1. IKI10201 03a-Boolean Algebra Bobby Nazief Semester-I 2005 - 2006 The materials on these slides are adopted from those in CS231’s Lecture Notes at UIUC, which is derived from Howard Huang’s work and developed by Jeff Carlyle.

2. 2 Road Map Boolean Algebra Logic Gates & Flip-flops Register-Transfer Design Finite-State Machines Binary Systems & Data Represent. Generalized FSM Sequential Design Techniques Logic Design Techniques Combinatorial Components Storage Components Processor Components 2 3 3 4 5 6 7 6 8 8 9

3. 3 Boolean Algebra

4. 4 Boolean values Earlier, we used electrical voltages to represent two discrete values 1 and 0, from which binary numbers can be formed. It’s also possible to think of voltages as representing two logical values, true and false. For simplicity, we often still write digits instead: – 1 (high) is true – 0 (low) is false We will use these values 1 and 0 as the elements of our Boolean System.

5. 5 Basic boolean operations There are two basic operations for logical values. AND (product) of two inputs OR (sum) of two inputs xy, or x  yx + y Operation: Expression: Truth table:

6. 6 Axiomatic definition of Boolean Algebra A Boolean algebra requires – A set of elements B, consisting of two elements (0 and 1) – Two binary operations OR and AND – The axioms below must always be true – Based on axiom #5, we can develop a unary (one-argument) operation NOT

7. 7 Basic theorems of Boolean Algebra In addition to the axioms, additional laws can be derived; they are called theorems of Boolean Algebra These theorems are useful in performing algebraic manipulations of Boolean expressions

8. 8 Proofing the theorems using axioms Idempotency: x + x = x Proof: x + x = (x + x)  1by identity = (x + x)  (x + x’)by complement = x + x  x’by distributivity = x + 0by complement = xby identity Idempotency: x  x = x Proof: x  x = (x  x) + 0by identity = (x  x) + (x  x’)by complement = x  (x + x’)by distributivity = x  1by complement = xby identity

9. 9 Duality principle The left and right columns of axioms are duals – exchange all ANDs with ORs, and 0s with 1s So are the theorems:

10. 10 Boolean Functions Computers take inputs and produce outputs, just like functions in math! Mathematical functions can be expressed in two ways: We can represent logical functions in two analogous ways too: – A finite, but non-unique Boolean expression. – A truth table, which will turn out to be unique and finite. f(x,y)= 2x + y = x + x + y = 2(x + y/2) =... An expression is finite but not unique A function table is unique but infinite

11. 11 Boolean expressions We can use these basic operations to form more complex expressions: f(x,y,z) = (x + y’)z + x’ Some terminology and notation: – f is the name of the function. – (x,y,z) are the input variables, each representing 1 or 0. Listing the inputs is optional, but sometimes helpful. – A literal is any occurrence of an input variable or its complement. The function above has four literals: x, y’, z, and x’. Precedence is important, but not too difficult. – NOT has the highest precedence, followed by AND, and then OR. – Fully parenthesized, the function above would be kind of messy: f(x,y,z) = (((x +(y’))z) + x’)

12. 12 Truth tables A truth table shows all possible inputs and outputs of a function. Remember that each input variable represents either 1 or 0. – Because there are only a finite number of values (1 and 0), truth tables themselves are finite. – A function with n variables has 2 n possible combinations of inputs. Inputs are listed in binary order—in this example, from 000 to 111. f(0,0,0)= (0 + 1)0 + 1= 1 f(0,0,1)= (0 + 1)1 + 1= 1 f(0,1,0)= (0 + 0)0 + 1= 1 f(0,1,1)= (0 + 0)1 + 1= 1 f(1,0,0)= (1 + 1)0 + 0= 0 f(1,0,1)= (1 + 1)1 + 0= 1 f(1,1,0)= (1 + 0)0 + 0= 0 f(1,1,1)= (1 + 0)1 + 0= 1 f(x,y,z) = (x + y’)z + x’

13. 13 Complement of a function The complement of a function always outputs 0 where the original function outputted 1, and 1 where the original produced 0. In a truth table, we can just exchange 0s and 1s in the output column(s) f(x,y,z)= x(y’z’ + yz)

14. 14 Complementing a function algebraically You can use DeMorgan’s law to keep “pushing” the complements inwards You can also take the dual of the function, and then complement each literal – If f(x,y,z) = x(y’z’ + yz)… – …the dual of f is x + (y’ + z’)(y + z)… – …then complementing each literal gives x’ + (y + z)(y’ + z’)… – …so f’(x,y,z) = x’ + (y + z)(y’ + z’) f(x,y,z)= x(y’z’ + yz) f’(x,y,z)= ( x(y’z’ + yz) )’[ complement both sides ] = x’ + (y’z’ + yz)’[ because (xy)’ = x’ + y’ ] = x’ + (y’z’)’ (yz)’[ because (x + y)’ = x’ y’ ] = x’ + (y + z)(y’ + z’)[ because (xy)’ = x’ + y’, twice]

15. 15 Algebraic manipulation We can now start doing some simplifications x’y’ + xyz + x’y = x’(y’ + y) + xyz[ Distributive; x’y’ + x’y = x’(y’ + y) ] = x’  1 + xyz[ Axiom 5; y’ + y = 1 ] = x’ + xyz[ Axiom 2; x’  1 = x’ ] = (x’ + x)(x’ + yz)[ Distributive ] = 1  (x’ + yz)[ Axiom 5; x’ + x = 1 ] = x’ + yz[ Axiom 2 ; x’  1 = x’]

16. 16 Canonical Forms Any boolean function that is expressed as a sum of minterms or as a product of maxterms is said to be in its canonical form.

17. 17 Minterms A minterm is a special product of literals, in which each input variable appears exactly once. A function with n variables has 2 n minterms (since each variable can appear complemented or not) A three-variable function, such as f(x,y,z), has 2 3 = 8 minterms: Each minterm is true for exactly one combination of inputs: x’y’z’x’y’zx’yz’x’yz xy’z’xy’zxyz’xyz MintermIs true when…Shorthand x’y’z’x=0, y=0, z=0m 0 x’y’zx=0, y=0, z=1m 1 x’yz’x=0, y=1, z=0m 2 x’yzx=0, y=1, z=1m 3 xy’z’x=1, y=0, z=0m 4 xy’zx=1, y=0, z=1m 5 xyz’x=1, y=1, z=0m 6 xyzx=1, y=1, z=1m 7

18. 18 Sum of minterms form Every function can be written as a sum of minterms, which is a special kind of sum of products form The sum of minterms form for any function is unique If you have a truth table for a function, you can write a sum of minterms expression just by picking out the rows of the table where the function output is 1 (1-minterm). f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’ = m 0 + m 1 + m 2 + m 3 + m 6 =  (0,1,2,3,6) f’ = xy’z’ + xy’z + xyz = m 4 + m 5 + m 7 =  (4,5,7) f’ contains all the minterms not in f

19. 19 Sum of minterms: practise F = x + yz, how to express this in the sum of minterms? = x(y + y’)(z + z’) + (x + x’)yz = xyz + xyz’ + xy’z + xy’z’ + xyz + x’yz = x’yz + xy’z’ + xy’z + xyz’ + xyz =  (3,4,5,6,7) or, convert the expression into truth-table and then read the minterms from the table

20. 20 Maxterms A maxterm is a sum of literals, in which each input variable appears exactly once. A function with n variables has 2 n maxterms The maxterms for a three-variable function f(x,y,z): Each maxterm is false for exactly one combination of inputs: x’ + y’ + z’x’ + y’ + zx’ + y + z’x’+ y + z x + y’ + z’x + y’ + zx + y + z’x + y + z MaxtermIs false when…Shorthand x + y + zx=0, y=0, z=0M 0 x + y + z’x=0, y=0, z=1M 1 x + y’ + zx=0, y=1, z=0M 2 x + y’ + z’x=0, y=1, z=1M 3 x’ + y + zx=1, y=0, z=0M 4 x’ + y + z’x=1, y=0, z=1M 5 x’ + y’ + zx=1, y=1, z=0M 6 x’ + y’ + z’x=1, y=1, z=1M 7

21. 21 Product of maxterms form Every function can be written as a unique product of maxterms If you have a truth table for a function, you can write a product of maxterms expression by picking out the rows of the table where the function output is 0 (0-maxterm). f = (x’ + y + z)(x’ + y + z’)(x’ + y’ + z’) = M 4 M 5 M 7 = ∏(4,5,7) f’ = (x + y + z)(x + y + z’)(x + y’ + z) (x + y’ + z’)(x’ + y’ + z) = M 0 M 1 M 2 M 3 M 6 = ∏(0,1,2,3,6) f’ contains all the maxterms not in f

22. 22 Product of maxterms: practise F = x’y’ + xz, how to express this in the product of maxterms? = (x’y’ + x)(x’y’ + z) = (x’ + x)(y’ + x)(x’ + z)(y’ + z) = (x + y’)(x’ + z)(y’ + z) = (x + y’ + zz’)(x’ + z + yy’)(xx’ + y’ + z) = (x + y’ + z)(x + y’ + z’)(x’ + y + z)(x’ + y’ + z)(x + y’ + z)(x’ + y’ + z) = (x + y’ + z)(x + y’ + z’)(x’ + y + z)(x’ + y’ + z) = ∏(2,3,4,6) or, convert the expression into truth-table and then read the minterms from the table

23. 23 Minterms and maxterms are related Any minterm m i is the complement of the corresponding maxterm M i For example, m 4 ’ = M 4 because (xy’z’)’ = x’ + y + z MaxtermShorthand x + y + zM 0 x + y + z’M 1 x + y’ + zM 2 x + y’ + z’M 3 x’ + y + zM 4 x’ + y + z’M 5 x’ + y’ + zM 6 x’ + y’ + z’M 7 MintermShorthand x’y’z’m 0 x’y’zm 1 x’yz’m 2 x’yzm 3 xy’z’m 4 xy’zm 5 xyz’m 6 xyzm 7

24. 24 Converting between canonical forms We can convert a sum of minterms to a product of maxterms In general, just replace the minterms with maxterms, using maxterm numbers that don’t appear in the sum of minterms: The same thing works for converting from a product of maxterms to a sum of minterms From beforef=  (0,1,2,3,6) andf’=  (4,5,7) = m 4 + m 5 + m 7 complementing(f’)’= (m 4 + m 5 + m 7 )’ sof = m 4 ’ m 5 ’ m 7 ’[ DeMorgan’s law ] = M 4 M 5 M 7 [ By the previous page ] = ∏(4,5,7) f=  (0,1,2,3,6) = ∏(4,5,7)

25. 25 Standard Forms Any boolean function that is expressed as a sum of products (SOP) or as a product of sums (POS), where each product-term or sum-term may require fewer than (n-1) operations, is said to be in its standard form. Standard forms are not unique, there can be several different SOPs and POSs for a given function. A SOP expression contains: – Only OR (sum) operations at the “outermost” level – Each term (implicant) must be a product of literals A POS expression contains: – Only AND (product) operations at the “outermost” level – Each term (implicate) must be a sum of literals f(x,y,z) = xy + x’yz + xy’z f(x,y,z) = (x’ + y’)(x + y’ + z’)(x’ + y + z’)

26. 26 Strategy for operator (literal) reduction

27. 27 Other Logic Operations We’ve already seen all the basic Boolean operations and the associated primitive logic gates. There is a few additional gates that are often used in logic design. – They are all equivalent to some combination of primitive gates. – But they have some interesting properties in their own right. NAND (NOT-AND) NOR (NOT-OR) XOR (eXclusive OR) (xy)’ = x’ + y’(x + y)’ = x’ y’x  y = x’y + xy’ Operation: Expressions: Truth table: