Boolean Algebra Basics PDF

‫ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜﺎﻧﻴﺔ‪ :‬ﺃﺳﺎﺳﻴﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ‪Basics of Boolean Algebra‬‬ ‫ﳏﺘﻮﻳﺎﺕ ﺍﻟﻮﺣﺪﺓ‬ ‫ﲤﻬﻴﺪ‬...

‫ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜﺎﻧﻴﺔ‪ :‬ﺃﺳﺎﺳﻴﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ‪Basics of Boolean Algebra‬‬ ‫ﳏﺘﻮﻳﺎﺕ ﺍﻟﻮﺣﺪﺓ‬ ‫ﲤﻬﻴﺪ‬ ‫ﺃﻫﺪﺍﻑ ﺍﻟﻮﺣﺪﺓ‬ ‫ﺍﳌﺘﻐﲑ ﺍﳌﻨﻄﻘﻲ )‪(Logical Variable‬‬ ‫‪.1‬‬ ‫ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ )‪(Logical Operations‬‬ ‫‪.2‬‬ ‫ﺗﻐﻴﲑ ﻋﺪﺩ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ )‪ (Fan-In‬ﻟﻠﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫‪.3‬‬ ‫ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ )‪(Logical Expression‬‬ ‫‪.4‬‬ ‫ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ )‪(Logic Circuit‬‬ ‫‪.5‬‬ ‫ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ )‪(Logic Diagram‬‬ ‫‪.6‬‬ ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ )‪(Truth Table‬‬ ‫‪.7‬‬ ‫ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ )‪(Boolean Algebra Theorems‬‬ ‫‪.8‬‬ ‫ﺍﺳﺘﺨﺪﺍﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰱ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫‪.9‬‬ ‫ﲤﻬﻴﺪ‬ ‫ﻣﺮﺣﺒﺎﹰ ﺑﻚ ﻋﺰﻳﺰﻱ ﺍﻟﺪﺍﺭﺱ ﰲ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺜﺎﻧﻴﺔ ﻣﻦ ﻣﻘﺮﺭ "ﺃﺳﺎﺳﻴﺎﺕ ﺍﻟﺘﺼﻤﻴﻢ ﺍﳌﻨﻄﻘﻲ"‪.‬ﺗﺘﻨﺎﻭﻝ ﻫﺬﻩ ﺍﻟﻮﺣﺪﺓ ﺃﺳﺎﺳﻴﺎﺕ‬ ‫ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ )‪ ،(Boolean Algebra‬ﻭ ﻫﻮ ﺟﱪ ﺍﳌﺘﻐﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‪.‬ﻭ ﺍﳌﺘﻐﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻫﻮ ﻧﻮﻉ ﺍﳌﺘﻐﲑﺍﺕ ﺍﻟﺬﻱ ﻳﺘﻢ‬ ‫ﺍﻟﺘﻌﺎﻣﻞ ﻣﻌﻪ ﰲ ﺍﻟﺪﻭﺍﺋﺮ ﺍﳌﻨﻄﻘﻴﺔ )‪.(Logic Circuits‬ﺣﻴﺚ ﻧﺘﻌﺮﻑ ﻋﻠﻰ ﺑﻌﺾ ﺍﳌﻔﺎﻫﻴﻢ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﱵ ﳓﺘﺎﺝ ﺇﻟﻴﻬﺎ ﰲ‬ ‫ﺩﺭﺍﺳﺘﻨﺎ ﻟﻸﺟﺰﺍﺀ ﺍﻟﺘﺎﻟﻴﺔ ﻣﻦ ﺍﳌﻘﺮﺭ‪.‬‬ ‫ﺃﻫﺪﺍﻑ ﺍﻟﻮﺣﺪﺓ‬ ‫ﻋﺰﻳﺰﻱ ﺍﻟﺪﺍﺭﺱ‪ ،‬ﺑﻌﺪ ﺩﺭﺍﺳﺔ ﻫﺬﻩ ﺍﻟﻮﺣﺪﺓ ﻳﻨﺒﻐﻲ ﺃﻥ ﺗﻜﻮﻥ ﻗﺎﺩﺭﺍﹰ ﻋﻠﻰ‪:‬‬ ‫ ﻛﺘﺎﺑﺔ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻭ ﺍﻟﺘﻌﺎﻣﻞ ﻣﻌﻬﺎ‪.‬‬ ‫ ﺇﻧﺸﺎﺀ ﺟﺪﺍﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻭ ﺍﺳﺘﺨﺪﺍﻣﻬﺎ‪.‬‬ ‫ ﻓﻬﻢ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ‪.‬‬ ‫ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻨﻈﺮﻳﺎﺕ‪.‬‬ ‫‪1‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪ -1‬ﺍﳌﺘﻐﲑ ﺍﳌﻨﻄﻘﻲ )‪(Logical Variable‬‬ ‫ﺍﳌﺘﻐﲑ ﺍﳌﻨﻄﻘﻲ ﻫﻮ ﺃﻱ ﻣﺘﻐﲑ ﳝﻜﻦ ﺃﻥ ﻳﺄﺧﺬ ﻗﻴﻤﺔ ﻭﺍﺣﺪﺓ ﻓﻘﻂ ﻣﻦ ﻗﻴﻤﺘﲔ‪.‬ﻣﺜﻼﹰ‪:‬‬ ‫ﺧﻄﺄ‬ ‫ﺃﻭ‬ ‫ﺻﻮﺍﺏ‬ ‫‪False‬‬ ‫ﺃﻭ‬ ‫‪True‬‬ ‫‪OFF‬‬ ‫ﺃﻭ‬ ‫‪ON‬‬ ‫‪0 Volts‬‬ ‫ﺃﻭ‬ ‫‪+5 Volts‬‬ ‫‪Low‬‬ ‫ﺃﻭ‬ ‫‪High‬‬ ‫ﺃﺳﻮﺩ‬ ‫ﺃﻭ‬ ‫ﺃﺑﻴﺾ‬ ‫‪Female‬‬ ‫ﺃﻭ‬ ‫‪Male‬‬ ‫ﻳﺮﻣﺰ ﻹﺣﺪﻯ ﺍﻟﻘﻴﻤﺘﲔ ﺑﺎﻟﺮﻣﺰ ‪ 1‬ﻭ ﻟﻠﻘﻴﻤﺔ ﺍﻷﺧﺮﻯ ﺑﺎﻟﺮﻣﺰ ‪... 0‬ﻓﺄﻱ ﻣﺘﻐﲑ ﻣﻨﻄﻘﻲ ﻻ ﳝﻜﻦ ﺃﻥ ﻳﺄﺧﺬ ﺇﻻ ﺇﺣﺪﻯ ﻫﺎﺗﲔ‬ ‫ﺍﻟﻘﻴﻤﺘﲔ‪ ،‬ﻭ ﻻ ﻳﻮﺟﺪ ﺃﻱ ﺍﺣﺘﻤﺎﻝ ﺛﺎﻟﺚ‪.‬ﻓﺈﺫﺍ ﻛﺎﻥ ‪ x‬ﻣﺘﻐﲑ ﻣﻨﻄﻘﻲ ﻓﺈﻧﻪ ﺇﻣﺎ ﺃﻥ ﻳﻜﻮﻥ ‪ x = 1‬ﺃﻭ ‪. x = 0‬‬ ‫‪ -2‬ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ )‪(Logical Operations‬‬ ‫ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻫﻲ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻟﱵ ﳝﻜﻦ ﺇﺟﺮﺍﺅﻫﺎ ﻋﻠﻰ ﺍﳌﺘﻐﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‪.‬ﺑﻌﺾ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺎﺕ ﻫﻲ ﻋﻤﻠﻴﺎﺕ‬ ‫ﺃﺳﺎﺳﻴﺔ‪ ،‬ﻭ ﻫﻲ ﻋﻤﻠﻴﺎﺕ ‪ NOT‬ﻭ ‪ AND‬ﻭ ‪ ،OR‬ﻭ ﺑﻌﻀﻬﺎ ﻋﻤﻠﻴﺎﺕ ﻏﲑ ﺃﺳﺎﺳﻴﺔ‪ ،‬ﻣﺜﻞ ﻋﻤﻠﻴﺎﺕ ‪ NAND‬ﻭ ‪ NOR‬ﻭ‬ ‫‪ ،XOR‬ﻭ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺎﺕ ﳝﻜﻦ ﺍﻟﺘﻌﺒﲑ ﻋﻨﻬﺎ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻷﺳﺎﺳﻴﺔ‪.‬‬ ‫‪ 1-2‬ﻋﻤﻠﻴﺔ ‪NOT‬‬ ‫ﻳﻄﻠﻖ ﻋﻠﻴﻬﺎ ﺃﻳﻀﺎﹰ ﻋﻤﻠﻴﺔ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ )‪ ،(Logical Inversion‬ﻭﻓﻴﻬﺎ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻋﺒﺎﺭﺓ ﻋﻦ ﻣﻌﻜﻮﺱ ﺍﻟﺪﺧﻞ‪ ،‬ﻓﺈﺫﺍ‬ ‫ﻛﺎﻥ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺎﹰ ‪ 1‬ﻓﺈﻥ ﺍﳋﺮﺝ ﻳﻜﻮﻥ ﻣﺴﺎﻭﻳﺎﹰ ‪ ،0‬ﻭ ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺎﹰ ‪ 0‬ﻓﺈﻥ ﺍﳋﺮﺝ ﻳﻜﻮﻥ ﻣﺴﺎﻭﻳﺎﹰ ‪.1‬ﻳﺮﻣﺰ‬ ‫ﻟﻠﻌﻤﻠﻴﺔ ﺑﻮﺿﺢ ﺧﻂ ﻓﻮﻕ ﺍﳌﺘﻐﲑ‪ ،‬ﳑﺎ ﻳﻌﲏ ﺃﻧﻪ ﻣﻌﻜﻮﺱ‪.‬‬ ‫‪x = NOT A‬‬ ‫‪x= A‬‬ ‫‪2‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻳﺴﻤﻰ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ )‪ ،(Truth Table‬ﻭﻫﻮ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻌﻤﻠﻴﺔ ‪ ،NOT‬ﻭ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‬ ‫ﻳﻮﺿﺢ ﲨﻴﻊ ﺍﺣﺘﻤﺎﻻﺕ ﺍﻟﺪﺧﻞ ﻭ ﺍﳋﺮﺝ ﺍﳌﻘﺎﺑﻞ ﻟﻜﻞ ﻣﻨﻬﺎ‪.‬‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫ﻻﺣﻆ ﺃﻥ ﺍﻟﺪﺧﻞ ﻫﻨﺎ ﻫﻮ ‪ A‬ﻭ ﺍﳋﺮﺝ ﻫﻮ ‪. x‬ﻭ ﺍﻟﺪﺧﻞ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻋﺒﺎﺭﺓ ﻋﻦ ﻣﺘﻐﲑ ﻭﺍﺣﺪ ﳝﻜﻦ ﺃﻥ ﻳﺄﺧﺬ ﻭﺍﺣﺪﺓ ﻣﻦ‬ ‫ﻗﻴﻤﺘﲔ‪ ،‬ﺇﻣﺎ ‪ 0‬ﺃﻭ ‪ ،1‬ﺃﻱ ﺃﻥ ﻫﻨﺎﻙ ﺍﺣﺘﻤﺎﻟﲔ ﻓﻘﻂ ﻟﻠﺪﺧﻞ‪.‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ )‪ (Logic Gate‬ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻲ ﺑﻮﺍﺑﺔ ‪ ،(NOT Gate) NOT‬ﺍﻟﱵ ﻳﻄﻠﻖ ﻋﻠﻴﻬﺎ ﺃﻳﻀﺎﹰ‬ ‫ﺍﻟﻌﺎﻛﺲ ﺍﳌﻨﻄﻘﻲ )‪.(Logic Inverter‬ﻭ ﳝﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﺃﻱ ﻣﻦ ﺍﻟﺸﻜﻠﲔ ﺍﻟﺘﺎﻟﻴﲔ ﰲ ﲤﺜﻴﻞ ﺑﻮﺍﺑﺔ ‪:NOT‬‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪ 2-2‬ﻋﻤﻠﻴﺔ ﺍﻟﺘﻜﺎﻓﺆ )‪(Equivalence‬‬ ‫ﰲ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻣﺴﺎﻭﻳﺎﹰ ﻟﻠﺪﺧﻞ‪ ،‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﻌﻼﻣﺔ ﺍﻟﺘﺴﺎﻭﻱ‬ ‫‪x= A‬‬ ‫ﻭﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻠﻌﻤﻠﻴﺔ ﻫﻮ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﺗﺴﻤﻰ ﺍﻟﻌﺎﺯﻝ )‪ ،(Buffer‬ﻭ ﻳﺘﻢ ﲤﺜﻴﻠﻬﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‪:‬‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪3‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪ 3-2‬ﻋﻤﻠﻴﺔ ‪AND‬‬ ‫ﰲ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻣﺴﺎﻭﻳﺎﹰ ‪ 1‬ﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ﲨﻴﻊ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺔ ‪ ،1‬ﻭ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻳﻜﻮﻥ ﻣﺴﺎﻭﻳﺎﹰ‬ ‫‪ 0‬ﺇﺫﺍ ﻛﺎﻥ ﺃﻱ ﻣﺘﻐﲑ ﻣﻦ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺎﹰ ‪.0‬ﻭ ﻳﺮﻣﺰ ﳍﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﺑﺄﻱ ﻣﻦ ﺍﻟﻄﺮﻕ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪x = A AND B‬‬ ‫‪x = A⋅ B‬‬ ‫‪x = AB‬‬ ‫)ﺍﻟﻄﺮﻳﻘﺔ ﺍﻷﺧﲑﺓ ﻫﻲ ﺍﻷﻛﺜﺮ ﺍﺳﺘﺨﺪﺍﻣﺎﹰ(‬ ‫ﻓﻴﻤﺎ ﻳﻠﻲ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﺒﻮﺍﺑﺔ ‪ AND‬ﲟﺪﺧﻠﲔ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﻻﺣﻆ ﺃﻧﻪ ﻧﻈﺮﺍﹰ ﻟﻮﺟﻮﺩ ﻣﺘﻐﲑﻳﻦ ﻟﻠﺪﺧﻞ ﻫﻨﺎ ﳘﺎ ‪ A‬ﻭ ‪ ، B‬ﻓﺄﻧﻪ ﺗﻮﺟﺪ ﺃﺭﺑﻌﺔ ﺍﺣﺘﻤﺎﻻﺕ ﻟﻠﺪﺧﻞ‪.‬ﻭ ﺍﻟﻘﺎﻋﺪﺓ ﺍﻟﻌﺎﻣﺔ ﰲ‬ ‫ﺟﺪﺍﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻫﻲ ﺃﻧﻪ ﺇﺫﺍ ﻛﺎﻥ ﻋﺪﺩ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻫﻮ ‪ N‬ﻓﺈﻥ ﻋﺪﺩ ﺍﺣﺘﻤﺎﻻﺕ ﺍﻟﺪﺧﻞ‪ ،‬ﺃﻱ ﻋﺪﺩ ﺃﺳﻄﺮ ﺟﺪﻭﻝ‬ ‫ﺍﻟﺼﻮﺍﺏ‪ ،‬ﻫﻮ ‪. 2 N‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻲ ﺑﻮﺍﺑﺔ ‪ ،AND‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ AND‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input AND Gate‬‬ ‫‪4‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻗﺪ ﻳﻜﻮﻥ ﻟﺒﻮﺍﺑﺔ ‪ AND‬ﺃﻛﺜﺮ ﻣﻦ ﻣﺪﺧﻠﲔ‪.‬ﻣﺜﻼﹰ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪C‬‬ ‫ﺑﻮﺍﺑﺔ ‪ AND‬ﺑﺜﻼﺛﺔ ﻣﺪﺍﺧﻞ‬ ‫)‪(3-Input AND Gate‬‬ ‫ﺗﺪﺭﻳﺐ ‪:1‬‬ ‫ﻗﻢ ﺑﺈﻧﺸﺎﺀ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ )‪ (Truth Table‬ﻟﺒﻮﺍﺑﺔ ‪ AND‬ﺑﺜﻼﺛﺔ ﻣﺪﺍﺧﻞ‪.‬‬ ‫‪ 4-2‬ﻋﻤﻠﻴﺔ ‪OR‬‬ ‫ﰲ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻣﺴﺎﻭﻳﺎﹰ ‪ 1‬ﺇﺫﺍ ﻛﺎﻥ ﺃﻱ ﻣﻦ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺎﹰ ‪ ،1‬ﻭ ﻳﻜﻮﻥ ﺍﳋﺮﺝ ﻳﻜﻮﻥ ﻣﺴﺎﻭﻳﺎﹰ ‪ 0‬ﺇﺫﺍ‬ ‫ﻛﺎﻧﺖ ﲨﻴﻊ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻣﺴﺎﻭﻳﺔ ‪.0‬ﻭ ﻳﺮﻣﺰ ﳍﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﺑﺄﻱ ﻣﻦ ﺍﻟﻄﺮﻳﻘﺘﲔ ﺍﻟﺘﺎﻟﻴﺘﲔ‬ ‫‪x = A OR B‬‬ ‫‪x = A+ B‬‬ ‫ﻓﻴﻤﺎ ﻳﻠﻲ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﺒﻮﺍﺑﺔ ‪ OR‬ﲟﺪﺧﻠﲔ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻲ ﺑﻮﺍﺑﺔ ‪ ،OR‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ OR‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input OR Gate‬‬ ‫‪5‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻗﺪ ﻳﻜﻮﻥ ﻟﺒﻮﺍﺑﺔ ‪ OR‬ﺃﻛﺜﺮ ﻣﻦ ﻣﺪﺧﻠﲔ‪.‬ﻣﺜﻼﹰ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪C‬‬ ‫‪D‬‬ ‫ﺑﻮﺍﺑﺔ ‪ OR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‬ ‫)‪(4-Input OR Gate‬‬ ‫ﺗﺪﺭﻳﺐ ‪:2‬‬ ‫ﻗﻢ ﺑﺈﻧﺸﺎﺀ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ )‪ (Truth Table‬ﻟﺒﻮﺍﺑﺔ ‪ OR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‪.‬‬ ‫‪ 5-2‬ﻋﻤﻠﻴﺔ ‪NAND‬‬ ‫ﻋﻤﻠﻴﺔ ‪ NAND‬ﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻦ ﻋﻤﻠﻴﺔ ‪ AND‬ﻣﺘﺒﻮﻋﺔ ﺑﻌﻤﻠﻴﺔ ‪ ،NOT‬ﺃﻱ ﺃ‪‬ﺎ ﻋﻤﻠﻴﺔ ‪ ،NOT AND‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺄﻱ‬ ‫ﻣﻦ ﺍﻟﻄﺮﻕ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪x = A NAND B‬‬ ‫‪x = A AND B‬‬ ‫‪x = A⋅ B‬‬ ‫‪x = AB‬‬ ‫‪x = A↑ B‬‬ ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻫﻮ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻌﻤﻠﻴﺔ ‪ ،NAND‬ﻭﻫﻮ ﻋﻜﺲ ﻋﻤﻠﻴﺔ ‪ AND‬ﻛﻤﺎ ﻫﻮ ﻣﺘﻮﻗﻊ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻰ ﺑﻮﺍﺑﺔ ‪ ،NAND‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ NAND‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input NAND Gate‬‬ ‫‪6‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻛﻔﺎﻳﺔ ﻋﻤﻠﻴﺔ ‪(Sufficiency of NAND) NAND‬‬ ‫ﺍﳌﻘﺼﻮﺩ ﺑﻜﻔﺎﻳﺔ ﻋﻤﻠﻴﺔ ‪ NAND‬ﻫﻮ ﺃﻥ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ )‪ (OR ،AND ،NOT‬ﳝﻜﻦ ﺇﺟﺮﺍﺅﻫﺎ‬ ‫ﲨﻴﻌﹰﺎ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪ ،NAND‬ﻭ ﺑﺎﻟﺘﺎﱄ ﳝﻜﻦ ﺑﻨﺎﺀ ﺃﻱ ﺩﺍﺋﺮﺓ ﻣﻨﻄﻘﻴﺔ ﺑﺎﻟﻜﺎﻣﻞ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪ NAND‬ﻓﻘﻂ‪.‬‬ ‫ﰲ ﺍﳉﺰﺀ ﺍﻟﺘﺎﱄ ﺳﻨﻮﺿﺢ ﻃﺮﻳﻘﺔ ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪.NAND‬‬ ‫ﻋﻤﻠﻴﺔ ‪:NOT‬‬ ‫ﳝﻜﻦ ﺃﻥ ﻧﻘﻮﻡ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺔ ‪ NAND‬ﻛﻌﺎﻛﺲ ﻣﻨﻄﻘﻲ ﺑﺮﺑﻂ ﲨﻴﻊ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ ﳍﺎ ﰲ ﻃﺮﻑ ﻭﺍﺣﺪ‬ ‫‪A‬‬ ‫‪A‬‬ ‫ﻭ ﳝﻜﻦ ﺃﻥ ﻧﺮﻣﺰ ﻟﺒﻮﺍﺑﺔ ‪ NAND‬ﺍﳌﺴﺘﺨﺪﻣﺔ ﻛﻌﺎﻛﺲ ﻣﻨﻄﻘﻲ ﺑﺒﻮﺍﺑﺔ ‪ NAND‬ﺑﻄﺮﻑ ﺩﺧﻞ ﻭﺍﺣﺪ‪ ،‬ﺃﻱ‬ ‫‪A‬‬ ‫‪A‬‬ ‫ﻋﻤﻠﻴﺔ ‪:AND‬‬ ‫ﳝﻜﻦ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ AND‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ NAND‬ﻣﺘﺒﻮﻋﺔ ﺑﻌﻤﻠﻴﺔ ﻋﻜﺲ ﻣﻨﻄﻘﻲ‬ ‫‪A‬‬ ‫‪A⋅ B‬‬ ‫‪B‬‬ ‫‪A⋅ B‬‬ ‫ﻋﻤﻠﻴﺔ ‪:OR‬‬ ‫ﳝﻜﻦ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ OR‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ NAND‬ﻣﺴﺒﻮﻗﺔ ﺑﻌﻤﻠﻴﺔ ﻋﻜﺲ ﻣﻨﻄﻘﻲ ﻟﻜﻞ ﻃﺮﻑ ﻣﻦ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ‬ ‫‪A‬‬ ‫‪A‬‬ ‫‪A⋅ B = A + B‬‬ ‫‪B‬‬ ‫‪B‬‬ ‫‪7‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪A⋅ B = A + B‬‬ ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﺍﻟﺘﺎﱄ ﻳﺜﺒﺖ ﺃﻥ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪B A⋅ B A⋅ B A+ B‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪ 6-2‬ﻋﻤﻠﻴﺔ ‪NOR‬‬ ‫ﻋﻤﻠﻴﺔ ‪ NOR‬ﻫﻲ ﻋﺒﺎﺭﺓ ﻋﻦ ﻋﻤﻠﻴﺔ ‪ OR‬ﻣﺘﺒﻮﻋﺔ ﺑﻌﻤﻠﻴﺔ ‪ ،NOT‬ﺃﻱ ﺃ‪‬ﺎ ﻋﻤﻠﻴﺔ ‪ ،NOT OR‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺄﻱ ﻣﻦ‬ ‫ﺍﻟﻄﺮﻕ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪x = A NOR B‬‬ ‫‪x = A OR B‬‬ ‫‪x = A+ B‬‬ ‫‪x = A↓ B‬‬ ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻫﻮ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻌﻤﻠﻴﺔ ‪ ،NOR‬ﻭ ﻫﻮ ﻋﻜﺲ ﻋﻤﻠﻴﺔ ‪ OR‬ﻛﻤﺎ ﻫﻮ ﻣﺘﻮﻗﻊ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻰ ﺑﻮﺍﺑﺔ ‪ ،NOR‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ NOR‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input NOR Gate‬‬ ‫‪8‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻛﻔﺎﻳﺔ ﻋﻤﻠﻴﺔ ‪(Sufficiency of NOR) NOR‬‬ ‫ﺍﳌﻘﺼﻮﺩ ﺑﻜﻔﺎﻳﺔ ﻋﻤﻠﻴﺔ ‪ NOR‬ﻫﻮ ﺃﻥ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ )‪ (OR ،AND ،NOT‬ﳝﻜﻦ ﺇﺟﺮﺍﺅﻫﺎ ﲨﻴﻌﺎﹰ‬ ‫ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪ ،NOR‬ﻭ ﺑﺎﻟﺘﺎﱄ ﳝﻜﻦ ﺑﻨﺎﺀ ﺃﻱ ﺩﺍﺋﺮﺓ ﻣﻨﻄﻘﻴﺔ ﺑﺎﻟﻜﺎﻣﻞ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪ NOR‬ﻓﻘﻂ‪.‬‬ ‫ﰲ ﺍﳉﺰﺀ ﺍﻟﺘﺎﱄ ﺳﻨﻮﺿﺢ ﻃﺮﻳﻘﺔ ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺎﺕ ‪.NOR‬‬ ‫ﻋﻤﻠﻴﺔ ‪:NOT‬‬ ‫ﳝﻜﻦ ﺃﻥ ﻧﻘﻮﻡ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺑﻮﺍﺑﺔ ‪ NOR‬ﻛﻌﺎﻛﺲ ﻣﻨﻄﻘﻲ ﺑﺮﺑﻂ ﲨﻴﻊ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ ﳍﺎ ﰲ ﻃﺮﻑ ﻭﺍﺣﺪ‬ ‫‪A‬‬ ‫‪A‬‬ ‫ﻭ ﳝﻜﻦ ﺃﻥ ﻧﺮﻣﺰ ﻟﺒﻮﺍﺑﺔ ‪ NOR‬ﺍﳌﺴﺘﺨﺪﻣﺔ ﻛﻌﺎﻛﺲ ﻣﻨﻄﻘﻲ ﺑﺒﻮﺍﺑﺔ ‪ NOR‬ﺑﻄﺮﻑ ﺩﺧﻞ ﻭﺍﺣﺪ‪ ،‬ﺃﻱ‬ ‫‪A‬‬ ‫‪A‬‬ ‫ﻋﻤﻠﻴﺔ ‪:OR‬‬ ‫ﳝﻜﻦ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ OR‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ NOR‬ﻣﺘﺒﻮﻋﺔ ﺑﻌﻤﻠﻴﺔ ﻋﻜﺲ ﻣﻨﻄﻘﻲ‬ ‫‪A‬‬ ‫‪A+ B‬‬ ‫‪B‬‬ ‫‪A+ B‬‬ ‫ﻋﻤﻠﻴﺔ ‪:AND‬‬ ‫ﳝﻜﻦ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ AND‬ﻋﻦ ﻃﺮﻳﻖ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ NOR‬ﻣﺴﺒﻮﻗﺔ ﺑﻌﻤﻠﻴﺔ ﻋﻜﺲ ﻣﻨﻄﻘﻲ ﻟﻜﻞ ﻃﺮﻑ ﻣﻦ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ‬ ‫‪A‬‬ ‫‪A‬‬ ‫‪A + B = A⋅ B‬‬ ‫‪B‬‬ ‫‪B‬‬ ‫‪9‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪A + B = A⋅ B‬‬ ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﺍﻟﺘﺎﱄ ﻳﺜﺒﺖ ﺃﻥ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪B A+ B A+ B‬‬ ‫‪A⋅ B‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﻭ ﺗﺘﻮﻓﺮ ﺑﻮﺍﺑﺎﺕ ‪ NAND‬ﻭ ﺑﻮﺍﺑﺎﺕ ‪ NOR‬ﺑﺄﻛﺜﺮ ﻣﻦ ﻣﺪﺧﻠﲔ‪ ،‬ﻣﺜﻠﻬﺎ ﰲ ﺫﻟﻚ ﻣﺜﻞ ﺑﻮﺍﺑﺎﺕ ‪ AND‬ﻭ ﺑﻮﺍﺑﺎﺕ ‪.OR‬‬ ‫‪ 7-2‬ﻋﻤﻠﻴﺔ ‪XOR‬‬ ‫‪ XOR‬ﻫﻮ ﺍﺧﺘﺼﺎﺭ ﻟﻌﺒﺎﺭﺓ ‪ ،Exclusive OR‬ﻭ ﺗﺴﻤﻰ ﻋﻤﻠﻴﺔ ﺍﻻﺧﺘﻼﻑ‪ ،‬ﺣﻴﺚ ﺃﻥ ﺍﳋﺮﺝ ﻳﺴﺎﻭﻱ ‪ 1‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺪﺧﻼﻥ‬ ‫ﳐﺘﻠﻔﲔ‪ ،‬ﻭ ﻳﺴﺎﻭﻱ ‪ 0‬ﺇﺫﺍ ﻛﺎﻧﺎ ﻣﺘﺸﺎ‪‬ﲔ‪.‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺈﺣﺪﻯ ﺍﻟﻄﺮﻳﻘﺘﲔ ﺍﻟﺘﺎﻟﻴﺘﲔ‬ ‫‪x = A XOR B‬‬ ‫‪x = A⊕ B‬‬ ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻫﻮ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻌﻤﻠﻴﺔ ‪XOR‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻰ ﺑﻮﺍﺑﺔ ‪ ،XOR‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ XOR‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input XOR Gate‬‬ ‫ﻭ ﳝﻜﻦ ﺍﻟﺘﻌﺒﲑ ﻋﻦ ﻋﻤﻠﻴﺔ ‪ XOR‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﻛﺎﻟﺘﺎﱄ‬ ‫‪A ⊕ B = AB + AB‬‬ ‫‪10‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪A ⊕ B = AB + AB‬‬ ‫ﻭ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﺍﻟﺘﺎﱄ ﻳﺜﺒﺖ ﺃﻥ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪AB AB‬‬ ‫‪AB + AB‬‬ ‫‪A⊕ B‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪ 8-2‬ﻋﻤﻠﻴﺔ ‪XNOR‬‬ ‫ﻫﻲ ﻣﻌﻜﻮﺱ ﻋﻤﻠﻴﺔ ‪ ،XOR‬ﻭ ﺗﺴﻤﻰ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺴﺎﻭﻱ‪ ،‬ﺣﻴﺚ ﺃﻥ ﺍﳋﺮﺝ ﻳﺴﺎﻭﻱ ‪ 1‬ﺇﺫﺍ ﻛﺎﻥ ﺍﻟﺪﺧﻼﻥ ﻣﺘﺴﺎﻭﻳﲔ‪ ،‬ﻭ‬ ‫ﻳﺴﺎﻭﻱ ‪ 0‬ﺇﺫﺍ ﻛﺎﻧﺎ ﳐﺘﻠﻔﲔ‪.‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺈﺣﺪﻯ ﺍﻟﻄﺮﻳﻘﺘﲔ ﺍﻟﺘﺎﻟﻴﺘﲔ‬ ‫‪x = A XNOR B‬‬ ‫‪x = A⊕ B‬‬ ‫ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻫﻮ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻌﻤﻠﻴﺔ ‪XNOR‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫ﺍﻟﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﻫﺬﻩ ﺍﻟﻌﻤﻠﻴﺔ ﻫﻰ ﺑﻮﺍﺑﺔ ‪ ،XNOR‬ﻭ ﻳﺮﻣﺰ ﳍﺎ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫ﺑﻮﺍﺑﺔ ‪ XNOR‬ﲟﺪﺧﻠﲔ‬ ‫)‪(2-Input XNOR Gate‬‬ ‫ﻭ ﳝﻜﻦ ﺍﻟﺘﻌﺒﲑ ﻋﻦ ﻋﻤﻠﻴﺔ ‪ XNOR‬ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﻛﺎﻟﺘﺎﱄ‬ ‫‪A ⊕ B = AB + AB‬‬ ‫ﺗﺪﺭﻳﺐ ‪:3‬‬ ‫‪A ⊕ B = AB + AB‬‬ ‫ﻗﻢ ﺑﺈﻧﺸﺎﺀ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﺍﻟﺬﻱ ﻳﺜﺒﺖ ﺃﻥ‬ ‫ﻭ ﲞﻼﻑ ﺑﻮﺍﺑﺎﺕ ‪ AND‬ﻭ ‪ OR‬ﻭ ‪ NAND‬ﻭ ‪ ،NOR‬ﻻ ﺗﺘﻮﻓﺮ ﺑﻮﺍﺑﺎﺕ ‪ XOR‬ﺃﻭ ﺑﻮﺍﺑﺎﺕ ‪ XNOR‬ﺑﺄﻛﺜﺮ ﻣﻦ‬ ‫ﻣﺪﺧﻠﲔ‪.‬‬ ‫‪11‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫‪ -3‬ﺗﻐﻴﲑ ﻋﺪﺩ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ )‪ (Fan-In‬ﻟﻠﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫ﰲ ﻛﺜﲑ ﻣﻦ ﺍﻷﺣﻴﺎﻥ ﻗﺪ ﺗﺘﻮﻓﺮ ﻟﻨﺎ ﺑﻮﺍﺑﺎﺕ ﻣﻨﻄﻘﻴﺔ ﺑﻌﺪﺩ ﻣﻦ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ )‪ (Fan-In‬ﺃﻛﱪ ﺃﻭ ﺃﻗﻞ ﳑﺎ ﳓﺘﺎﺝ ﺇﻟﻴﻪ‪.‬‬ ‫ﺳﻨﻮﺿﺢ ﰲ ﻫﺬﺍ ﺍﳉﺰﺀ ﺍﻷﺳﺎﻟﻴﺐ ﺍﳌﺨﺘﻠﻔﺔ ﺍﻟﱵ ﳝﻜﻨﻨﺎ ﺇﺗﺒﺎﻋﻬﺎ ﻟﺘﻐﻴﲑ ﻋﺪﺩ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ ﻟﻠﺒﻮﺍﺑﺔ ﺍﳌﻨﻄﻘﻴﺔ ﺑﺎﻟﺰﻳﺎﺩﺓ ﺃﻭ‬ ‫ﺑﺎﻟﻨﻘﺼﺎﻥ‪.‬‬ ‫ﺗﻘﻠﻴﻞ ﻋﺪﺩ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ‪:‬‬ ‫ﻳﺘﻢ ﺫﻟﻚ ﺑﺮﺑﻂ ﻃﺮﻑ ﺍﻟﺪﺧﻞ ﺍﻟﺰﺍﺋﺪ ﺑﺄﺣﺪ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ ﺍﳌﺴﺘﺨﺪﻣﺔ‪ ،‬ﻣﺜﻼﹰ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A⋅B‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A+B‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪A⋅B⋅C‬‬ ‫‪A‬‬ ‫‪A‬‬ ‫ﰲ ﺍﳊﺎﻟﺔ ﺍﻷﻭﱃ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺑﻮﺍﺑﺔ ‪ AND‬ﺑﺜﻼﺛﺔ ﻣﺪﺍﺧﻞ ﻛﺒﻮﺍﺑﺔ ‪ AND‬ﲟﺪﺧﻠﲔ‪ ،‬ﻭ ﺫﻟﻚ ﺑﺎﻟﺘﺨﻠﺺ ﻣﻦ ﻃﺮﻑ ﺍﻟﺪﺧﻞ‬ ‫ﺍﻟﺜﺎﻟﺚ ﻏﲑ ﺍﳌﺮﻏﻮﺏ ﻓﻴﻪ ﺑﺮﺑﻄﻪ ﺑﺄﺣﺪ ﻃﺮﰲ ﺍﻟﺪﺧﻞ ﺍﳌﺴﺘﺨﺪﻣﲔ‪.‬ﻭ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﺜﺎﻧﻴﺔ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺑﻮﺍﺑﺔ ‪ OR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‬ ‫ﻛﺒﻮﺍﺑﺔ ‪ OR‬ﲟﺪﺧﻠﲔ‪.‬‬ ‫ﻛﻤﺎ ﳝﻜﻦ ﺃﻥ ﻳﺘﻢ ﺍﻟﺘﺨﻠﺺ ﻣﻦ ﻃﺮﻑ ﺍﻟﺪﺧﻞ ﺍﻟﺰﺍﺋﺪ ﺑﻮﺿﻊ ﺍﻟﻘﻴﻤﺔ ﺍﳌﻨﻄﻘﻴﺔ ‪ 1‬ﰲ ﻃﺮﻑ ﺍﻟﺪﺧﻞ ﺍﻟﺰﺍﺋﺪ ﰲ ﺑﻮﺍﺑﺎﺕ ‪ AND‬ﻭ‬ ‫‪ ،NAND‬ﻭ ﻭﺿﻊ ﺍﻟﻘﻴﻤﺔ ﺍﳌﻨﻄﻘﻴﺔ ‪ 0‬ﰲ ﻃﺮﻑ ﺍﻟﺪﺧﻞ ﺍﻟﺰﺍﺋﺪ ﰲ ﺑﻮﺍﺑﺎﺕ ‪ OR‬ﻭ ‪ ،NOR‬ﻣﺜﻼﹰ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A⋅B‬‬ ‫‪1‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪0‬‬ ‫‪A+B‬‬ ‫‪0‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪A ⋅B⋅C‬‬ ‫‪1‬‬ ‫‪A‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪A‬‬ ‫‪0‬‬ ‫‪12‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻫﻨﺎ ﺃﻳﻀﺎﹰ ﰲ ﺍﳊﺎﻟﺔ ﺍﻷﻭﱃ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺑﻮﺍﺑﺔ ‪ AND‬ﺑﺜﻼﺛﺔ ﻣﺪﺍﺧﻞ ﻛﺒﻮﺍﺑﺔ ‪ AND‬ﲟﺪﺧﻠﲔ‪ ،‬ﻭ ﺫﻟﻚ ﺑﺎﻟﺘﺨﻠﺺ ﻣﻦ ﻃﺮﻑ‬ ‫ﺍﻟﺪﺧﻞ ﺍﻟﺜﺎﻟﺚ ﻏﲑ ﺍﳌﺮﻏﻮﺏ ﻓﻴﻪ ﺑﻮﺿﻊ ﺍﻟﻘﻴﻤﺔ ﺍﳌﻨﻄﻘﻴﺔ ‪ 1‬ﻓﻴﻪ‪.‬ﻭ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﺜﺎﻧﻴﺔ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺑﻮﺍﺑﺔ ‪ OR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‬ ‫ﻛﺒﻮﺍﺑﺔ ‪ OR‬ﲟﺪﺧﻠﲔ‪ ،‬ﻭ ﺫﻟﻚ ﺑﻮﺿﻊ ﺍﻟﻘﻴﻤﺔ ﺍﳌﻨﻄﻘﻴﺔ ‪ 0‬ﰲ ﻃﺮﰲ ﺍﻟﺪﺧﻞ ﺍﻟﺰﺍﺋﺪﻳﻦ‪.‬‬ ‫ﺯﻳﺎﺩﺓ ﻋﺪﺩ ﺃﻃﺮﺍﻑ ﺍﻟﺪﺧﻞ‪:‬‬ ‫ﻳﺘﻢ ﺫﻟﻚ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺃﻛﺜﺮ ﻣﻦ ﺑﻮﺍﺑﺔ ﻭﺍﺣﺪﺓ ﻭ ﺍﺳﺘﺨﺪﺍﻡ ﺧﺮﺝ ﺍﻟﺒﻮﺍﺑﺔ ﺍﻷﻭﱃ ﻛﺪﺧﻞ ﻟﻠﺒﻮﺍﺑﺔ ﺍﻟﺜﺎﻧﻴﺔ‪ ،‬ﻣﺜﻼﹸ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A⋅ B ⋅ C‬‬ ‫‪C‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A+ B + C + D‬‬ ‫‪C‬‬ ‫‪D‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪A⊕ B ⊕ C ⊕ D‬‬ ‫‪C‬‬ ‫‪D‬‬ ‫ﰲ ﺍﳊﺎﻟﺔ ﺍﻷﻭﱃ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺑﻮﺍﺑﱵ ‪ ،AND‬ﻛﻞ ﻣﻨﻬﻤﺎ ﲟﺪﺧﻠﲔ‪ ،‬ﻛﻮﺍﺑﺔ ‪ AND‬ﺑﺜﻼﺛﺔ ﻣﺪﺍﺧﻞ‪.‬ﻭ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﺜﺎﻧﻴﺔ‬ ‫ﺍﺳﺘﺨﺪﻣﻨﺎ ﺛﻼﺛﺔ ﺑﻮﺍﺑﺎﺕ ‪ ،OR‬ﻛﻞ ﺑﻮﺍﺑﺔ ﻣﻨﻬﺎ ﲟﺪﺧﻠﲔ‪ ،‬ﻛﺒﻮﺍﺑﺔ ‪ OR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‪.‬ﻭ ﰲ ﺍﳊﺎﻟﺔ ﺍﻟﺜﺎﻟﺜﺔ ﺍﺳﺘﺨﺪﻣﻨﺎ ﺛﻼﺛﺔ‬ ‫ﺑﻮﺍﺑﺎﺕ ‪ ،XOR‬ﻛﻞ ﺑﻮﺍﺑﺔ ﻣﻨﻬﺎ ﲟﺪﺧﻠﲔ‪ ،‬ﻛﺒﻮﺍﺑﺔ ‪ XOR‬ﺑﺄﺭﺑﻌﺔ ﻣﺪﺍﺧﻞ‪.‬‬ ‫‪ -4‬ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ )‪(Logical Expression‬‬ ‫ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﳎﻤﻮﻋﺔ ﻣﻦ ﺍﳌﺘﻐﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﳌﺮﺗﺒﻄﺔ ﻣﻊ ﺑﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﺑﻌﻤﻠﻴﺎﺕ ﻣﻨﻄﻘﻴﺔ‪.‬ﻣﺜﻞ‬ ‫‪x = A+ B ⋅ C‬‬ ‫ﻳﺘﻜﻮﻥ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻫﻨﺎ ﻣﻦ ﺃﺭﺑﻌﺔ ﻣﺘﻐﲑﺍﺕ ﻫﻲ ‪ A‬ﻭ ‪ B‬ﻭ ‪ C‬ﻭ ‪ ، x‬ﺗﺮﺑﻂ ﺑﻴﻨﻬﺎ ﻋﻤﻠﻴﺎﺕ ‪ NOT‬ﻭ ‪ AND‬ﻭ ‪OR‬‬ ‫ﻭ ﻋﻤﻠﻴﺔ ﺍﻟﺘﻜﺎﻓﺆ )=(‪.‬‬ ‫‪13‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺃﺳﺒﻘﻴﺔ ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ )‪:(Operation Precedence‬‬ ‫ﻳﺘﻢ ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ﺑﺎﻟﺘﺮﺗﻴﺐ ﺍﻟﺘﺎﱄ‪:‬‬ ‫‪ -1‬ﻋﻤﻠﻴﺔ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ‪NOT‬‬ ‫‪ -2‬ﻋﻤﻠﻴﺔ ‪AND‬‬ ‫‪ -3‬ﻋﻤﻠﻴﺔ ‪OR‬‬ ‫ﻓﻔﻲ ﺍﻟﺘﻌﺒﲑ ﺃﻋﻼﻩ‪ ،‬ﻣﺜﻼﹰ‪ ،‬ﻳﺘﻢ ﺃﻭﻻﹰ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ﻟﻠﻤﺘﻐﲑﻳﻦ ‪ B‬ﻭ ‪ C‬ﺍﻭﻻﹰ‪ ،‬ﰒ ﻋﻤﻠﻴﺔ ‪ AND‬ﺑﲔ ‪ B‬ﻭ‬ ‫‪ ، C‬ﻭ ﺃﺧﲑﺍﹰ ﻋﻤﻠﻴﺔ ‪.OR‬‬ ‫ﰲ ﺣﺎﻟﺔ ﻇﻬﻮﺭ ﻋﺪﺓ ﻋﻤﻠﻴﺎﺕ ﻣﺘﺴﺎﻭﻳﺔ ﻣﻦ ﺣﻴﺚ ﺍﻷﺳﺒﻘﻴﺔ ﰲ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻳﺘﻢ ﺇﺟﺮﺍﺅﻫﺎ ﺑﺎﻟﺘﺮﺗﻴﺐ ﻣﻦ ﺍﻟﻴﺴﺎﺭ ﻟﻠﻴﻤﲔ‪.‬‬ ‫ﳝﻜﻦ ﺍﺳﺘﺨﺪﺍﻡ ﺍﻷﻗﻮﺍﺱ ﻟﻠﺘﺤﻜﻢ ﰲ ﺗﺮﺗﻴﺐ ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ‪ ،‬ﺣﻴﺚ ﺃﻥ ﺍﻷﻗﻮﺍﺱ ﳍﺎ ﺍﻷﺳﺒﻘﻴﺔ ﺍﻟﻌﻠﻴﺎ‪ ،‬ﺃﻱ ﺃﻥ ﻣﺎ ﺑﲔ ﺍﻷﻗﻮﺍﺱ‬ ‫ﻳﺘﻢ ﺣﺴﺎﺑﻪ ﺩﺍﺋﻤﺎﹰ ﺃﻭﻻﹰ‪.‬ﻣﺜﻼﹰ ﺇﺫﺍ ﻗﻤﻨﺎ ﰲ ﺍﻟﺘﻌﺒﲑ ﺍﻟﺴﺎﺑﻖ ﺑﺈﺿﺎﻓﺔ ﻗﻮﺳﲔ ﻛﺎﻟﺘﺎﱄ‬ ‫‪x = ( A + B) ⋅ C‬‬ ‫ﻓﺈﻧﻪ ﻳﺘﻢ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ‪ OR‬ﺍﳌﻮﺟﻮﺩﺓ ﺑﲔ ﺍﻟﻘﻮﺳﲔ ﻗﺒﻞ ﻋﻤﻠﻴﺔ ‪ ،AND‬ﻭ ﺫﻟﻚ ﻋﻠﻰ ﺍﻟﺮﻏﻢ ﻣﻦ ﺃﻥ ﻋﻤﻠﻴﺔ ‪ AND‬ﳍﺎ ﺃﺳﺒﻘﻴﺔ‬ ‫ﺃﻋﻠﻰ ﻣﻦ ﻋﻤﻠﻴﺔ ‪.OR‬ﻭ ﺍﻟﺴﺒﺐ ﰲ ﺫﻟﻚ ﻫﻮ ﻭﺟﻮﺩ ﻋﻤﻠﻴﺔ ‪ OR‬ﻣﺎ ﺑﲔ ﺍﻟﻘﻮﺳﲔ‪.‬ﺣﻴﺚ ﻳﺘﻢ ﺃﻭﻻﹰ ﺣﺴﺎﺏ ﻣﺎ ﺑﲔ‬ ‫ﺍﻟﻘﻮﺳﲔ‪ ،‬ﻓﻴﺘﻢ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ﻟﻠﻤﺘﻐﲑ ‪ ، B‬ﰒ ﻋﻤﻠﻴﺔ ‪ OR‬ﺑﲔ ‪ A‬ﻭ ‪ ، B‬ﻭ ﺑﻌﺪ ﺍﻻﻧﺘﻬﺎﺀ ﻣﻦ ﺍﻷﻗﻮﺍﺱ ﻳﺘﻢ‬ ‫ﺇﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺧﺎﺭﺟﻬﺎ‪ ،‬ﻓﻴﺘﻢ ﺇﺟﺮﺍﺀ ﻋﻤﻠﻴﺔ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ﻟﻠﻤﺘﻐﲑ ‪ ، C‬ﰒ ﻋﻤﻠﻴﺔ ‪ AND‬ﳌﺎ ﺑﲔ ﺍﻟﻘﻮﺳﲔ ﻭ ‪. C‬‬ ‫‪ -5‬ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ )‪(Logic Circuit‬‬ ‫ﳝﻜﻦ ﲤﺜﻴﻞ ﺃﻱ ﺗﻌﺒﲑ ﻣﻨﻄﻘﻲ ﺑﺪﺍﺋﺮﺓ ﻣﻨﻄﻘﻴﺔ‪ ،‬ﺣﻴﺚ ﻧﻨﻈﺮ ﻟﻠﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﳌﻮﺟﻮﺩﺓ ﺑﺎﻟﺘﻌﺒﲑ ﻭ ﻧﻘﻮﻡ ﺑﺮﺑﻂ ﺍﻟﺒﻮﺍﺑﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﺗﻠﻚ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺑﺎﻷﺳﻠﻮﺏ ﺍﳌﻨﺎﺳﺐ‪.‬ﻣﺜﻼ‪ ً،‬ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪x = A+ B ⋅ C‬‬ ‫ﳝﻜﻦ ﲤﺜﻴﻠﻪ ﺑﺎﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫‪A‬‬ ‫‪x‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪14‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻭ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪x = ( A + B) ⋅ C‬‬ ‫ﳝﻜﻦ ﲤﺜﻴﻠﻪ ﺑﺎﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x‬‬ ‫‪C‬‬ ‫‪ -6‬ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ )‪(Logic Diagram‬‬ ‫ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﳐﻄﻂ ﻣﺒﺴﻂ ﻳﻮﺿﺢ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻟﻠﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻭ ﻣﺴﻤﻴﺎ‪‬ﺎ ﻭ ﻣﺘﻐﲑﺍﺕ ﺍﳋﺮﺝ ﻭﻣﺴﻤﻴﺎ‪‬ﺎ‪ ،‬ﺑﺎﻹﺿﺎﻓﺔ‬ ‫ﺇﱃ ﺍﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﺪﺍﻝ ﻋﻠﻰ ﻭﻇﻴﻔﺘﻬﺎ‪.‬‬ ‫ﻣﺜﻼﹰ‪ ،‬ﻛﻼ ﺍﻟﺪﺍﺋﺮﺗﲔ ﺍﳌﻨﻄﻘﻴﺘﲔ ﺃﻋﻼﻩ ﳝﻜﻦ ﲤﺜﻴﻠﻬﻤﺎ ﺑﺎﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ ﺍﻟﺘﺎﱄ‪:‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫ﺍﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ‬ ‫‪x‬‬ ‫‪C‬‬ ‫ﻭ ﻧﻘﻮﻡ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﳌﺨﻄﻄﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻛﺒﺪﻳﻞ ﻟﻠﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﺍﳌﻔﺼﻠﺔ ﻛﻨﻮﻉ ﻣﻦ ﺍﻟﺘﺒﺴﻴﻂ‪ ،‬ﻭ ﺫﻟﻚ ﻋﻨﺪﻣﺎ ﻻ ﻧﻜﻮﻥ ﲝﺎﺟﺔ‬ ‫ﻟﻠﺘﻔﺎﺻﻴﻞ ﺍﻟﺪﺍﺧﻠﻴﺔ ﻟﻠﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‪.‬ﻛﻤﺎ ﰲ ﺍﻟﺪﻭﺍﺋﺮ ﺍﳌﻌﻘﺪﺓ ﺍﳌﻜﻮﻧﺔ ﻣﻦ ﻋﺪﺩ ﻣﻦ ﺍﻟﺪﻭﺍﺋﺮ ﺍﻟﺼﻐﲑﺓ ﺍﳌﺮﺑﻮﻃﺔ ﻣﻊ ﺑﻌﻀﻬﺎ‬ ‫ﺍﻟﺒﻌﺾ‪ ،‬ﺣﻴﺚ ﻧﻘﻮﻡ ﺑﺘﻤﺜﻴﻞ ﺗﻠﻚ ﺍﻟﺪﻭﺍﺋﺮ ﺍﻟﺼﻐﲑﺓ ﲟﺨﻄﻄﺎ‪‬ﺎ ﺍﳌﻨﻄﻘﻴﺔ‪.‬‬ ‫‪ -7‬ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ )‪(Truth Table‬‬ ‫ﻋﺒﺎﺭﺓ ﻋﻦ ﺟﺪﻭﻝ ﻳﻮﺿﺢ ﲨﻴﻊ ﺍﺣﺘﻤﺎﻻﺕ ﺍﻟﺪﺧﻞ ﻟﻠﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻭ ﻗﻴﻢ ﺍﳋﺮﺝ ﺍﳌﻘﺎﺑﻞ ﻟﻜﻞ ﻣﻨﻬﺎ‪.‬ﻣﺜﻼﹰ‪ ،‬ﻹﻧﺸﺎﺀ ﺟﺪﻭﻝ‬ ‫ﺻﻮﺍﺏ ﻟﻠﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪x = A+ B ⋅ C‬‬ ‫ﻧﺒﺪﺃ ﺑﺘﺤﺪﻳﺪ ﻋﺪﺩ ﺍﻟﺼﻔﻮﻑ ﻭ ﻋﺪﺩ ﺍﻷﻋﻤﺪﺓ ﰲ ﺍﳉﺪﻭﻝ‪.‬ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻫﻲ ‪ A‬ﻭ ‪ B‬ﻭ ‪ ، C‬ﻭ ﻋﺪﺩﻫﺎ ‪ ،3‬ﺃﻱ ﻋﻦ‬ ‫ﻋﺪﺩ ﺍﺣﺘﻤﺎﻻﺕ ﺍﻟﺪﺧﻞ ﻫﻮ ‪ ، 2 3 = 8‬ﻭ ﻫﻮ ﻋﺪﺩ ﺃﺳﻄﺮ )ﺻﻔﻮﻑ( ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‪.‬ﺃﻣﺎ ﻋﻦ ﺍﻷﻋﻤﺪﺓ ﻓﻨﺤﺘﺎﺝ ﻋﻤﻮﺩﺍﹰ‬ ‫‪15‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻟﻜﻞ ﻣﺘﻐﲑ ﻣﻦ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻭ ﻋﻤﻮﺩﺍﹰ ﻟﻜﻞ ﻣﺘﻐﲑ ﻣﻦ ﻣﺘﻐﲑﺍﺕ ﺍﳋﺮﺝ‪.‬ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻋﺪﺩﻫﺎ ‪ ،3‬ﻛﻤﺎ ﺫﻛﺮﻧﺎ ﻣﻦ‬ ‫ﻗﺒﻞ‪ ،‬ﻭ ﻫﻨﺎﻙ ﻣﺘﻐﲑ ﺧﺮﺝ ﻭﺍﺣﺪ ﻫﻮ ‪ ، x‬ﺃﻱ ﳓﺘﺎﺝ ﺇﱃ ﺃﺭﺑﻌﺔ ﺃﻋﻤﺪﺓ ﳌﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ ﻭ ﻣﺘﻐﲑﺍﺕ ﺍﳋﺮﺝ‪.‬ﻛﻤﺎ ﳓﺘﺎﺝ ﺇﱃ‬ ‫ﺃﻋﻤﺪﺓ ﺇﺿﺎﻓﻴﺔ ﻹﺟﺮﺍﺀ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﺣﻴﺚ ﳓﺘﺎﺝ ﻋﻤﻮﺩﺍﹰ ﻹﳚﺎﺩ ‪ ، B‬ﻭ ﻋﻤﻮﺩﺍﹰ ﺁﺧﺮ ﻹﳚﺎﺩ ‪ ، C‬ﻛﻤﺎ ﳓﺘﺎﺝ ﻋﻤﻮﺩﺍﹰ‬ ‫ﻹﳚﺎﺩ ‪ ، B ⋅ C‬ﻭ ﺃﺧﲑﺍﹰ ﳓﺘﺎﺝ ﻋﻤﻮﺩﺍﹰ ﻹﳚﺎﺩ ‪ ، A + B ⋅ C‬ﻭ ﻫﻮ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻧﻔﺲ ﻋﻤﻮﺩ ﺍﳋﺮﺝ ‪. x‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫= ‪C B⋅C x‬‬ ‫‪A+ B ⋅ C‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫ﻭ ﺑﺎﳌﺜﻞ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻟﻠﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪x = ( A + B) ⋅ C‬‬ ‫ﻫﻮ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪A+ B x = ( A + B) ⋅ C‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺭﺳﻢ ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‪ ،‬ﻭ ﺃﻛﻤﻞ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‪ ،‬ﰒ ﺍﺭﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻟﻠﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + AB‬‬ ‫ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‬ ‫‪A‬‬ ‫‪B‬‬ ‫ﺍﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ‬ ‫‪y‬‬ ‫‪C‬‬ ‫‪16‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪ABC AB ABC + AB x = ABC + AB‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪x = ABC + AB‬‬ ‫‪C‬‬ ‫ﺗﺪﺭﻳﺐ ‪:4‬‬ ‫ﺍﺭﺳﻢ ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‪ ،‬ﻭ ﺃﻛﻤﻞ ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‪ ،‬ﰒ ﺍﺭﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻟﻜﻞ ﺗﻌﺒﲑ ﻣﻦ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ‪:‬‬ ‫‪x = A( B + C ) -1‬‬ ‫‪y = AB( A + C ) -2‬‬ ‫‪z = AB + C D -3‬‬ ‫‪ -8‬ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ )‪(Boolean Algebra Theorems‬‬ ‫ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﻫﻮ ﺟﱪ ﺍﳌﺘﻐﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﻭ ﺍﳍﺪﻑ ﺍﻷﺳﺎﺳﻲ ﻣﻦ ﺩﺭﺍﺳﺘﻨﺎ ﻟﻨﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﻫﻮ ﺍﺳﺘﺨﺪﺍﻡ ﺗﻠﻚ‬ ‫ﺍﻟﻨﻈﺮﻳﺎﺕ ﰲ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‪.‬‬ ‫ﻟﻜﻞ ﻧﻈﺮﻳﺔ )‪ (Theorem‬ﻣﻦ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﻧﻈﺮﻳﺔ ﻣﻘﺎﺑﻠﺔ ﺃﻭ ﻣﻨﺎﻇﺮﺓ ﳍﺎ )‪.(Dual Theorem‬ﻭ ﻟﻠﺤﺼﻮﻝ‬ ‫ﻋﻠﻰ ﺍﻟﻨﻈﺮﻳﺔ ﺍﳌﻘﺎﺑﻠﺔ ﻷﻱ ﻧﻈﺮﻳﺔ ﻧﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﺍﻟﺘﺒﺪﻳﻼﺕ ﺍﻟﺘﺎﻟﻴﺔ ﰲ ﺍﻟﻨﻈﺮﻳﺔ ﺍﻷﺻﻠﻴﺔ‪:‬‬ ‫ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻱ ‪ 0‬ﺑـ ‪1‬‬ ‫ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻱ ‪ 1‬ﺑـ ‪0‬‬ ‫ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻱ ﻋﻤﻠﻴﺔ ‪ AND‬ﺑﻌﻤﻠﻴﺔ ‪OR‬‬ ‫ ﺍﺳﺘﺒﺪﺍﻝ ﺃﻱ ﻋﻤﻠﻴﺔ ‪ OR‬ﺑﻌﻤﻠﻴﺔ ‪AND‬‬ ‫‪17‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻭﻋﻤﻮﻣﺎﹰ ﳝﻜﻦ ﺇﺛﺒﺎﺕ ﺻﺤﺔ ﺃﻱ ﻧﻈﺮﻳﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺟﺪﺍﻭﻝ ﺍﻟﺼﻮﺍﺏ‪.‬ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ﻳﻮﺿﺢ ﺍﻟﻨﻈﺮﻳﺎﺕ ﺍﻷﺳﺎﺳﻴﺔ ﺍﳌﺴﺘﺨﺪﻣﺔ ﰲ‬ ‫ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﳌﻘﺎﺑﻠﺔ‬ ‫ﺍﻟﻨﻈﺮﻳﺔ‬ ‫ﺍﺳﻢ ﺍﻟﻨﻈﺮﻳﺔ‬ ‫‪A= A‬‬ ‫‪A= A‬‬ ‫ﻋﻜﺲ ﺍﻟﻌﻜﺲ‬ ‫‪A⋅ 0 = 0‬‬ ‫‪A+ 1 = 1‬‬ ‫ﺍﻟﻌﻤﻠﻴﺎﺕ ﻣﻊ ‪ 1‬ﻭ ‪0‬‬ ‫‪A⋅ 1 = A‬‬ ‫‪A+ 0 = A‬‬ ‫‪A⋅ A = A‬‬ ‫‪A+ A = A‬‬ ‫ﺍﳌﺘﻐﲑ ﻣﻊ ﻧﻔﺴﻪ‬ ‫‪A⋅ A = 0‬‬ ‫‪A+ A = 1‬‬ ‫ﺍﳌﺘﻐﲑ ﻣﻊ ﻋﻜﺴﻪ‬ ‫‪A⋅ B = B ⋅ A‬‬ ‫‪A+ B = B + A‬‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻹﺑﺪﺍﻟﻴﺔ‬ ‫) ‪( A ⋅ B) ⋅ C = A ⋅ ( B ⋅ C‬‬ ‫) ‪( A + B) + C = A + ( B + C‬‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﺠﻤﻴﻌﻴﺔ‬ ‫) ‪A + B ⋅ C = ( A + B) ⋅ ( A + C‬‬ ‫‪A ⋅ (B + C) = A⋅ B + A⋅ C‬‬ ‫ﺍﻟﻨﻈﺮﻳﺔ ﺍﻟﺘﻮﺯﻳﻌﻴﺔ‬ ‫‪A ⋅ ( A + B) = A‬‬ ‫‪A+ A⋅ B = A‬‬ ‫ﺍﻻﻣﺘﺼﺎﺹ ﺃﻭ ﺍﻻﺑﺘﻼﻉ‬ ‫‪A ⋅ ( A + B) = A ⋅ B‬‬ ‫‪A+ A⋅ B = A+ B‬‬ ‫‪A ⋅ B = A+ B‬‬ ‫‪A+ B = A⋅ B‬‬ ‫ﺩﻱ ﻣﻮﺭﻏﺎﻥ )‪(De Morgan‬‬ ‫‪ -9‬ﺍﺳﺘﺨﺪﺍﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰱ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‬ ‫ﺍﳍﺪﻑ ﻣﻦ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻫﻮ ﺗﺒﺴﻴﻂ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﺃﻱ ﺗﻘﻠﻴﻞ ﻋﺪﺩ ﺍﻟﺒﻮﺍﺑﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﺪﺍﺧﻠﺔ ﰱ ﺑﻨﺎﺋﻬﺎ‪ ،‬ﻭ ﺫﻟﻚ‬ ‫ﻟﺘﻘﻠﻴﻞ ﺗﻜﻠﻔﺘﻬﺎ‪.‬ﻛﻤﺎ ﻳﻌﺘﱪ ﺗﻘﻠﻴﻞ ﺗﻔﺮﻉ ﺍﻟﺪﺧﻞ ﻟﻠﺒﻮﺍﺑﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﳌﺴﺘﺨﺪﻣﺔ ﰱ ﺑﻨﺎﺀ ﺍﻟﺪﺍﺋﺮﺓ ﻧﻮﻋﺎﹰ ﻣﻦ ﺍﻟﺘﺒﺴﻴﻂ ﺃﻳﻀﺎﹰ‪.‬‬ ‫ﺳﻨﻌﺮﺽ ﰲ ﻫﺬﺍ ﺍﳉﺰﺀ ﻋﺪﺩﺍﹰ ﻣﻦ ﺍﻷﻣﺜﻠﺔ ﻟﺘﻮﺿﻴﺢ ﻃﺮﻳﻘﺔ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻨﻈﺮﻳﺎﺕ‪.‬ﻭ ﻧﺬﻛﺮ ﺍﻟﻘﺎﺭﺉ‬ ‫ﺑﻀﺮﻭﺭﺓ ﺍﻟﺘﺪﺭﺏ ﻋﻠﻰ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺒﺴﻴﻂ ﺑﻔﻬﻢ ﺍﻷﻣﺜﻠﺔ ﺟﻴﺪﺍﹰ ﻭ ﺇﻋﺎﺩﺓ ﺣﻠﻬﺎ ﻭ ﺣﻞ ﺍﻟﺘﺪﺭﻳﺒﺎﺕ‪.‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰱ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + AB‬‬ ‫ﰒ ﺍﺭﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻗﺒﻞ ﺍﻟﺘﺒﺴﻴﻂ ﻭ ﺑﻌﺪﻩ‪.‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫‪y = ABC + AB‬‬ ‫)‪y = ( A+ B + C ) ⋅ ( A+ B‬‬ ‫ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫)‪y = ( A + B + C ) ⋅ ( A + B‬‬ ‫ﻋﻜﺲ ﺍﻟﻌﻜﺲ‬ ‫‪y = A+ ( B + C ) ⋅ B‬‬ ‫ﺍﻟﺘﻮﺯﻳﻌﻴﺔ‬ ‫‪y = A+ C B‬‬ ‫ﺍﻻﺑﺘﻼﻉ‬ ‫‪18‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺣﻞ ﺁﺧﺮ‪:‬‬ ‫‪y = ABC + AB‬‬ ‫)‪y = A ⋅ ( BC + B‬‬ ‫ﺍﻟﺘﻮﺯﻳﻌﻴﺔ‬ ‫)‪y = A ⋅ (C + B‬‬ ‫ﺍﻻﺑﺘﻼﻉ‬ ‫‪y = A+ C B‬‬ ‫ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫‪y = A+ C B‬‬ ‫ﻋﻜﺲ ﺍﻟﻌﻜﺲ‬ ‫ﺍﻟﺪﺍﺋﺮﺓ ﻗﺒﻞ ﺍﻟﺘﺒﺴﻴﻂ‪:‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪y‬‬ ‫‪C‬‬ ‫ﺍﻟﺪﺍﺋﺮﺓ ﺑﻌﺪ ﺍﻟﺘﺒﺴﻴﻂ‪:‬‬ ‫‪A‬‬ ‫‪y‬‬ ‫‪B‬‬ ‫‪C‬‬ ‫ﻻﺣﻆ ﺃﻥ ﺍﻟﺪﺍﺋﺮﺓ ﻗﺒﻞ ﺍﻟﺘﺒﺴﻴﻂ ﻣﻜﻮﻧﺔ ﻣﻦ ‪ 6‬ﺑﻮﺍﺑﺎﺕ‪ ،‬ﻭ ﺑﻌﺪ ﺍﻟﺘﺒﺴﻴﻂ ﺃﺻﺒﺤﺖ ﻣﻜﻮﻧﺔ ﻣﻦ ‪ 4‬ﺑﻮﺍﺑﺎﺕ ﻓﻘﻂ‪.‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰱ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = A( A + B) + C + CB‬‬ ‫ﰒ ﺍﺭﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ ﻗﺒﻞ ﺍﻟﺘﺒﺴﻴﻂ ﻭ ﺑﻌﺪﻩ‪.‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫‪y = A( A + B) + C + CB‬‬ ‫‪y = AB + C + CB‬‬ ‫ﺍﻻﺑﺘﻼﻉ‬ ‫‪y = AB + C + B‬‬ ‫ﺍﻻﺑﺘﻼﻉ‬ ‫‪y = AB + B + C‬‬ ‫ﺍﻹﺑﺪﺍﻟﻴﺔ‬ ‫‪y = B+C‬‬ ‫ﺍﻻﺑﺘﻼﻉ‬ ‫‪19‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺍﻟﺪﺍﺋﺮﺓ ﻗﺒﻞ ﺍﻟﺘﺒﺴﻴﻂ‬ ‫‪A‬‬ ‫‪B‬‬ ‫‪y‬‬ ‫‪C‬‬ ‫ﺍﻟﺪﺍﺋﺮﺓ ﺑﻌﺪ ﺍﻟﺘﺒﺴﻴﻂ‬ ‫‪B‬‬ ‫‪y‬‬ ‫‪C‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰲ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫ﳍﺬﺍ ﺍﳌﺜﺎﻝ ﺃﳘﻴﺔ ﺧﺎﺻﺔ‪ ،‬ﻭ ﺫﻟﻚ ﻧﻈﺮﺍﹰ ﺇﱃ ﺃﻥ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻳﻈﻬﺮ ﰲ ﺻﻮﺭﺓ ﳑﻴﺰﺓ ﺗﺴﻤﻰ ﺻﻮﺭﺓ ﳎﻤﻮﻉ ﺍﳊﺪﻭﺩ ﺍﻟﺼﻐﺮﻯ‬ ‫)‪.(Sum of minterms‬ﻭ ﰲ ﻫﺬﻩ ﺍﻟﺼﻮﺭﺓ ﻳﺘﻜﻮﻥ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ ﻣﻦ ﳎﻤﻮﻋﺔ ﻣﻦ ﺍﳊﺪﻭﺩ ﺍﳌﺮﺗﺒﻄﺔ ﻣﻊ ﺑﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ‬ ‫ﺑﻌﻤﻠﻴﺎﺕ ‪.OR‬ﻭ ﻳﺴﻤﻰ ﻛﻞ ﺣﺪ ﻣﻨﻬﺎ ﺑﺎﳊﺪ ﺍﻷﺻﻐﺮ )‪.(minterm‬ﻭ ﺍﳊﺪ ﺍﻷﺻﻐﺮ ﺗﻈﻬﺮ ﻓﻴﻪ ﲨﻴﻊ ﻣﺘﻐﲑﺍﺕ ﺍﻟﺪﺧﻞ‬ ‫ﻣﺮﺗﺒﻄﺔ ﻣﻊ ﺑﻌﻀﻬﺎ ﺍﻟﺒﻌﺾ ﺑﻌﻤﻠﻴﺎﺕ ‪ ،AND‬ﻭ ﻳﻜﻮﻥ ﺑﻌﺾ ﻫﺬﻩ ﺍﳌﺘﻐﲑﺍﺕ ﻣﻌﻜﻮﺳﺎﹰ ﻭ ﺑﻌﻀﻬﺎ ﺍﻵﺧﺮ ﻏﲑ ﻣﻌﻜﻮﺱ‪.‬‬ ‫ﻟﺘﺒﺴﻴﻂ ﻫﺬﺍ ﺍﻟﻨﻮﻉ ﻣﻦ ﺍﻟﺘﻌﺒﲑﺍﺕ ﻧﺒﺤﺚ ﻋﻦ ﺍﻟﺘﺸﺎ‪‬ﺎﺕ ﻣﺎ ﺑﲔ ﺍﳊﺪﻭﺩ‪.‬ﻭ ﺍﳊﺪﺍﻥ ﺍﳌﺘﺸﺎ‪‬ﺎﻥ ﳘﺎ ﺣﺪﻳﻦ ﻳﺘﻔﻘﺎﻥ ﰱ ﻛﻞ‬ ‫ﺷﻲﺀ ﻋﺪﺍ ﻣﺘﻐﲑ ﻭﺍﺣﺪ ﻳﻈﻬﺮ ﰱ ﺃﺣﺪﳘﺎ ﻣﻌﻜﻮﺳﺎﹰ ﻭ ﰲ ﺍﻵﺧﺮ ﺑﺪﻭﻥ ﻋﻜﺲ‪.‬ﻣﺜﻼﹰ‪ ،‬ﰲ ﺍﻟﺘﻌﺒﲑ ﺃﻋﻼﻩ ﺍﳊﺪ ﺍﻷﻭﻝ ‪ABC‬‬ ‫ﻳﺸﺒﻪ ﺍﳊﺪ ﺍﻟﺜﺎﱐ ‪ ، ABC‬ﺣﻴﺚ ﻳﺘﻔﻖ ﺍﳊﺪﺍﻥ ﰲ ﻛﻞ ﺷﻲﺀ ﻋﺪﺍ ﺍﳌﺘﻐﲑ ‪ C‬ﺍﻟﺬﻱ ﻳﻈﻬﺮ ﰲ ﺍﳊﺪ ﺍﻷﻭﻝ ﻣﻌﻜﻮﺳﺎﹰ ﻭ ﰲ ﺍﳊﺪ‬ ‫ﺍﻟﺜﺎﱐ ﺑﺪﻭﻥ ﻋﻜﺲ‪.‬ﻭ ﺑﻨﻔﺲ ﺍﻟﻄﺮﻳﻘﺔ ﻳﺘﺸﺎﺑﻪ ﺍﳊﺪﺍﻥ ﺍﻟﺜﺎﻟﺚ ‪ ABC‬ﻭ ﺍﻟﺮﺍﺑﻊ ‪ ، ABC‬ﺣﻴﺚ ﻳﺘﻔﻘﺎﻥ ﰲ ﻛﻞ ﺷﻲﺀ ﻋﺪﺍ‬ ‫ﺍﳌﺘﻐﲑ ‪ A‬ﺍﻟﺬﻱ ﻳﻈﻬﺮ ﰲ ﺍﳊﺪ ﺍﻟﺜﺎﻟﺚ ﻣﻌﻜﻮﺳﺎﹰ ﻭ ﰲ ﺍﳊﺪ ﺍﻟﺮﺍﺑﻊ ﺑﺪﻭﻥ ﻋﻜﺲ‪.‬‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫ﻻﺣﻆ ﺃﻥ ﺍﻹﺧﺘﻼﻑ ﻣﺎ ﺑﲔ ﺍﳊﺪﻳﻦ ﺍﳌﺘﺸﺎ‪‬ﲔ ﳚﺐ ﺃﻥ ﻳﻜﻮﻥ ﰲ ﻣﺘﻐﲑ ﻭﺍﺣﺪ ﻓﻘﻂ ﻭ ﻻ ﳚﻮﺯ ﺃﻥ ﻳﻜﻮﻥ ﰲ ﺃﻛﺜﺮ ﻣﻦ ﻣﺘﻐﲑ‪.‬‬ ‫‪20‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺑﻌﺪ ﺇﳚﺎﺩ ﺍﻟﺘﺸﺎ‪‬ﺎﺕ ﻣﺎ ﺑﲔ ﺍﳊﺪﻭﺩ ﻧﻘﻮﻡ ﲜﻤﻊ ﻛﻞ ﺣﺪﻳﻦ ﻣﺘﺸﺎ‪‬ﲔ ﰲ ﺣﺪ ﻭﺍﺣﺪ ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﺍﻟﻌﺎﻣﻞ ﺍﳌﺸﺘﺮﻙ ﻣﺎ ﺑﲔ‬ ‫ﺍﳊﺪﻳﻦ‪ ،‬ﺃﻣﺎ ﺍﳌﺘﻐﲑ ﺍﳌﺨﺘﻠﻒ ﻓﻴﺘﻢ ﺍﺧﺘﺼﺎﺭﻩ‪.‬‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫)‪y = AB(C + C ) + BC ( A + A‬‬ ‫ﺑﺈﺧﺮﺍﺝ ﺍﻟﻌﺎﻣﻞ ﺍﳌﺸﺘﺮﻙ ﰲ ﻛﻞ ﺣﺪﻳﻦ ﻣﺘﺸﺎ‪‬ﲔ‬ ‫)‪y = AB(1) + BC (1‬‬ ‫ﲜﻤﻊ ﺍﳌﺘﻐﲑ ﻣﻊ ﻋﻜﺴﻪ‬ ‫‪y = AB + BC‬‬ ‫ﺑﺎﻟﻌﻤﻠﻴﺎﺕ ﻣﻊ ‪1‬‬ ‫ﻻﺣﻆ ﰲ ﺍﳌﺜﺎﻝ ﺍﻟﺴﺎﺑﻖ ﻭﺟﻮﺩ ﺗﺸﺎﺑﻪ ﺇﺿﺎﰲ ﺑﲔ ﺍﳊﺪﻭﺩ‪ ،‬ﺣﻴﺚ ﺃﻥ ﺍﳊﺪ ﺍﻟﺜﺎﱐ ‪ ABC‬ﻳﺸﺒﻪ ﺍﳊﺪ ﺍﻟﺜﺎﻟﺚ ‪ ، ABC‬ﻭ ﻟﻜﻦ‬ ‫ﱂ ﻧﻜﻦ ﰲ ﺣﺎﺟﺔ ﻻﺳﺘﺨﺪﺍﻡ ﻫﺬﺍ ﺍﻟﺘﺸﺎﺑﻪ ﰲ ﻋﻤﻠﻴﺔ ﺍﻟﺘﺒﺴﻴﻂ‪.‬‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰲ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + ABC + ABC + ABC + ABC‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫ﺍﻟﺘﻌﺒﲑ ﻫﻨﺎ ﰲ ﺻﻮﺭﺓ ﳎﻤﻮﻉ ﺍﳊﺪﻭﺩ ﺍﻟﺼﻐﺮﻯ‪ ،‬ﻟﺬﻟﻚ ﻧﺒﺤﺚ ﻋﻦ ﺍﻟﺘﺸﺎ‪‬ﺎﺕ ﻣﺎ ﺑﲔ ﺍﳊﺪﻭﺩ‪.‬‬ ‫ﺍﳊﺪ ﺍﻷﻭﻝ ﻳﺸﺒﻪ ﺍﳊﺪ ﺍﻟﺜﺎﱐ‪ ،‬ﻭ ﺍﳊﺪ ﺍﻟﺮﺍﺑﻊ ﻳﺸﺒﻪ ﺍﳊﺪ ﺍﳋﺎﻣﺲ‪ ،‬ﻭ ﺍﳊﺪ ﺍﻟﺜﺎﻟﺚ ﻳﺸﺒﻪ ﺍﳊﺪ ﺍﻷﻭﻝ‪.‬‬ ‫‪y = ABC + ABC + ABC + ABC + ABC‬‬ ‫ﻧﻼﺣﻆ ﻫﻨﺎ ﻭﺟﻮﺩ ﻣﺸﻜﻠﺔ ﺗﺘﻤﺜﻞ ﰲ ﺃﻥ ﺍﳊﺪ ﺍﻷﻭﻝ ﻳﺘﺸﺎﺑﻪ ﰲ ﻧﻔﺲ ﺍﻟﻮﻗﺖ ﻣﻊ ﻛﻞ ﻣﻦ ﺍﳊﺪﻳﻦ ﺍﻟﺜﺎﱐ ﻭ ﺍﻟﺜﺎﻟﺚ‪.‬ﰲ ﻣﺜﻞ‬ ‫ﻫﺬﻩ ﺍﳊﺎﻻﺕ ﻧﻘﻮﻡ ﺑﺘﻜﺮﺍﺭ ﺍﳊﺪ ﺍﻷﻭﻝ )ﻣﺴﺘﺨﺪﻣﲔ ﻧﻈﺮﻳﺔ ﺍﳌﺘﻐﲑ ﻣﻊ ﻧﻔﺴﻪ( ﲝﻴﺚ ﻳﺘﻢ ﲨﻌﻪ ﻣﻊ ﻛﻼ ﺍﳊﺪﻳﻦ ﺍﻟﺜﺎﱐ ﻭ‬ ‫ﺍﻟﺜﺎﻟﺚ‪.‬‬ ‫‪y = ABC + ABC + ABC + ABC + ABC‬‬ ‫‪y = ABC + ABC + ABC + ABC + ABC + ABC‬‬ ‫ﺑﺘﻜﺮﺍﺭ ﺍﳊﺪ ﺍﻷﻭﻝ‬ ‫‪y = AB + AC + AB‬‬ ‫ﲜﻤﻊ ﻛﻞ ﺣﺪﻳﻦ ﻣﺘﺸﺎ‪‬ﲔ‬ ‫‪y = AB + AB + AC‬‬ ‫ﺑﺎﻟﻨﻈﺮﻳﺔ ﺍﻹﺑﺪﺍﻟﻴﺔ‬ ‫‪y = B + AC‬‬ ‫ﲜﻤﻊ ﺍﳊﺪﻳﻦ ﺍﳌﺘﺸﺎ‪‬ﲔ‬ ‫‪21‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰲ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫ﻧﻼﺣﻆ ﺃﻥ ﻣﺎ ﺃﺳﻔﻞ ﺧﻂ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ﺍﳋﺎﺭﺟﻲ ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﺗﻌﺒﲑ ﰲ ﺻﻮﺭﺓ ﳎﻤﻮﻉ ﺍﳊﺪﻭﺩ ﺍﻟﺼﻐﺮﻯ‪ ،‬ﻟﺬﻟﻚ ﻧﺒﺤﺚ‬ ‫ﻋﻦ ﺍﻟﺘﺸﺎ‪‬ﺎﺕ ﻣﺎ ﺑﲔ ﺍﳊﺪﻭﺩ‪.‬‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫‪y = BC + AB‬‬ ‫ﲜﻤﻊ ﻛﻞ ﺣﺪﻳﻦ ﻣﺘﺸﺎ‪‬ﲔ‬ ‫)‪y = ( BC ) ⋅ ( AB‬‬ ‫ﺑﻨﻈﺮﻳﺔ ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫)‪y = ( B + C ) ⋅ ( A + B‬‬ ‫ﺑﻨﻈﺮﻳﺔ ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫ﻣﺜﺎﻝ‪:‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰲ ﺗﺒﺴﻴﻂ ﺍﻟﺘﻌﺒﲑ ﺍﳌﻨﻄﻘﻲ‬ ‫‪y = ABC + ABC + ABC‬‬ ‫ﺍﳊﻞ‪:‬‬ ‫ﻧﻼﺣﻆ ﺃﻥ ﻣﺎ ﺃﺳﻔﻞ ﺧﻂ ﺍﻟﻌﻜﺲ ﺍﳌﻨﻄﻘﻲ ﺍﳋﺎﺭﺟﻲ ﻫﻮ ﻋﺒﺎﺭﺓ ﻋﻦ ﺗﻌﺒﲑ ﰲ ﺻﻮﺭﺓ ﳎﻤﻮﻉ ﺍﳊﺪﻭﺩ ﺍﻟﺼﻐﺮﻯ‪ ،‬ﻟﺬﻟﻚ ﻧﺒﺤﺚ‬ ‫ﻋﻦ ﺍﻟﺘﺸﺎ‪‬ﺎﺕ ﻣﺎ ﺑﲔ ﺍﳊﺪﻭﺩ‪.‬‬ ‫‪y = ABC + ABC + ABC‬‬ ‫‪y = ABC + ABC + ABC‬‬ ‫‪y = ABC + ABC + ABC + ABC‬‬ ‫ﺑﺘﻜﺮﺍﺭ ﺍﳊﺪ ﺍﻟﺜﺎﻟﺚ‬ ‫‪y = BC + AB‬‬ ‫ﲜﻤﻊ ﻛﻞ ﺣﺪﻳﻦ ﻣﺘﺸﺎ‪‬ﲔ‬ ‫)‪y = B(C + A‬‬ ‫ﺑﺄﺧﺬ ﺍﻟﻌﺎﻣﻞ ﺍﳌﺸﺘﺮﻙ‬ ‫)‪y = B + (C + A‬‬ ‫ﺑﻨﻈﺮﻳﺔ ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫‪y = B+CA‬‬ ‫ﺑﻨﻈﺮﻳﺔ ﺩﻱ ﻣﻮﺭﻏﺎﻥ‬ ‫‪22‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  ‫ﺗﺪﺭﻳﺐ ‪:5‬‬ ‫ﺍﺳﺘﺨﺪﻡ ﻧﻈﺮﻳﺎﺕ ﺍﳉﱪ ﺍﻟﺒﻮﻟﻴﺎﱐ ﰲ ﺗﺒﺴﻴﻂ ﻛﻞ ﻣﻦ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫‪A = x + xyz + x yz + xw + x w + x y‬‬ ‫‪-1‬‬ ‫‪B = ( x + y + xy)( x + y) xy‬‬ ‫‪-2‬‬ ‫)‪C = ( x + y + x y)( xy + xz + yz‬‬ ‫‪-3‬‬ ‫ﺍﳋﻼﺻﺔ‬ ‫ﻋﺰﻳﺰﻱ ﺍﻟﺪﺍﺭﺱ‪ ،‬ﺗﻌﻠﻤﻨﺎ ﰲ ﻫﺬﻩ ﺍﻟﻮﺣﺪﺓ ﺑﻌﺾ ﺍﳌﻬﺎﺭﺍﺕ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﱵ ﺳﻨﺤﺘﺎﺝ ﺇﻟﻴﻬﺎ ﰲ ﺩﺭﺍﺳﺘﻨﺎ ﻟﺒﻘﻴﺔ ﺍﳌﻘﺮﺭ‪.‬ﺣﻴﺚ ﺗﻌﺮﻓﻨﺎ‬ ‫ﻋﻠﻰ ﺍﻟﻌﻤﻠﻴﺎﺕ ﺍﳌﻨﻄﻘﻴﺔ ﺍﳌﺨﺘﻠﻔﺔ ﻭ ﺍﻟﺒﻮﺍﺑﺎﺕ ﺍﻟﱵ ﺗﻘﻮﻡ ﺑﺈﺟﺮﺍﺀ ﺗﻠﻚ ﺍﻟﻌﻤﻠﻴﺎﺕ‪ ،‬ﻭ ﺗﻌﻠﻤﻨﺎ ﻛﻴﻔﻴﺔ ﻛﺘﺎﺑﺔ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻭ‬ ‫ﺍﻟﺘﻌﺎﻣﻞ ﻣﻌﻬﺎ ﻭ ﺍﻟﻘﻴﺎﻡ ﺑﺘﺒﺴﻴﻄﻬﺎ ﺑﺎﺳﺘﺨﺪﺍﻡ ﺍﻟﻨﻈﺮﻳﺎﺕ‪ ،‬ﻛﻤﺎ ﺗﻌﻠﻤﻨﺎ ﻛﻴﻔﻴﺔ ﺇﻧﺸﺎﺀ ﺟﺪﺍﻭﻝ ﺍﻟﺼﻮﺍﺏ ﻭ ﺑﻨﺎﺀ ﺍﻟﺪﻭﺍﺋﺮ ﺍﳌﻨﻄﻘﻴﺔ‪.‬‬ ‫ﶈﺔ ﻣﺴﺒﻘﺔ ﻋﻦ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺘﺎﻟﻴﺔ‬ ‫ﰲ ﺍﻟﻮﺣﺪﺓ ﺍﻟﺘﺎﻟﻴﺔ ﺳﻨﺘﻨﺎﻭﻝ ﺍﳋﻄﻮﺍﺕ ﺍﳌﺘﺒﻌﺔ ﰲ ﺗﺼﻤﻴﻢ ﺍﻟﺪﻭﺍﺋﺮ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﺍﺑﺘﺪﺍﺀﺍﹰ ﻣﻦ ﲢﺪﻳﺪ ﻣﻮﺍﺻﻔﺎﺕ ﺍﻟﺪﺍﺋﺮﺓ‪ ،‬ﰒ ﻛﺘﺎﺑﺔ‬ ‫ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ ﻟﻠﺪﺍﺋﺮﺓ ﰲ ﺍﻟﺼﻮﺭﺓ ﺍﳌﻨﺎﺳﺒﺔ‪ ،‬ﻓﺘﺒﺴﻴﻂ ﺗﻠﻚ ﺍﻟﺘﻌﺒﲑﺍﺕ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﻭﺧﺘﺎﻣﺎﹰ ﺑﻨﺎﺀ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‪ ،‬ﺇﻣﺎ ﺑﺎﺳﺘﺨﺪﺍﻡ‬ ‫ﺍﻟﺒﻮﺍﺑﺎﺕ ﺍﻷﺳﺎﺳﻴﺔ ﺍﻟﺜﻼﺙ ‪ NOT‬ﻭ ‪ AND‬ﻭ ‪ ،OR‬ﺃﻭ ﺑﺎﺳﺘﺨﺪﺍﻡ ﻧﻮﻉ ﻭﺍﺣﺪ ﻣﻦ ﺍﻟﺒﻮﺍﺑﺎﺕ )‪ NAND‬ﺃﻭ ‪.(NOR‬‬ ‫ﺇﺟﺎﺑﺎﺕ ﺍﻟﺘﺪﺭﻳﺒﺎﺕ‬ ‫ﺗﺪﺭﻳﺐ ‪:1‬‬ ‫‪A‬‬ ‫‪B‬‬ ‫=‪C x‬‬ ‫‪A⋅ B ⋅ C‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪0‬‬ ‫‪0‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪1‬‬ ‫‪23‬‬ ‫‪PDF created with pdfFactory Pro trial version www.pdffactory.com‬‬  :2 ‫ﺗﺪﺭﻳﺐ‬ A B C D x = A+ B + C + D 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 1 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 :3 ‫ﺗﺪﺭﻳﺐ‬ A B A B AB AB AB + AB A⊕ B 0 0 1 1 0 1 1 1 0 1 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 1 0 0 1 0 1 1 :4 ‫ﺗﺪﺭﻳﺐ‬ ‫ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‬ -1 A B x C 24 PDF created with pdfFactory Pro trial version www.pdffactory.com  ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‬ A B C B B + C A( B + C ) x = A( B + C ) 0 0 0 1 1 0 1 0 0 1 1 1 0 1 0 1 0 0 0 0 1 0 1 1 0 1 0 1 1 0 0 1 1 1 0 1 0 1 1 1 1 0 1 1 0 0 0 0 1 1 1 1 0 1 1 0 ‫ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‬ A B x C ‫ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‬ -2 A B y C ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‬ A B C A C AB A+ C y = AB( A + C ) 0 0 0 1 1 0 1 0 0 0 1 1 0 0 0 0 0 1 0 1 1 1 1 1 0 1 1 1 0 1 0 0 1 0 0 0 1 0 1 0 1 0 1 0 0 0 1 0 1 1 0 0 1 0 1 0 1 1 1 0 0 0 1 0 25 PDF created with pdfFactory Pro trial version www.pdffactory.com  ‫ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﻄﻘﻴﺔ‬ A y B C ‫ﺍﳌﺨﻄﻂ ﺍﳌﻨﻄﻘﻲ‬ -3 A B z C D ‫ﺟﺪﻭﻝ ﺍﻟﺼﻮﺍﺏ‬ A B C D B D AB C D AB + C D y = AB + C D 0 0 0 0 1 1 0 0 0 1 0 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0 1 1 0 0 0 1 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0 1 0 1 0 1 0 0 0 0 0 1 0 1 1 0 0 1 0 1 1 0 0 1 1 1 0 0 0 0 0 1 1 0 0 0 1 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 1 1 1 1 1 0 1 0 1 1 1 0 1 0 1 0 1 1 0 0 0 1 0 0 0 1 1 1 0 1 0 0 0 0 0 1 1 1 1 0 0 1 0 1 1 0 1 1 1 1 0 0 0 0 0 1 26 PDF created with pdfFactory Pro trial version www.pdffactory.com  :5 ‫ﺗﺪﺭﻳﺐ‬ A= x+ y -1 B=0 -2 C = xy + x yz -3 ‫ﺍﳌﺼﺎﺩﺭ ﻭ ﺍﳌﺮﺍﺟﻊ‬ Fredrick J. Hill & Gerald R. Peterson, “Introduction to Switching Theory & Logical Design”, Third Edition, John Wiley & Sons, 1981 27 PDF created with pdfFactory Pro trial version www.pdffactory.com