Boolean Algebra and Digital Circuits

1. Boolean Algebra and Digital Circuits Reading: Chapter 8 (138-162) from the text book

2. Defn of a Boolean Algebra A Boolean algebra consists of: • a set B={0, 1}, • 2 binary operations on B(denoted by + & ×), • a unary operation on B(denoted by '), such that : 0 + 0 = 0 0 × 0 = 0 1 + 0 = 1 0 × 1 = 0 0 + 1 = 1 1 × 0 = 0 1 + 1 = 1 1 × 1 = 1 0’=1 and 1’=0.

3. Rules of a Boolean Algebra The following axioms (‘rules’) are satisfied for all elements x, y& zof B: (1) x + y = y + x (commutative axioms) x× y = y × x (2) x + (y + z) = (x + y) + z (associative axioms) x × (y × z) = (x × y) × z (3) x × (y + z) = (x × y) + (x × z) x + (y × z) = (x + y) × (x + z) (distributive axioms) (4) x + 0 = x x× 1 = x (identity axioms) (5) x + x' = 1 x × x' = 0 (inverse axioms)

4. Duality • To form the dual of an expression, replace all + operators with × operators, all × operators with + operators, all 1’s with 0’s, and all 0’s with 1’s. • The principle of duality says that if an expression is valid in Boolean algebra, the dual of that expression is also valid.

5. Duality Exercise: Form the dual of the expression a + (bc) = (a + b)(a + c) Solution: Following the replacement rules… a(b + c) = ab + ac • Take care not to alter the location of the parentheses if they are present.

6. Laws of Boolean Algebra • In addition to the laws given by the axioms of Boolean Algebra, we canshow the following laws x'' = x (double complement) x + x = x x× x = x (idempotent ) (x + y)' = x' × y' (x × y)' = x' + y' (de Morgan’s laws) x + 1 = 1 x × 0 = 0 (annihilation) x + (x × y) = x x× (x + y) = x (absorption) 0' = 1 1' = 0 (complement)

7. Exercise Simplify the Boolean expression (x' × y) + (x × y) Solution: (x' × y) + (x × y) = (y × x') + (y × x) (commutative) = y × (x' + x) (distributive) = y × (x + x') (commutative) = y × 1 (inverse) = y (identity) Thus (x' × y) + (x × y) = y

8. Digital Circuits • The circuitry in a digital computer operates with signals that can take only 2 values ‘on/off’ (i.e. 0/1). • We’ll use the particular Boolean Algebra where Bhas just the 2 elts 0 & 1, and where • Boolean addition corresponds to parallel switch contacts:

9. Boolean Addition

10. Boolean Multiplication • Boolean multiplication corresponds to series switch contacts:

11. Boolean Notation • This means that in effect we’ll be employing Boolean Algebra notation. • The truth tables can be rewritten as

12. Notational Short-cuts We will employ short-cuts in notation: • In ‘multiplication’ we’ll omit the symbol ×, & write xyfor x × y (just as in ordinary algebra) (2) The associative law says that x + (y + z) = (x + y) + z So we’ll write this as simply x + y + z, because the brackets aren’t necessary.

13. Notational Short-cuts Similarly, write the product of 3 terms as xyz (3) In ordinary algebra, the expression x + y × z means x + (y × z),because of the convention that multiplication takes precedence over addition. e.g.x + yz means x + (y × z),and not (x + y) × z Similarly, ab + cd means (a × b) + (c × d)

14. Logic Circuit . . . . Inputs Outputs Digital Circuits • A digital circuit (or logic gate circuit) is an electronic device for carrying out digital computations (e.g. addition of 2 numbers) • It accepts 1 or more inputs, each of which has 2 possible states (0 for ‘off’ & 1 for ‘on’) • For each possible combination of inputs, one or more outputs are produced

15. A Y Symbol Input Output Describing Circuit Functionality: Inverter Truth Table • Truth table completely specifies outputs for all input combinations. • The above circuit is an inverter. • An input of 0 is inverted to a 1. • An input of 1 is inverted to a 0.

16. A Y B The AND Gate • This is an AND gate. • So, if the two inputs signals are asserted (high (ON)) the output will also be asserted (ON). • Otherwise, the output will be asserted (low (OFF)). Truth Table Input 1 input2 output

17. A Y B The OR Gate • This is an OR gate. • So, if at least one of the two input signals is asserted (ON), then output will be asserted (ON). • Otherwise, the output will be asserted (low (OFF)). Input 1 Input 2 Output

18. Consider Three-input Gate 3 Input ORGate

19. x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 F 0 0 0 0 1 0 1 1 x y F = x(y+z’) z y+z’ z’ F = x(y+z’) Boolean Functions • Boolean algebra deals with binary variables and logic operations. • Function results in binary 0 or 1

20. G 0 0 0 1 0 0 1 1 x xy y G = xy +yz z yz Boolean Functions x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 xy 0 0 0 0 0 0 1 1 yz 0 0 0 1 0 0 0 1 We will learn how to transform between expression and truth table.

21. G 0 0 0 1 0 0 1 1 Truth Table to Expression Converting a truth table to an expression • Each row with output of 1becomes a product term • Sumproduct terms together to have the Boolean function. x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 Any Boolean Expression can be represented in sum of products form! xyz + xyz’ + x’yz

22. x x x x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 G 0 0 0 1 0 0 1 1 G x x x x x x x z y G = xyz + xyz’ + x’yz Equivalent Representations of Circuits • Number of 1’s in truth table outputcolumn equals AND terms for Sum-Of-Products (SOP)

23. Reducing Boolean Expressions • Is this the smallest possible implementation of this expression? No! • Use Boolean Algebra rules to reduce complexity while preserving functionality. • Step 1: Use idempotent law (a + a = a). So xyz + xyz’ + x’yz = xyz + xyz + xyz’ + x’yz G = xyz + xyz’ + x’yz

24. Reducing Boolean Expressions • Step 2: Use distributive law a(b + c) = ab + ac. So xyz+xyz+xyz’+x’yz=xy(z + z’) +yz(x + x’) • Step 3: Use Inverse law (a + a’ = 1). So xy(z + z’)+yz(x + x’) = xy.1 +yz.1 • Step 4: Use Identity law (a . 1 = a). So xy +yz=xy.1 +yz.1= xyz + xyz’ + x’yz

25. x 0 0 0 0 1 1 1 1 y 0 0 1 1 0 0 1 1 z 0 1 0 1 0 1 0 1 G 0 0 0 1 0 0 1 1 Reduced Hardware Implementation • Reducedequation requires less hardware! • Same function implemented! x x G x x x y z G = xyz + xyz’ + x’yz = xy + yz

26. y 0 1 x x’y’ x’y 0 x y F 0 0 1 0 1 1 1 0 0 1 1 0 y 1 xy’ xy 0 1 x 1 1 0 1 0 0 Karnaugh maps • Alternate way of representing Boolean function • All rows of truth table represented with a square • Each square represents a minterm

27. Karnaugh maps • Easy to convert between truth table, K-map, and SOP. • Unoptimized form: number of 1’s in K-map equals number of minterms (products) in SOP. • Optimized form: reduced number of minterms F(x,y) = x’y + x’y’ = x’

28. B 0 1 A 0 0 1 F=AB+A’B 1 1 0 BC 00 01 11 10 A 0 0 1 0 1 1 1 1 1 1 Karnaugh Maps • A Karnaugh map is a graphical tool for assisting in the general simplification procedure. • Two variable maps. B 0 1 A 0 0 1 F=AB+AB+AB 1 1 1 A B F C • Three variable maps. 0 0 0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1 + F=AB’C’+ABC+ABC +ABC + A’B’C + A’BC’

29. y 0 1 x 1 1 0 1 0 0 Rules for K-Maps • We can reduce functions by circling1’s in the K-map. • Each circle represents minterm reduction. • Following circling, we can deduce minimized and-or form. F(x,y) = x’y + x’y’ = x’

30. Rules for K-Maps Rules to consider • Every cell containing a 1 must be included at least once. • The largest possible “power of 2rectangle” must be enclosed.

31. B B 0 1 0 1 A A 0 0 1 0 0 1 F=AB+A’B 1 1 0 1 1 1 BC 00 01 11 10 A 0 0 1 0 1 1 1 1 1 1 Karnaugh Maps • A Karnaugh map is a graphical tool for assisting in the general simplification procedure. • Two variable maps. F=AB+AB+AB F=A+B • Three variable maps. F=A+BC+BC F=AB’C’+ABC+ABC +ABC + A’B’C + A’BC’

32. More Karnaugh Map Examples a a Examples 0 1 0 1 b b 0 1 1 0 0 1 1 0 0 1 0 1 g = b' f = a ab ab c 00 01 11 10 c 00 01 11 10 0 0 0 1 0 0 0 1 1 0 0 1 1 1 1 0 0 1 1 1 cout = ab + bc + ac f = a 1. Circle the largest groups possible. 2. Group dimensions must be a power of 2. 3. Remember what circling means!