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CSE-221 Digital Logic Design (DLD)

CSE-221 Digital Logic Design (DLD). Lecture-5: Canonical and Standard forms and Integrated Circuites. CANONICAL AND STANDARD FORMS Minterms and Maxterms

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CSE-221 Digital Logic Design (DLD)

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  1. CSE-221Digital Logic Design (DLD) Lecture-5: Canonical and Standard forms and Integrated Circuites

  2. CANONICAL AND STANDARD FORMS Minterms and Maxterms A binary variable may appear either in its normal form (x) or in its complement form (x ). Now consider two binary variables x and y combined x and y combined with an AND operation. Since each variable may appear in either form, there are four possible combinations: xy, xy, xy, and xy. Each of these four AND terms is called a minterm , or a standard product . In a similar fashion , n variables forming an OR term, with each variable being primed or unprimed , provide 2n possible combinations called maxterms ,or standard sums.

  3. x y z Term Designation Term Designation 0 0 0 xyzm0 x+y+z M0 0 0 1 xyz m1 x+y+z M1 0 1 0 xyzm2 x+y+z M2 0 1 1 xyz m3 x+y+zM3 1 0 0 xyzm4 x+y+z M4 1 0 1 xyz m5 x+y+zM5 1 1 0 xyzm6 x+y+z M6 1 1 1 xyz m7 x+y+zM7 Minterms and Maxterms for Three Binary Variables MINTERM MAXTERMS Functions of Three Variables one of these minterms results in f1 = 1, we have f1 = xyz + xyz+xyz = m1 +m4 +m7 Similarly, it may be easily verified that f2=xyz+xyz+ xyz+xyz = m3+m5+m6+m7

  4. The complement of f1 is read as f1 = xyz + xyz+ xyz+xyz+ xyz If we take the complement of f1, we obtain the function f1 f1 = (x+y+z)(x+y+z)(x+y+z)(x+y+z) =M0.M2. M3. M5. M6 Similarly, it is possible to read the expression for f2 from the table: f2 =(x+y+z)(x+y+z)( x+y+z) (x+y+z) = M0M1M2M4

  5. SUM OF MINTERMS It was previously stated that for n binary variables, one can obtain 2n distinct minterms, and that any Boolean function can be expressed as a sum of minterms . Example Express the Boolean function F = A + BC in a sum of minterms. The function has three variables, A, B, and C. The first term A is missing two variables ; therefore: A= A(B+B)= AB +AB This function is still missing one variable: A = AB(C+C)+AB(C+C) = ABC +ABC +ABC +ABC The second term BC is missing one variable: BC = BC(A+A) =ABC+ABC Combing all terms, we have: F = A+BC = ABC +ABC +ABC+ABC +ABC F = ABC+ABC +ABC +ABC +ABC = m1+ m4+ m5 + m6+ m7

  6. It is sometimes convenient to express the Boolean function, when in its sum of minterms, in the following short notation: F(A, B, C) =  (1, 4, 5, 6, 7) The summation symbol  stands for the ORing of terms; the numbers following it are the minterms of the function. The letters in parenthesis following F form a list of the variables in the order taken when the minterm is inverted to an AND term. F= A+BC Truth Table for F = A+BC

  7. PRODUCT OF MAXTERMS Example:- Express the Boolean function F = xy +xz in a product of maxterm form. First, convert the function into OR terms using the distributive law: F= xy +xz = (xy+x)(xy+z) = (x+x)(y+x)(x+z)(y+z) = (x+y)(x+z)(y+z) The function has three variable: x, y, and z, Each OR term is missing one variable; therefore: x+y = x+y+zz= (x+y+z)(x+y+z) x+z= x+z+yy= (x+y+z)(x+y+z) y+z= y+z+xx= (x+y+z)(x+y+z) Combing all the term and removing those that appear more than once, we finally obtain: F = (x+y+z)(x+y+z)(x+y+z)(x+y+z) = M0M2M4M5 A convenient way to express this function is as follows: F(x, y, z) =  (0, 2, 4, 5) The product symbol, , denotes the ANDing of maxterms; the numbers are the maxterms of the function.

  8. Conversion between Canonical Forms The complement of a function expressed as the sum of minterms equals the sum of minterms missing from the original function. This is because the original function is expressed by those minterms that make the function equal to 1, whereas its complement is a 1 for those minterms that the function is a 0. As an example, consider the function F(A, B, C) = (1, 4, 5, 6, 7) The has a complement that can be expressed as F(A, B, C) =  (0, 2, 3) = m0+ m2+ m3 Now, if we take the complement of F by DeMorgan’s theorem, we obtain F in a different form: F= (m0+ m2+ m3)  = m0 . m2 . m3 = M0 M2 M3 =  (0, 2, 3) The last conversion follows from the definition of minterms and maxterms as shown in Table 2-3. From the table, it is clear that the following relation holds true: mj = Mj That is the maxterm with subscript j is a complement of the minterm with the same subscript j, and vice versa.

  9. F1 = y +xy +xyz F2 = x(y +z)(x +y+z) F3 = AB +C(D+E) , F3 = AB +C(D+E)= AB +CD+ CE.

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