數(shù)字系統(tǒng)設(shè)計(jì)與pld應(yīng)用第二章-資料下載頁
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【正文】 1D0E N D = 1S T A R T = 1 ?C R = 0 , C A = 1C B 0 = C B 1 = 1C C = 1Bi= 1 ?C M 0 = 1C M 1 = 1C M 0 = 1C B 0 = 1i = 4 ?NYYNNYS3S2S1S000 01 10 11 (2) 推導(dǎo)激勵(lì)函數(shù) (3) 推導(dǎo)輸出方程 E N D = 1S T A R T = 1 ?C R = 0 , C A = 1C B 0 = C B 1 = 1C C = 1Bi= 1 ?C M 0 = 1C M 1 = 1C M 0 = 1C B 0 = 1i = 4 ?NYYNNY1 11 00 10 0END = Q1 Q0 CR = Q1 Q0 CA = CB1 = Q1 Q0 CB0 = Q1 Q0 + Q1 Q0 = Q0 CC = Q1 Q0 CM1 = Q1 Q0 Bi CM0 = Q1 Q0 Bi + Q1 Q0 (4) 畫邏輯電路圖 101 E N0123S T A R T0111M U Xi = 4Q0Q1 30G 7 4 1 5 3C PC 11 DC 11 D7 4 7 4110123BiE N DC RC A ( C B 1 )C B 0C M 1C M 0C CE ND X01Q0Q1? ?G0300 7 4 1 3 911例 212： 根據(jù)乘法器的 ASM圖 , 以多 D觸發(fā)器為核心 ,采用“一對(duì)一”的狀態(tài)分配方法 , 設(shè)計(jì)四位乘法器的控制單元 . (1) 狀態(tài)分配 共 4個(gè)狀態(tài)，故需 4個(gè)觸發(fā)器 令 : S0—— Q0(0000)， S1—— Q1(1100)， S2—— Q2(1010)， S3—— Q3(1001) E N D = 1S T A R T = 1 ?C R = 0 , C A = 1C B 0 = C B 1 = 1C C = 1Bi= 1 ?C M 0 = 1C M 1 = 1C M 0 = 1C B 0 = 1i = 4 ?NYYNNYS3S2S1S0(2) 列出 ASM圖的次態(tài)表 次態(tài) 現(xiàn)態(tài) 輸入 S0 S0 START S3 i=4 S1 S0 START S2 S1 S3 i=4 S3 S2 (3) 推導(dǎo)激勵(lì)方程 D0 = Q0 START + Q3(i=4) D1 = Q0 START D2 = Q1 + Q3 i=4 D3 = Q2 次態(tài) 現(xiàn)態(tài) 輸入 S0 S0 START S3 i=4 S1 S0 START S2 S1 S3 i=4 S3 S2 (4) 推導(dǎo)輸出方程 E N D = 1S T A R T = 1 ?C R = 0 , C A = 1C B 0 = C B 1 = 1C C = 1Bi= 1 ?C M 0 = 1C M 1 = 1C M 0 = 1C B 0 = 1i = 4 ?NYYNNYS3S2S1S0END = Q0 CR = Q1 CA = CB1 = Q1 CB0 = Q1 + Q3 CC = Q2 CM1 = Q2 Bi CM1 = Q2 Bi + Q3 (5) 畫出邏輯電路圖 謝謝觀看 /歡迎下載 BY FAITH I MEAN A VISION OF GOOD ONE CHERISHES AND THE ENTHUSIASM THAT PUSHES ONE TO SEEK ITS FULFILLMENT REGARDLESS OF OBSTACLES. BY FAITH I BY FAITH
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