1 / 20

D e MORGAN’s LAWS

D e MORGAN’s LAWS. De Morgan’s Laws provide an easy way to find the inverse of a boolean expression: (X + Y)’ = X’ Y’ (X Y)’ = X’ + Y’ An easy way to remember this is that each TERM is complemented, and that OR’s become AND’s; AND’s become OR’s.

tovi
Download Presentation

D e MORGAN’s LAWS

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. DeMORGAN’s LAWS De Morgan’s Laws provide an easy way to find the inverse of a boolean expression: (X + Y)’ = X’ Y’ (X Y)’ = X’ + Y’ An easy way to remember this is that each TERM is complemented, and that OR’s become AND’s; AND’s become OR’s. (Easy to prove this via a truth table, see textbook p. 47.) The complement of the product is the sum of complements The complement of the sum is the product of complements. -- Can easily generalize to n variables using above rules

  2. APPLYING DeMORGAN’S LAW Apply De Morgan’s Law to a more complex expression: (AB + C’D)’ = (AB)’ (C’D)’ = (A’+B’)(C + D’) Note that De Morgan’s law was applied twice. Another example:[(A’ + B)C’]’ = (A’ + B)’ + (C’)’ = (A’)’ (B)’ + C = AB’ + C

  3. NAND, NOR GATES Why do we care about De Morgan’s Law? There are two other gate types that produce the complement of a boolean function! A B Y0 0 10 1 11 0 11 1 0 A B Y0 0 10 1 01 0 01 1 0 (AB)’ (A+B)’ A A B B NAND NOR

  4. NAND, NOR (cont.) NAND (NOT AND) - can be thought of as an AND gate followed by an inverter. A (AB)’ A AB (AB)’ B B NAND NOR (NOT OR) - can be thought as an OR gate followed by an inverter. A+B A (A+B)’ A (A+B)’ B B NOR

  5. Actually…. In the real world, an AND gate is made from an NAND gate followed by an inverter. An OR gate is made from a NOR gate followed by an inverter. A A (AB)’ AB AB B B NAND AND (A+B)’ A A+B A+B A B B NOR OR

  6. What is this logic function in SOP form? A (AB)’ B F = ((AB)’ (CD)’)’ C D (CD)’ Lets use De Morgan’s Law F = ((AB)’ (CD)’)’ = ((AB)’)’ + ((CD)’)’ = AB + CD An interesting result…………… SOP Form

  7. NAND-NAND form = AND-OR form A (AB)’ B F = ((AB)’ (CD)’)’ C D (CD)’ Same logic function A (AB) B F = AB + CD C D (CD)

  8. nand nor and or not DeMORGAN’S THEOREM • Symbolic DeMorgan’s duals exist for all gate primitives

  9. DeMORGAN’S THEOREM (cont’d) Can save you time in evaluating/designing combinatorial logic – Align bubbles and non-bubbles whenever possible

  10. SHORTCUTS for MULTIPLYING A short cut theorem for Distribution (Multiplication) (X + Y) (X’ + Z) = XZ + X’Y Only works when you have a variable (X) and its complement (X’). To PROVE this, lets do the distribution the long way. (X + Y)(X’ + Z) = X X’ + XZ + X’Y + YZ Redundant by consensus theorem p.40 #17 0 = 0 + XZ + X’Y = XZ + X’Y

  11. EXAMPLE: CONVERT to SOP FORM (p. 51) (A+B+C’) (A+B+D) (A+B+E) (A+D’+E)(A’+C) Let X=A+B, Y=C’, Z=D (X+Y)(X+Z) = X+YZ (#8D p.40) =(A+B+C’D) (A+B+E) (A+D’+E) (A’+C) Let X=A, Y=D’+E, Z=C,  (X+Y)(X’+Z) = XZ+X’Y (#16 p.40) =(A+B+C’D)(A+B+E) (AC+A’(D’+E)) =(A+B+C’D)(A+B+E) (AC+A’D’+A’E)) by distr. law Let X=A+B, Y=C’D, Z=E,  (X+Y)(X+Z) = X+YZ =(A+B+C’DE) (AC+A’D’+A’E)) Mult. out by distr. law and eliminate terms such asAA’D’=AAC+AA’D’+AA’E+ABC+A’BD’+A’BE +C’DEAC+C’DEA’D+C’DEA’E = AC + ABC + A’BD + A’BE + A’C’DE Let X=AC, Y=B X+XY = X (#10 p.40) = AC + A’BD’ + A’BE + A’C’DE

  12. EXAMPLE: CONVERT to POS FORM (p. 51) AC + A’BD’ + A’BE + A’C’DE (A’ is common to three terms) = AC + A’(BD’ + BE + C’DE) by distr. law Let X=A,X’=A’, Y=BD’ + BE + C’DE, Z=CXZ+X’Y=(X+Y)(X’+Z) =(A+BD’+BE+C’DE)(A’+C)=(A+C’DE+BD’+BE)(A’+C) -re-arranging = A+C’DE +B(D’+E ) (A’+C) by distr. law Let X= A+C’DE, Y=B, Z=D’+E X+YZ=(X+Y)(X+Z) (8D) = (A+B+C’DE)(A+C’DE +D’+E)(A’+C) But E+C’DE = E (using X+XY=X) so C’DE is redundant = (A+B+C’DE) (A+D’+E)(A’+C) Let X=A+B, Y=C’, Z=DEX+YZ=(X+Y)(X+Z) =(A+B+C’)(A+B+DE) (A+D’+E)(A’+C) Let X=A+B, Y=D, Z=EX+YZ=(X+Y)(X+Z) =(A+B+C’) (A+B+D)(A+B+E) (A+D’+E)(A’+C)

  13. A F = A  B B EXCLUSIVE-OR FUNCTION One last gate type is the XOR Gate (Exclusive OR gate). XOR A B Y0 0 00 1 11 0 11 1 0 XOR gate is common in logic circuits that do binary addition/subtraction. Note that:F = A  B= A’B + AB’ = (A+B) (AB)’ -- see eqns. 3-17 to 3-24 p. 52 for XOR theorems

  14. EQUIVALENCE FUNCTION (º) Equivalence is the complement of XOR. º A F = (A  B)’= A º B A B Y0 0 10 1 01 0 01 1 1 B º A º B A -- alternate symbol B Note: In order to simplify expressions containing AND and OR as well as XOR and º, it is usually desirable to eliminate XOR and º by subst. defns. in terms of AND and OR.

  15. POSITIVE and NEGATIVE LOGIC • General Concept • Positive Logic • High Voltage => Logic 1 • Low Voltage => Logic 0 • Negative Logic • High Voltage => Logic 0 • Low Voltage => Logic 1

  16. POSITIVE and NEGATIVE LOGIC (cont’d) • Implication • Positive Logic High Voltage => Logic 1 Low Voltage => Logic 0 Voltage A B F Low Low Low Low High Low High Low Low High High High Logic A B F 0 0 0 0 1 0 1 0 0 1 1 1 - Equivalent gate: AND

  17. POSITIVE and NEGATIVE LOGIC (cont’d 2) • Implication • Negative Logic High Voltage => Logic 0 Low Voltage => Logic 1 Voltage A B F Low Low Low Low High Low High Low Low High High High Logic A B F 1 1 1 1 0 1 0 1 1 0 0 0 - Equivalent gate: OR

  18. POSITIVE and NEGATIVE LOGIC (cont’d 3) • Implication • Positive Logic High Voltage => Logic 1 Low Voltage => Logic 0 Voltage A B F Low Low Low Low High High High Low High High High High Logic A B F 0 0 0 0 1 1 1 0 1 1 1 1 - Equivalent gate: OR

  19. POSITIVE and NEGATIVE LOGIC (cont’d 4) • Implication • Negative Logic High Voltage => Logic 0 Low Voltage => Logic 1 Voltage A B F Low Low Low Low High High High Low High High High High Logic A B F 1 1 1 1 0 0 0 1 0 0 0 0 - Equivalent gate: AND -- Negative logic Thm. p. 56 (skip proof)

  20. What do you need to know? • De Morgan’s Laws • Duality (covered in previous presentation) • NAND, NOR gates • Multiplying Out and Factoring Expressions • XOR Gate, Equivalence Gate • Positive and Negative Logic

More Related