Computer - Science - Notes - CH07 - Boolean - Algebra - Solutionrider
Computer - Science - Notes - CH07 - Boolean - Algebra - Solutionrider
Computer - Science - Notes - CH07 - Boolean - Algebra - Solutionrider
Truth table:
A truth table is composed of one column for each input variable (for example, A and B), and one
final column for all of the possible results of the logical operation that the table is meant to represent (for
example, A XOR B). Each row of the truth table therefore contains one possible configuration of the input
variables (for instance, A = true B = false), and the result of the operation for those values.
Logical Operators:
In Algebraic function e use +,-,*,/ operator but in case of Logical Function or Compound statement we
use AND,OR & NOT operator.
Example: He prefers Computer Science NOT IP.
er
3. AND
id
NOT Operator—Operates on single variable. It gives the complement value of
nr
variable.
tio
X X
0 1
1 0
lu
OR Operator -It is a binary operator and denotes logical Addition operation and is represented by
”+” symbol
so
0+ 0 = 0
0 + 1 = 1
1 + 0 = 1
1 + 1 =1
X Y X+Y
0 0 0
0 1 1
1 0 1
1 1 1
AND Operator – AND Operator performs logical multiplications and symbol is (.) dot.
0.0=0
0.1=0
1.0=0
1.1=1
solutionrider
101
thesolutionrider.blogspot.com
solutionrider
Truth table:
X Y X.Y
0 0 0
0 1 0
1 0 0
1 1 1
A logic gate is an physical device implementing a Boolean function, that is, it performs a logical operation
on one or more logic inputs and produces a single logic output. Gates also called logic circuits.
Or
A gate is simply an electronic circuit which operates on one or more signals to produce an output signal.
NOT gate (inverter):The output Q is true when the input A is NOT true, the output is the inverse of the
input:
Q = NOT A
A NOT gate can only have one input. A NOT gate
is also called an inverter.
er
id
nr
AND gate
tio
OR gate
The output Q is true if input A OR input B is true (or
both of them are true): Q = A OR B
An OR gate can have two or more inputs, its output is
true if at least one input is true.
Boolean algebra consists of fundamental laws that are based on theorem of Boolean algebra. These
fundamental laws are known as basic postulates of Boolean algebra. These postulates states basic relations
in boolean algebra, that follow:
0 + 0 = 0
I If X != 0 then x=1 and If X!=1 then x=0 0 + 1 = 1
II OR relations(logical addition) 1 + 0 = 1
1 + 1 = 1
102
thesolutionrider.blogspot.com
solutionrider
0.0 = 0
III AND relations (logical multiplication) 0.1 = 0
1.0 = 0
1.1 = 1
IV Complement Rules 0 1, 1 0
Principal of Duality
This principal states that we can derive a Boolean relation from another Boolean relation by
performing simple steps. The steps are:-
1. Change each AND(.) with an OR(+) sign
2. Change each OR(+) with an AND(.) sign
3. Replace each 0 with 1 and each 1 with 0
e.g
0+0=0 then dual is 1.1=1
1+0=1 then dual is 0.1=0
er
1. Properties of 0 and 1
(a) 0+X=X id
(b) 1+X=1
nr
(c) 0.X=0
(d) 1.X=X
2. Indempotence Law
tio
(a) X+X=X
(b) X.X=X
lu
3. Involution Law
so
(X) = X
4. Complementarity Law
(a) X + X=1
(b) X. X=0
5. Commutative Law
(a) X+Y=Y+X
(b) X.Y=Y.X
6. Associative Law
(a) X+(Y+Z)=(X+Y)+Z
(b) X(YZ)=(XY)Z
7. Distributive Law
(a) X(Y+Z)=XY_XZ
(b) X=YZ=(X+Y)(X+Z)
8. Absorption Law
(a) X+XY= X
(b) X(X+Y)=X
103
thesolutionrider.blogspot.com
solutionrider
solutionrider
Demorgan’s Theorem
A mathematician named DeMorgan developed a pair of important rules regarding group complementation
in Boolean algebra.
Demorgan’s First Theorem
er
id
nr
Derivation of Boolean expression:-
Minterm :
tio
missing in second term), multiply that term with (missing term +complement( missing term) )factor e.g. if
Y is missing multiply with Y+Y” )
3. Expand the expression .
4. Remove all duplicate terms and we will have minterm form of an expression.
Example: Convert X + Y
X + Y = X.1 + Y.1
=X.(Y+Y”) + Y(X+X”)
=XY + XY”+XY+X”Y
=XY+XY”+XY
104
thesolutionrider.blogspot.com
Shorthand Minterm notation:
Since all the letters must appear in every product, a shorthand notation has been developed that saves
actually writing down the letters themselves. To form this notation, following steps are to be followed:
1. First of all, Copy original terms
2. Substitute 0s for barred letters and 1s for nonbarred letters
3. Express the decimal equivalent of binary word as a subscript of m.
Rule1. Find Binary equivalent of decimal subscript e.g.,for m6 subscript is 6, binary equivalent of 6 is
110.
Rule 2. For every 1s write the variable as it is and for 0s write variables complemented form i.e., for 110 t
is XYZ. XYZ is the required minterm for m6.
maxterm:
A maxterm is a sum of all the literals (with or without the bar) within the logic system. Boolean
Expression composed entirely either of Minterms or Maxterms is referred to as Canonical Expression.
Canonical Form:
Canonical expression can be represented is derived from
(i) Sum-of-Products(SOP) form
(ii) Product-of-sums(POS) form
Sum of Product (SOP)
1. Various possible input values
er
2. The desired output values for each of the input combinations
id
nr
tio
1.Algebraic method:This method makes use of Boolean postulates, rules and theorems to simplify the
expression.
Example No. 1: Reduce the expression XY X XY .
105
thesolutionrider.blogspot.com
solutionrider
er
id
nr
tio
lu
Karnaugh Maps:
Karnaugh map or K Map is a graphical display of the fundamental product in a truth table.
For example:
Put a 1 in the box for any minterm that appears in the SOP expansion.
Basic idea is to cover the largest adjacent blocks you can whose side length is some power of 2.
Blocks can “wrap around” the edges.
For example, the first K-map here represents x y x y x ( y y ) x . (since y+y’=1)
The second K-map, similarly, shows x y x y ( x x ) y y
106
thesolutionrider.blogspot.com
solutionrider
Remember, group together adjacent cells of 1s, to form largest possible rectangles of sizes that are
powers of 2. Notice that you can overlap the blocks if necessary.
Sum Of Products Reduction using K- Map
er
id
nr
tio
lu
so
For reducing the expression first mark Octet, Quad, Pair then single.
• Pair: Two adjacent 1’s makes a pair.
• Quad: Four adjacent 1’s makes a quad.
• Octet: Eight adjacent 1’s makes an Octet.
• Pair removes one variable.
• Quad removes two variables.
• Octet removes three variables.
Reduction of expression: When moving vertically or horizontally in pair or a quad or an octet it can be
observed that only one variable gets changed that can be eliminated directly in the expression.
For Example
In the above Ex
Step 1 : In K Map while moving from m7 to m15 the variable A is changing its state Hence it can be
removed directly, the solution becomes B.CD = BCD. This can be continued for all the pairs, Quads, and
Octets.
solutionrider 107
thesolutionrider.blogspot.com
solutionrider
Step 2 : In K map while moving from m0 to m8 and m2 to m10 the variable A is changing its state. Hence
B’ can be taken similarly while moving from m0 to m2 and m8 to m10 the variable C is changing its state.
Hence D’ can be taken; the solution becomes B’.D’
The solution for above expression using K map is BCD + B’D’.
Example1: Reduce the following Boolean expression using K-Map:
F(P,Q,R,S)=Σ(0,3,5,6,7,11,12,15)
Soln:
This is 1 quad, 2pairs & 2 lock
Quad(m3+m7+m15+m11) reduces to RS
Pair(m5+m7) reduces to P‟QS
Pair (m7+m6) reduces to P‟QR
Block m0=P‟Q‟R‟S‟
M12=PQR‟S‟
hence the final expressions is F=RS + P‟QS + P‟QR + PQR‟S‟ + P‟Q‟R‟S‟
Example2: Reduce the following Boolean expression using K-Map:
F(A,B,C,D)=Π(0,1,3,5,6,7,10,14,15)
Soln:
Reduced expressions are as follows:
For pair 1, (A+B+C)
For pair 2, (A‟+C‟+D)
For Quad 1, (A+D‟)
er
For Quad 2, (B‟+C‟)
Hence final POS expression will be id
More about Gates:
nr
NAND gate (NAND = Not AND)
This is an AND gate with the output inverted, as shown
tio
108
thesolutionrider.blogspot.com
solutionrider
solutionrider
Q = (A AND B) OR (NOT A AND NOT B) EX-NOR gates can only have 2 inputs.
Summary truth tables
The summary truth tables below show the output states for all types of 2-input and 3-input gates.
er
id
nr
tio
a) X+XY=X b) X(X+Y)=X
b) Verify X’.Y+X.Y’=(X’+Y’).(X+Y) algebraically.
Ans. LHS= X’Y + XY’
= (X’+X) (X’+Y’) (Y+X) (Y+Y’)
= 1.(X’+Y’) (X+Y).1
= (X’+Y’) (X+Y)
= RHS, hence proved
c) Write the equivalent Boolean Expression F for the following circuit diagram :
solutionrider 109
thesolutionrider.blogspot.com
solutionrider
Ans.: A’B+AB+B’C
d) If F(P,Q,R,S) = ∏ (3,4,5,6,7,13,15) , obtain the simplified form using K-Map.
Ans.:
.
Reduction of groups following the reduction rule :
Quad1 = M4.M5.M6.M7
= P+Q’
Quad2 = M5.M7.M13.M15
er
= Q’+S’
Pair = M3.M7
= P+R’+S’
id
nr
Therefore POS of F(P,Q,R,S) = (P+Q’)(Q’+S’)(P+R’+S’)
tio
e) F(a,b,c,d)=Σ(0,2,4,5,7,8,10,12,13,15)
F(a,b,c,d)=B1+B2+B3
lu
B1=m0+m4+m12+m8==c’d’
B2=m5+m7+m13+m15=bd
so
B3=m0+m2+m8+m10=b’d’
F(a,b,c,d)=c’d’+bd+b’d’
f) Write the equivalent Boolean expression for the following logic circuit:
X Y Z F
0 0 0 0
0 0 1 1
0 1 0 0
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 1
Express in the product of sums form, the Boolean function F(X,Y,Z), the truth table for which is
given below:
solutionrider
110
thesolutionrider.blogspot.com
solutionrider
5. Write the equivalent Boolean Expression F for the following circuit diagram : 2
er
id
nr
tio
lu
6 Write the equivalent Boolean Expression F for the following circuit diagram : 2
so
7. Convert the following Boolean expression into its equivalent Canonical Sum of Product
Form((SOP)
(X’+Y+Z’).(X’+Y+Z).(X’+Y’+Z).(X’+Y’+Z’) 1
8. Convert the following Boolean expression into its equivalent Canonical Product of Sum form (POS):
A.B’.C + A’.B.C +A’.B.C’ 1
9. Draw a Logical Circuit Diagram for the following Boolean expression:
A.(B+C’) 2
10. Write the equivalent Boolean Expression F for the following circuit diagram: 2
solutionrider 111
thesolutionrider.blogspot.com
solutionrider
er
15 Obtain a simplified form for a boolean expression
F(U,V,W,Z)= π (0,1,3,5,6,7,10,14,15)
id
16 . Reduce the following boolean expression using K-Map
nr
F(A,B,C,D) = Σ(5,6,7,8,9,12,13,14,15)
tio
lu
so
solutionrider
112
thesolutionrider.blogspot.com