Function Simplification

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Function Simplification (1/2)  Why simplify?  Simpler expression uses less logic gates.  Thus: cheaper, less power, faster (sometimes).  Simplification techniques:  Algebraic Simplification.  simplify symbolically using theorems/postulates.  requires skill but extremely open-ended.  Karnaugh Maps.  diagrammatic technique using ‘Venn-like diagram’.  easy for humans (pattern-matching skills).  simplified standard forms.  limited to not more than 6 variables.

Function Simplification (2/2)  Simplification techniques:  Quine-McCluskey tabulation technique.  tabulation technique based on certain ‘cancellation theorems’.  simplified standard forms.  tedious, repetitive step-by-step technique.  boring to humans BUT suitable for computers.  larger variables possible, but computationally intensive.

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Algebraic Simplification (1/5)  Algebraic simplification aims to minimise (i) number of literals, and (ii) number of terms  But sometimes conflicting.  Let’s aim at reducing the number of literals.

Algebraic Simplification (2/5)  Example: (x+y).(x+y').(x'+z)(6 literals) = (x.x+x.y'+x.y+y.y').(x'+z)(assoc.) = (x+x.(y'+y)+0).(x'+z)(idemp.,assoc., compl.) = (x+x.(1)+0).(x'+z)(complement) = (x+x+0).(x'+z)(identity 1) = (x).(x'+z)(idemp, identity 0) = (x.x'+x.z)(assoc.) = (0+x.z)(complement) = x.z (identity 0) Number of literals reduced from 6 to 2.

Algebraic Simplification (3/5)  Find minimal SOP and POS expressions of f(x,y,z) = x'.y.(z + y'.x) + y'.z x'.y.(z+y'.x) + y'.z = x'.y.z + x'.y.y'.x + y'.z(distributivity) = x'.y.z y'.z(complement, null element 0) = x'.y.z + y'.z(identity 0) = x'.z + y’.z(absorption) = (x' + y').z(distributivity) Minimal SOP of f = x'.z + y'.z (Two 2-input AND gates & One 2-input OR gate) Minimal POS of f = (x' + y').z (One 2-input OR gate & One 2-input AND gate)

Algebraic Simplification (4/5)  Find minimal SOP expression of f(a,b,c,d) = a.b.c + a.b.d + a'.b.c' + c.d + b.d‘ a.b.c + a.b.d + a'.b.c' + c.d + b.d' = a.b.c + a.b + a'.b.c' + c.d + b.d' (absorption) = a.b.c + a.b + b.c' + c.d + b.d' (absorption) = a.b + b.c' + c.d + b.d' (absorption) = a.b + c.d + b.(c' + d') (distributivity) = a.b + c.d + b.(c.d)' (DeMorgan) = a.b + c.d + b (absorption) = b + c.d (absorption) Number of literals reduced form 13 to 3.

Algebraic Simplification (5/5)  Difficulty – needs good algebraic manipulation skills.  Advantage – very open-ended (to your desired form!)

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Half Adder (1/2)  Half-Adder is a circuit which adds two single bits (called X,Y) together, to produce a result of two bits (called C, S).  A black-box representation of this circuit is: Truth table representation is: Half adder X Y (X+Y) S C

Half Adder (2/2)  In sum-of-minterms form: C = X.Y S = X'.Y + X.Y'  Algebraic simplification could simplify S to: S = X'.Y + X.Y' = X  Y  Giving: XYXY S C

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Introduction to K-maps  Systematic method to obtain simplified sum-of-products (SOPs) Boolean expressions.  Objective: Fewest possible terms/literals.  Diagrammatic technique based on a special form of Venn diagram.  Advantage: Easy with visual aid.  Disadvantage: Limited to 5 or 6 variables.

Outline  Function Simplification  Algebraic Simplification  Half Adder  Introduction to K-maps  Venn Diagrams

Venn Diagrams (1/2)  Venn diagram to represent the space of minterms.  Example of 2 variables (4 minterms): a.b' a'.b a'.b' a.b a b

Venn Diagrams (2/2)  Each set of minterms represents a Boolean function. Examples: { a.b, a.b' }  a.b + a.b' = a.(b+b') = a { a'.b, a.b }  a'.b + a.b = (a'+a).b = b { a.b }  a.b { a.b, a.b', a'.b }  a.b + a.b' + a'.b = a + b { }  0 { a'.b',a.b,a.b',a'.b }  1 a.b' a'.b a'.b' a.b a b