3-7 Other Two-level Implementations

NAND and NOR logic implementation the most important from a practical point of view Wired logic 1) the open-collector TTL NAND gates a wired-AND logic Chap.10, ? Fig.10-11 ? Fig.3-22(a) F = (AB)’(CD)’= (AB + CD)’ AND-OR-INVERT function

2) the NOR output of ECL gates a wired-OR logic ? Fig.3-22(b) F = (A+B)’+ (C+D)’= ((A+B)(C+D))’ OR-AND-INVERT function Nondegenerate Forms Two-level combinations of gates AND, OR, NAND, NOR 16 possible combinations of two-level forms

8 degenerate forms degenerate to a single operation Ex) AND-AND ==> AND of all input variables OR-OR ==> OR AND-NAND ==> NAND OR-NOR ==> NOR NAND-OR ==> AND-NAND ==> NAND NAND-NOR ==> AND-AND ==> AND NOR-AND ==> OR-NOR ==> NOR NOR-NAND ==> OR-OR ==> OR

8 nondegerate forms an implementation in sum of products or product of sums dual AND-OR OR-AND basic forms NAND-NAND NOR-NOR the most practical forms NOR-OR NAND-AND OR-NAND AND-NOR AND-OR-INVERT implementation NAND-AND and AND-NOR equivalent forms of the AND-OR-INVERT

function ? Fig.3-23 F = (AB+CD+E)’ (a) AND-NOR sum of product + invert (b) AND-NOR (c) NAND-AND OR-AND-INVERT implementation OR-NAND and NOR-OR equivalent forms of the OR-AND-INVERT

function ? Fig.3-24 F = ((A+B)(C+D)E)’ (a) OR-NAND product of sums + invert (b) OR-NAND (c) NOR-OR Tabular Summary and Example Table 3-4 implementation ? with other Two-level forms AND-NOR and NAND-AND OR-NAND and NOR-OR AND-OR-INVERT form ---------- -------- F’ F F’= sum of products form for 0’s in the map

∑ OR-NAND and NOR-OR OR-AND-INVERT form ---------- -------- F’ F F’= product of sums forms F = sum of products form for 1’s in the map Ex 3-11 ? Implement the function of Fig.3-19(a) ( F(x,y,z) = ∑ (0,6) )

with the 4 two-level forms in Table 3-4 ? F’= x’y + xy’+ z for 0’s in the map F = (x’y + xy’+ z)’ in AND-OR-INVERT form AND-NOR and NAND-AND implementation ? Fig.3-25(a) F = x’y’z’+ xyz’for 1’s in the map F’= (x + y + z)(x’+ y’+ z) F = ((x + y + z)(x’+y’+z))’ in OR-AND-INVERT form

∑ OR-NAND and NOR-OR implementation ? Fig.3-25(b) 3-8 Don’t-Care Conditions don’t care condition the unspecified minterms whose outputs can be either 0 or 1, marked with X ==> used for further simplification Ex 3-12 Simplify the Boolean Function F(w,x,y,z) = ∑

∑ (1,3,7,11,15) d(w,x,y,z) = (0,2,5) ? Fig.3-26 (a) F = yz + w’x’ (b) F = yz + w’z the smallest literals and smallest terms The choice between 0 or 1 of don’t care term X depends on the simplicity Ex 3-12 F(w,x,y,z) = yz + w’x’ =

∑ (1,3,5,7,11,15) Fig.3-26(b) (0,1,2,3,7,11,15) F(w,x,y,z) = yz + w’z Both expressions include minterms 1,3,7,11 and 15 of minterm ‘1’ ? Fig.3-26 F’= z’+ wy’ for 0’s F(w,x,y,z) = (z’+ wy’)’ = z(w’+ y) ? (1,3,5,7,11,15) Fig.3-26(b)

3-9 The Tabulation Method The map method of simplification convenient for 3,4,5 numbers of variable for reasonable selection of adjacent squares disadvantage rely on the recognition ability of certain patterns through trial-and-error procedure

The tabulation mathod Quine-Macluskey method a guaranteed specific step-by-step procedure for a simplified standard-form expression suitable for machine computation for problems with many variables tedious for human use prone to mistakes because of routine and monotonous process

The tabula method 1) find by an exhaustive search all the terms(prime implicants ) that are candidates for inclusion in the simplified function 2) choose among the prime implicants for an expression with the least number of literals 3-10 Determination of prime implicants

Find the prime implicants by using a matching process 1) compare each minterm with every other minterm 2) If two minterms differ in only one variable, that variable is removed (a term with one less literal is found) 3) repeat until no further elimination of literals

∑ The prime implicants The remaining terms and all the terms that did not match during the process Ex 3-13 Simplify the following Boolean function by using the tabulation method F = ∑ (0,1,2,8,10,11,14,15) ? step 1 : Table 3-5(a) minterms of 1’s group into 5 sections

The 1st section, the number with no 1’s The 2nd section, the number with one 1’s The 3rd section, the number with two 1’s The 4th section, the number with three 1’s The 5th section, the number with four 1’s ? step 2 : Table 3-5(b) combine two minterms that differ by only one variable (compare the next section down only)

In the same section more than one variable differs 두 숫자가 같지 않으면, minterm1이 1일 때, minterm2가 0이고 minterm1이 0일 때, minterm2가 1인 두쌍이 반드시 하나 이상있음 예) minterm1 : 1 0 0 1 1 0 0 minterm2 : 0 1 1 0 0 1 0 ----- ---- ----- remove the unmatched variable of (a) place a check mark to show that they have been used step 3 : The terms of column (b) has only 3 variables

? Table 3-5(c) with only 2 variables no check mark of 000- because 000- does not match with any other term step 4 : the prime implicants the unchecked terms in the table because each checked terms are changed into a simpler from in the subsequent column w’x’y’(000-) in column(b) x’z’(-0-0) and wy(1-1-) in column(c)

A simplified expression the sum of the prime implicants F = w’x’y’+ x’z’+ wy The map method, ? Fig.3-27 The sum of prime implicants ≠ the expression with minimum number of terms Ex 3-14 Comparing with decimal number ( in stead of binary )

When two minterms differ by only one position, the difference is a power of 2 Ex) 1 1 0 1(25) - 1 0 0 1(17) ---------------- 0 1 0 0(8 = 23) Table 3-6 ? Example 3-13 with Decimal Notation a) Group the minterms b) Write two numbers whose difference is a power of 2(1,2,4,8,16...)

Mark two numbers at right of (a) The number in the ( ) the position of - c) Compare only those terms with the same number in the ( ) 0,2(2) 쭽 8 쭾 =====> 0,2,8,10(2,8) 8,10(2) 0,8(8) 2 쭾 =====> 0,2,8,10(2,8) 2,10(8) 10,11(1) 4 쭾 =====> 10,11,14,15(1,4) 14,15(1) 10,14(4) 1 쭾 =====> 10,11,14,15(1,4) 11,15(4)

The prime implicants terms not checked in the table Conversion to binary 0,1(1) ==> 0000,0001 000- 0,2,8,10(2,8) ==> 0000 0010 00-0 1000 1010 10-0 -0-0 Ex 3-14

∑ Determine the prime implicants of the function F(w,x,y,z) = (1,4,6,7,8,9,11,15) Table 3-7 ? The prime implicants x’y’z, w’xz’, w’xy, xyz, wyz, wx’ F = x’y’z + w’xz’+ w’xy + xyz + wyz + wx’ not necessarily minimum number of terms implicants marked in Table 3-8 ?

The map method F = x’y’z + w’xz’+ xyz + wx’ 3-11 Selection of Prime Implicants Place X’s in each row to show the composition of minterms that make the prime implicants Choose a minimum set of prime implicants that cover all the minterms in the function Ex 3-15 Minimize the function of Example 3-14 ?

Table 3-8 Prime Implicant Table x’y’z is from 1,9 marked with X. Inspect for column containing only a single X 1, 4, 8, 10 Essential prime implicants x’y’z, w’xz’, wx’ prime implicants with a single X should be included

Choose the prime implicants ( xyz ) which cover the other minterms than those covered by essential prime implicants marked in Table 3-8 ? F = x’y’z + w’xz’+ xyz + wx’ The tabulation process with 1’s the sum of products form The tabulation process with 0’s the product of sums form A function with don’t-care conditions the don’t-care terms included in the list of minterms for the determination of prime implicants

allows the derivation of prime implicants with the least number of literals not included in the prime implicant table ( because the don’t-care terms do not have to be covered ) 3-12 Concluding Remarks Boolean-function simplification minimization of the number of literals of Boolean function in standard forms ( restriction ) sum of products form product of sums form map method tabulation method

From Fig.3-15 ? standard forms ==> no more than two levels of gates nonstandard forms ==> more than the Gray-code sequence for the maps a change in only one bit between adjacent squares ? Variations of the three-variable map Fig.3-29 Variations of the four-variable map Fig.3-30 (a) very popular ? (b) original Veitch diagram modified to (a) by Karnaugh

The tabulation method error-prone in comparing numbers over long lists step-by-step procedure suitable for computer mechanization The map method for more than 5 variables, not the best simplified expression For the functions with multiple outputs, further simplification is possible because of common gates an extension of the tabulation method for multiple-output circuits practically important only if a computer program of the method is available (because of tediousness for human manipulation)