Table 2.8 Boolean Expressions for the 16 Functions of Two Variables Boolean Operator Symbol Name Comments Functions Binary constant FO = 0 Null F1 =xy x'y AND x and y F2 = x y' Inhibition x, but not y F3=x Transfer F4 =x'y ylx Inhibition y, but not x F5= y Transfer y F6 =x y'+x'y x® y Exclusive- x or y, but not OR both F7=x+ y x+y OR х огу F8 = (x+ y)' x y NOR Not-OR F9 =x y +x'y'(x y )' Equivalence x equals y F 10 = y' y' Complement Not y F 11 = x + y' xCy Implication If y, then x F 12 = x' x' Complement Not x F 13 = x'+y x >y Implication If x, then y F 14 = (x y)' x t y NAND Not-AND Binary constant F 15 = 1 Identity

Just answer three, four and that's OK

Consider the minterm expansion for f=Σm(1, 3, 4, 5, 6, 7, 9, 11) with 4 variables. (a) Draw its Karnaugh map. (b) Write down expression for all the prime implicants. (c) Write down expression for all the essential prime implicants. (d) Write down the minimum sum of product expression for f.

Use the procedure described in slides chapter6-part1 (proof 2, pages 8-13) to convert the following finite automaton to a regular expression.

Implement the following Boolean function F, using the two-level forms of logic (a) NAND-AND, (b) AND-NOR, (c) OR-NAND, and (d) NOR-OR: F(W, X, Y, Z) = ∑ (2, 3, 4, 8, 9, 10, 12, 14)

The correct statements are: Group of answer choices A DPDA must halt. A NDPDA must halt. A NDPDA without ε-transitions must halt. A DPDA or NDPDA without ε-transitions must halt in at most |w| steps.

Using Karnaugh Maps, simplify the following logic expression: T = ABCD + ABCD + ABCD + ABCD + ABCD + ABCD + ABCD The "Don't Care" states are: ABCD, ĀBCD, ABCD and ABCD.

A combinational circuit is described by this Boolean Expression: F(A,B,C,D) = 2m(0,2,4,8,9) i) Minimise the Boolean Expression given above using a Karnaugh map. ii) Given Don't Care conditions d = Em (10,11,12), use a separate Karnaugh map to demonstrate how you would further simplify the expression in part i). iii) Use Switching Algebra to show how you would implement the function obtained in part ii) by using NAND gates only. Provide a circuit diagram in your answer. Show all your working.

For certain functions which cannot be simplified beyond the sum of minterms, for such functions, the sum of minterms representation has the same number of literals as any othersum of products representation of this function.Check if this condition is satisfied for the following functions:f (x, y, z) = x ⊕ y ⊕ z f (x, y, z) = x + y + zxyf (x, y, z) = xyz + x′y′z

Given the Boolean functions F1 and F2, show that- The boolean function E = F1 +F2 contains the sum of mintemrs of F1 and F2- The boolean function G = F1*F2 contains only the minterms that are common to F1 and F2