Lecture 3: Incompletely Specified Functions and K Maps CSE 140: - - PowerPoint PPT Presentation

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Lecture 3: Incompletely Specified Functions and K Maps CSE 140: - - PowerPoint PPT Presentation

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Lecture 3: Incompletely Specified Functions and K Maps 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 1

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Lecture 3: Incompletely Specified Functions and K Maps

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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  • Literals

xi or xi’

  • Product Term x2x1’x0
  • Sum Term

x2 + x1’ + x0

  • Minterm of n variables: A product of n literals in

which every variable appears exactly once.

  • Maxterm of n variables: A sum of n literals in which

every variable appears exactly once.

Definitions

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Id a b f (a, b) 0 0 0 1 1 0 1 0 2 1 0 1 3 1 1 X: don’t care

For example: 1) The input does not happen. 2) The input happens, but the

  • utput is ignored.

Example:

  • Decimal number 0… 9 uses 4 bits. (1,1,1,1) does not happen.
  • Situations where the output of a switching function can be

either 0 or 1 for a particular combination of inputs

  • This is specified by a don’t care in the truth table

Incompletely Specified Function

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Consider a circuit that produces the difference between two inputs only if the second input is less than or equal to the first input. Let’s describe the truth table:

Id a b g(a,b,c) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

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How to completely specify the truth table in canonical form? We have three types of output which divides the input space into three sets: Onset F: All the input conditions for which the output is 1 Offset R: All the input conditions for which the output is 0 Don’t care D: All the input conditions for which the output is a ‘don’t care’ Id a b g(a,b,c) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

Which of the following is needed to express a Boolean function?:

  • A. Any one of F, R, D
  • B. Any two of F, R, D
  • C. All three sets:F, R, D
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Boolean Expressions for onset, offset and Don’t care

F (a, b) = Σm(2)= m(2) R(a, b) = Σm(0,3) D= m(1) Id a b g(a,b,c) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

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Q: Is F (a, b) =R (a, b) for the given truth table?

Id a b g(a,b,c) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

  • A. Yes
  • B. No
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Q: Is F (a, b) =R (a, b) for the given truth table?

Id a b g(a,b,c) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

  • A. Yes
  • B. No because the output

contains don’t cares

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Q: Is F (a, b) =R (a, b) for the given truth table?

Id a b g(a,b,c) 0 0 0 0 1 0 1 1 2 1 0 1 3 1 1 0

  • A. Yes
  • B. No
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Q: Is F (a, b) =R (a, b) for the given truth table?

Id a b g(a,b,c) 0 0 0 0 1 0 1 1 2 1 0 1 3 1 1 0

  • A. Yes
  • B. No
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PI Q: True or False: For a given digital combinational circuit, the value of the output can be one of three states

  • A. True: It can be 0, 1 or don’t care.
  • B. False: It can only be 0 or 1.

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PI Q: True or False: For a given digital combinational circuit, the value of the output can be one of three states

  • A. True: 0, 1 or don’t care
  • B. False: The don’t care is not a separate state,

rather it is used in the truth table to indicate that the output can be either 0 or 1

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Id a b f (a, b) 0 0 0 0 1 0 1 X 2 1 0 1 3 1 1 0

Q: Which of the following assumptions would result in an implementation with the fewest gates?

Reducing Incompletely Specified Functions

  • A. f(0,1) = 1
  • B. f(0,1) = 0
  • C. Neither A or B
  • D. Both A and B
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Id a b f (a, b) 0 0 0 0 1 0 1 0 2 1 0 1 3 1 1 X

Q: Which of the following assumptions would result in an implementation with the fewest gates?

Reducing Incompletely Specified Functions

  • A. f(1,1) = 1
  • B. f(1,1) = 0
  • C. Neither A or B
  • D. Both A and B
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Id a b f (a, b) 0 0 0 0 1 0 1 0 2 1 0 1 3 1 1 X

Don’t care set is important because it allows us to minimize the function

Reducing Incompletely Specified Functions

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Implementation

Specification è Schematic Diagram Net list, Switching expression Obj min cost è Search in solution space (max performance) Cost: wires, gates è Literals, product terms, sum terms We want to minimize # of terms, and # of literals

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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 f(0,0) 1 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

ID A B f(A,B) minterm 1 1 1 A’B 2 1 1 AB’ 3 1 1 1 AB

An example of 2-variable function, f(A,B)

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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

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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 f(0,0) 1 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) 1 1 1 2 1 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) 1 1 1 2 1 1 3 1 1 1

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Two Variable K-maps

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)

# possible 2-variable functions: For 2 variables as inputs, we have 4=22 entries. Each entry can be 0 or 1. Thus we have 16=24 possible functions. f(a,b) a b

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Representation of k-Variable Func.

  • Boolean Expression
  • Truth Table
  • Cube
  • K Map
  • Binary Decision

Diagram

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(0,1,1,1) (0,1,1,0) (0,0,0,0) (0,0,0,1) (1,0,0,1) (1,1,1,1) (1,1,0,1) (1,0,0,0) (0,0,1,0) (1,1,1,0) (0,0,1,1) (1,0,1,1) (0,1,0,1) (1,0,1,0)

A cube of 4 variables: (A,B,C,D)

D C B A

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ID A B C f(A,B,C) f(0,0,0) 1 1 f(0,0,1) 2 1 f(0,1,0) 3 1 1 f(0,1,1) 4 1 f(1,0,0) 5 1 1 f(1,0,1) 6 1 1 f(1,1,0) 7 1 1 1 f(1,1,1)

(A,B)

(0,0) (0,1) (1,0) (1,1) C=0 f(0,0,0) f(0,1,0) f(1,0,0) f(1,1,0) C=1 f(0,0,1) f(0,1,1) f(1,0,1) f(1,1,1)

K Map 1

(A,B)

(0,0) (0,1) (1,1) (1,0) C=0 f(0,0,0) f(0,1,0) f(1,1,0) f(1,0,0) C=1 f(0,0,1) f(0,1,1) f(1,1,1) f(1,0,1)

Q: Which of the following is a valid K’Map representation of the truth table ?

K Map 2 A: K’ Map 1 B: K’ Map 2 C: Both K Maps 1 and 2 D: Neither K Map 1 or 2

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ID A B C f(A,B,C) f(0,0,0) 1 1 f(0,0,1) 2 1 f(0,1,0) 3 1 1 f(0,1,1) 4 1 f(1,0,0) 5 1 1 f(1,0,1) 6 1 1 f(1,1,0) 7 1 1 1 f(1,1,1)

(A,B)

(0,0) (0,1) (1,0) (1,1) C=0 f(0,0,0) f(0,1,0) f(1,0,0) f(1,1,0) C=1 f(0,0,1) f(0,1,1) f(1,0,1) f(1,1,1)

K Map 1

(A,B)

(0,0) (0,1) (1,1) (1,0) C=0 f(0,0,0) f(0,1,0) f(1,1,0) f(1,0,0) C=1 f(0,0,1) f(0,1,1) f(1,1,1) f(1,0,1)

Q: Which of the following is a valid K’map representation of the truth table ?

K Map 2 A: K’ map 1 B: K’ map 2: Although map 1 is a correct mapping of the truth table it is not useful in getting a reduced expression because terms in adjacent cells cannot be combined C: Both K maps 1 and 2 D: Neither K map 1 or 2

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Corresponding three variable K-map

0 2 6 4 1 3 7 5

c = 1

1 1 1 1 0 0 0 0 (0,0) (0,1) (1,1) (1,0)

c = 0 Gray code

Id a b c f (a,b,c) 0 0 0 0 1 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 0

Truth table

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Corresponding three variable K-map

0 2 6 4 1 3 7 5

b = 1 c = 1 a = 1

1 1 1 1 0 0 0 0 (0,0) (0,1) (1,1) (1,0)

c = 0 Gray code f(a,b,c) = c’

Id a b c f (a,b,c) 0 0 0 0 1 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 0 6 1 1 0 1 7 1 1 1 0

Truth table

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Karnaugh Maps (K-Maps)

  • Boolean expressions can be minimized by combining

terms

  • K-maps minimize equations graphically

C 00 01 1 Y 11 10 AB 1 1 C 00 01 1 Y 11 10 AB ABC ABC ABC ABC ABC ABC ABC ABC B C 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 Y

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  • Circle 1’s in adjacent squares
  • Find rectangles which correspond to product

terms in Boolean expression

K-map

C 00 01 1 Y 11 10 AB 1 1

B C 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 Y

y(A,B)=A’B’C’+A’B’C= A’B’(C’+C)=A’B’

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Corresponding K-map

0 2 6 4 1 3 7 5

c = 1

0 1 X 1 0 0 1 1 (0,0) (0,1) (1,1) (1,0)

c = 0

Id a b c f (a,b,c) 0 0 0 0 0 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 X 7 1 1 1 1

Another 3-Input truth table

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Corresponding K-map

0 2 6 4 1 3 7 5

b = 1 c = 1 a = 1

0 1 X 1 0 0 1 1 (0,0) (0,1) (1,1) (1,0)

c = 0 f(a,b,c) = a + bc’

Id a b c f (a,b,c) 0 0 0 0 0 1 0 0 1 0 2 0 1 0 1 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 X 7 1 1 1 1

Another 3-Input truth table

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Corresponding K-map

0 2 6 4 1 3 7 5

c = 1

1 X 0 1 1 0 0 1 (0,0) (0,1) (1,1) (1,0)

c = 0

Id a b c f (a,b,c,d) 0 0 0 0 1 1 0 0 1 1 2 0 1 0 X 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0

3 Input Truth table with Don’t cares PI Q: In the above case the minimal Boolean expression is obtained when the don’t care term is:

  • A. Included i.e. Circled in with one or more minterms that evaluate to 1
  • B. Excluded when grouping the minterms that evaluate to one
  • C. Either approach gives the minimal expression
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Corresponding K-map

0 2 6 4 1 3 7 5

c = 1

1 X 0 1 1 0 0 1 (0,0) (0,1) (1,1) (1,0)

c = 0

Id a b c f (a,b,c,d) 0 0 0 0 1 1 0 0 1 1 2 0 1 0 X 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0

3 Input Truth table with Don’t cares PI Q: In the above case the minimal Boolean expression is obtained when the don’t care term is:

  • A. Included i.e. Circled in with one or more minterms that evaluate to 1
  • B. Excluded when grouping the minterms that evaluate to one
  • C. Either approach gives the minimal expression
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Corresponding K-map

0 2 6 4 1 3 7 5

b = 1 c = 1 a = 1

1 X 0 1 1 0 0 1 (0,0) (0,1) (1,1) (1,0)

c = 0 f(a,b,c) = b’

Id a b c f (a,b,c,d) 0 0 0 0 1 1 0 0 1 1 2 0 1 0 X 3 0 1 1 0 4 1 0 0 1 5 1 0 1 1 6 1 1 0 0 7 1 1 1 0

3 Input Truth table with Don’t cares

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Proof of Consensus Theorem using K Maps

Consensus Theorem: A’B+AC+BC=A’B+AC

C 00 01 1 Y 11 10 AB 1 1 1 1 C 00 01 1 Y 11 10 AB ABC ABC ABC ABC ABC ABC ABC ABC B C 1 1 1 1 A 1 1 1 1 1 1 1 1 1 1 1 1 Y

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Reading

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[Harris] Chapter 2, 2.7