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Lecture 2: Combinational Logic

CSE 140: Components and Design Techniques for Digital Systems

Spring 2014

CK Cheng, Diba Mirza Dept. of Computer Science and Engineering

University of California, San Diego

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Outline •  What is a combinational circuit? •  Combinational Logic

1.  Scope 2.  Specification : Boolean algebra, truth tables 3.  Synthesis: circuits

PI Q: What is a combinational circuit?

A. A circuit whose output depends only on current inputs

B. A circuit that has memory of its past states C. A circuit that contains any logic gate

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What is a combinational circuit?

•  Output depends only on current inputs •  No memory •  Example: Circuit that adds to binary numbers

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Digital Circuits Operate on Binary Values

•  Typically consider only two discrete values: 1’s and 0’s –  1, TRUE, HIGH –  0, FALSE, LOW

•  1 and 0 can be represented by anything (typically voltage) 5

Basic Combinational Logic: Gates

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A

B Y=

Basic Combinational Logic: Gates

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A

B

Y= AB

A B Y 0 0 0 0 1 0 1 0 0 1 1 1

AND

Truth Table Equivalence between: – Symbol (Logic circuit) – Truth table – Boolean Equation

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Two-input operator AND (˄, · )

A B Y 0 0 0 0 1 0 1 0 0 1 1 1

AND

A 1 A

A B Y 0 0 0 0 1 1 1 0 1 1 1 1

OR

A 1 1

A 0 0

A 0 A

A Y 0 1 1 0

NOT

Boolean algebra and switching functions:

Operators and Digital Logic Gates

0 dominates in AND 0 blocks the output 1 passes signal A

1 dominates in OR 1 blocks the output 0 passes signal A

Two-input operator OR (˅, + )

One-input operator NOT (Complement, ` )

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Starting with basic gates we want to build more complex logic circuits

Problem Specification:

a b c d e

Synthesis

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Combinational Logic: Scope •  Description

–  Language: e.g. C Programming, Verilog, VHDL –  Boolean algebra –  Truth table

•  Design –  Schematic Diagram –  Inputs, Gates, Nets, Outputs

•  Goal –  Validity: correctness, turnaround time –  Performance: power, timing, cost –  Testability: yield, diagnosis, robustness

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•  Introduced binary variables •  Introduced the three fundamental

logic operations: AND, OR, and NOT.

•  Born to working class parents •  Taught himself mathematics and

joined the faculty of Queen’s College in Ireland.

•  Wrote An Investigation of the Laws of Thought (1854)

Boolean Algebra: George Boole

George Boole: 1815 - 1864

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Boolean Algebra Similar to regular algebra but defined on on a set (B) with only three basic ‘logical’ operations: 1.  Intersection: AND (2-input) , Operator: . 2.  Union: OR (2 input), Operator: + 3.  Complement: NOT ( 1 input) Operator:

These operations satisfy the following laws: •  Commutative laws: a+b=b+a, a·b=b·a •  Distributive laws: a+(b·c)=(a+b)·(a+c), a·(b+c)=a·b+a·c •  Identity laws: a+0=a, a·1=a •  Complement laws: a+a’=1, a·a’=0

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Switching Algebra •  Boolean Algebra: multiple-valued logic, i.e. each variable

have multiple values. •  Switching Algebra: binary logic, i.e. each variable can be

either 1 or 0. •  Boolean Algebra ≠ Switching Algebra

BB

Boolean Algebra

Switching Algebra

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Starting with basic gates we want to build more complex logic circuits

Problem Specification: Words Truth table Boolean (switching) expression Reduced Boolean expression

a b c d e

Synthesis

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Example: Half Adder

a b carry sum

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

Truth Table a

b

Sum

Carry

Specifying Logic Problems: Truth tables

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a b carry sum

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

Truth Table

How do we go from Truth table to a Boolean Equation?

Sum (a, b) = a’b + ab’ Carry (a, b) = ab

Switching Expressions or Boolean Equation:

Switching Function

Need Some Definitions •  Complement: variable with a bar over it A, B, C •  Literal: variable or its complement A, A, B, B, C, C •  Implicant: product of literals ABC, AC, BC •  Implicate: sum of literals (A+B+C), (A+C), (B+C) •  Minterm: product that includes all input variables ABC, ABC, ABC •  Maxterm: sum that includes all input variables (A+B+C), (A+B+C), (A+B+C) 17

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Logic circuit vs. Boolean Algebra Expression

Schematic (circuit) Diagram: 5 primary inputs 4 components 9 signal nets 12 pins

ab + cd a b

c d

e

cd

ab

y=e (ab+cd)

Boolean Algebra: 5 literals 4 operators

Obj: min #terms min #literals

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A B carry sum Min term Max term

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

Specifying input combinations as Boolean expressions

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A B carry sum Min term Max term

0 0 0 0 A’B’ A+B 0 1 0 1 A’B A+B’ 1 0 0 1 AB’ A’+B 1 1 1 0 AB A’+B’

Sum of Product Canonical Form

I.  sum(A,B) = II.  carry(A,B)=

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PI Q: Is the expression for sum obtained in the previous slide expected and correct?

A.  No, it is not expected and it is not correct. It should have been S= A+B

B.  It might not be expected but it is correct

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A B carry sum Min term Max term

0 0 0 0 m0 M0

0 1 0 1 m1 M1 1 0 0 1 m2 M2 1 1 1 0 m3 M3

Sum of Product Canonical Form

I.  sum(A,B) =

II.  carry(A,B)=

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A B carry sum Min term Max term

0 0 0 0 A’B’ A+B 0 1 0 1 A’B A+B’ 1 0 0 1 AB’ A’+B 1 1 1 0 AB A’+B’

Product of Sum Canonical Form

I.  sum(A,B) =

II.  carry(A,B)=

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A B carry sum Min term Max term

0 0 0 0 m0 M0

0 1 0 1 m1 M1 1 0 0 1 m2 M2 1 1 1 0 m3 M3

Product of Sum Canonical Form

I.  sum(A,B) = II.  carry(A,B)=

PI Q: The SOP form seems to be more compact than the POS form. Is there a situation where we would prefer the latter over the former to express the switching function?

A.  Yes, when the output is 1 more often than 0 B.  Yes, when the output is 0 more often than 1 C.  No, we always prefer the SOP form because its

more compact

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PI Q: How can we check if our switching expression in POS and SOP canonical forms is correct?

A.  We can’t – just listen to the professor and follow B.  Regenerate the truth table from the switching

expression C.  Derive one canonical form from the other using

Boolean Algebra

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Re-deriving the truth table

Switching Expressions: Sum (A,B) = A’B + AB’ Carry (A, B) = AB

Ex:

Sum (0,0) = 0’.0 + 0.0’ = 0 + 0 = 0 Sum (0,1) = 0’1 + 0.1’ = 1 + 0 = 1 Sum (1,1) = 1’1 + 1.1’ = 0 + 0 = 0

a

b

sum a b

carry

Logic circuit for half adder:

A B carry sum

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

Truth Table

The SOP and POS forms don’t usually give the optimal circuit or the simplest Boolean expression of the switching function

To optimize the circuit, simplify the Boolean expression using: 1.  Boolean Algebra axioms and theorems 2.  Karnaugh Maps (next lecture)

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1.  B = 0 if B not equal to 1 2.  0’ = 1 3.  1.1 = 1 4.  0.1 = 0 5.  a+0=a, a.1=a Identity law 6.  a+a’=1, a.a’=0 Complement law

Axioms of Boolean Algebra

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I.  Commutative Law: A + B = B +A AB = BA

II. Distributive Law A(B+C) = AB + AC

A+BC = (A+B)(A+C)

A B C

A C

A B

A B C

A C

A B

Theorems of Boolean Algebra

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III Associativity (A+B) + C = A + (B+C) (AB)C = A(BC)

C A B

A B C

C A B

A B C

IV: Consensus Theorem: AB+B’C+AC = AB+B’C

PI Q: Which term in AB’+AC+BC can be deleted? A.  AB’ B.  AC C.  BC D.  None of the above

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Proof of consensus Theorem using Boolean Algebra

AB+B’C+AC = AB+B’C

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V. DeMorgan’s Theorem

•  Y = AB = A + B

•  Y = A + B = A B

AB Y

AB Y

AB Y

AB Y

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Bubble Pushing •  Pushing bubbles backward (from the output) or forward

(from the inputs) changes the body of the gate from AND to OR or vice versa.

•  Pushing a bubble from the output back to the inputs puts bubbles on all gate inputs.

•  Pushing bubbles on all gate inputs forward toward the output puts a bubble on the output and changes the gate body.

AB Y A

B Y

AB YA

B Y

Example of transforming circuits using bubble pushing

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Shannon’s Expansion

•  Shannon’s expansion assumes a switching algebra system

•  Divide a switching function into smaller functions •  Pick a variable x, partition the switching function

into two cases: x=1 and x=0 –  f(x,y,z,…)= xf(x=1,y,z,…) + x’f(x=0,y,z,…)

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Shannon’s expansion

•  Shannon’s expansion: –  f(x)=xf(1)+x’f(0)

–  f(x,y)=xf(1,y)+x’f(0,y) –  f(x,y,z,…)=xf(1,y,z,…)+x’f(0,y,z,…)

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Shannon’s Expansion: Example

PIQ: f(x,y,z)=xf(?,y,z)+x’f(?,y,z) A.  ?=0 B.  ?=1

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Shannon’s Expansion: Example

PI Q f(x,y)=(x+f(?,y))(x’+f(?,y)) A. ?=0 B. ?=1.

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Proof of Shannon’s Expansion

f(x,y)=(x+f(0,y))(x’+f(1,y)) {Enumerative induction}

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Shannon’s Expansion

Reduction of Boolean Expression •  AB+AC+B’C =AB+B’C

•  (A+B)(A+C)(B’+C) =(A+B)(B’+C)

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To prove the reduction using (1)  Boolean algebra, (2)  Logic simulation and (3)  Shannon’s expansion (exercise)

Summary: Switching Algebra and Karnaugh Map

•  Shannon’s expansion and consensus theorem are used for logic optimization

•  Shannon’s expansion divides the problem into smaller functions

•  Consensus theorem finds common terms when we merge small functions

•  Karnaugh map mimics the above two operations in two dimensional space as a visual aid.

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We are in a position to build a circuit to do n-bit Binary Addition

5 + 7

1 2 Carry Sum

1 1 1 1 0 1 + 1 1 1

1 1 0 0

Carryout Sums

Carry bits 5 7 12

Binary Addition: Hardware

•  Half Adder: Two inputs (a,b) and two outputs (carry, sum).

•  Full Adder: Three inputs (a,b,c) and two outputs (carry, sum).

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Full Adder

Id a b cin carry sum

0 0 0 0 0 0 1 0 0 1 0 1 2 0 1 0 0 1 3 0 1 1 1 0 4 1 0 0 0 1 5 1 0 1 1 0 6 1 1 0 1 0 7 1 1 1 1 1

Truth Table

a

b

Sum

Carry

Cin

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Minterm and Maxterm

Id a b cin carryout

0 0 0 0 0 a+b+c 1 0 0 1 0 a+b+c’ 2 0 1 0 0 a+b’+c 3 0 1 1 1 a’ b c 4 1 0 0 0 a’+b+c 5 1 0 1 1 a b’c 6 1 1 0 1 a b c’ 7 1 1 1 1 a b c minterm

maxterm

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f1(a,b,c) = a’bc + ab’c + abc’ + abc f2(a,b,c) = (a+b+c)(a+b+c’)(a+b’+c)(a’+b+c)

f1(a, b, c) = m3 + m5 + m6 + m7 = Sm(3,5,6,7) f2(a, b, c) = M0M1M2M4 = PM(0, 1, 2, 4)

PI Q: Is f1 = f2? A. Yes B. No

PIQ: Reduce using Boolean algebra theorems Carry(A,B,C)=A’BC+AB’C+ABC’+ABC

A. A’BC+AB’C+AB B. A’BC+AC+AB C. AB+AC+CA D. ABC E.  Cannot be reduced further

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A’BC+AB’C+ABC’+ABC

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f1(a,b,c) = a’bc + ab’c + abc’ + abc = ab +c(b +a)

Circuit for Full Adder Carry out

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Next Lecture Reading

[Harris] Chapter 2, Section 2.5, 2.7