The document discusses Boolean functions and simplification methods. It recaps Boolean algebra and other tools used for logic system analysis and synthesis. It then explains how to write Boolean functions in canonical/standard forms using variables, minterms and maxterms. Examples are provided to demonstrate conversion between sum of products and product of sums forms. The document also discusses handling cases where some variables are missing. Finally, it introduces Karnaugh maps as a method to minimize Boolean expressions of 2, 3 or 4 variables without using Boolean algebra theorems. Implementation of functions using NAND/NOR gates in sum of products or product of sums form is also briefly covered.

1. Boolean Functions &
Simplification Methods
Fundamentals of Automation Engineering I
ES1111

2. Recapitulating
• We have seen that –
• Boolean algebra (postulates and theorems) can be used as
tool for analysis and synthesis of logic systems.
• Other tools used are truth tables, schematic diagrams and
timing diagrams.
• Today, we will learn in this week –how to write Boolean
functions in two standard forms.
• And K map to minimize Boolean Expression

3. Truth tables
x y f
0 0 0
0 1 1
1 0 1
1 1 0
Example of function
dependent on two inputs
x y z F
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 0
Example of function
dependent on four inputs
u v w x F
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0
Example of function
dependent on inputs

4. Canonical/Standard forms: 3 variables
Variables Minterms Maxterms
x y z Term Designation Term Designation
0 0 0 x´y´z´ m0 x+y+z MO
0 0 1 x´y´z m1 x+y+z´ M1
0 1 0 x´yz´ m2 x+y´+z M2
0 1 1 x´yz m3 x+y´+z´ M3
1 0 0 xy´z´ m4 x´+y+z M4
1 0 1 xy´z m5 x´+y+z´ M5
1 1 0 xyz´ m6 x´+y´+z M6
1 1 1 xyz m7 x´+y´+z´ M7
4
Eg : F=m2+m1 +m5

6. Conversion between Canonical forms
X Y Z F Min-
term
Max
-
term
0 0 0 0 M0
0 0 1 1 m1
0 1 0 0 M2
0 1 1 0 M3
1 0 0 1 m4
1 0 1 0 M5
1 1 0 0 M6
1 1 1 1 m7
F= x’y’z+ xy’z’+xyz= m1 + m4 + m7
F= (x+y+z)(x+y’+z)(x+y’+z’)(x’+y+z')(x’+y’+z)
= M0M2M3M5M6
Sum Of Products (SOP) : look for 1’s in output,
OR the minterms
Product Of Sum (POS): look for 0’s in output
AND the maxterms

7. To do
Write the functions in slide 3 in
• (a) SOP form
• (b) POS form
• Ans
(a) SOP form F= x’y+xy’= m1+m2
(b) POS form F=(x+y)(x’+y’) = M0. M3
x y f
0 0 0
0 1 1
1 0 1
1 1 0

8. More Practice
x y z F
0 0 0 0
0 0 1 1
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 1
1 1 0 0
1 1 1 0
u v w x F
0 0 0 0 0
0 0 0 1 1
0 0 1 0 1
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0
SOP form ??? m1+m2+m5
POS form ???
(M0)(M3)(M4)(M6)(M7)
SOP form ???
POS form ???

9. What if some variables are missing?
• F= A+B’C
We can use Boolean algebra and write the function as
F=A(B+B’)(C+C’)+(A+A’)B’C
= ABC+AB’C+ABC’+AB’C’+AB’C’+AB’C+A’B’C
=ABC+AB’C+ABC’+AB’C’+A’B’C
= m7+m5+m6+m4+m1
= 𝑚(1,4,5,6,7)

11. • Canonical SoP form
• Canonical SoP form means Canonical Sum of Products form. In this
form, each product term contains all literals. So, these product terms
are nothing but the min terms. Hence, canonical SoP form is also
called as sum of min terms form.
• Standard SoP form means Standard Sum of Products form. In this
form, each product term need not contain all literals. So, the product
terms may or may not be the min terms. Therefore, the Standard
SoP form is the simplified form of canonical SoP form.
• Similarly for POS forms
• A literal is a single variable within a term which may or may not be
complemented.

13. K Map (Karnaugh Map)
• In many digital circuits and practical problems we need to find
expression with minimum variables.
• We can minimize Boolean expressions of 2, 3, 4 variables very
easily using K-map without using any Boolean algebra
theorems.
• K-map can take two forms Sum of Product (SOP) and Product
of Sum (POS) according to the need of problem. K-map is table
like representation but it gives more information than TRUTH
TABLE

14. Gray Code: binary numeral system such that two successive
values differ in only one bit (binary digit)

15. K Map (Karnaugh Map)

16. 3 variable Kmap
m0 m1 m3
1
m2
1
m4
1
m5
1
m7 m6
yz 00 01 11 10
x
0
1
F(x,y,z)=∑(2,3,4,5)
xy’
x’y
• One square represents one minterm, giving a term with 3 literals.
• Two adjacent squares can be combined to a single term with 2 literals
• Four adjacent squares can be combined to a single terms with 1 literal.
• Eight adjacent squares encompass the whole map and can be
combined to form a function that is always 1.

18. 4 Variable K map

20. Prime Implicants and Essential PI (see
pdf)

22. 4 variable K map (Example)
a= F1 (A,B,C,D) =∑(0,2,3,5,6, 7,8,9) + d ∑(10,11,12,13,14,15)

24. Q1
F= x1’x3’x4+ …+…

26. Implementation of Function in SOP form
Implement the Boolean Function with NAND gates
F=xy’+x’y+z
x
x’
y
y’
z
x’
y’
x
z
y
z
x
y'
x'
y
Stage 1 Stage 2 Stage 3
DeMorgan’s th. 5 b)
(AB)’ =A’+B’

27. Implementation of function POS form
F=(A+B)(C+D)E
A
B
C
D
E
A
C
B
E
D
A
C
D
E
B

29. A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 0
0 1 1 1 0
1 0 0 0 1
1 0 0 1 1
1 0 1 0 1
1 0 1 1 1
1 1 0 0 1
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0