數字邏輯設計及應用教學課件：4-4 冒險 總結

1、 class exersise Using Karnaugh maps, find a minimal sum-of-products expression for each of the following logic function F. Indicate the distinguished 1-cells in each map. F = A,B,C,D ( 0, 1, 2, 3, 4, 5, 7, 14, 15 )Using Karnaugh maps, find a minimal sum-of-products expression for each of the followi

2、ng logic function F. Indicate the distinguished 1-cells in each map.F = A,B,C,D ( 0, 1, 2, 3, 4, 5, 7, 14, 15 )1、填圖2、圈組 找奇異“1”單元 圈質主蘊含項 圈其它的13、讀圖CDAB00 01 11 1000011110111111111F(A,B,C,D) = AB + AC + AD + ABCUsing Karnaugh maps, find a minimal sum-of-products expression for each of the following log

3、ic function F. Indicate the distinguished 1-cells in each map.F = A,B,C,D ( 0, 1, 2, 3, 4, 5, 7, 14, 15 )1、填圖2、圈組 找奇異“1”單元 圈質主蘊含項 圈其它的13、讀圖CDAB00 01 11 1000011110111111111F(A,B,C,D) = AB + AC + AD + ABCNot the distinguished 1-cells ExampleCDAB00 01 11 1000011110111111CDAB00 01 11 100001111011Given t

4、wo prime implicants P and Q in a reduced map, P is said to eclipse(重疊) Q. (P220)(written P Q) if P covers at least all the 1-cells covered by Q. (P220) BCD is a secondary essential prime implicant .BCDCDAB00 01 11 1000011110111111沒有奇異“1”單元沒有質主蘊含項CDAB00 01 11 1000011110111111Example1.Prime Implicant

5、Theorem : A minimal sum is a sum of prime implicants.2. A minimal sum is not the sum of all the prime-implicants.3. the sum of all the prime-implicants of a logic function is called the complete sum.（完全和） “Dont-Care” Input Combinations ( “無關”輸入組合) (P222)有時組合電路的輸出和某些輸入組合無關F = A,B,C,D(1,2,3,5,7) + d(1

6、0,11,12,13,14,15)CDAB00 01 11 1000011110dddddd11111F = AD + BCADBCd 集（d-set）Example 11. Using Karnaugh maps, find a minimal sum-of-products expression for each of the following logic function F. F=ABCD+ABC+ABCD+BCDCDAB00 01 11 1000011110111111Example 2F2= A,B,C,D(1,2,3,6,7,9,11,13,15) 111111111CDAB0

7、0 01 11 10000111100000000Example 31. Using Karnaugh maps, find a minimal products expression for each of the following logic function F. F=(AB)(B+D)(C+D)(A+C+D)(B+C+D)Example 4what is the relation between the fonction F and fonction G?F=AB+AD+BC+CDG=(ABD+AC+BCD)Example 5 if F1=ACD+ABD+BCD+ACD F2=(AC

8、D+BC+ACD),what is the F1F2,F1+F2,F1 F2?Example 6Z1=AB+BD+BCD+ABC,please express the Z1 with the minimal NAND-NAND expresstion.please express the Z1 with the minimal NOR-NOR expresstion.please express the Z1 with the minimal AOI expresstion.Z1=AB+BD+BCD+ABCCDAB00 01 11 10000111101111111111Z1=AB+BD+AC

9、D最小和：NAND-NAND expresstionZ1=（AB+BD+ACD）Z1=AB+BD+BCD+ABCCDAB00 01 11 10000111100Z1=AB+BD+ACD最小和：NAND-NAND expresstionZ1=(（AB+BD+ACD）)最小積？00000Z1=(A+B+C).(A+C+D).(A+B+D)(A+B+D)NOR-NOR expresstionZ1=(（(A+B+C).(A+C+D).(A+B+D)(A+B+D)Z1=AB+BD+BCD+ABCCDAB00 01 11 10000111101Z1=AB+BD+ACD最小和：NAND-NAND expre

10、sstionZ1=(（AB+BD+ACD）)反函數Z1？11111Z1=ABC+ACD+ABD+ABDAOI expresstionZ1=(ABC+ACD+ABD+ABD)4.4 Timing Hazards(定時冒險) steady-state behavior & stransient behavior 穩(wěn)態(tài)特性 和 瞬態(tài)特性 (P224)Circuit delayhazard （冒險）AAAFFglitch尖峰AA hazard is said to exist when a circuit has the possibility of producing such a glitch.

11、4.4.1 Static Hazards 靜態(tài)冒險 (P225)static-1 hazard 靜態(tài)-1型冒險static-0 hazard 靜態(tài)-0型冒險主要存在于“與或”電路中AFAFSteady state is 1. F = (AA) = A+ASteady state is 0. F = (A+A) = AA主要存在于“或與”電路中Example:when WYZ=001, F=X ;There are three passes from X to F.Dynamic hazard4.4.2 Finding Static Hazards Using Maps 利用卡諾圖發(fā)現靜態(tài)冒險

12、(P226)ZXY00 01 11 10011 11 1若卡諾圖中，圈與圈之間有相切現象，則可能出現靜態(tài)冒險。消除冒險的方法： 引入額外項乘積項覆蓋冒險的輸入對。F = XZ + YZ + XYstatic-1 hazard Example Indicate whether or not existence or of the static hazards occurs ?ZXY00 01 11 10011 11 1ZXY00 01 11 10011 11 1Eliminate the hazart 補充：競爭冒險（清華教材）1&AAY111AY2AAY1Y2競爭：門電路兩個輸入信號同時向相反

13、的邏輯電平跳變。若后繼負載電路是一個對脈沖敏感的電路，這種尖峰脈沖可能使負載電路發(fā)生誤動作。競爭冒險：由于競爭而在電路輸出端可能產生尖峰脈沖檢查競爭冒險現象的方法 只要輸出端的邏輯函數在一定條件下能簡化成Y = A + AY = AA或則可判定存在競爭冒險如：Y = AB +AC當 B = C = 1 時，Y = A + A ，存在競爭冒險又如：Y = ( A + B ) ( B + C )當 A = C = 0 時，Y = BB ，存在競爭冒險 采用計算機輔助分析手段 用實驗來檢查電路輸出端是否產生尖峰脈沖消除競爭冒險現象的方法 接入濾波電容尖峰脈沖一般都很窄，輸出端并接一個很小的濾波電容，足以將其幅度削弱到門電路的閾值電壓以下。增加了輸出電壓波形的上升時間和下降時間，使波形變壞不是一個好辦法1&AAY1C f消除競爭冒險現象的方法 引入選通脈沖 修改邏輯設計Y = AB + AC = AB + AC + BC增加冗余項消除冒險（可以利用卡諾圖）1&AAY1PAAY1P第四章 小結4.1 開關代數公理、定理摩根定理對偶、反演邏輯函數的標準表示法真值表積之和、和之積標準項n 變量最小項（最大項）4.2 組合電路分析4.3 組合電路綜合4.5 定時冒險(電路的描述與設計) Exampl