Find the sum-of-products expansions of these Boolean functions. a) F(x, y, z)=x+y+z b) F(x, y, z

Find the sum-of-products expansions of these Boolean...

Key Concepts

Boolean Algebra

Boolean algebra is a branch of algebra that deals with binary variables and logical operations. It forms the mathematical basis for digital circuit design and computer programming by defining rules for operations like AND, OR, and NOT, and it provides a framework for simplifying and manipulating Boolean expressions.

Sum-of-Products Expansion

A sum-of-products expansion is a form of Boolean expression where several product (AND) terms are OR'ed together. Each product term represents a conjunction of literals (variables or their complements) that, when evaluated, contribute to the overall function. This form is fundamental in digital logic design and is commonly used for implementation with logic gates.

Minterms

Minterms are specific types of product terms in Boolean functions that correspond to exactly one row in the truth table where the function is true. Each minterm includes every variable in the function, either in its direct or complemented form, and serves as a building block for the sum-of-products form.

Boolean Function Simplification

Boolean function simplification involves using Boolean identities and laws, such as the distributive, associative, and De Morgan's laws, to reduce a complex Boolean expression into a simpler or more canonical form. Simplification is critical in reducing the number of logic gates required in circuit design and improving efficiency.

Transcript

00:01 We are given boolean functions and we're asked to find the sum of product expansions of these boolean functions.

00:10 In part a, we're given the function f of x, y, z equals x plus y plus c.

00:29 Determine the sum of products expansion of f, you can either use boolean identities or construct a table that determines all value of f.

00:39 This would be a pretty large table.

00:43 Instead, i'm just going to use boolean identities.

00:46 So first of all, x plus y plus z, by the identity laws, is the same as x times 1 plus y times 1 plus z times 1.

01:00 And this, again, by the identity laws, is equal to x times 1 plus y times 1 times 1 plus z times 1, plus z times 1.

01:15 Off parentheses because of associativity.

01:25 And then using the unit property of boolean variables, this is equal to x times y plus y bar times z plus z bar.

01:53 Now to do a similar thing with the other two terms.

01:56 So plus y times x plus x bar times z plus z bar.

02:02 And plus z times x plus x bar times y plus y bar next i'll use the distributive property for boolean variables so this is xy plus xy bar times z plus z bar plus y x plus y x plus yx plus y x bar times z plus z plus z bar plus x x plus x x bar times y plus y plus y bar then i'll use the distributive law again for these three terms and i get x y z plus x y bar z technically i'm using the distributive law twice here but i'm going to skip some steps to get faster plus x y z bar plus x y bar plus xy bar z bar plus y x z plus y x plus y x bar z plus y x bar plus y x bar plus x bar y plus x bar y bar.

04:13 Next i'll use the commutative law for multiplication.

04:19 And so we get all of our terms x, y, z plus x, y, bar, z, plus x, y, bar, z, plus x, y, z, plus x, y, z again, plus x bar y z, plus x bar, plus x bar, y, z, plus x, y, z bar, plus x, x bar y z bar plus x y z plus x bar y z plus x bar y z plus x y bar z plus x bar y bar z plus x bar y bar z and finally i will use the idempotent law for fully in algebras so we have x plus x times y times z plus x times y times z is x times y times z adding another x to my time z is still x times y times z then we add the x y bar z we add these with the other x y bar z we still have x y bar z then we have our x y z bar we add this with the other x y z bar we still have x y b bar then we have x y bar z bar then we have x y bar z bar adding this with the other x y bar z bar we still have x y bar z bar we still have x y bar z bar adding this the other one, we still have x bar y z.

06:15 Then we have x bar y z bar, and this appears to be the only one of this kind.

06:35 And then we have x bar y bar z.

06:51 And so this is the sum of product formula for part a.

06:59 Okay, in part b, we are given the boolean function f of x y z equals x plus z times y, again, to determine the sum of product expansion of this function, you could use boolean identities or construct a table that determines all values of the function.

07:34 However, constructing the table would take quite a bit of effort, and i don't think it would be particularly enlightening about this function.

07:43 So instead, let's use boolean identities.

07:49 So first, i'll use the distributive law.

07:53 This is xy plus zy.

08:00 Then i'll use the identity law.

08:02 This gives us xy times 1 plus zy times 1...

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