(X + X’) = XY + X’Z + XYZ + X’YZ = XY(Z + 1) + X’Z (Y + 1)
§ Dual form:
(X+Y)(X’ + Z)(Y + Z) = (X+Y)(X’ + Z)
§ Minterm expression:
Sum of products where F = 1
§ Maxterm expression
Product of sum where F = 0
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Problem #1
§ Find the consensus term in each expression and delete it: § abc’d + a’be + bc'de
Let X = a , Y = bc‘d, Z = be , then can be rewritten as : XY + X‘Z + YZ bc‘de to be deleted à abc‘d + a‘be
§ (x'+y+z)(x+w)(y+z+w)
Dual form of consensus form is (X+Y)(X'+Z)(Y+Z) à (y+z+w) to be deleted à (x’ + y + z)(x + w)
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Problem #2
§ Simplify each expression by only using consensus theorem:
§ bc'd'+abc'+ac'd+ab'd+a'bd’
bc’d’ can be deleted and keep abc’ + a’bd’ à abc’ + ac’d + ab’d + a’bd’ ac’d can be deleted and keep abc’ + ab’d à abc’ + ab’d + a’bd’
§ (b'+c'+d')(b+c+d)(a+b+c)(a'+c+d)
(b + c + d) can be deleted and keep (a+b+c)(a’+c+d) à (b'+c'+d') (a+b+c)(a’+c+d)
§ a'b’c + abd + a’cde + bcde + a'bde
bcde can be deleted and keep abd + a’cde à a’b’c + abd + a’cde + a’bde à a’cde can be deleted and keep a’b’c + a’bde à a’b’c + abd + a’bde
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Problem #3
§ Simplify each expression to a sum of three terms :
§ a'bcd+a'bc'd+b'ef+cde'g+a'def+a'b'ef
a'bd[c+c'] + ef[b'(1+a')] +efa'd + cde'g =a'[b]d + ef[b'] + [efa'd] (consensus of 1st two terms) + cde'g = a'bd + efb'+cde'g ; a'bd+b'ef+cde'g
§ w'x'y'+w'xz'+((x+y+w'z)(x'+z'+wy'))’
w’x’y’ + w’xz’ + (x+y+w’z)’ + (x’ + z’ + wy’)’ à w’x’y’ + w’xz’ + x’y’(w’z)’ + xz(wy’)’ à w’x’y’ + w’xz’ + x’y’(w + z’) + xz(w’ + y) à x’y’(w’ + w + z’) + w’xz’ + w’xz + xyzà x’y’ + w’x(z+z’) + xyz à x'y’ + w’x + xyz
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Problem #4
§ Simplify F = a’b ⊕
⊕ bc bc ⊕ ab ⊕ ⊕ ab ⊕ b’c b’c’
b(a' ⊕ a) ⊕ (bc ⊕ b'c’) à (b ⊕ bc) ⊕ b’c’ à b(1 ⊕ c) ⊕ b'c’ à bc' ⊕ b'c’ à c'(b ⊕ b') = c'
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Problem #5
§ Factor Z = abc+de+acf+ad'+ab'e' and simplify it to form (x+x)(x+x)(x+x+x+x)
where each x represents a literal. Then express Z as a minimum sum of products in the form xx + xx+ xx+ xx a(bc+cf+b'e'+d')+de à (a+d)(a+e)(bc + cf +b'e'+d' +d)(bc +cf +b'e'+d' +e)à (a+d)(a+e)(bc + cf + b'e' + d' + e) à (e + e’b’ = e + b’) (a+d)(a+e)(bc+cf+b'+d’+e) à (bc + b’ = c + b’) (a+d)(a+e)(c+cf +b' +d’+e) à (a+d)(a+e)(c+d'+b'+e) (a + de)(c + d’ + b’ + e) à ac + ad’ + ab’ + ae + de + dec + deb’ à (ae consensus term of ad’ + de) ac+ad'+ab’+de(1 + b’) à ac+ad'+ab'+de
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Problem #6
§ statement always true? if a + b = c then ad' + bd' = cd’
YES
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Problem #7
§ A network has three 1-bit inputs and two 1-bit outputs. The output
signals b0, b1 represent a binary number which is equal to the number of input signals which are zero
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x y z b1 b0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Minterm b0 = m(0,3,5,6) Minterm b1 = m(0,1,2,4) Maxterm b0 = M(1,2,4,7) Maxterm b1 = M(3,5,6,7)
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