[英]How to simplify Logic Gate - mux simulation
當嘗試制作一個多路復用器時,您如何去做:
Not(in=a, out=nota);
Not(in=b, out=notb);
Not(in=sel, out=notsel);
And(a=a, b=b, out=aAndb);
And(a=a, b=notb, out=aAndNotb);
And(a=nota, b=b, out=bAndNota);
And(a=aAndb, b=sel, out=aAndBAndSel);
And(a=aAndb, b=notsel, out=aAndBAndNotSel);
And(a=aAndNotb, b=notsel, out=aAndNotBAndNotSel);
And(a=bAndNota, b=sel, out=bAndNotaAndSel);
Or(a=aAndBAndSel, b=aAndBAndNotSel, out=o1);
Or(a=o1, b=aAndNotBAndNotSel, out=o2);
Or(a=o2, b=bAndNotaAndSel, out=out);
對此:
Nand(a=sel, b=sel, out=notsel);
Nand(a=a, b=notsel, out=asel);
Nand(a=b, b=sel, out=bnotsel);
Nand(a=asel, b=bnotsel, out=out);
我的回答總是很長,我不確定您將如何尋找更優雅的解決方案。
MUX Truth Table Answers
| a | b | sel | out |
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 0 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
寫下您擁有的表達式的真值表,使用卡諾圖將其最小化,然后使用布爾代數獲得第二個表達式。 (通過加倍不適用並適用摩根法律)
簡化為:
BS + AS'
或使用其他符號:
(B & S) ∨ (A & ~S)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.