Prof. Sin-Min Lee Department of Computer Science CS147 Lecture 6 POS, K-map and Multiplexer Mid 1 Revision Prof. Sin-Min Lee Department of Computer Science

Unsimplifying expressions xy + y’z + xz = (xy  1) + (y’z  1) + (xz  1) = (xy  (z’ + z)) + (y’z  (x’ + x)) + (xz  (y’ + y)) = (xyz’ + xyz) + (x’y’z + xy’z) + (xy’z + xyz) = xyz’ + xyz + x’y’z + xy’z You can also convert the expression to a sum of minterms with Boolean algebra. Apply the distributive law in reverse to add in missing variables. Very few people actually do this, but it’s occasionally useful. In both cases, we’re actually “unsimplifying” our example expression. The resulting expression is larger than the original one! But having all the individual minterms makes it easy to combine them together with the K-map.

K-maps from truth tables You can also fill in the K-map directly from a truth table. The output in row i of the table goes into square mi of the K-map. Remember that the rightmost columns of the K-map are “switched.”

The Sum-of-Products (SOP) Form When two or more product terms are summed by Boolean addition

Conversion of a General Expression to SOP Form Any logic expression can be change into SOP form by applying Boolean Algebra techniques Example: Try This:

The Standard SOP Form Multiply:

The Products-of-Sum (POS) Form When two or more sum terms are multiplied.

The Standard POS Form Rule 12! Add:

Converting SOP to Truth Table Examine each of the products to determine where the product is equal to a 1. Set the remaining row outputs to 0.

Converting POS to Truth Table Opposite process from the SOP expressions. Each sum term results in a 0. Set the remaining row outputs to 1.

Converting from Truth Table to SOP and POS Inputs Output A B C X 1 POS:

The 3-Variable K-Map

The 4-Variable K-Map

K-Map SOP Minimization A 1 is placed on the K-Map for each product term in the expression. Each 1 is placed in a cell corresponding to the value of a product term

Example: Map the following standard SOP expression on a K-Map: Solution:

Example: Map the following standard SOP expression on a K-Map: Solution:

Exercise: Map the following standard SOP expression on a K-Map:

Answer:

K-Map Simplification of SOP Expressions A group must contain either 1, 2, 4, 8 or 16 cells. Each cell in group must be adjacent to one or more cells in that same group but all cells in the group do not have to be adjacent to each other Always include the largest possible number 1s in a group in accordance with rule 1 Each 1 on the map must be included in at least one group. The 1s already in a group can be included in another group as long as the overlapping groups include noncommon 1s Goal: To maximize the size of the groups and to minimize the number of groups

Example: Group the 1s in each K-Maps

Determining the minimum SOP Expression from the Map Groups the cells that have 1s. Each group of cells containing 1s create one product term composed of all variables that occur in only one form (either uncomplemented or complemented) within the group. Variable that occurs both uncomplemented and complemented within the group are eliminated. These are called contradictory variables.

Example: Determine the product term for the K-Map below and write the resulting minimum SOP expression 1

Example: Use a K-Map to minimize the following standard SOP expression

Example: Use a K-Map to minimize the following standard SOP expression

Mapping Directly from a Truth Table

Don’t Care (X) Conditions A situation arises in which input variable combinations are not allowed Don’t care terms either a 1 or a 0 may be assigned to the output

Exercise: Use K-Map to find the minimum SOP from 1 2

SOP

POS

MUX Enable 2n Data Inputs Data Output n Input Select

(Only one O/P asserted at any time Remember the 2 – 4 Decoder? Sel(3) S1 Sel(2) Sel(1) S0 Sel(0) Mutually Exclusive (Only one O/P asserted at any time

Z Y X F 0 F=0 1 1 F=1 1 F= X´ 0 F= X 1 X´ X F Z Y

Y X Z F 0 F=Z 1 0 F=0 1 F= Z´ 1 F= 1 Z Z´ 1 F Y X

Y X Z F X .Y ´=0 1 X⊕Y=0 X⊕Y=1 X .Y ´=1

MUX Example (cont.) When A=B=0, F=C When A=0, B=1, F=C 1 When A=B=0, F=C When A=0, B=1, F=C When A=1, B=0, F=C When A=B=1, F=C’

Implementation using MUXs: Now we implement the output f1 using an 8x1MUX. Selection inputs to the MUX are x2 y1 y2 (that is, if x2y1y2=001, input I1 of the MUX is selected). The implementation table is tabulated as

Combinational circuit implementation using MUX We can use Multiplexers to express Boolean functions also. Expressing Boolean functions as MUXs is more efficient than as decoders. First n-1 variables of the function used as selection inputs; last variable used as data inputs. If last variable is called Z, then each data input has to be Z, Z’, 0, or 1.