Digital Logic Design (CSNB163)
Digital Logic Design (CSNB163)Module 51Combinations of Binary
VariablesIn Module 3, you have learned the concept of Boolean
Algebra which consists of binary variables and binary operator.A
binary variable x, can either appear in its:normal form (x) or
complement form (x) If there are two binary variables (e.g. x and
y) beings the inputs to a binary operator (e.g. AND), then there
are basically 22 number of possible combinations (e.g. xy, xy, xy,
xy),Thus, 2n possible combinations where n is the total number of
input variable. Boolean Algebra RepresentationA Boolean Algebraic
expression can appear in any of the following forms:
standard form
canonical form
Sum of Product (SOP)Product of Sum (POS)Sum of MintermsProduct
of MaxtermsStandard Vs Canonical Form of Boolean AlgebraThe major
difference between standard and canonical form of Boolean Algebra
expression is in terms of the input binary variables between
terms.
CriteriaStandard FormCanonical Forminput binary variables
between termsNot necessarily consistentMust be
consistentExampleF(a, b, c) = a + bcF(a, b, c) = abc + abc + abc +
abcStandard Boolean Algebra (Sum of Product)Sum of Product (SOP) is
a Boolean expression containing AND terms ( also known as product
term) whereby the AND terms may contain one or more input binary
variables and are combined by OR operationsExample:
F(x, y, z) = xy + yz + xyzcontain AND terms with one or more
input binary variables AND terms are combined through OR
operationsStandard Boolean Algebra (Product of Sum)Product of Sum
(POS) is a Boolean expression containing OR terms ( also known as
sum term) whereby the OR terms may contain one or or more input
binary variables and are combined by AND operationsExample:
F(x, y, z) = (x+y ) . (y+z ). (x+y+z)contain OR terms with one
or more input binary variables AND terms are combined through AND
operationsCanonical Boolean Algebra (Sum of Minterms)Sum of
Minterms is a Boolean expression containing AND terms ( also known
as minterms) whereby the minterms must contain similar input binary
variables and are combined by OR operationsExample:
F(x, y, z) = xyz+ xyz+ xyzcontain minterms with similar input
binary variables minterms are combined through OR
operationsCanonical Boolean Algebra (Sum of Minterms cont.)Each
input binary variable in a minterm can be associated with a binary
value, whereby:normal form = 1 (e.g. x = 1) complement form = 0
(e.g. x = 0) E.g. If given F1:
the equivalence
Thus, output is onlyassociated with binary 1
xyzmintermdesignationF1000xyzm00001xyzm10010xyzm20011xyzm30100xyzm41101xyzm51110xyzm60111xyzm71F1
= xyz+ xyz+ xyzF1 = m4 + m5 + m7 = (m4, m5, m7)Canonical Boolean
Algebra (Sum of Minterms) (Example1)Given the following truth
table:
Derive the Sum of Minterms Boolean expression for Y.Draw the
circuit diagram.
ABY000011101111Canonical Boolean Algebra (Sum of Minterms)
(Example1)Answer =
ABY000011101111Y= AB + AB + AB
Standard Boolean Algebra (Product of Maxterms)Product of
Maxterms is a Boolean expression containing OR terms ( also known
as maxterms) whereby the maxterms must contain similar input binary
variables and are combined by AND operationsExample:
contain maxterms with similar input binary variables maxterms
are combined through AND operationsF2 = (x+y+z) (x+y+z)
(x+y+z)Canonical Boolean Algebra (Product of Maxterms cont.)Each
input binary variable in a maxterm can be associated with a binary
value, whereby:normal form = 0 (e.g. x = 0) complement form = 1
(e.g. x = 1) If given F2:
the equivalence
Thus, output is onlyassociated with binary 0
xyzmaxtermdesignationF2000x+y+zM01001x+y+zM10010x+y+zM21011x+y+zM30100x+y+zM41101x+y+zM51110x+y+zM60111x+y+zM71F2
= (x+y+z) (x+y+z) (x+y+z)F2 = M1 . M3 . M6 = (M1, M3, M6)Canonical
Boolean Algebra (Product of Maxterms) (Example1)Given the following
truth table:
Derive the Product of Maxterms Boolean expression for Y.Draw the
circuit diagram.
CBAY00010011010101111000101111011110Canonical Boolean Algebra
(Product of Maxterms) (Example1)Answer =
Y= (C+B+A) (C + B + A)CBAY00010011010101111000101111011110
Sum of Minterms Product of MaxtermsAn Algebraic Boolean
expression in Sum of Minterms can be converted into its Product of
Maxterms (and vice versa)
Sum of MintermsProduct of MaxtermsThis can be proven
throughTheorems of Boolean Algebra!Sum of Minterms Product of
Maxterms (Example1)Given the following truth table:
Derive the Sum of Minterms Boolean expression for Y.Derive the
Product of Maxterms Boolean expression for Y.Prove that both
expressions are equal.
xyZY00000010010001111000101111011111Sum of Minterms Product of
Maxterms (Example1)Answer:
xyzYmM0000m0M00010m1M10100m2M20111m3M31000m4M41011m5M51101m6M61111m7M7Sum
of MintermsY1 = m3 + m5 + m6 + m7 = xyz + xyz + xyz + xyzProduct of
MaxtermsY2 = M0.M1.M2.M4 = (x+y+z)(x+y+z) (x+y+z) (x+y+z)Sum of
Minterms Product of Maxterms (Example1)Answer:
xyzYmM0000m0M00010m1M10100m2M20111m3M31000m4M41011m5M51101m6M61111m7M7Proving:Sum
of Minterms = Product of MaxtermsY1= m0+m1+m2+m4Y1 = (Y1)=
(m0+m1+m2+m4)Y1=M0.M1.M2.M4 = Y2Y1 = m0+m1+m2 +m4 = xyz+xyz+xyz+xyz
Y1=(Y1)= (xyz+xyz+xyz+xyz )via Involution Y1=(xyz) . (xyz) . (xyz)
. (xyz) via DeMorgansY1=(x+y+z).(x+y+z).(x+y+z).(x+y+z)via
DeMorgansY1=(x+y+z).(x+y+z).(x+y+z).(x+y+z)=Y2 via
InvolutionConversion between Boolean Algebra RepresentationsWe can
convert standard form of Boolean Algebraic expression into
canonical form (and vice versa). For example:Convert Sum of Product
into sum of minterms (and vice versa)Convert Product of Sum into
product of maxterms (and vive versa)Conversion is carried out by
manipulating the basic postulates, properties and theorems of
Boolean Algebra taught in Module 4.
Convert Standard Form into Canonical Form (Example1)Convert the
following Boolean Function
to its canonical Sum of Minterms form.
F(a, b, c) = a + bca + bc= a(1) + bc via Identity Element= a(b +
b) + (bc) via Complement= ab + ab + bcvia Distributive=(ab + ab)(1)
+ (bc)(1)via Identity Element=(ab + ab)(c+c) + bc(a + a) via
Complement=abc + abc + abc + abc + abc + abc via Distributive=abc +
abc + abc + abc + abcvia IdempotencyConvert Standard Form into
Canonical Form (Example2)Convert the following Boolean Function
to its canonical Product of Maxterms form.
F(x, y, z) = xy +xz(1st step get the Sum of Minterms)F (x, y,
z)= xy + xz= xy(1) + xz(1) via Identity Element= xy(z+z) + xz(y+y)
via Complement= xyz + xyz+ xyz +xyzvia Distributive= m7 + m6 + m3+
m1Sum of MintermsConvert Standard Form into Canonical Form
(Example2)Convert the following Boolean Function
to its canonical Product of Maxterms form.
F(x, y, z) = xy +xz(2nd step convert Sum of Minterms into
Product of Maxterms)F(x,y,z)= m7+m6+m3+m1 Sum of Minterms Thus F
(x, y, z)= m5+m4+m2+m0= xyz + xyz+xyz+xyzF(x,y,z)={F (x, y, z)}
=(xyz + xyz+xyz+xyz) via Involution=(xyz) . (xyz).(xyz)+(xyz) via
DeMorgans=(x+y+z).(x+y+z).(x+y+z).(x+y+z)via
DeMorgans=(x+y+z).(x+y+z).(x+y+z).(x+y+z)via Involution= M5 +
M4+M2+M0 Product of MaxtermsDigital Logic Design (CSNB163)End of
Module 5
