【正文】 1 0 0 0 1 0 1 0 1 0 0 1 0 1 1 0 1 1 1 0 2121 BBABBAALA ??????? ? ?21_________ _____ 21_____ _ BBABBALALB ?????? 用邏輯代數(shù)證明下列不等式 (a) BABAA ??? 由交換律 ))(( CABABCA ???? ，得 BABAAABAA ?????? ))(( (b) ACABCABCBAA B C ???? ACABBCA CBCACBCBBCACABCBAA B C ???? ??????? )( )()( (c) ? ? ECDAEDCCDACBAA ??????? ? ?ECDAECDCDAEDCCDAAEDCCDACBAA??????????????_ _ _ _ _)( 用代數(shù)法化簡下列等式 (a) )( ABCAB ? ABABA B CABCAB ???? )( (b) ))(( BABA? BABABA ?? ))(( (c) )(_______ CBBCA ? CABCCBCACBABCBCBACBBCA???????????? ))(()(_ _ _ _ _ _ _ (d) BCCBBCAAB CA ???? _ __ __ CABCCBBCAA B CA ?????? _____ (e) _ __ __ _ ___ __ __ _ __ ___ __ __ _ __ __ BABABAAB ??? 0_ __ _ __ _ ___ __ _ __ _ __ __ _ __ _ __ __ __ _ __ _ __ _ ?????? AABABABAAB (f) ____ __ __ __ ____ __ __ __ ____ __ __ __ ____ __ __ __ __ __ __ __ ___ __ __ ___ __ __ __ ____ __ __ __ __ )()()()( BABABABA ????? BABABABBABABABABBABABABABABABABA??????????????????)())(()()()()()()()()(_ _ __ _ _ __ _ __ _ _ _ _ __ _ _ __ _ _ __ _ __ _ _ ____ _ __ _ _ _ _ ___ _ __ _ _ _ _ ___ _ __ _ _ _ _ ___ _ __ _ _ _ _ __ _ _ __ _ _ __ _ __ _ _ __ _ __ _ _ _ _ ___ _ __ _ _ _ _ __ (g) ))(( CBACBA ???? BACBACBA ?????? ))(( (h) CBAA B CCBACBA ???? CACBCBACBCBAACBAA B CCBACBA ????? ??????? (i) _____________________________)( BAAB ?? BABABABAABBAAB ?????????? ))(()()( ____ _____ _________ ______ ____ ______ (j) BACAABCB ??? CAACBCAA B CBBACAA B CB ????????? (k) CBA B C DDBCA B DDA B C ???? DBCBABDACABDCADACBDBCBA B DA B CDCBA B DA B CCDCBA B DA B CCBA B C DDBCA B DDA B C???????????????????????????)()()()((l) __ ____ ____ __ ____ ____ __ ____ ____ __ __ __ ______ ____ __ __ CABCBBCAAC ??? BCBCBCACBABCCBABCAA B CCBACBBCAACCABCBBCAAC????????????????????)())(()()()(_ _ ___ _ __ _ _ __ _ __ _ __ _ _ __ _ __ _ __ _ _ __ _ _ _ _ __ _ __ _ __ _ _ __ _ _ (m) ______ __ __ __ ____ __ __ __ ____ __ __ __ __ __ __ ____ __ __ __ __ )( BABAAB CBA ??? 0)()()(__ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ ___ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ __ _ _ __ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ __ _ _ __ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _ _ _ _ _ _ __ _ _ _ _ _ _ _ _ _???????????????ACBAACBABAABBCBABABAA B CBA 將下列各式轉(zhuǎn)換成與 – 或形式 (a) __________________ DCBA ??? (1)當 0________ ??BA ， 1__________??DC 時，真值為 1。 解： (1) 微處理器屬于超大規(guī)模； (2) IC 計算器屬于大規(guī)模； (3) IC 加法器屬于中規(guī)模； (4) 邏輯門屬于小規(guī)模； (5) 4 兆位存儲器 IC 屬于甚大規(guī)模。 將下列每一二進制數(shù)轉(zhuǎn)換為十六進制碼： (1) 101001B (2) 解： (1) 101001B=29H (2) = 將下列十進制轉(zhuǎn)換為十六進制數(shù)： (1) 500D (2) 59D (3) (4) 解： (1) 500D=1F4H (2) 59D=3BH (3) = (4) = 將下列十六進制數(shù)轉(zhuǎn)換為二進制數(shù)： (1) (2) 解： (1) =10 0011 0101B (2) =1010 0000 0100 0001B 將下列十六進制數(shù)轉(zhuǎn)換為十進制數(shù)： (1) (2) 解： (1) = (2) = 解 ： (1) LSTTL 驅(qū)動同類門 mAIOL 8(max) ? mAIIL (max) ? ?? mAmAN OL mAIOH (max) ? mAIIH (m ax) ? ?? mAmAN OH N=20 (2) LSTTL 驅(qū)動基本 TTL門 mAIOL 8(max) ? mAIIL (max) ? ?? mAmANOL mAIOH (max) ? mAIIH (m ax) ? ?? mAmAN OH N=5 解： EDBCABEDBCABL ???????? __ _ __ __ ___ _ __ __ _ ____ _ ___ _ _ 解： B=0 時，傳輸門開通， L=A； B=1 時，傳輸門關(guān)閉， A 相當于經(jīng)過 3 個反相器到達輸出 L， L= A A B L 0 0 0 0 1 1 1 0 1 1 1 0 所以， BABABAL ???? 解 ： C ， __________ BCCB ? D ， __________ DEDE ? __________ DEBC? ， __ ____ ___ ____ ______ _____ ____ ____