‫ﺣﻞّ اﻣﺘﺤﺎن ﻣﻴﺎنﺗﺮم آزﻣﻮنﭘﺬﻳﺮي‬
‫‪.1‬‬
‫ﺑﺎﺳﻤﻪ ﺗﻌﺎﻟﻲ‬
‫ﻧﻴﻢﺳﺎل اول ‪88-89‬‬
‫اﻟﻒ )‪ 2‬ﻧﻤﺮه( در ﺷﻜﻞ ﻣﻘﺎﺑﻞ‪ ،‬آﻳﺎ راﺑﻄﻪاي )ﻣﻌـﺎدل ﺑـﻮدن ﻳـﺎ ﻏﻠﺒـﻪ اﺷـﻜﺎل( ﺑـﻴﻦ دو اﺷـﻜﺎل‬
‫‪ A2/1‬و ‪ B2/1‬وﺟﻮد دارد؟ ﭼﺮا؟ ‪ A2/1‬ﻳﻌﻨﻲ ‪.A2 stuck-at-one‬‬
‫ج‪ -‬اﻳﻦ دو اﺷﻜﺎل‪ ،‬ﻃﺒﻖ ﻣﺜـﺎل ‪ 4.11‬ﻛﺘـﺎب از ﻧﻈـﺮ ‪ functional‬ﻣﻌﺎدﻟﻨـﺪ زﻳـﺮا ﺑـﻪ ازاي ﻫـﺮ دو‪،‬‬
‫ﺧﺮوﺟﻲ ﻋﺒﺎرت اﺳﺖ از‪C = A + B :‬‬
‫ب )‪ 2‬ﻧﻤﺮه( ﺑﺎ ﻓﺮض ‪ ،single stuck-at fault‬ﺑﺮاي ﻣﺪار ﻣﻘﺎﺑﻞ ﺑﺎ اﺳﺘﻔﺎده از رواﺑﻂ ﻣﻌﺎدل ﺑﻮدن اﺷﻜﺎلﻫﺎ )‪ ،(fault equivalence‬ﻟﻴﺴﺖ ﻛﻠﻴـﻪ‬
‫اﺷﻜﺎلﻫﺎﻳﻲ را ﻛﻪ ﺑﺮاي ﺗﻮﻟﻴﺪ ﺑﺮدار آزﻣﻮن ﻻزم دارﻳﻢ ﺑﻪ دﺳﺖ آورﻳﺪ )‪.(equivalence collapsed set‬‬
‫ﺟﻮاب‪ :‬در ﺷﻜﻞ ﻣﻘﺎﺑﻞ‪ ،‬اﺷﻜﺎلﻫﺎﻳﻲ ﻛﻪ ﻫﺎﺷﻮر ﺧـﻮردهاﻧـﺪ ﺣـﺬف ﻣـﻲﺷـﻮﻧﺪ و ﻓﻘـﻂ ‪12‬‬
‫اﺷﻜﺎل ﺑﺎﻗﻲ ﻣﻲﻣﺎﻧﺪ‪ :‬اﻟﺒﺘﻪ‪ ،‬ﻃﺒﻖ اﻟﻒ‪ A2/1 ،‬ﻧﻴﺰ ﻗﺎﺑﻞ ﺣﺪف اﺳـﺖ‪ .‬ﻫـﻢﭼﻨـﻴﻦ ‪ A1/1‬و‬
‫‪ B1/1‬ﻣﻌﺎﻟﻨﺪ ﻛﻪ ﻓﻘﻂ ﻳﻜﻲ را ﻧﮕﻪ ﻣﻲدارﻳﻢ‪.‬‬
‫ج )‪ 2‬ﻧﻤﺮه( ﻣﺠﻤﻮﻋﻪ ﺣﺪاﻗﻞ ﺑﺮدارﻫﺎي ﺗﺴﺖ ﺑﺮاي ﻛﺸﻒ ﺗﻤﺎم اﺷﻜﺎلﻫﺎ را ﺑﻪ دﺳـﺖ‬
‫آورﻳﺪ‪.‬‬
‫ج‪ -‬ﺑﺮاي ﻛﺸﻒ ﺗﻤﺎم اﺷﻜﺎلﻫﺎ‪ ،‬ﺑﺎﻳﺪ ﻫﺮ ﭼﻬﺎر ﺑﺮدار ﺗﺴﺖ ﻣﻤﻜﻦ را ﺑﻪ ﻣﺪار اﻋﻤﺎل ﻛﻨﻴﻢ‪ .‬اﻟﺒﺘﻪ اﮔﺮ ﻧﻘﺎط ﻣﻴﺎﻧﻲ ﻣﻬﻢ ﻧﺒﻮد‪ ،‬ﺑﺮدار ‪ 00‬ﺿﺮوري ﻧﺒﻮد‪ ،‬وﻟﻲ اﻳﻦ ﺑﺮدار‪،‬‬
‫ﺗﻨﻬﺎ ﺑﺮداري اﺳﺖ ﻛﻪ اﺷﻜﺎل ‪ A1/1‬و ‪ B1/1‬را ﻛﺸﻒ ﻣﻲﻛﻨﺪ‪.‬‬
‫‪ .2‬ﺑﺮاي ﺷﻜﻞ ﻣﻘﺎﺑﻞ‪:‬‬
‫اﻟﻒ‪ 3) -‬ﻧﻤﺮه( ﺗﺴﺖ ﺟﺎﻣﻊ )‪ (exhaustive‬و ﺷﺒﻪ ﺟـﺎﻣﻊ )‪ (pseudo-exhaustive‬ﺑـﻪ ﭼﻨـﺪ‬
‫ﺑﺮدار ﺗﺴﺖ ﻧﻴﺎز دارد؟ ﻣﻨﻈﻮر از ﺗﺴﺖ ﺷﺒﻪ ﺟﺎﻣﻊ آن اﺳﺖ ﻛﻪ ﺗﻤﺎم ﺧﺮوﺟﻲﻫﺎي ﻣـﺪار ﺑـﻪ ﻃـﻮر ﺟـﺎﻣﻊ‬
‫ﺗﺴﺖ ﺷﻮﻧﺪ‪.‬‬
‫ب‪ 4) -‬ﻧﻤﺮه( ﻓﺮض ﻛﻨﻴﺪ ﻣﻲﺧﻮاﻫﻴﻢ ﻫﺮ ﭼﻬﺎر ‪ sub-circuit‬ﻣﺪار )داﺧﻞ ﺧﻂ ﭼﻴﻦ( را ﻃﻮري ﺗﺴﺖ ﻛﻨﻴﻢ ﻛﻪ ﺗﻤﺎم وروديﻫﺎي ﻣﻤﻜﻦ آن ‪ sub-circuit‬ﺑﻪ‬
‫آن ا‪‬ﻋﻤﺎل ﺷﻮد )اﻳﻦ روش ‪ cell-fault model‬ﻧﺎم دارد(‪ .‬ﻛﻤﺘﺮﻳﻦ ﺗﻌﺪاد ﺑﺮدار ﺗﺴﺖ ﺑﺮاي اﻳﻦ ﻣﻨﻈﻮر را ﺑﻪ دﺳﺖ آورﻳﺪ و ﺑﺮدارﻫﺎ را ﺑﻨﻮﻳﺴﻴﺪ‪.‬‬
‫ﺣﻞ‪ :‬اﻟﻒ‪ -‬ﺗﺴﺖ ﺟﺎﻣﻊ‪26 = 64 :‬‬
‫ﺗﺴﺖ ﺷﺒﻪ ﺟﺎﻣﻊ‪ :‬ﺑﺮاي ‪ X‬ﻧﻴﺎز ﺑﻪ ‪ 25 = 32‬ﺑﺮدار‪ ،‬و ﺑﺮاي ‪ Y‬ﻧﻴﺎز ﺑﻪ ‪ 24 = 16‬ﺑﺮدار دارﻳﻢ‪ .‬ﺑﻪ ﺳﺎدﮔﻲ ﻣﻲﺗﻮان اﻳﻦ ﺗﺴﺖﻫﺎ را ﻃﻮري اﻋﻤﺎل ﻛﺮد ﻛﻪ ﻫـﻢﭘﻮﺷـﺎﻧﻲ‬
‫داﺷﺘﻪ ﺑﺎﺷﻨﺪ‪ ،‬ﻟﺬا ﺟﻤﻌﺎً ﻫﻤﺎن ‪ 32‬ﺑﺮدار ﻧﻴﺎز اﺳﺖ‪.‬‬
‫‪ Sub-circuit TUW has 3 inputs and requires 23 = 8 tests, while all the other sub-circuits have just 2‬ب‪-‬‬
‫‪inputs and require 4 tests each. We must also propagate the test responses from the internal sub-circuits‬‬
‫‪TUW and V to at least one of the 2 primary outputs. For example, we can propagate all the responses of‬‬
‫‪TUW to primary output Y by setting F= 1. This will also apply 2 of its 4 tests, namely WF= 01 and 11, to‬‬
‫‪sub-circuit (AND gate) Y. We then require two additional tests that make WF= 10 and 10 to complete the‬‬
testing of sub-circuit Y. This leads to 10 tests overall. These 10 patterns can easily be overlapped with
the tests needed by sub-circuits V and X, which must be propagated to primary output X, and so are
largely independent of the tests being propagated to output Y.
We can test the entire circuit with just 9 tests if we propagate one of the responses of TUW to
output X and the other responses to output Y. This is the minimum possible, since the 8 patterns that
exhaustively test TUW produce only one 0 on line W, so at least one additional test pattern is needed to
ensure that sub-circuit Y has both 00 and 01 applied to it.
‫ ﺑـﺎ ذﻛـﺮ دﻟﻴـﻞ ﺑﻴـﺎن ﻛﻨﻴـﺪ آﻳـﺎ ﻫـﻴﺞ اﺷـﻜﺎل‬،CK ‫ ﻧﻤﺮه( ﺑﻪ ﻓﺮض ﺳﺎﻟﻢ ﺑـﻮدن ﺧـﻂ‬3)
.3
‫( در اﻳــﻦ ﻣــﺪار وﺟــﻮد دارد ﻛــﻪ ﻣــﺎﻧﻊ از ﻣﻘــﺪاردﻫﻲ اوﻟﻴــﻪ‬SSF) ‫ﭼﺴــﺒﻴﺪﮔﻲ ﺗﻜــﻲ‬
‫( ﺧﺮوﺟﻲ ﮔﺮدد؟‬initialization)
‫ ﻟـﺬا‬.‫ ﺑﺎﺷـﺪ‬Q ‫ ﻣﺴـﺘﻘﻞ از‬Q+ ‫ ﺑﺎﻳﺪ‬،‫ ﺑﺮاي ﻣﻘﺪاردﻫﻲ اوﻟﻴﻪ‬،‫ ﺑﺎﺷﺪ‬Q = X ‫ اﮔﺮ‬،‫ ﻣﻲﺑﺎﺷﺪ‬Q+ = (Q + A)(A’ + B) ‫ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ اﻳﻦ ﻛﻪ‬:‫ﺟﻮاب‬
‫ ﻣـﺎﻧﻊ از‬A sa0 ‫ ﺑﺪﻳﻬﻲ اﺳﺖ ﻛﻪ‬.‫ ﻣﻲﺗﻮان ﺧﺮوﺟﻲ ﻣﺪار ﺳﺎﻟﻢ را ﻣﻘﺪاردﻫﻲ ﻛﺮد‬B ‫ و ﺑﺎ اﻧﺘﺨﺎب‬Q+ = B :‫ ﺑﺎﺷﺪ ﻛﻪ ﺧﻮاﻫﻴﻢ داﺷﺖ‬A = 1 ‫ﺑﺎﻳﺪ‬
‫ ﻫـﻴﺞ اﺷـﻜﺎل ﺗﻜـﻲ دﻳﮕـﺮي ﻣـﺎﻧﻊ از‬.‫ ﺧﻮاﻫـﺪ ﻣﺎﻧـﺪ‬X ‫ ﻫﻤﻮاره‬،‫ ﺑﺎﺷﺪ‬Q = X ‫ و اﮔﺮ‬Q+ = Q :‫ﻣﻘﺪاردﻫﻲ اوﻟﻴﻪ ﺧﻮاﻫﺪ ﺷﺪ زﻳﺮا ﺧﻮاﻫﻴﻢ داﺷﺖ‬
‫ ﻛـﻪ‬Q+ = Q(A’ + B) :‫ ﺧـﻮاﻫﻴﻢ داﺷـﺖ‬A ‫ ﺑﻮدن ﺷﺎﺧﻪي ﺑﺎﻻﻳﻲ ﺧﺎرج ﺷـﺪه از‬stuck-at-0 ‫ ﻣﺜﻼً در ﺻﻮرت‬،‫ﻣﻘﺪاردﻫﻲ اوﻟﻴﻪ ﻧﺨﻮاﻫﺪ ﺷﺪ‬
.‫ ﺧﺮوﺟﻲ را ﺑﻪ ﺻﻔﺮ ﻣﻘﺪاردﻫﻲ ﻛﺮد‬B = 0 ‫ و‬A = 1 ‫ ﻣﻲﺗﻮان ﺑﺎ‬،Q ‫ﻣﺴﺘﻘﻞ از ﻣﻘﺪار‬
‫ ﺷـﺒﻴﻪﺳـﺎزي اﺳـﺘﻨﺘﺎﺟﻲ را ﺑـﺮاي ﻟﻴﺴـﺖ‬abc = 111 ‫ ﺑـﺮاي ورودي‬،‫ ﻧﻤﺮه( در ﻣﺪار ﻣﻘﺎﺑﻞ‬4) .٤
:‫اﺷﻜﺎل زﻳﺮ اﻧﺠﺎم دﻫﻴﺪ‬
{a/0, a/1, b/0, b/1, c/0, c/1, d/1, e/1, h/1, k/1, p/1, m/0, n/0, q/1}
:‫ﺣﻞ‬