1

Representation of Logic CircuitsRepresentation of Logic Circuits

EE 208 – Logic Design

Chapter 2

Sohaib Majzoub

2

Binary Logic and GatesBinary Logic and Gates

• Binary variables take on one of two values.• Logical operators operate on binary values and

binary variables.• Basic logical operators are the logic functions AND,

OR and NOT.• Logic gates implement logic functions.• Boolean Algebra: a useful mathematical system for

specifying and transforming logic functions.• We study Boolean algebra as a foundation for

designing and analyzing digital systems!

3

Binary VariablesBinary Variables

• Recall that the two binary values have different names:– True/False– On/Off– Yes/No– 1/0

• We use 1 and 0 to denote the two values.

4

Logical OperationsLogical Operations

• The three basic logical operations are:– AND – OR– NOT

• AND is denoted by a dot (·).

• OR is denoted by a plus (+).

• NOT is denoted by an overbar ( ¯ ), or a single quote mark (') after,

5

Chapter 2 - Part 1 5

Logic Diagrams and ExpressionsLogic Diagrams and Expressions

• Boolean equations, truth tables and logic diagrams describe the same function!• Truth tables are unique; expressions and logic diagrams are not. This gives

flexibility in implementing functions.

X

Y F

Z

Logic Diagram

Equation

ZY’ X F

Truth Table

11 1 1

11 1 0

11 0 1

11 0 0

00 1 1

00 1 0

10 0 1

00 0 0

X Y Z Z X F Y’

6

1.

3.

5.

7.

9.

11.

13.

15.

17.

Commutative

Associative

Distributive

DeMorgan’s

2.

4.

6.

8.

X . 1 X=

X . 0 0=

X . X X=

0=X . X’

BooleanBoolean AlgebraAlgebra• An algebraic structure defined on a set of at least two elements,

B, together with three binary operators (denoted +, · and ) that satisfies the following basic identities:

10.

12.

14.

16.

X + Y Y + X=

(X + Y) Z+ X + (Y Z)+=

X(Y + Z) XY XZ+=

(X + Y)’ X’ . Y’=

XY YX=

(XY) Z X(Y Z)=

X + YZ (X + Y) (X + Z)=

(X . Y)’ X’ + Y’=

X + 0 X=

+X 1 1=

X + X X=

1=X + X’

(X’)’ = X

7

Chapter 2 - Part 1 7

Boolean Operator PrecedenceBoolean Operator Precedence

The order of evaluation in a Boolean expression is:

1. Parentheses2. NOT3. AND4. OR

Consequence: Parentheses appear around OR expressions Example: F = A(B + C)(C + D’)

8

Standard Representation of Logic FunctionsStandard Representation of Logic Functions

• Truth Table– List all possible combinations of input and output– Not convenient for large # of variables ( > 5)

Need more convenient alternative.

X Y Z F

0 0 0 00 0 1 10 1 0 00 1 1 01 0 0 11 0 1 01 1 0 01 1 1 1

9Canonical FormsCanonical Forms

• What are Canonical Forms?

• Minterms and Maxterms

• Index Representation of Minterms and Maxterms

• Sum-of-Minterm (SOM) Representations

• Product-of-Maxterm (POM) Representations

• Representation of Complements of Functions

• Conversions between Representations

10Minterms: Sum Of Products (SOP)Minterms: Sum Of Products (SOP)X Y

F1

1000

0 00 11 01 1

F2

0100

F3

0010

F4

0001

F5

0110

MintermX’Y’X’YXY’XY

F6

0111

Minterm: A logical product term that contains ALL variables of a logic expression only once, either complemented (if corresponding logic is 0) or not complemented (if corresponding logic is 1).

Canonical Sum of Products (SOP): A logical sum of all minterms of logic expression for which the output is 1.

m0

m1

m2

m3

F1 = X’Y’ = m0 = X,Y(0)

F2 = X’Y = m1 = X,Y(1)

F3 = XY’ = m2 = X,Y(2)

F4 = XY = m3 = X,Y(3)

F5 = m1 + m2 = X,Y(1,2)

F6 = m1 + m2 + m3 = X,Y(1,2,3)

11Maxterms: Product Of Sums (POS)Maxterms: Product Of Sums (POS)X Y

F1

0111

0 00 11 01 1

F2

1011

F3

1101

F4

1110

F5

0110

MaxtermX+YX+Y’X’+YX’+Y’

F6

0111

Maxterm: A logical sum term that contains ALL variables of a logic expression only once, either complemented (if corresponding logic is 1) or not complemented (if corresponding logic is 0).

Canonical Products of Sums (POS): A logical product of all maxterms of logic expression for which the output is 0.

M0

M1

M2

M3

F1 = X+Y = M0 = X,Y(0)

F2 = X+Y’ = M1 = X,Y(1)

F3 = X’+Y = M2 = X,Y(2)

F4 = X’+Y’ = M3 = X,Y(3)

F5 = M0 . M3 = X,Y(0,3)

F6 = M0 = X,Y(0)

12

• Yes!

F5 = m1 + m2 = X,Y(1,2) = X’Y + XY’

= M0.M3 = X,Y(0,3) = (X+Y).(X’+Y’) = XX’ + XY’ + X’Y + YY ’ = X’Y + XY’

SOP = POSSOP is NOT POS’

Are SOP and POS Equivalent?Are SOP and POS Equivalent?

X Y 0 00 11 01 1

F5

0110

13

x y z index m1 + m4 + m7 = F1

0 0 0 0 0 + 0 + 0 = 0

0 0 1 1 1 + 0 + 0 = 1

0 1 0 2 0 + 0 + 0 = 0

0 1 1 3 0 + 0 + 0 = 0

1 0 0 4 0 + 1 + 0 = 1

1 0 1 5 0 + 0 + 0 = 0

1 1 0 6 0 + 0 + 0 = 0

1 1 1 7 0 + 0 + 1 = 1

Minterm Function ExampleMinterm Function Example

• Example: Find F1 = m1 + m4 + m7

• F1 = x’ y’ z + x y’ z’ + x y z

14

Minterm Function ExampleMinterm Function Example

• F(A, B, C, D, E) = m2 + m9 + m17 + m23

• F(A, B, C, D, E) =

15Maxterm Function ExampleMaxterm Function Example

• Example: Implement F1 in Maxterms: F1 = M0 · M2 · M3 · M5 · M6

F1 = (x+y+z).(x+y’+z).(x+y’+z’).(x’+y+z’).(x’+y’+z)

x y z i M0 M2 M3 M5 M6 = F1 0 0 0 0 0 1 1 1 = 0 0 0 1 1 1 1 1 1 1 = 1 0 1 0 2 1 0 1 1 1 = 0 0 1 1 3 1 1 0 1 1 = 0 1 0 0 4 1 1 1 1 1 = 1 1 0 1 5 1 1 1 0 1 = 0 1 1 0 6 1 1 1 1 0 = 0 1 1 1 7 1

1 1 1 1 = 1

1

16

Maxterm Function ExampleMaxterm Function Example

• • F(A, B,C,D) =

14 11 8 3 M M MM)D,C,B,A(F

17Canonical Sum of MintermsCanonical Sum of Minterms

• Any Boolean function can be expressed as a Sum of Minterms.– For the function table, the minterms used are the

terms corresponding to the 1's– For expressions, expand all terms first to explicitly

list all minterms. Do this by “ANDing” any term missing a variable v with a term (v+v’).

• Example: Implement f = x + x’.y’ as a sum of minterms.

First expand terms: f = x(y+y’)+x’y’Then distribute terms: f = xy +xy’ +x’y’

Express as sum of minterms: f = m3 + m2 + m0

18Another SOM ExampleAnother SOM Example

• Example: F = A + B’ . C• There are three variables, A, B, and C which we

take to be the standard order.• Expanding the terms with missing variables:

F = A(B + B’)(C + C’) + (A + A’) B’ C

= ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C

= ABC + ABC’ + AB’C + AB’C’ + A’B’C• Collect terms (removing all but one of duplicate

terms): • Express as SOM:

F = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7

19Shorthand SOM FormShorthand SOM Form

• From the previous example, we started with:

F = A + B’ . C• We ended up with:

F = m1+m4+m5+m6+m7

• This can be denoted in the formal shorthand:

• Note that we explicitly show the standard variables in order and drop the “m” designators.

)7,6,5,4,1()C,B,A(F m

20

Canonical Product of MaxtermsCanonical Product of Maxterms• Any Boolean Function can be expressed as a

Product of Maxterms (POM).– For the function table, the maxterms used are the terms

corresponding to the 0's.– For an expression, expand all terms first to explicitly list all

maxterms. Do this by first applying the second distributive law , “ORing” terms missing variable v with a term equal to (v.v’) and then applying the distributive law again.

• Example: Convert to product of maxterms:f(x,y,z) = x + x’.y’

Apply the distributive law: x + x’.y’ = (x+x’).(x+y’)= 1.(x+y’) = x+y’Add missing variable z: x+y’ +z.z’ =(x+y’+z).(x+y’+z’)

Express as POM: f = M2 · M3

21

• Convert to Product of Maxterms:f(A,B,C) = A.C’+B.C+A’.B’

• Use x + y z = (x+y)·(x+z) with x= (A.C’ + B.C), y=A’ and z=B’ to get:

f = (A.C’+B.C+A’)(A.C’+B.C+B’)• Then use x + x’.y = x + y to get:

f = (C’+B.C+A’).(A.C’+C+B’)and a second time to get:

f = (C’+B+A’).(A+C+B’)• Rearrange to standard order:

f = (A’+B+C’).(A+B’+C) to give f = M5 · M2

Another POM ExampleAnother POM Example

22

Function ComplementsFunction Complements

• The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical forms.

• Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices.

• Example: Given

F(x,y,z) m(1,3,5,7)

F '(x,y,z) m(0,2,4,6)

F '(x,y,z) M (1,3,5,7)

23ConversionConversion BetweenBetween FormsForms• To convert between sum-of-minterms and product-

of-maxterms form (or vice-versa) we follow these steps:– Find the function complement by swapping terms in the

list with terms not in the list.– Change from products to sums, or vice versa.

• Example: Given F as before:• Form the Complement: • Then use the other form with the same indices –

this forms the complement again, giving the other form of the original function:

)7,5,3,1()z,y,x(F m

F '(x,y,z) m(0,2,4,6)

)6,4,2,0()z,y,x(F M

24

Chapter 2 - Part 1 24

• Standard Sum-of-Products (SOP) form: equations are written as an OR of AND terms

• Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms

• Examples:– SOP: ABC+A’B’C+B– POS: (A+B).(A+B’+C’).C

• These “mixed” forms are neither SOP nor POS– (A.B+C).(A+C)– A.B.C’+A.C.(A+B)

StandardStandard FormsForms

25

Combinational Circuit AnalysisCombinational Circuit Analysis

• Determine the output for certain input

• Algebraically manipulate logic expressions

• Transform algebraic expression into canonical form (SOP or POS)

26Simplification of CircuitsSimplification of Circuits

• Example, Simplify the following function

F = [((W.X’)’.Y)’+(W’+X+Y’)’+(W+Z)’]’

(a)Using algebra

(b)Using Circuits

(c) More and better techniques in coming lectures

27Analysis ExampleAnalysis Example

XYZ

F

Truth Table X Y Z F

0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1

28Analysis ExampleAnalysis Example

XYZ

F

Truth Table 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 1

29Analysis Example (cntd.)Analysis Example (cntd.)

XYZ

F

Find an expression for F

F =

30Analysis Example (cntd.)Analysis Example (cntd.)

XYZ

F

Find an expression for F

F = (X + Y’).Z + X’ . Y . Z’

31Analysis Example (cntd.)Analysis Example (cntd.)

Find canonical SOP of F (need to expand)

F = (X + Y’).Z + X’ . Y . Z’

F = X .Z + Y’.Z + X’ . Y . Z’ (SOP)

F = X .Y.Z+ X .Y’.Z+ X.Y’.Z +X’.Y’.Z + X’.Y. Z’

F = X .Y.Z+ X .Y’.Z+ X’.Y’.Z +X’.Y.Z’

(Canonical SOP)

F = X,Y,Z(1,2,5,7) (Canonical SOP)

2-level AND-OR Logic

32Analysis Example (cntd.)Analysis Example (cntd.)

Canonical of SOP, F = X .Y.Z+ X .Y’.Z+ X’.Y’.Z + X’.Y. Z’

X’ YZ’

F

X’ Y’Z

X YZ

X Y’Z

X’ YZ’

F

Y’

Z

X

Z

SOP, 2-levels AND-OR LogicF = X .Z+ Y’.Z + X’.Y. Z’

Canonical of SOP is more expensive to represent a boolean function, more gates and more inputs

33Analysis Example (cntd.)Analysis Example (cntd.)

Find canonical POS of F

F = (X + Y’).Z + X’ . Y . Z’

Remember DL:

a.b+c.d.e=(a+c)(a+d)(a+e)(b+c)(b+d)(b+e)?

a = (X+Y’); b = Z; c = X’, d=Z’; e = Y;

F= (X+Y’+X’).(X+Y’+Y).(X+Y’+Z’).(Z+X’).(Z+Y). (Z+Z’)

F =1.1.(X+Y’+Z’).(Z+X’).(Z+Y).1

F =(X+Y’+Z’).(Z+X’).(Z+Y) (POS, 2-level OR-AND logic) not POS canonical yet

34Analysis Example (cntd.)Analysis Example (cntd.)

F =(X+Y’+Z’).(X’+Z).(Y+Z)

F =(X+Y’+Z’).(X’+Y.Y’+Z).(X.X’+Y+Z)

Again DL: a = X’+Z and b.c= Y.Y’

a + bc = (a+b)(a+c)

F =(X+Y’+Z’).(X’+Y+Z).(X’+Y’+Z).(X+Y+Z) .(X’+Y+Z)

F =(X’+Y’+Z) .(X’+Y+Z). (X+Y’+Z’). (X+Y+Z)

F =X,Y,Z(0,3,4,6)

Canonical POS, 2-level OR-AND logic

35Design ExampleDesign Example

• Alarm System: the alarm will sound-off if:

- The panic button is pushed

- OR alarm is enabled AND someone broke-in

- Alarm locks are on door, window, & garage.

Output : Alarm Sound

Input: Panic Button (P), enable (E), Secure (S), where S = W.D.G (window, door, and garage)

Assume that:

S is active-low ( 0 if activated, 1 if deactivated)

P is active-high (1 if activated, 0 if deactivated)

36Design ExampleDesign Example

A = P + E.S’

S = W.G.D

E D’

AEG’

E W’

P

37NAND GatesNAND Gates

• Use less transistors, cheaper to implement, less power consumption etc.

• How to convert a logic circuit to NAND-NAND circuit:

- Find 2-level AND-OR (SOP) implementation

- Replace all gates by NAND gates

- convert single literal inputs to 2nd NAND level gate

38NAND-NAND circuitsNAND-NAND circuits

• Example

E D’

AEG’

E W’

P

E D’

AEG’

E W’

P

E D’

AEG’

E W’

P

39NOR GatesNOR Gates

• To convert a logic circuit to NOR-NOR circuit:

- Find 2-level OR-AND (POS) implementation

- Replace all gates by NOR gates

- Invert single literal inputs to 2nd level NOR gate