1 / 11

Lecture 14: Obtaining Canonical Representations of Functions

Lecture 14: Obtaining Canonical Representations of Functions. PROF. INDRANIL SENGUPTA DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING. Algebraic Procedure to Obtain Canonical s-o-p. Examine each term of a given sum-of-products expression; if it is not a minterm , go to the next step.

patriciafletcher
Download Presentation

Lecture 14: Obtaining Canonical Representations of Functions

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Lecture 14: Obtaining Canonical Representations of Functions PROF. INDRANIL SENGUPTA DEPARTMENT OF COMPUTER SCIENCE AND ENGINEERING

  2. Algebraic Procedure to Obtain Canonical s-o-p • Examine each term of a given sum-of-products expression; if it is not a minterm, go to the next step. • For all missing variable xi, multiply the term by (xi’ + xi). • Multiply out all products and eliminate redundant product terms. Example: f (a, b, c) = a.b’ + b + a.b.c = a.b’ (c + c’) + b (a + a’)(c + c’) + a.b.c = a.b’.c + a.b’.c’ + a.b.c + a.b.c’ + a’.b.c + a’.b.c’ + a.b.c = a.b’.c + a.b’.c’ + a.b.c + a.b.c’ + a’.b.c + a’.b.c’

  3. Algebraic Procedure to Obtain Canonical p-o-s • Examine each term of a given product-of-sums expression; if it is not a maxterm, go to the next step. • For all missing variable xi, add the term xi’ xi • Obtain the sum terms, and eliminate redundant terms. Example: f (a, b, c) = a’ (b’ + c) = (a’ + bb’ + cc’) (b’ + c + aa’) = (a’ + b + c)(a’ + b + c’)(a’ + b’ + c)(a’ + b’ + c’)(a + b’ + c)(a’ + b’ + c) = (a’ + b + c)(a’ + b + c’)(a’ + b’ + c)(a’ + b’ + c’)(a + b’ + c)

  4. Transforming One Form to Another • Double complement the given function and apply De Morgan’s theorem. • Rule to be followed: • Complement of a sum of true minterms is the same as the sum of the false minterms. • Example: f (a, b, c) = a’.b’.c’ + a’.b’.c + a’.b.c’ + a.b.c’ + a.b.c f = (f’)’ = [(a’.b’.c’ + a’.b’.c + a’.b.c’ + a.b.c’ + a.b.c)’]’ = [a’.b.c + a.b.c’ + a.b’.c]’ (Consider remaining minterms) = (a + b’ + c’)(a’ + b’ + c)(a’ + b + c)

  5. Canonical s-o-p from the Truth Table • Consider rows of the truth table for which the output is 1. • For each such row, form a minterm. • If the input variable is 0, the corresponding variable will appear in complemented form in the minterm. • If the input variable is 1, the corresponding variable will appear in uncomplementedform in the minterm. • Take the sum of all such minterms. • We get the canonical sum-of-products expression.

  6. Example

  7. Converting to Gate Level Realization

  8. Canonical p-o-s from the Truth Table • Consider rows of the truth table for which the output is 0. • For each such row, form a maxterm. • If the input variable is 0, the corresponding variable will appear in uncomplemented form in the maxterm. • If the input variable is 1, the corresponding variable will appear in complemented form in the maxterm. • Take the product of all such maxterms. • We get the canonical product-of-sums expression.

  9. Example

  10. Converting to Gate Level Realization

  11. END OF LECTURE 14

More Related