eTEACH presentation accessible player instructionseTEACH presentation accessible player instructionsThis eTEACH player requires that you have Flash 8 or above installed. If you do not, download the player from Adobe.
The next major heading marks the start of the presentation. Each next lower level heading thereafter marks the beginning of a slide. An audio control button that plays the audio track for the slide follows the contents and notes for each slide in the presentation. Pressing this button after the audio playback toggles pausing and resuming the playback. Before moving off the play audio button, complete listening to the audio track for the slide or pause the audio. Control + Shift + L increases volume. Control + Shift + Q decreases volume.
There is a table of contents for the presentation following the presentation. The table of contents can be used to access sections of the presentation in random order. Type Control + Shift + C to go to the table of contents.
When links to external resources are present in the presentation, a list of these resource links follows the table of contents. When following a link to a resource, you should remember which slide you were on. When you return to the main page, your screen-reader may not return you to that slide. You can use the table of contents or just click through the headings to return to the correct slide.
This presentation contains resources. Click here to proceed immediately to the resource list. A link to resources follows each slide.
If the presentation contains one or more quizzes, a button to take you to each quiz follows the slide at which the quiz is to occur. Each question in the quiz begins with a level three heading. The answer choices will be presented in a list. If your instructor has supplied a hint for the questions, it will follow the answers and is formatted as an anchor. The correct answer to the question follows the list of answers and the hint (if included). The correct answer is also formatted as an anchor.
Chapter 2 - Part 1 - PPT - Mano & Kime - 2nd EdSlide 1 Lecture 6 Canonical FormsLecture 6 Canonical Forms
Slide 2 OutlineOutline What are Canonical Forms? Minterms and Maxterms Index Representation of Minterms and Maxterms Sum-of-Minterm (SOM) Representations Product-of-Maxterm (POM) Representations Representation of Complements of Functions Conversions between Representations
Slide 3 Canonical FormsCanonical Forms It is useful to specify Boolean functions in a form that: Allows comparison for equality. Has a correspondence to the truth tables Canonical Forms in common usage: Sum of Minterms (SOM) Product of Maxterms (POM)
Slide 4 MintermsMinterms Minterms are AND terms with every variable present in either true or complemented form. Given that each binary variable may appear normal (e.g., x) or complemented (e.g., ), there are 2n minterms for n variables. Example: Two variables (X and Y)produce 2 x 2 = 4 combinations: (both normal) (X normal, Y complemented) (X complemented, Y normal) (both complemented) Thus there are four minterms of two variables. Y X XY Y X Y X x
Slide 5 MaxtermsMaxterms Maxterms are OR terms with every variable in true or complemented form. Given that each binary variable may appear normal (e.g., x) or complemented (e.g., x), there are 2n maxterms for n variables. Example: Two variables (X and Y) produce 2 x 2 = 4 combinations: (both normal) (x normal, y complemented) (x complemented, y normal) (both complemented) Y X + Y X + Y X + Y X +
Slide 6 Maxterms and MintermsExamples: Two variable minterms and maxterms. The index above is important for describing which variables in the terms are true and which are complemented. Maxterms and Minterms
Slide 7 Standard OrderStandard Order Minterms and maxterms are designated with a subscript The subscript is a number, corresponding to a binary pattern The bits in the pattern represent the complemented or normal state of each variable listed in a standard order. All variables will be present in a minterm or maxterm and will be listed in the same order (usually alphabetically) Example: For variables a, b, c: Maxterms: (a + b + c), (a + b + c) Terms: (b + a + c), a c b, and (c + b + a) are NOT in standard order. Minterms: a b c, a b c, a b c Terms: (a + c), b c, and (a + b) do not contain all variables
Slide 8 Purpose of the IndexPurpose of the Index The index for the minterm or maxterm, expressed as a binary number, is used to determine whether the variable is shown in the true form or complemented form. For Minterms: 1 means the variable is Not Complemented and 0 means the variable is Complemented . For Maxterms: 0 means the variable is Not Complemented and 1 means the variable is Complemented .
Slide 9 Index Example in Three VariablesIndex Example in Three Variables Example: (for three variables) Assume the variables are called X, Y, and Z. The standard order is X, then Y, then Z. The Index 0 (base 10) = 000 (base 2) for three variables). All three variables are complemented for minterm 0 ( ) and no variables are complemented for Maxterm 0 (X,Y,Z). Minterm 0, called m0 is . Maxterm 0, called M0 is (X + Y + Z). Minterm 6 ? Maxterm 6 ? Z , Y , X Z Y X m6 = X Y Z M6 = (X + Y + Z)
m6 = X Y Z M6 = (X + Y + Z)
Slide 10 Minterm and Maxterm RelationshipReview: DeMorgan's Theorem and Two-variable example: and Thus M2 is the complement of m2 and vice-versa. Since DeMorgan's Theorem holds for n variables, the above holds for terms of n variables giving: and Thus Mi is the complement of mi. Minterm and Maxterm Relationship y x y · x + = y x y x × = + y x M 2 + = y x· m 2 = i m M = i i i M m =
Slide 11 Function Tables for BothFunction Tables for Both Minterms of Maxterms of 2 variables 2 variables Each column in the maxterm function table is the complement of the column in the minterm function table since Mi is the complement of mi. x y m 0 m 1 m 2 m 3 0 0 1 0 0 0 0 1 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 1 x y M 0 M 1 M 2 M 3 0 0 0 1 1 1 0 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 0
Slide 12 ObservationsObservations In the function tables: Each minterm has one and only one 1 present in the 2n terms (a minimum of 1s). All other entries are 0. Each maxterm has one and only one 0 present in the 2n terms All other entries are 1 (a maximum of 1s). We can implement any function by "ORing" the minterms corresponding to "1" entries in the function table. These are called the minterms of the function. We can implement any function by "ANDing" the maxterms corresponding to "0" entries in the function table. These are called the maxterms of the function. This gives us two canonical forms: Sum of Minterms (SOM) Product of Maxterms (POM) for stating any Boolean function.
Slide 13 Minterm Function Examplex y z index m1 + m4 + m7 = F1 0 0 0 0 0 + 0 + 0 = 0 0 0 1 1 1 + 0 + 0 = 1 0 1 0 2 0 + 0 + 0 = 0 0 1 1 3 0 + 0 + 0 = 0 1 0 0 4 0 + 1 + 0 = 1 1 0 1 5 0 + 0 + 0 = 0 1 1 0 6 0 + 0 + 0 = 0 1 1 1 7 0 + 0 + 1 = 1 Minterm Function Example Example: Find F1 = m1 + m4 + m7 F1 = x y z + x y z + x y z
Slide 14 Maxterm Function ExampleMaxterm Function Example Example: Implement F1 in maxterms: F1 = M0 · M2 · M3 · M5 · M6 ) z y z)·(x y ·(x z) y (x F 1 + + + + + + = z) y x )·( z y x ·( + + + + x y z i M0 M2 M3 M5 M6 = F1 0 0 0 0 0 1 1 1 = 0 0 0 1 1 1 1 1 1 1 = 1 0 1 0 2 1 0 1 1 1 = 0 0 1 1 3 1 1 0 1 1 = 0 1 0 0 4 1 1 1 1 1 = 1 1 0 1 5 1 1 1 0 1 = 0 1 1 0 6 1 1 1 1 0 = 0 1 1 1 7 1 1 1 1 1 = 1 1
Slide 15 Canonical Sum of MintermsCanonical Sum of Minterms Any Boolean function can be expressed as a Sum of Minterms. For the function table, the minterms used are the terms corresponding to the 1's For expressions, expand all terms first to explicitly list all minterms. Do this by ANDing any term missing a variable v with a term ( ). Example: Implement as a sum of minterms. First expand terms: Then distribute terms: Express as sum of minterms: f = m3 + m2 + m0 y x x f + = y x ) y y ( x f + + = y x y x xy f + + = v v +
Slide 16 Another SOM ExampleAnother SOM Example Example: There are three variables, A, B, and C which we take to be the standard order. Expanding the terms with missing variables: F = A(B + B )(C + C ) + (A + A ) B C = ABC + ABC + AB C + AB C + AB C + A B C Collect terms (removing all but one of duplicate terms): = ABC + ABC + AB C + AB C + A B C Express as SOM: = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7 C B A F + =
F = A(B + B )(C + C ) + (A + A ) B C = ABC + ABC + AB C + AB C + AB C + A B C = ABC + ABC + AB C + AB C + A B C = m7 + m6 + m5 + m4 + m1 = m1 + m4 + m5 + m6 + m7
Slide 17 Shorthand SOM FormShorthand SOM Form From the previous example, we started with: We ended up with: F = m1+m4+m5+m6+m7 This can be denoted in the formal shorthand: Note that we explicitly show the standard variables in order and drop the m designators. C B A F + =
Slide 18 Canonical Product of MaxtermsCanonical Product of Maxterms Any Boolean Function can be expressed as a Product of Maxterms (POM). For the function table, the maxterms used are the terms corresponding to the 0's. For an expression, expand all terms first to explicitly list all maxterms. Do this by first applying the second distributive law , ORing terms missing variable v with a term equal to and then applying the distributive law again. Example: Convert to product of maxterms: Apply the distributive law: Add missing variable z: Express as POM: f = M2 · M3 y x x ) z , y , x ( f + = y x ) y (x 1 ) y )(x x (x y x x + = + × = + + = + ( ) z y x ) z y x ( z z y x + + + + = × + + v v × A+BC = (A+B)(A+C)
Slide 19 Another POM ExampleConvert to Product of Maxterms: Use x + y z = (x+y)·(x+z) with , and to get: Then use to get: and a second time to get: Rearrange to standard order, to give f = M5 · M2 Another POM Example B A C B C A C) B, f(A, + + = B z = ) B C B C )(A A C B C (A f + + + + = y x y x x + = + ) B C C )(A A BC C ( f + + + + = ) B C )(A A B C ( f + + + + = C) B )(A C B A ( f + + + + = A y C), B (A x = + = C
Slide 20 Function ComplementsFunction Complements The complement of a function expressed as a sum of minterms is constructed by selecting the minterms missing in the sum-of-minterms canonical forms. Alternatively, the complement of a function expressed by a Sum of Minterms form is simply the Product of Maxterms with the same indices. Example: Given ) 7 , 5 , 3 , 1 ( ) z , y , x ( F m S = ) 6 , 4 , 2 , 0 ( ) z , y , x ( F m S = ) 7 , 5 , 3 , 1 ( ) z , y , x ( F M P =
Slide 21 Conversion Between FormsConversion Between Forms To convert between sum-of-minterms and product-of-maxterms form (or vice-versa) we follow these steps: Find the function complement by swapping terms in the list with terms not in the list. Change from products to sums, or vice versa. Example:Given F as before: Form the Complement: Then use the other form with the same indices this forms the complement again, giving the other form of the original function: ) 6 , 4 , 2 , 0 ( ) z , y , x ( F m S =
Slide 22 Standard FormsStandard Sum-of-Products (SOP) form: equations are written as an OR of AND terms Standard Product-of-Sums (POS) form: equations are written as an AND of OR terms Examples: SOP: POS: These mixed forms are neither SOP nor POS Standard Forms B C B A C B A + + C · ) C B (A · B) (A + + + C) (A C) B (A + + B) (A C A C B A + +
Slide 23 Standard Sum-of-Products (SOP)Standard Sum-of-Products (SOP) A sum of minterms form for n variables can be written down directly from a truth table. Implementation of this form is a two-level network of gates such that: The first level consists of n-input AND gates, and The second level is a single OR gate (with fewer than 2n inputs). This form often can be simplified so that the corresponding circuit is simpler.
Slide 24 Standard Sum-of-Products (SOP)A Simplification Example: Writing the minterm expression: F = A'B'C + AB'C' + AB'C + ABC' + ABC Simplifying: F = A B C + A (B C + B C + B C + B C) = A B C + A (B + B) (C + C) = A B C + A.1.1 = A B C + A = B C + A Simplified F contains 3 literals compared to 15 in minterm F Standard Sum-of-Products (SOP)
F = A B C + A (B C + B C + B C + B C) = A B C + A (B + B) (C + C) = A B C + A.1.1 = A B C + A = B C + A
Slide 25 AND/OR Two-level Implementation of SOP ExpressionAND/OR Two-level Implementation of SOP Expression The two implementations for F are shown below it is quite apparent which is simpler!
Slide 26 SOP and POS ObservationsSOP and POS Observations The previous examples show that: Canonical Forms (Sum-of-minterms, Product-of-Maxterms), or other standard forms (SOP, POS) differ in complexity Boolean algebra can be used to manipulate equations into simpler forms. Simpler equations lead to simpler two-level implementations Questions: How can we attain a simplest expression? Is there only one minimum cost circuit? The next lecture will deal with these issues.
Slide 27 SummarySummary What are Canonical Forms? Minterms and Maxterms Index Representation of Minterms and Maxterms Sum-of-Minterm (SOM) Representations Product-of-Maxterm (POM) Representations Representation of Complements of Functions Conversions between Representations
Table of ContentsThe object immediately following this sentence is the media player.