代数法化简逻辑函数 代数法化简逻辑函数
\u903b\u8f91\u51fd\u6570\u7684\u4ee3\u6570\u5316\u7b80\u6cd5AC + A'BC = C(A+A'B) = C(A+B), \u6ce8\uff1aa+a'b = a+b
(AC+A'BC)' + B'C + ABC'
=(C(A+B))' + B'C + ABC'
=C'+(A+B)' + B'C + ABC' \u6ce8\uff1a(ab)' = a' + b'
=C' + ABC' + B'C + (A+B)'
=C'(1+AB) + B'C + A'B' \u6ce8\uff1a(a+b)' = a'b'
=C' + B'C + A'B'
=C' + B' + A'B' \u6ce8\uff1aa + a'b = a + b
=C' + B'(1+A')
=C' + B'
\u90a3\u4e48\uff0c\u539f\u5f0f
=(C'+B')'
=(C')'*(B')' \u6ce8\uff1a(a+b)' = a'b'
=C*B
=BC
Y = AB' + BD + DCE + DA'
= AB' + (A+A')*BD + DCE + DA' \u6ce8\uff1aA+A'=1
= AB' + ABD + A'BD + DCE + DA'
= AB' + ABD + A'BD + DA' + DCE
= A*(B'+BD) + A'D*(B+1) + DCE \u6ce8\uff1aB+1=1
= A*(B'+D) + A'D + DCE \u6ce8\uff1aB'+BD=B'+D
= AB' + AD + A'D + DCE
= AB' + (A+A')*D + DCE \u6ce8\uff1aA+A'=1
= AB' + D + DCE
= AB' + D(1+CE) \u6ce8\uff1a1+CE=1
= AB' + D
逻辑函数,是一类返回值为逻辑值true或逻辑值false的函数。true:代表判断后的结果是真的,正确的,也可以用1表示;false:代表判断后的结果是假的,错误的,也可以用0表示。按一定逻辑规律进行运算的代数。与普通代数不同,布尔代数中的变量是二元值的逻辑变量。
绛旓細L=(A+B)'(A+B'C)+(A+C)B =(A'B')(A+B'C)+AB+BC =AA'B'+A'B'B'C+AB+BC =0+A'B'C+AB+BC =C(B+A'B')+AB =C(B+A')+AB =AB+A'C+BC銆
绛旓細绛旀锛歒=AC'+ABC+ACD'+CD(A+1)=AC'+ABC+(ACD'+ACD)+CD =AC'+ABC+AC(D'+D)+CD =AC'+ABC+AC+CD =AC'+AC(B+1)+CD =A(C'+C)+CD =A+CD
绛旓細AC + A'BC = C(A+A'B) = C(A+B), 娉細a+a'b = a+b (AC+A'BC)' + B'C + ABC'=(C(A+B))' + B'C + ABC'=C'+(A+B)' + B'C + ABC' 娉細(ab)' = a' + b'=C' + ABC' + B'C + (A+B)'=C'(1+AB) + B'C + A'B' 娉細(a+b)' =...
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绛旓細鍥炵瓟锛氱敤A'琛ㄧず闈濧, F=AC'+ABC+AC =AC'+AC =A(C'+C) =A.
绛旓細1锛欶=AB'+B+BCD =AB'+B(1+CD)=AB'+B=A+B 2锛欶=A+A'B+AB+A'B'=A+(A'+A)B+A'B'=A+B+A'B'=A+B'+B=1 3锛欶=AB+AD+B'D'+AC'D' =A(B+D)+B'D'+AC'D'=A+B'D'+AC'D'=A(1+C'D')+B'D'=A+B'D'
绛旓細閫昏緫鍑芥暟鐨勫寲绠灏辨槸浣夸竴涓渶鍒濈殑閫昏緫鍑芥暟缁忚繃鍖栫畝鍚庡緱鍒板紡涓殑鈥滀笌鈥濋」锛屸滄垨鈥濋」椤规暟鏈灏戯紝鑰屾瘡椤逛腑鐨勫彉閲忔暟涔熸渶灏戙備粠鑰屼娇缁勬垚鐨勯昏緫鐢佃矾鏈绠(閫昏緫闂ㄦ暟鍜屾瘡闂ㄧ殑杈撳叆绔暟鏈灏)銆備笁銆侀昏緫鍑芥暟鐨浠f暟娉曞寲绠 浠f暟娉曟槸鍒╃敤閫昏緫浠f暟宸ュ叿鏉ヨ揪鍒颁娇寮忓瓙绠鍖栫殑鐩殑銆傚寲绠渚濇嵁锛氶昏緫浠f暟瀹氬緥銆佸父鐢ㄥ叕寮忋佸拰...
绛旓細涓烘柟渚夸功鍐欙紝鐢ˋ'琛ㄧず鈥滈潪A鈥滷=AB'C'+A'B'+A'D+C+BD =B'(AC'+A')+A'D+C+BD ...鐢卞垎閰嶅緥锛屽緱AB'C'+A'B'=B'(AC'+A')=B'(A'+C')+A'D+C+BD ...鐢卞惛鏀跺緥锛屽緱A'+AC'=A'+C'=A'B'+B'C'+A'D+C+BD =A'B'+A'D+BD+B'+C ...鐢卞惛鏀跺緥锛屽緱...
绛旓細F=AB+AC+BC+CD+D=AB+AC+BC+D(C+1)=AB+AC+BC+D F2=AB+A'C+B'C+C'D+D'=AB+A'C+B'C+C'+D'=AB+A'+B'+C'+D'=B+A'+B'+C'+D'=1 F4=A'B'C'+A+B+C=(A+B+C)'+A+B+C=1
绛旓細F=AB+ABD+A'C+A'CE =AB(1+D)+A'C(1+E)=AB+A'C