COE 211

Lab 3 Gate-Level Minimization

By.Eng. Ohoud Almohammadi

Exercise 3.2❖ Simplify the following Boolean functions, using three-

variable maps: (c) F(x, y, z) = ∑(0, 1,6, 7)

Exercise 3.2❖ (d) F(x,y,z) = ∑(0,1,3,4,5)

Exercise 3.6❖ Simplify the following Boolean expressions, using four-

variable maps:

❖ (a)A'B'C'D'+ AC'D'+ B'CD'+ A'BCD + BC'D

Exercise 3.7❖ Simplify the following Boolean expressions, using four-

variable maps: (a) w'z + xz + x'y + wx'z

Prime Implicants❖ Any single 1 or group of 1's that can be combined together on a Karnaugh map of the function

F represents a product term which is called an IMPLICANT.

❖ A PRIME IMPLICANT is a product term that cannot be combined with another term to eliminate a variable.

❖ A single 1 is a prime implicant if it is not adjacent to any other 1’s.

❖ Two adjacent 1's form a prime implicant if they are not contained in a group of four adjacent 1’s.

❖ Four adjacent 1's form a prime implicant if they are not contained in a group of eight adjacent 1’s.

❖ The minimum sum-of-products expression for a function consists of some (BUT NOT NECESSARILY ALL) of the prime implicants of a function

❖ If a minterm in a square is covered by only one prime implicant, that prime implicant is said to be essential.

Exercise 3.9❖ Find all the prime implicants for the following Boolean

functions, and determine which are essential:

❖ (a)F(w,x,y,z) = ∑(0,2,4,5,6,7,8,10,13,15)

Don 't-care Minterm❖ A don 't-care minterm is a combination of variables whose

logical value is not specified. Such a minterm cannot be marked with a 1 in the map. because it would require that the function always be a 1 for such a combination. Likewise. putting a 0 on the square requires the function to be 0. To distinguish the don'r-cere condition from l's and 0’s an X is used. Thus, an X inside a square in the map indicates that we don't care whether the value of 0 or 1 is assigned to F for the particular minterm.

Exercise 3.15❖ Simplify the following Boolean function F, together with

the don’t-care conditions d, and then express the simplified function in sum-of-minterms form: (a) F(x,y,z) = ∑(0,1,4,5,6)

❖ F=1

NAND Circuits❖ The NAND gate is.said to be a universal gate because any digital

system can be implemented with it.

❖ TO show that any Boolean function can be implemented with NAND gates, we need only show that the logical operations of AND and OR. and complement can be obtained with NAND alone.

❖ TheAND operation requires. two NAND gates. The first produces the NAND operation and the second inverts the logical sense of the signal. The OR operation is achieved through a NAND gate with additional inverters in each input.

NAND Implementation

Exercise 3.16❖ Simplify the following functions, and implement them

with two-level NAND gate circuits:

❖ (a) F(A, B, C, D) = AC’D’ + A’C + ABC + AB’C + A’C’D’