Lecture 4: Simplification using Boolean Algebra K Maps ...€¦ · Lecture 4: Simplification using...

Lecture 4: Simplification using Boolean Algebra, K Maps,

Incompletely Specified Functions

CSE 140: Components and Design Techniques for Digital Systems

Diba Mirza

Dept. of Computer Science and Engineering University of California, San Diego

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What you should know at the end of this lecture….

1.  How to simplify combinational circuits using Boolean Algebra axioms and theorems

2.  How to simplify combinational circuits using Karnaugh maps

3.  Why Karnaugh maps are preferred in simplifying circuits

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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) = (x+f(0)). (x’+f(1))

–  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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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: Example

f(x,y,z)=xf(0,y,z) + x’f(1,y,z) Is the above equation correct? A. Yes B. No

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Shannon’s Expansion Decompose the switching function into min terms f(x,y)=xf(1,y)+x’f(0,y)

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Shannon’s Expansion Decompose the switching function into max terms f(x,y)=(x+f(0,y)). (x’+ f(1,y))

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

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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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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 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

Building the simplest possible circuit

We can get the simplest circuit in two ways: 1. Truth table –> Canonical POS/SOP –> Most simplified switching function -> Simplest Circuit

•  This can get difficult when we have many inputs variables

2. Truth table or Canonical POS/SOP ->Karnaugh map-> Simplest Circuit

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Implementation: Specification => Logic Diagram

Flow 1: 1.  Specification 2.  Truth Table 3.  Sum of Products or Product of Sums

canonical form 4.  Reduced expression using Boolean

Algebra 5.  Schematic Diagram of Two Level Logic

Flow 2: 1. Specification 2. Truth Table 3. Karnaugh Map (truth table in two dimensional space) 4. Reduce using K’Maps 5. Reduced expression (SOP or POS) 6. Schematic Diagram of Two Level Logic

Karnaugh Map: A 2-dimensional truth table

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Truth Table vs. Karnaugh Map

ID A B f(A,B)

0 0 0 f(0,0)

1 0 1 f(0,1)

2 1 0 f(1,0)

3 1 1 f(1,1)

2-variable function, f(A,B)

A=0 A=1

B=0 f(0,0) f(1,0)

B=1 f(0,1) f(1,1)

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Truth Table vs. Karnaugh Map

ID A B f(A,B)

0 0 0 0

1 0 1 1

2 1 0 1

3 1 1 1

2-variable function, f(A,B)

A=0 A=1

B=0 0 1

B=1 1 1

Boolean expr:

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Function can be represented by sum of minterms: f(A,B) = A’B+AB’+AB

This is not optimal however! We want to minimize the number of literals and terms.

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To minimize the number of literals and terms. We factor out common terms – A’B+AB’+AB= A’B+AB’+AB+AB =(A’+A)B+A(B’+B)=B+A Hence, we have

f(A,B) = A+B

How can we guarantee the most reduced expression was reached?

•  Boolean expressions can be minimized by combining terms •  K-maps minimize equations graphically

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ID A B f(A,B)

0 0 0 f(0,0)

1 0 1 f(0,1)

2 1 0 f(1,0)

3 1 1 f(1,1)

A=0 A=1

B=0 A’B’ AB’

B=1 A’B AB

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K-Map: Truth Table in 2 Dimensions

A = 0 A = 1

B = 0 B = 1

0 2

1 3

0 1 1 1

ID A B f(A,B)

0 0 0 0

1 0 1 1

2 1 0 1

3 1 1 1

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K-Map: Truth Table in 2 Dimensions

A = 0 A = 1

B = 0 B = 1

0 2

1 3

0 1 1 1

A’B

AB’

AB

f(A,B) = A + B

ID A B f(A,B)

0 0 0 0

1 0 1 1

2 1 0 1

3 1 1 1

Review 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) 27

Next Lecture…

•  Quiz (multiple choice based on material covered so far)

•  Bring your clickers for the quiz, paper submissions will not be accepted

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