Code 100 - 1410 Words

Running head: Executive Management Project Code 100 Chamberlain College of Nursing EX630: Executive Practicum Professor Valda Upenieks 6/16/2012 The Master of Science Nursing student with the collaboration of my nurse executive mentor Mrs. Darville created a project that is valuable for the facility and a rewarding experience for me as a graduate student. My personal practicum experience took place at St Elizabeth’s Hospital, which is a small 90 bed community hospital located in Gonzales, La. St Elizabeth’s Hospital is a part of the Franciscan Missionaries of Our Lady Health System which was organized in 1984 to unite with three existing hospitals in Louisiana which were already a part of the system†¦show more content†¦The RRT intervenes before a crisis, which the Code 100 will be to activate the RRT, as a result lives are saved and adverse events are prevented. Data collection from the initiative has proven to be a beneficial tool; it has decreased code blue calls and has produced positive outcomes (Rankins, 2006). The IHI initiative included clinicians in collaboration with hospital and nurse leads to implement best practice protocols to prevent unnecessary deaths. The RRT deployed at first sign of patient decline does save lives (Richard, 2005). Objectives Code 100 implementation will prevent avoidable inpat ient deaths that are not an expected or outcome of a patient’s hospital stay (MHA, 2012). Code 100 which is the first section of a larger project that will require the staff nurses to document whenever they care a physician especially during the day shift and document all the signs and symptoms the patients’ exhibit and the intervention performed. The nurses will have a form to complete and the data will be collected to track the number of incidents and the outcome. The facility will put in place a Rapid Response Team in the coming weeks which will be activated when a nurse on the floor observe signs of deterioration from a patient to prevent the declination of the patient’s status. Mock Code will also be implemented to ensure the staff is ready when a Code 100 is calledShow MoreRelatedProcedures And Procedures Of Medical Coding871 Words   |  4 Pagesdetailed process in which health care professionals use to code an insurance claim, to submit for reimburseme nt of services. In the following scenario, my task would be to explain the process and steps I took, to code patients cases using ICD-9. When coding it’s very important to code correctly to avoid delays in reimbursement and proper reimbursement. The following will explain the steps I took to perform proper medical coding. 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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

Comparing poems Essay Example For Students

Comparing poems Essay The poem Presents from my aunts in Pakistan is about a half English, half Pakistani Girl who was born in Pakistan and brought up in England. Her Aunts in Pakistan sent her some traditional Pakistani clothes. In private she loves the clothes but she doest what to show her friends because she thinks that her friend wont like them.  The poets thoughts and feelings change throughout the poem. In lines 1-19 the girl is happy and grateful glistening like an orange split open (line 4) this is an optimistic image, the presents seem to be exhilarating, stunning and full of assurance. In lines 20-26 the girl feels she cant wear these clothes because she is half-English, unlike Aunt Jamila.(Lines 25 and 26). Lines 27-38 show two sides to her. One side by her saying I wanted my parents camel skin lamp. (Line 27) where she is attracted to but at the same time repelled from to consider the cruelty (line 29) but then in line 32 back to attracted to the lamp to marvel at the colours (line 32).  Lines 40-44 is about her in her bed room trying to show off her clothes but her friend isnt impressed asked to see my weekend clothes. (Line 42-43)  Lines 50-end are meant to be about the girl searching from somewhere she belongs And I was there with no fixed nationality she doesnt believe that she belongs to any fixed nationality. Staring through the fret work (line 68) she is too far away from the re-establishing Pakistan. There seems to be no set structural form to this poem. The stanzas are split up in which ever way the poet feels like in broken rhythm. The poem is written in free verse so there is no rhyme.  The use of similes and metaphors are mostly used in lines (lines 1-39) where the content is describing how beautiful the clothes are. The poet uses colours and fruit to enforce a mood across to the reader of the poem.  I liked this poem because of the way she used colours and fruit to show the way in which the girl in the poem is feeling. I also liked how she describes the all the clothes. Charlotte ONeils Song  The poem Charlotte ONeils Song is about a girl who is a servant in a mansion. She talks about her life scrubbedscrappedwashed and how she is getting bored of doing everything. The girl Charlotte ONeil, according to ships records, is a 17-year-old girl sailing to New Zealand to be a general servant in a rich mans home.  At the end she says I wont be there anymore. and it seems she slams the door and shes gone.  The mood of this poem is quite anger at the beginning and tired of doing the same job day in day out and at the end she glad she gone but know youre on your own my dear. The structure to this poem is not set up in stanzas with set syllables and number of lines. The rhyme starts of steady but the further you go into the poem the angrier she gets and the more lose the rhymes scheme goes. There are no similies or metaphors because charlotte is trying to keep you focused on what shes saying so you wont be distracted by the comparisons making what she has gone through less graphic.  I like this poem because there are no adjectives in the poem so it looks a little bare in places but she keeps it tighter by increasing the anger and letting the rhyme scheme slip to show this. I also liked the way in which she concluded the piece by saying and you can open your own front door.  Nothings changed  This poem is about the apartheid in district six, Cape Town, Africa. A man is walking through the field the fields just out side of dictrict six. He then walks into district six no board says it is, but my feet know he walks up to a white s only Inn and says theres a guard at the gatepost to hold the true Africans out. .ub8501c962e615083cf19e8c7356dab35 , .ub8501c962e615083cf19e8c7356dab35 .postImageUrl , .ub8501c962e615083cf19e8c7356dab35 .centered-text-area { min-height: 80px; position: relative; } .ub8501c962e615083cf19e8c7356dab35 , .ub8501c962e615083cf19e8c7356dab35:hover , .ub8501c962e615083cf19e8c7356dab35:visited , .ub8501c962e615083cf19e8c7356dab35:active { border:0!important; } .ub8501c962e615083cf19e8c7356dab35 .clearfix:after { content: ""; display: table; clear: both; } .ub8501c962e615083cf19e8c7356dab35 { 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; } .ub8501c962e615083cf19e8c7356dab35:active , .ub8501c962e615083cf19e8c7356dab35:hover { opacity: 1; transition: opacity 250ms; webkit-transition: opacity 250ms; background-color: #2C3E50; } .ub8501c962e615083cf19e8c7356dab35 .centered-text-area { width: 100%; position: relative ; } .ub8501c962e615083cf19e8c7356dab35 .ctaText { border-bottom: 0 solid #fff; color: #2980B9; font-size: 16px; font-weight: bold; margin: 0; padding: 0; text-decoration: underline; } .ub8501c962e615083cf19e8c7356dab35 .postTitle { color: #FFFFFF; font-size: 16px; font-weight: 600; margin: 0; padding: 0; width: 100%; } .ub8501c962e615083cf19e8c7356dab35 .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; } .ub8501c962e615083cf19e8c7356dab35:hover .ctaButton { background-color: #34495E!important; } .ub8501c962e615083cf19e8c7356dab35 .centered-text { display: table; height: 80px; padding-left : 18px; top: 0; } .ub8501c962e615083cf19e8c7356dab35 .ub8501c962e615083cf19e8c7356dab35-content { display: table-cell; margin: 0; padding: 0; padding-right: 108px; position: relative; vertical-align: middle; width: 100%; } .ub8501c962e615083cf19e8c7356dab35:after { content: ""; display: block; clear: both; } READ: Rhoda"s Diary The Withered Arm by Thomas Hardy EssayThe second half of the poem is describes the way hes feeling about the whites that are better than him .he walks up to the window and without out looking already knows what hes going to see. Before I see them, there will be crush ice white glass then he takes us down the street to working mans he backs away from the window and has a burning desire to smash the clear panes in any way which he can.  The poet feeling and thoughts develop as the significantly during the course of the poem. From when the poem begins hes walking thought the fields and although the terrain is harsh and hard he feels comfortable there. He then walks down into district six and his anger begins to be revealed. Talks about his hands, bones, lungs and hot, white inwards turning anger of my eyes. This surggests that he is beginning he is angry a bit but as the poem progresses the anger enhances with it.  The three poems are about different cultures. I liked the nothings changed poem because it has the most effect on me. The way in which the poet unravels the plot gradually shows how much nothing has changed.