Geriatrics and Long-term Care Essay Example for Free

Geriatrics and Long-term Care Essay Though her sons and daughters check in on her all of the time, they are not there 24 hours a day. She does not want to have something happen and no one find out until several hours or days go by. She is very active in the community and church and I expect that she will remain so, even after moving into the assisted living facilities. This report seeks to uncover long-term care/housing programs and services provided to older adults. I will focus on the mission and services provided. From this report, I expect to gain an understanding of long-term care options and the differences amongst them, so it will be useful for my aging parents. Introduction There are 1,065,502 people (15. 1%), aged 60 and above, in Virginia (U. S. Census Bureau, 2000). There are 216,588 households with those aged 65 and over living alone with 565,204 households that contain individuals aged 65 and above (U. S. Census Bureau, 2000). The second number, 565,204 does not state that the household contains only 65 and over individual or could be an older person living with their son or daughters family. If taken into the later context that leaves 283,728 elderly that lives other than their own home or with another family member. This results in a huge demand for housing of the elderly on this country. With the onset of the baby boomers coming of old age, it is necessary that the government looks at ways to handle the increased need for housing of the elderly. Determining long-term care options Not everyone will need a long-term care option when they age. In fact currently most of the elderly remain in their residence. The questions elderly must ask themselves if considering long-term care are many. Elderly may consider long term care if they have a physical or mental disability, chronic illness, terminal illness or if they are not able to care for themselves. Everyone will need to make their own decision when it comes time, but having the information about what services and programs are available will make the decision much easier. Long-Term Care Options There are seven types of licensed care services and facilities; Home Care Services, Community Based Care Services, Adult Day Care Centers, Continuing Care Retirement Communities (CCRCs), Assisted Living Facilities, Nursing Homes and Senior Housing. Which one to choose depends on factors that include, cost, insurance, health needs, medical condition of the person and value for services provided? Home care services Home care services are broken down into skilled care and home support. Skilled care is provided under direction of a physician and administered by registered nurses, physical, speech and occupational therapists. Home support provides shopping, meal preparation and light housekeeping, to include bathing and dressing. Other home support services provided include counseling and social work services. Home care services allow older and disabled persons to remain in a familiar environment while maintaining their independence and security. Home care is designed for elderly and disabled people that do not need nursing home care, but needs assistance with day-to-day health and personal needs. The cost of home care is often less expensive than hospital and nursing home care. Home care service can: †¢ Preserve independence and security; †¢ Allow recipient to remain home; †¢ Relieve stress for recipient and family members or caregivers; and †¢ Prevent unnecessary hospital or nursing home bills. Community-Based Care Services If you have the ability to transport yourself to social activities, health appointments or go out for meals, community-based care services are probably not needed. However, for the elderly who cannot get out on their own, or have family that can provide transportation, community-based care services can help. Community-based organizations are broadly made up of an all volunteer staff. Services offered differ amongst organizations. Adult Day Care Centers Provides social interaction and meals in a protected environment, thus allowing those take care of the elderly person time-away. Social interaction includes activities such as physical exercise, games, trips, art and music. Some adult care programs offer medical services, such as help taking medications or checking blood pressure. In the United States of America 1,141 of 3,141 counties lack enough space for adult day care (Shellenbarger, 2002). Transportation to and from adult daycare is sometimes offered by the adult day care center. Currently cost of adult day care averages about $60 a day. Though it sounds expensive it is cheap compared to home-care, assisted living and nursing home care. Adult day care is covered through Medicaid and those insurance companies that offer long-term care policies. Continuing Care Retirement Communities Continuing Care Retirement Communities (CCRCs) offer independent living in a cottage setting to skilled nursing care and in between. The services can be all-inclusive, modified to meet the residents needs or in a fee-for-service. CCRCs promote wellness, independence and socialization in a residential environment. The idea behind a CCRC is that elderly can stay in one place rather than moving from one long-term care option to the next. Example, your parents move into senior housing, then assisted living area, then for further care move into a nursing home, all of which are located in the same complex. Vice paying monthly for rent and services provided, elderly pay a fee or endowment to be part of the CCRC. Assisted Living Facilities Assisted living facilities are broken down into independent, residential or assisted living facilities. Independent and residential living facilities provide minimal assistance for those elderly with minor limitations. Assisted living provides more assistance for those elderly that need help due to major limitations. Services offered include oversight, health care services and help with daily living activities. Assisted living facilities are one of the fastest growing long-term care options available today. In 2000, there were only about 1,000,000, aged 65 and above living in assisted living or residential care (Munn, Hanson, Zimmerman, Sloane, Mitchell, 2006). Since then assisted living facilities have blossomed to over 36,000 licensed facilities providing for more than 9,000,000 residents (Hernandez, 2005-2006). The boom is due to affordability and the homelike living arrangements offered. Nursing Homes When family can no longer take of their elderly member that is injured or disabled, home cares is the preferred option, but if there are no availabilities, then nursing homes are appropriate. In 2000, 1,557,800, aged 65 and above lived in nursing homes (Munn, Hanson, Zimmerman, Sloane, Mitchell, 2006). Most elderly that utilize nursing homes are recovering from illness or injury. Nursing homes also provide hospice care for those terminal elderly; provide rehabilitation; or maintain care for those elderly with chronic health care needs. Nursing homes provide around the clock care for those recovering from illness and injury. The homes are for those elderly that need more medical attention than social gathering. Nursing homes also provide personal care in the form of bathing, dressing and going to the bathroom for the elderly. Senior Housing This is for those elderly that do not need long-term care, but live in a home that is not considered safe. Senior housing often is apartments that have been adapted for the elderly and include railing in bathrooms, wider hallways and raised outlets. Optional services provided include meals, housekeeping and social activities. Choosing the right long-term care option With the long-term care options, mentioned above, selecting which program or service is right can be overwhelming. Think about what it is that you mother, father, or both want in their older years. Ask the elderly what it is that they want? What are their needs? Do they need help with chores? Do they prefer small facilities, certain location, special living conditions? What is their financial situation? Will you be paying for long-term care or is insurance providing coverage? Is Medicare or Medicaid involved? Research available long-term care facilities near your residence first. Ask questions about what services/programs that are/are not provided, if they are insured, costs and vacancies. Check on fee’s for special care services or if extra fees are charged for services. Think about waiting list, if you find the one you want, but are not needed immediately or near term. Visit and tour the facility with your loved ones. Is it friendly, does it look clean, are people happy, what activities do the have for residents. How many staff is available, come back during different times of the day on different days. What the rooms are like are any options available? What are the residents allowed/not allowed doing? Is there a schedule for anything, eating, naps, bedtime, and social visits? How often are your loved ones checked on? Does the staff conduct regular care training? Ask other residents what their opinion of the long-term care facility? Check with the Better Business Bureau to check for any complaints filed. Conclusion In conclusion, though there are many choices for long-term care, it will come down to the individual, the elderly, the elderly family or a combination to determine which type of care will best fit there needs. There is no one fits all long-term care facility, such as there is no specific type of elderly person. Everyone age’s differently and each will have their own very specific needs when it comes time to make a decision on long-term care. Some will choose to remain in their home while others may choose a community setting. However, one thing is known and that is the elderly population is increasing in the United States. This will lead to a higher demand for long-term care services in the future. In the past families tended to take care of their own and their parents, whereas now the elderly will have to start looking at looking at for themselves, though some will be taken care of by their children. In the end, follow your instincts. Choose a place that treats your parents with respect and makes them feel comfortable. References Bolda, E. J. (2006). Community Partnerships for Older Adults: Meeting the Housing Challenge. Generations, 29(4), 61-63.

Origin of Paper :: Essays

The word paper comes from the Greek term for the ancient Egyptian writing material called papyrus, which was formed from beaten strips of papyrus plants. Papyrus was produced as early as 3000 BC in Egypt, and sold to ancient Greece and Rome. The establishment of Great library at Alexandria put a drain on the supply of Papyrus, so According to the Roman Varro, Pliny's Natural History records (xiii.21), parchment was invented under the patronage of Eumenes of Pergamum, to build his rival libray at Permagum. parchment or vellum, made of processed sheepskin or calfskin, replaced papyrus, as the papyrus plant requires subtropical conditions to grow. In China, documents were ordinarily written on bone or bamboo, making them very heavy and awkward to transport. Silk was sometimes used, but was normally too expensive to consider. Indeed, most of the above materials were rare and costly. While the Chinese court official Cai Lun is widely regarded to have first described the modern method of papermaking (inspired from wasps and bees) from wood pulp in AD 105, the 2006 discovery of specimens bearing written characters in north-west China's Gansu province suggest that paper was in use by the ancient Chinese military more than 100 years before Cai in 8 BCE [1]. Archà ¦ologically however, true paper without writing has been excavated in China dating from the 2nd-century BCE. In America, archaeological evidence indicates that paper was invented by the Mayas no later than the 5th century AD.[1] Called Amatl, it was in widespread use among Mesoamerican cultures until the Spanish conquest. In small quantities, traditional Maya papermaking techniques are still practiced today. Paper is considered to be one of the Four Great Inventions of Ancient China. It spread slowly outside of China; other East Asian cultures, even after seeing paper, could not figure out how to make it themselves. Instruction in the manufacturing process was required, and the Chinese were reluctant to share their secrets. The paper was thin and translucent, not like modern western paper, and thus only written on one side. Books were invented in India, of Palm leaves (where we derive the name leaf for a sheet of a book). The technology was first transferred to Korea in 604 and then imported to Japan by a Buddhist priest, Dam Jing (曇å ¾ ´) from Goguryeo, around 610, where fibres (called bast) from the mulberry tree were used. After further commercial trading and the defeat of the Chinese in the Battle of Talas, the invention spread to the Middle East, Production was started in Baghdad, where the arabs invented a method to make a thicker sheet of paper.

Simplification of Switching Function

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 C hecking 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. 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 Karnaugh 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 2C+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 DAB 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 Simplif ied 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 †¦ Exam ple: 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 minterms: {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 onl y 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 function 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 ABEssential prime implicants (minimum cover) 24 Don’t Care Conditions â₠¬ ¢ †¢ †¢ †¢ 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 Plo tting 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 CConvert 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 31Plotting K-map from SOP expression â₠¬ ¢ †¢ 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 ABAB 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 cellsY = 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 Simplification of Switching Function 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 C hecking 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. 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 Karnaugh 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 2C+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 DAB 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 Simplif ied 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 †¦ Exam ple: 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 minterms: {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 onl y 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 function 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 ABEssential prime implicants (minimum cover) 24 Don’t Care Conditions â₠¬ ¢ †¢ †¢ †¢ 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 Plo tting 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 CConvert 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 31Plotting K-map from SOP expression â₠¬ ¢ †¢ 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 ABAB 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 cellsY = 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

Facts About Hypertension The Silent Killer - 831 Words

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