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Simplification of Switching Function Free Essays

EEN1036 Digital Logic Design Chapter 4 part I Simplification of Switching Function 1 Objective s s s s Simplifying logic circuit Minimization using Karnaugh map Using Karnaugh map to obtain simplified SOP and POS expression Five-variable Karnaugh map 2 Simplifying Logic Circuits †¢ †¢ †¢ A A Boolean expression for a logic circuit may be reduced to a simpler form The simplified expression can then be used to implement a circuit equivalent to the original circuit Consider the following example: B C A B C + A BC Y AB C + AB C Y = A B C + A BC + AB C + AB C 3 Continue †¦ Checking for common factor: Y = A B C + A BC + AB C + AB C = A C ( B + B ) + AB (C + C ) Reduce the complement pairs to ‘1’ Y = A C ( B + B ) + AB (C + C ) = A C + AB Draw the circuit based on the simplified expression A B C Y 4 Continue †¦ †¢ A Consider another logic circuit: B C Y Y = C( A + B + C ) + A + C Convert to SOP expression: Y = C( A + B + C ) + A + C = AC + B C + AC Checking for common factor: Y = A(C + C ) + B C = A + BC 5 Continue †¦ †¢ †¢ Simplification of logic circuit algebraically is not always an easy task The following two steps might be useful: i. The original expression is convert into the SOP form by repeated application of DeMorgan’s theorems and multiplication of terms ii. The product terms are then checked for common factors, and factoring is performed wherever possible 6 Continue †¦ †¢ Consider the truth table below: A 0 0 0 0 1 B 0 0 1 1 0 C 0 1 0 1 0 Y 0 0 1 0 0 Minterm Boolean expression: Simplify to yield: Y = A BC + ABC + AB C Y = BC ( A + A) + AB C = BC + AB C 1 0 1 1 1 1 0 1 1 1 1 0 †¢ If minterms are only differed by one bit, they can be simplified, e. We will write a custom essay sample on Simplification of Switching Function or any similar topic only for you Order Now g. A BC ABC 7 Continue †¦ †¢ More example: A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 Y 0 1 1 0 0 1 1 0 Minterm Boolean expression: Y = A B C + A BC + AB C + ABC Minterms 1 and 5, 2 and 6 are only differ by one bit: Y = B C ( A + A) + BC ( A + A) = BC + B C A B C Y 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 1 0 1 0 Minterm Boolean expression: Y = A B C + A BC + AB C + ABC Checking and factoring minterms differed by only by one bit: Y = A C ( B + B ) + AC ( B + B ) = A C + AC = C ( A + A) =C 8 Continue †¦ †¢ †¢ †¢ Though truth table can help us to detect minterms which are only differed by one bit, it is not arranged in a proper way A Karnaugh map (K-map) is a tool, which help us to detect and simplify minterms graphically It is a rearrangement of the truth table where each adjacent cell is only differed by one bit By looping adjacent minterms, it is similar to grouping the minterms with a single bit difference on the truth table 9 K arnaugh Map †¢ †¢ A K-map is just a rearrangement of truth table, so that minterms with a single-bit difference can be detected easily Figure below shows 4 possible arrangement of 3-variable K-map A BC 0 0 01 1 11 3 10 2 C AB 00 0 01 2 11 6 10 4 0 1 4 5 7 6 0 1 1 3 7 5 AB C 0 0 1 1 BC A 0 0 1 4 00 01 2 3 00 01 1 5 11 6 7 11 3 7 10 4 5 10 2 6 10 Continue †¦ †¢ Figure below show two possible arrangement of 4variable K-map CD AB 00 0 01 1 11 3 10 2 AB 00 CD 01 4 11 12 10 8 00 01 4 5 7 6 00 0 01 1 5 13 9 11 12 13 15 14 11 3 7 15 11 10 8 9 11 10 10 2 6 14 10 †¢ Notice that the K-map is labeled so that horizontally and vertically adjacent cells differ only by one bit. 11 Continue †¦ †¢ The K-map for both SOP and POS form are shown below: C D C D CD C D AB AB AB 0 1 3 2 C+D C+ D C + D C +D A +B 0 1 3 2 4 5 7 6 A+B A+B A +B 4 5 7 6 12 13 15 14 12 13 15 14 AB 8 9 11 10 8 9 11 10 SOP form (minterm) POS form (maxterm) †¢ †¢ The simplified SOP expression can be obtained by properly combining those adjacent cells which contains ‘1’ This process of combining adjacent minterms is known as 12 looping Continue †¦ †¢ †¢ Each loop of minterms will form a group which can be represented by a product term When a variable appears in both complemented and uncomplemented form within a group, that variable is eliminated from the product term C D C D CD C D AB AB AB AB 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 group 2 group 1: C D( AB + AB ) = AC D group 2: AB(C D + CD ) = ABD Simplified SOP expression: Y = AC D + ABD 13 group 1 Continue †¦ †¢ Consider another K-map: C D C D CD C D AB AB AB AB 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 group 1 C D C D CD C D AB AB AB AB 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 14 group 1: ( A B + AB )(C D + CD ) = BD Simplified SOP expression: Y = BD group 1: C D ( A B + A B + AB + AB ) = C D Simplified SOP expression: Y = CD group 1 From truth table to K-map †¢ The content of each cell can be directly plot on the Kmap according to the truth table Consider the following example: 0 1 2 3 4 5 6 7 A 0 0 0 0 1 1 1 1 B C Y 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0 B C B C BC B C A A 1 0 1 1 0 3 1 2 0 4 0 5 0 7 1 6 AB BC Simplified SOP expression: Y = A B + BC 15 Continue †¦ †¢ Consider the following 4-variable K-map A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C D Y 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 1 1 1 C D C D CD C D AB AB AB 0 0 0 1 1 1 0 1 0 0 1 0 3 0 0 0 ACD 2 4 5 7 6 12 13 15 14 AB 0 8 9 11 0 10 ABD Simplified SOP expression: Y = A C D + ABD 16 Continue †¦ †¢ Some guidelines: i. Construct K-map and fill it according to the truth table ii. Only loop cells in the power of 2, i. e. 2 cells, 4 cells, 8 cells and so on iii. Always start by looping the isolated minterms iv. Look for minterms which are adjacent to only one minterm and loop them together v. Proceed on to loop the largest possible groups, from eight minterms (octet), 4 minterms (quad) to 2 minterms (pair) vi. Obtain the product term for each group vii. The sum of these product terms will be the simplified SOP expression 17 Continue †¦ Example: a. Obtain the simplify SOP expression for the truth table: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Y 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 1 C D C D CD C D AB AB AB AB A B CD 0 0 0 0 0 0 1 1 0 1 0 1 1 3 1 0 0 0 2 4 5 7 6 12 13 15 14 8 9 1 11 10 BD ACD Simplified SOP expression: Y = A B CD + ACD + BD 18 Continue †¦ b. Obtain the simplify SOP expression from the K-map: ACD C D C D CD C D AB AB ABC 0 0 1 0 1 1 1 0 0 1 1 1 ACD 0 1 0 0 A BC AB AB Simplified SOP expression: Y = A C D + A BC + ACD + ABC 19 Continue †¦ c. Obtain the simplify SOP expression from the K-map: alternative solution: C D C D CD C D AB AB AB C D C D CD C D AB A CD 0 0 0 0 AC D 0 0 1 1 1 1 0 1 0 0 0 0 AB D 0 0 0 0 AC D 0 0 1 1 1 1 0 1 0 0 0 0 B CD A CD AB AB AB AB Y = A CD + AC D + AB D Y = A CD + AC D + B CD 20 General Terminology for Logic Minimization †¢ †¢ Here, we define four terms to provide the basis for general function minimization techniques These terms are implicant, prime implicant, essential prime implicant and cover We refer to the K-map below in explaining each term B C B C BC B C A A 1 0 1 1 3 2 1 4 1 5 1 7 6 †¢ †¢ An implicant is a product term that could be used to cover minterms of the function In the K-map above, there are 11 implicants: 5 m interms: {A B C , A BC , AB C , AB C , ABC} 5 group of two adjacent minterms: {AB , AC , A C , B C , BC} 1 group of four adjacent minterms:{C} 21 Continue †¦ †¢ †¢ †¢ A prime implicant is an implicant that is not part of any other mplicant In the K-map, there are two prime implicant: C and AB An essential prime implicant is a prime implicant that covers at least one minterm that is not covered by any other prime implicants Prime implicant AB is essential as it is the only prime implicant that covers minterm 4 Prime implicant C is also essential as it is the only prime implicant that covers minterm 1, 3 and 7 A cover of a function is a set of prime implicants for which each minterm of the function is contained in (covered by) at least one prime implicant All essential prime implicants must be used in any cover of a function 22 †¢ †¢ †¢ Continue †¦ †¢ †¢ For the K-map above, the set of implicants { AB , C} represents a cover of the fun ction A minimum cover contains the minimum number of prime implicants which contains all minterm in the function Consider the 4-variable K-map below: C D C D CD C D AB AB AB AB 1 1 Prime implicants †¢ C D C D CD C D AB AB AB 1 1 1 1 1 1 1 1 1 AB AB AB AB C D C D CD C D 1 1 1 1 Minimum cover 1 1 1 1 1 1 1 1 1 1 1 1 AB Essential prime implicants 23 Continue †¦ †¢ Consider another K-map C D C D CD C D AB AB AB AB 1 1 1 1 1 1 1 Prime implicants C D C D CD C D AB 1 1 1 1 1 1 1 1 1 1 AB AB 1 AB Essential prime implicants (minimum cover) 24 Don’t Care Conditions †¢ †¢ †¢ †¢ Some logic circuit will have certain input conditions whereby the output is unspecified This is usually because these input conditions would never occur In other words, we â€Å"don’t care† whether the output is HIGH or LOW Consider the following example: An air conditioning system has two inputs, C and H: – C will be ‘1’ if temperature is too cold (below 15 °C) Otherwise, it will be ‘0’ – H will be ‘1’ if temperature is too hot (above 25 °C) Otherwise, it will be ‘0’ – Output Y will be ‘1’ if temperature is too cold or too hot. If the temperature is acceptable, Y will be ‘0’ 25 Continue †¦ As there are two inputs, there are 4 possible logical conditions: C 0 0 1 1 H 0 1 0 1 Y 0 1 1 X meaning just nice too hot too cold ? Input condition C = 1, H = 1 has no real meaning, as it is impossible to be too hot and too cold at the same time We put a ‘X’ at the output corresponds to this input condition as this input condition cannot occur 26 K-map and Don’t Care Term †¢ Don’t care term, ‘X’ can be treated as ‘0’ or ‘1’ since they cannot occur In K-map, we can choose the don’t care term as ‘0’ or ‘1’ to our advantage A B C D Y 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 X 0 1 0 0 1 0 1 0 1 X 0 1 1 0 0 0 1 1 1 X 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 X 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 X C D C D CD C D AB AB AB 0 1 1 0 1 X 1 0 X X X X 0 0 1 0 AB Simplified Boolean expression: Y = AB + BC + A D 27 More examples †¦ C D C D CD C D AB AB AB AB C D C D CD C D AB AB AB AB 1 1 X 1 0 1 X 1 0 0 X X 0 1 X X 1 0 X 1 0 0 X 0 0 0 X X 1 X X Y = C D + BC + BD + A C D C D CD C D AB AB AB Y = B D + CD C D C D CD C D AB AB AB 0 0 1 0 1 X 1 1 0 1 X 0 0 0 0 0 1 1 X 0 1 X X 1 0 1 X X 0 0 X X 28 AB AB Y = ABC + C D + BD Y = A C + BD + AD Plotting function in Canonical Form †¢ Logic function may be expressed in many forms, ranging from simple SOP/POS expression to more complex expressions However, each of them has a unique canonical SOP/POS form If a Boolean expression is expressed in canonical form, it can be readily plotted on the K-map Consider the following Boolean expression: †¢ †¢ †¢ Y = ABC + B C Convert to canonical SOP expression: Y = ABC + B C ( A + A) = ABC + A B C + AB C 29 Continue †¦ Y = ABC + A B C + AB C Plotting the canonical SOP expression onto K-map B C B C BC B C A A 1 1 0 0 BC 0 0 0 1 AC Simplified SOP expression: Y = B C + AC †¢ Consider plotting the following Boolean expression on K-map: Y = C ( A ? B) + A + B 30 Continue †¦ First, convert to SOP expression Y = C ( A ? B) + A + B = C ( AB + A B) + A B = AB C + A BC + A B (C + C ) = AB C + A BC + A B C + A B C B C B C BC B C A A 1 0 AB 1 1 1 0 BC 0 0 AC ?Y = A B + B C + A C 31 Plotting K-map from SOP expression †¢ †¢ It is sometime too tedious to convert a Boolean expression to its canonical SOP form Consider the following Boolean expression: Y = AB (C + D )(C + D ) + A + B Convert to SOP form: Y = ( AB C + AB D )(C + D ) + A B = AB C D + AB CD + A B Convert to canonical form: Y = AB C D + AB CD + A B (C + C )( D + D) = AB C D + AB CD + ( A B C + A B C )( D + D) = AB C D + AB CD + A B C D + A B C D + A B CD + A B CD 32 Continue †¦ Y = AB C D + AB CD + A B C D + A B C D + A B CD + A B CD Plot the minterm on K-map: C D C D CD C D AB AB AB 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 AB AB B CD BC D Simplified SOP expression: Y = B C D + B CD + A B 33 Continue †¦ †¢ †¢ †¢ †¢ †¢ Boolean expression can be plotted on to the K-map from its SOP form Product terms with four variables are the minterms and correspond to a single cell on the K-map Product term with three variables corresponds to a loop of two adjacent minterms Product term with only two variables is a quad (a loop of four adjacent minterms) Product term with a single variable is an octet (a loop of eight adjacent minterms) 1 cell 2 cells Y = A + BC + B CD + ABCD 4 cells 8 cells 34 Continue †¦ †¢ Consider the previous example: Y = AB C D + AB CD + A B minterms 4 cells †¢ †¢ †¢ Both minterms are directly plotted on the K-map The loop which corresponds to A B is drawn on the K-map The cells inside the loops are filled with ‘1’ C D C D CD C D AB AB AB AB 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 AB AB C D A B CD 35 Continue †¦ †¢ Consider the following Boolean expression: Y = ( A + B )( AC + D ) Convert to SOP form: Y = AC + AD + ABC + BD Plot the SOP onto K-map C D C D CD C D AB AB AB AB AC BD C D C D CD C D AB AB ill cells in loops with ‘1’ 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 1 36 ABC AB AB AD Continue †¦ Obtain the simplified SOP expression from K-map: C D C D CD C D AB AB AB AB 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 1 Simplified SOP expression: Y = AC + AD + BD 37 Continue †¦ Example: Redesign the logic circuit below from its simplified SOP expression: A B C D Z Z = ( B + D )( B + D ) + B(CD + A D ) 38 Continue †¦ Z = ( B + D )( B + D ) + B(CD + A D ) = B + D + B + D + BCD + A BD = BD + B D + BCD + A BD C D C D CD C D AB AB AB 1 1 0 1 0 1 1 0 0 1 1 0 1 1 0 1 AB Z = BD + B D + A B 39 How to cite Simplification of Switching Function, Papers Simplification of Switching Function Free Essays EEN1036 Digital Logic Design Chapter 4 part I Simplification of Switching Function 1 Objective s s s s Simplifying logic circuit Minimization using Karnaugh map Using Karnaugh map to obtain simplified SOP and POS expression Five-variable Karnaugh map 2 Simplifying Logic Circuits †¢ †¢ †¢ A A Boolean expression for a logic circuit may be reduced to a simpler form The simplified expression can then be used to implement a circuit equivalent to the original circuit Consider the following example: B C A B C + A BC Y AB C + AB C Y = A B C + A BC + AB C + AB C 3 Continue †¦ Checking for common factor: Y = A B C + A BC + AB C + AB C = A C ( B + B ) + AB (C + C ) Reduce the complement pairs to ‘1’ Y = A C ( B + B ) + AB (C + C ) = A C + AB Draw the circuit based on the simplified expression A B C Y 4 Continue †¦ †¢ A Consider another logic circuit: B C Y Y = C( A + B + C ) + A + C Convert to SOP expression: Y = C( A + B + C ) + A + C = AC + B C + AC Checking for common factor: Y = A(C + C ) + B C = A + BC 5 Continue †¦ †¢ †¢ Simplification of logic circuit algebraically is not always an easy task The following two steps might be useful: i. The original expression is convert into the SOP form by repeated application of DeMorgan’s theorems and multiplication of terms ii. The product terms are then checked for common factors, and factoring is performed wherever possible 6 Continue †¦ †¢ Consider the truth table below: A 0 0 0 0 1 B 0 0 1 1 0 C 0 1 0 1 0 Y 0 0 1 0 0 Minterm Boolean expression: Simplify to yield: Y = A BC + ABC + AB C Y = BC ( A + A) + AB C = BC + AB C 1 0 1 1 1 1 0 1 1 1 1 0 †¢ If minterms are only differed by one bit, they can be simplified, e. We will write a custom essay sample on Simplification of Switching Function or any similar topic only for you Order Now g. A BC ABC 7 Continue †¦ †¢ More example: A 0 0 0 0 1 1 1 1 B 0 0 1 1 0 0 1 1 C 0 1 0 1 0 1 0 1 Y 0 1 1 0 0 1 1 0 Minterm Boolean expression: Y = A B C + A BC + AB C + ABC Minterms 1 and 5, 2 and 6 are only differ by one bit: Y = B C ( A + A) + BC ( A + A) = BC + B C A B C Y 0 0 0 1 0 0 1 0 0 0 1 1 1 1 1 1 0 0 1 1 0 1 0 1 0 1 1 0 1 0 1 0 Minterm Boolean expression: Y = A B C + A BC + AB C + ABC Checking and factoring minterms differed by only by one bit: Y = A C ( B + B ) + AC ( B + B ) = A C + AC = C ( A + A) =C 8 Continue †¦ †¢ †¢ †¢ Though truth table can help us to detect minterms which are only differed by one bit, it is not arranged in a proper way A Karnaugh map (K-map) is a tool, which help us to detect and simplify minterms graphically It is a rearrangement of the truth table where each adjacent cell is only differed by one bit By looping adjacent minterms, it is similar to grouping the minterms with a single bit difference on the truth table 9 K arnaugh Map †¢ †¢ A K-map is just a rearrangement of truth table, so that minterms with a single-bit difference can be detected easily Figure below shows 4 possible arrangement of 3-variable K-map A BC 0 0 01 1 11 3 10 2 C AB 00 0 01 2 11 6 10 4 0 1 4 5 7 6 0 1 1 3 7 5 AB C 0 0 1 1 BC A 0 0 1 4 00 01 2 3 00 01 1 5 11 6 7 11 3 7 10 4 5 10 2 6 10 Continue †¦ †¢ Figure below show two possible arrangement of 4variable K-map CD AB 00 0 01 1 11 3 10 2 AB 00 CD 01 4 11 12 10 8 00 01 4 5 7 6 00 0 01 1 5 13 9 11 12 13 15 14 11 3 7 15 11 10 8 9 11 10 10 2 6 14 10 †¢ Notice that the K-map is labeled so that horizontally and vertically adjacent cells differ only by one bit. 11 Continue †¦ †¢ The K-map for both SOP and POS form are shown below: C D C D CD C D AB AB AB 0 1 3 2 C+D C+ D C + D C +D A +B 0 1 3 2 4 5 7 6 A+B A+B A +B 4 5 7 6 12 13 15 14 12 13 15 14 AB 8 9 11 10 8 9 11 10 SOP form (minterm) POS form (maxterm) †¢ †¢ The simplified SOP expression can be obtained by properly combining those adjacent cells which contains ‘1’ This process of combining adjacent minterms is known as 12 looping Continue †¦ †¢ †¢ Each loop of minterms will form a group which can be represented by a product term When a variable appears in both complemented and uncomplemented form within a group, that variable is eliminated from the product term C D C D CD C D AB AB AB AB 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 group 2 group 1: C D( AB + AB ) = AC D group 2: AB(C D + CD ) = ABD Simplified SOP expression: Y = AC D + ABD 13 group 1 Continue †¦ †¢ Consider another K-map: C D C D CD C D AB AB AB AB 0 0 0 0 0 1 1 0 0 1 1 0 0 0 0 0 group 1 C D C D CD C D AB AB AB AB 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 14 group 1: ( A B + AB )(C D + CD ) = BD Simplified SOP expression: Y = BD group 1: C D ( A B + A B + AB + AB ) = C D Simplified SOP expression: Y = CD group 1 From truth table to K-map †¢ The content of each cell can be directly plot on the Kmap according to the truth table Consider the following example: 0 1 2 3 4 5 6 7 A 0 0 0 0 1 1 1 1 B C Y 0 0 1 0 1 1 1 0 1 1 1 0 0 0 0 0 1 0 1 0 1 1 1 0 B C B C BC B C A A 1 0 1 1 0 3 1 2 0 4 0 5 0 7 1 6 AB BC Simplified SOP expression: Y = A B + BC 15 Continue †¦ †¢ Consider the following 4-variable K-map A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C D Y 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 0 1 0 1 0 0 1 1 0 0 0 0 0 1 1 1 0 0 1 1 1 C D C D CD C D AB AB AB 0 0 0 1 1 1 0 1 0 0 1 0 3 0 0 0 ACD 2 4 5 7 6 12 13 15 14 AB 0 8 9 11 0 10 ABD Simplified SOP expression: Y = A C D + ABD 16 Continue †¦ †¢ Some guidelines: i. Construct K-map and fill it according to the truth table ii. Only loop cells in the power of 2, i. e. 2 cells, 4 cells, 8 cells and so on iii. Always start by looping the isolated minterms iv. Look for minterms which are adjacent to only one minterm and loop them together v. Proceed on to loop the largest possible groups, from eight minterms (octet), 4 minterms (quad) to 2 minterms (pair) vi. Obtain the product term for each group vii. The sum of these product terms will be the simplified SOP expression 17 Continue †¦ Example: a. Obtain the simplify SOP expression for the truth table: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 A 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 B 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1 C 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 D 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 Y 0 0 1 0 0 1 0 1 0 0 0 1 0 1 0 1 C D C D CD C D AB AB AB AB A B CD 0 0 0 0 0 0 1 1 0 1 0 1 1 3 1 0 0 0 2 4 5 7 6 12 13 15 14 8 9 1 11 10 BD ACD Simplified SOP expression: Y = A B CD + ACD + BD 18 Continue †¦ b. Obtain the simplify SOP expression from the K-map: ACD C D C D CD C D AB AB ABC 0 0 1 0 1 1 1 0 0 1 1 1 ACD 0 1 0 0 A BC AB AB Simplified SOP expression: Y = A C D + A BC + ACD + ABC 19 Continue †¦ c. Obtain the simplify SOP expression from the K-map: alternative solution: C D C D CD C D AB AB AB C D C D CD C D AB A CD 0 0 0 0 AC D 0 0 1 1 1 1 0 1 0 0 0 0 AB D 0 0 0 0 AC D 0 0 1 1 1 1 0 1 0 0 0 0 B CD A CD AB AB AB AB Y = A CD + AC D + AB D Y = A CD + AC D + B CD 20 General Terminology for Logic Minimization †¢ †¢ Here, we define four terms to provide the basis for general function minimization techniques These terms are implicant, prime implicant, essential prime implicant and cover We refer to the K-map below in explaining each term B C B C BC B C A A 1 0 1 1 3 2 1 4 1 5 1 7 6 †¢ †¢ An implicant is a product term that could be used to cover minterms of the function In the K-map above, there are 11 implicants: 5 m interms: {A B C , A BC , AB C , AB C , ABC} 5 group of two adjacent minterms: {AB , AC , A C , B C , BC} 1 group of four adjacent minterms:{C} 21 Continue †¦ †¢ †¢ †¢ A prime implicant is an implicant that is not part of any other mplicant In the K-map, there are two prime implicant: C and AB An essential prime implicant is a prime implicant that covers at least one minterm that is not covered by any other prime implicants Prime implicant AB is essential as it is the only prime implicant that covers minterm 4 Prime implicant C is also essential as it is the only prime implicant that covers minterm 1, 3 and 7 A cover of a function is a set of prime implicants for which each minterm of the function is contained in (covered by) at least one prime implicant All essential prime implicants must be used in any cover of a function 22 †¢ †¢ †¢ Continue †¦ †¢ †¢ For the K-map above, the set of implicants { AB , C} represents a cover of the fun ction A minimum cover contains the minimum number of prime implicants which contains all minterm in the function Consider the 4-variable K-map below: C D C D CD C D AB AB AB AB 1 1 Prime implicants †¢ C D C D CD C D AB AB AB 1 1 1 1 1 1 1 1 1 AB AB AB AB C D C D CD C D 1 1 1 1 Minimum cover 1 1 1 1 1 1 1 1 1 1 1 1 AB Essential prime implicants 23 Continue †¦ †¢ Consider another K-map C D C D CD C D AB AB AB AB 1 1 1 1 1 1 1 Prime implicants C D C D CD C D AB 1 1 1 1 1 1 1 1 1 1 AB AB 1 AB Essential prime implicants (minimum cover) 24 Don’t Care Conditions †¢ †¢ †¢ †¢ Some logic circuit will have certain input conditions whereby the output is unspecified This is usually because these input conditions would never occur In other words, we â€Å"don’t care† whether the output is HIGH or LOW Consider the following example: An air conditioning system has two inputs, C and H: – C will be ‘1’ if temperature is too cold (below 15 °C) Otherwise, it will be ‘0’ – H will be ‘1’ if temperature is too hot (above 25 °C) Otherwise, it will be ‘0’ – Output Y will be ‘1’ if temperature is too cold or too hot. If the temperature is acceptable, Y will be ‘0’ 25 Continue †¦ As there are two inputs, there are 4 possible logical conditions: C 0 0 1 1 H 0 1 0 1 Y 0 1 1 X meaning just nice too hot too cold ? Input condition C = 1, H = 1 has no real meaning, as it is impossible to be too hot and too cold at the same time We put a ‘X’ at the output corresponds to this input condition as this input condition cannot occur 26 K-map and Don’t Care Term †¢ Don’t care term, ‘X’ can be treated as ‘0’ or ‘1’ since they cannot occur In K-map, we can choose the don’t care term as ‘0’ or ‘1’ to our advantage A B C D Y 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 1 1 X 0 1 0 0 1 0 1 0 1 X 0 1 1 0 0 0 1 1 1 X 1 0 0 0 0 1 0 0 1 0 1 0 1 0 0 1 0 1 1 X 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 X C D C D CD C D AB AB AB 0 1 1 0 1 X 1 0 X X X X 0 0 1 0 AB Simplified Boolean expression: Y = AB + BC + A D 27 More examples †¦ C D C D CD C D AB AB AB AB C D C D CD C D AB AB AB AB 1 1 X 1 0 1 X 1 0 0 X X 0 1 X X 1 0 X 1 0 0 X 0 0 0 X X 1 X X Y = C D + BC + BD + A C D C D CD C D AB AB AB Y = B D + CD C D C D CD C D AB AB AB 0 0 1 0 1 X 1 1 0 1 X 0 0 0 0 0 1 1 X 0 1 X X 1 0 1 X X 0 0 X X 28 AB AB Y = ABC + C D + BD Y = A C + BD + AD Plotting function in Canonical Form †¢ Logic function may be expressed in many forms, ranging from simple SOP/POS expression to more complex expressions However, each of them has a unique canonical SOP/POS form If a Boolean expression is expressed in canonical form, it can be readily plotted on the K-map Consider the following Boolean expression: †¢ †¢ †¢ Y = ABC + B C Convert to canonical SOP expression: Y = ABC + B C ( A + A) = ABC + A B C + AB C 29 Continue †¦ Y = ABC + A B C + AB C Plotting the canonical SOP expression onto K-map B C B C BC B C A A 1 1 0 0 BC 0 0 0 1 AC Simplified SOP expression: Y = B C + AC †¢ Consider plotting the following Boolean expression on K-map: Y = C ( A ? B) + A + B 30 Continue †¦ First, convert to SOP expression Y = C ( A ? B) + A + B = C ( AB + A B) + A B = AB C + A BC + A B (C + C ) = AB C + A BC + A B C + A B C B C B C BC B C A A 1 0 AB 1 1 1 0 BC 0 0 AC ?Y = A B + B C + A C 31 Plotting K-map from SOP expression †¢ †¢ It is sometime too tedious to convert a Boolean expression to its canonical SOP form Consider the following Boolean expression: Y = AB (C + D )(C + D ) + A + B Convert to SOP form: Y = ( AB C + AB D )(C + D ) + A B = AB C D + AB CD + A B Convert to canonical form: Y = AB C D + AB CD + A B (C + C )( D + D) = AB C D + AB CD + ( A B C + A B C )( D + D) = AB C D + AB CD + A B C D + A B C D + A B CD + A B CD 32 Continue †¦ Y = AB C D + AB CD + A B C D + A B C D + A B CD + A B CD Plot the minterm on K-map: C D C D CD C D AB AB AB 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 AB AB B CD BC D Simplified SOP expression: Y = B C D + B CD + A B 33 Continue †¦ †¢ †¢ †¢ †¢ †¢ Boolean expression can be plotted on to the K-map from its SOP form Product terms with four variables are the minterms and correspond to a single cell on the K-map Product term with three variables corresponds to a loop of two adjacent minterms Product term with only two variables is a quad (a loop of four adjacent minterms) Product term with a single variable is an octet (a loop of eight adjacent minterms) 1 cell 2 cells Y = A + BC + B CD + ABCD 4 cells 8 cells 34 Continue †¦ †¢ Consider the previous example: Y = AB C D + AB CD + A B minterms 4 cells †¢ †¢ †¢ Both minterms are directly plotted on the K-map The loop which corresponds to A B is drawn on the K-map The cells inside the loops are filled with ‘1’ C D C D CD C D AB AB AB AB 1 0 0 1 1 0 0 0 1 0 0 1 1 0 0 0 AB AB C D A B CD 35 Continue †¦ †¢ Consider the following Boolean expression: Y = ( A + B )( AC + D ) Convert to SOP form: Y = AC + AD + ABC + BD Plot the SOP onto K-map C D C D CD C D AB AB AB AB AC BD C D C D CD C D AB AB ill cells in loops with ‘1’ 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 1 36 ABC AB AB AD Continue †¦ Obtain the simplified SOP expression from K-map: C D C D CD C D AB AB AB AB 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 1 Simplified SOP expression: Y = AC + AD + BD 37 Continue †¦ Example: Redesign the logic circuit below from its simplified SOP expression: A B C D Z Z = ( B + D )( B + D ) + B(CD + A D ) 38 Continue †¦ Z = ( B + D )( B + D ) + B(CD + A D ) = B + D + B + D + BCD + A BD = BD + B D + BCD + A BD C D C D CD C D AB AB AB 1 1 0 1 0 1 1 0 0 1 1 0 1 1 0 1 AB Z = BD + B D + A B 39 How to cite Simplification of Switching Function, Essay examples

Outcome

Outcome-Based Education Essay Outcome-based Education Denied Joan M. Montana, R. N. Outcome-based education Is the trend In todays curriculum. It Is currently favored Internationally In countries such as Canada, South Africa, New Zealand, and united States (Milan, 2004, cited by Butler, 2004). Outcome-based education encompasses learning at the students pace to achieve a desired outcome. It is defined by Chary (2003) as a method of curriculum design and teaching that focuses on what students can actually do after they are taught. Basically, it puts emphasis on measuring what the students have learned and understand after a rouser by not merely giving examinations but rather, activities that assess critical thinking of the students. In this light, outcome-based education is beneficial to our educational system In providing professional and competent workers. Biggs and Tang (2007) stated that It Is beneficial to the students since the learning outcome tell them precisely not only what they are supposing to be learning, but how and to what standard. In the development of outcome-based education, educators studied the work of Carroll and Bloom. Carroll said that it was inappropriate to fix the time for study and expect variable learning results from students (Davis, 2003). Bloom developed Carols thinking into the notion of mastery learning, in which a fixed level of performance was to be achieved by students by changing the relationship between time and learning (Davis, 2003). From this, the underlying principle of outcome-based education was achieved in a sense that the learning became fixed and the time to achieve the learning became the variable. The primary aim of outcome-based education Is to facilitate desired changes within the learners, by Increasing knowledge, developing skills and/or positively Influencing attitudes, values and Judgment (Butler, 2004). As a whole, It targets the cognitive, affective and psychosomatic domains of learning, thus, facilitating better understanding and retention of the topics learned. As stated by Killeen (2000), outcome-based learning is underpinned by three basic premises: All students can learn and succeed, but not all in the same time or in the same way. Successful learning promotes even more successful learning. * Schools (and teachers) control the conditions that determine whether or not students are successful at school learning. Spade and his colleagues (cited by Lawson ; Easels-Williams, 2007) identifies four organizing principles of OBOE, namely: (1) clarity of focus; (2) designing back; (3) high expectation for all students; and (4) capability of the teachers to provide expanded opportunities to allow for achievement of outcomes In a variety of ways. In clarity of focus, the learning outcomes broad and specific must be clearly Identified for the students; and all teaching and learning actively must be aligned with these outcomes. In designing back, the curriculum content should flow clearly from the most general valued outcomes, to the related, ore special outcomes, to class lesson actively. In Nell expectations Tort all students, it requires that successful and challenging learning experiences and achievement of high standards be part of learning for all students. In the fourth principle, it is suggested that different learners may take different routes, and different amounts of time or different numbers of attempts, to achieve the same outcome. Moreover, visionary elements, proposed by Spade (cited by Lawson ; Easels-Williams, 2007) emphasized the need for educational leaders to engage in empowerment thinking, visionary thinking and future-focused thinking that looks to the world as it should be in the future. Killeen (2000) acknowledged the significance of using outcomes to guide instructional planning. He claimed that there are three major steps in an OBOE system. These steps includes: deciding on the outcomes that students are to achieve; deciding how to assist students to achieve those outcome and deciding how to determine when students have achieved the outcomes. In addition, teaching strategies for OBOE is said to vary in numerous ways. Whatever approach to teaching will be used, it is important for educators to keep the following points in mind: * The main focus should be on learning rather than teaching Students cannot not learn when if they do not think * Thinking is facilitated and encouraged by the processes that were being used with the contents as well as by the content itself. The subject does not exist in isolation -educators should have to help students make links to other subject. * Educators have the main responsibility to help students learn how to learn. On the other hand, students also have responsibilities for learning in an OBOE system. Cookbook (1997, as cited by Killeen, 2000) suggest that learners are responsible for th eir own learning and progress. The principle acknowledges the fact that learning is ultimately a personal and an internal event. Students must be motivated if they are to learn. .u5125200dd93ef216613ea39f29f6407e , .u5125200dd93ef216613ea39f29f6407e .postImageUrl , .u5125200dd93ef216613ea39f29f6407e .centered-text-area { min-height: 80px; position: relative; } .u5125200dd93ef216613ea39f29f6407e , .u5125200dd93ef216613ea39f29f6407e:hover , .u5125200dd93ef216613ea39f29f6407e:visited , .u5125200dd93ef216613ea39f29f6407e:active { border:0!important; } .u5125200dd93ef216613ea39f29f6407e .clearfix:after { content: ""; display: table; clear: both; } .u5125200dd93ef216613ea39f29f6407e { display: block; transition: background-color 250ms; webkit-transition: background-color 250ms; width: 100%; opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #95A5A6; } .u5125200dd93ef216613ea39f29f6407e:active , .u5125200dd93ef216613ea39f29f6407e:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .u5125200dd93ef216613ea39f29f6407e .centered-text-area { width: 100%; position: relative ; } .u5125200dd93ef216613ea39f29f6407e .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .u5125200dd93ef216613ea39f29f6407e .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .u5125200dd93ef216613ea39f29f6407e .ctaButton { background-color: #7F8C8D!important; color: #2980B9; border: none; border-radius: 3px; box-shadow: none; font-size: 14px; font-weight: bold; line-height: 26px; moz-border-radius: 3px; text-align: center; text-decoration: none; text-shadow: none; width: 80px; min-height: 80px; background: url(https://artscolumbia.org/wp-content/plugins/intelly-related-posts/assets/images/simple-arrow.png)no-repeat; position: absolute; right: 0; top: 0; } .u5125200dd93ef216613ea39f29f6407e:hover .ctaButton { background-color: #34495E!important; } .u5125200dd93ef216613ea39f29f6407e .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .u5125200dd93ef216613ea39f29f6407e .u5125200dd93ef216613ea39f29f6407e-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .u5125200dd93ef216613ea39f29f6407e:after { content: ""; display: block; clear: both; } READ: What is culture? Argumentative EssayKilleen (2000) summarized three main points in the importance of motivating: students need to know why they are learning whatever they are learning, they need to see some value in this learning, and they need to believe that they can be successful. Outcome-based education allows the students to learn independently at heir own pace, having an outcome to achieve at the end of the learning process. The teachers are merely facilitators of learning guiding and assisting students to understand information and to transform it into their own personal knowledge. Outcome-based education, as a trend in the present curriculum, has its advantages and disadvantages. Mostly, students would benefit from this system. As stated by Razorblades and Uncharismatic (2013), OBOE encourages self-directed learning and allow the students to have a meta-cognitive understanding of the educational program and their role in that process. It also helps students to become aware of what they should be learning, aware of what they are actually learning, and aware of the control that they have over their own learning (Killeen, 2000). In contrast to its benefits, critics of OBOE pointed out several disadvantages. One is stated by Razorblades and Uncharismatic (2013) that OBOE conflicts with the wonderful, unpredictable voyages of exploration that characterize learning through discovery and inquiry. Another is that it is too technical or mechanical or inflexible such that innovation and creativity of the teachers would be killed. Furthermore, it emphasizes minimum levels of achievement and thus encourages mediocrity. However, these can De met Day OBOE In ten sense Tanat teachers would a De addle to practice Innovation Ana creativity in developing teaching strategies that would help the students achieve the outcomes. In terms of mediocrity, OBOE emphasizes the principle of high expectations for all the students wherein the teacher must set the criteria for the outcomes. It can be said that one of the major disadvantage of OBOE is the burden to the teachers in terms of findings ways to help the students arrive at the desired outcomes. In inclusion, outcome based education upgrades the educational system in terms of student involvement in the learning process, and promotes higher level of functioning after the learning process.