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Chapter 3: Karnaugh Maps. Uchechukwu Ofoegbu Temple University. Riddle.

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## Chapter 3: Karnaugh Maps

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**Chapter 3: Karnaugh Maps**Uchechukwu Ofoegbu Temple University**Riddle**Three people check into a hotel. They pay $30 to the manager and go to their room. The manager suddenly remembers that the room rate is $25 and gives $5 to the bellboy to return to the people. On the way to the room the bellboy reasons that $5 would be difficult to share among three people so he pockets $2 and gives $1 to each person. Now each person paid $10 and got back $1. So they paid $9 each, totalling $27. The bellboy has $2, totalling $29. Where is the missing $1?**Introduction to Karnaugh Maps**• If implemented correctly, they almost always produce a minimum solution. • They are more straightforward that algebraic manipulations • They generally produce SOPs, but POS can be generated from their complements if required.**Two-variable Karnaugh maps**A 0 1 B B’ A’ A 0 A’B’ AB’ B 1 A’B AB A 0 1 B 0 m0 m2 1 m1 m3**Three-variable Karnaugh maps**Product terms corresponding to groups of two.**Implicants**• An implicant of a function is a product term. • From the point of view of the map, an implicant is a rectangle of 1, 2, 4, 8, . . . (any power of 2) 1’s. That rectangle may not include any 0’s. • Example: • f = A’B’C’D’+A’B’CD+A’BCD+AB’CD+ABC’D’+ABC’D+ABCD AB 00 01 11 10 CD 00 1 1 01 1 1 1 1 11 1 10**Implicants**• The implicants of f are: Minterms (1 implicant) ABCD ABCD ABCD ABCD ABCD ABCD ABCD Groups of 2 ACD BCD ACD BCD ABC’ ABD Groups of 4 CD**Prime and Essential Prime Implicants**• Prime Implicant: • an implicant that (from the point of view of the map) is not fully contained in any one other implicant. • Essential Prime Implicant: • a prime implicant that includes at least one 1 that is not included in any other prime implicant. AB 00 01 11 10 CD 00 1 1 01 1 1 1 1 11 1 10**Minimum SOP Expressions From Karnaugh Maps**• Find all essential prime implicants. • Circle them on the map and mark the minterm(s) that make them essential with an asterisk (*). • 2. Find enough other prime implicants to cover the function. Do this using two criteria: • a. Choose a prime implicant that covers as many new 1’s (that is, those not already covered by a chosen prime implicant). • b. Avoid leaving isolated uncovered 1’s. • The main idea is • To Have all ones covered • To Have as few terms as possible • To have several rectangles with more 1’s and few rectangles with less 1’s**Example**f = w’x’y’z’+w’xy’z’+ w’xy’z+ w’xyz+ wx’y’z’+ w’xyz+ wxy’z’+ wxyz AB 00 01 11 10 CD * * * 00 1 1 1 1 unnecessary 1 01 1 1 * 11 1 10 f = y’z’+wyz+w’xz**In Groups**f = b + a' c**Don’t Cares**• Prime implicant • A rectangle of 1, 2, 4, 8, . . . 1’s or X’s not included in any one larger rectangle. • From the point of view of finding prime implicants, X’s (don’t cares) are treated as 1’s. • Essential prime implicant • A prime implicant that covers at least one 1 not covered by any other prime implicant (as always). • Don’t cares (X’s) do not make a prime implicant essential.**Example**f = Σm(1,7,10,11,13) + Σd(5,8,15) AB 00 01 11 10 CD 00 x Use don’t cares to get as many minterms in each tem as possible 1 x 01 1 1 x 1 11 10 1 F = BD + ACD + ABC**In groups**• For the following problem, find the minimum SOP expression within the options given • h(a,b,c) = Σm(0,1,5) + d(3,4,6,7) • h = a'b' + c + a • h = a + c + b’ • h = c + b’ • h = b’ • h = c**Example**F = A’B’C’+A’BC’+ABC’+ABC; G = A’B’C+A’BC+ABC’+ABC AB AB 00 01 11 10 00 01 11 10 C C 0 0 1 1 1 1 1 1 1 1 1 1 F = A’C’+AB G = A’C+AB**Example**F = AB+ABC G = AB +BC F = AB+ABC G = AB +ABC**Example**f = ab + bc g = ab + ac f = ab + abc g = ab + abc**Example**F = AC +ACD+ABCG = AC+ACD+ABC**NAND, NOR**• Many electronic systems automatically invert gates • Easier to fabricate with electronic components • Basic gates used in integrated circuits (IC) digital logic families. • NAND gate • universal gate • Could be used to construct any logic gate**NAND gates.**Alternate symbol for NAND. Symbols for NOR gate.**NAND Gate Implementation**When we have a circuit consisting of AND and OR gates such that the output of the circuit comes from an OR, the inputs to all OR gates come either from a system input or from the output of an AND, and the inputs to all AND gates come either from a system input or from the output of an OR. All gates are replaced by NAND gates, and any input coming directly into an OR is complemented.**Example**Try: g = wx(y+z)+x’y**NOR Gate Implementation**When we have a circuit consisting of AND and OR gates such that the output of the circuit comes from an AND, the inputs to all OR gates come either from a system input or from the output of an AND, and the inputs to all AND gates come either from a system input or from the output of an OR. All gates are replaced by NOR gates, and any input coming directly into an AND is complemented.**Example**Try: g = (x+y’)(x’+y)(x’+z)**XOR and XNOR**A xor B is 1 if a = 1 or b is 1 and 0 if both are 1 or 0; Develop a truth table for XOR**Homework**• 1-17 • 20 • 21 • 22 • 23

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