Altifigence Academy

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NAND, NOR and inversion symbols

Track the scope of inversion and connect equivalent symbols with De Morgan’s laws.

Chưa có bản dịch. Nội dung bài học gốc được hiển thị. (English)

The output circle complements the whole result

NAND complements AND; NOR complements OR:

YNAND=ab,YNOR=a+bY_{NAND}=\overline{ab},\qquad Y_{NOR}=\overline{a+b}

For input order 00, 01, 10, 11, NAND gives 1, 1, 1, 0 and NOR gives 1, 0, 0, 0. An inverted output is still a valid logic signal.

An inversion circle changes the truth table
NAND · Truth table
aby
001
011
101
110

Move inversions with De Morgan’s laws

ab=a+b,a+b=ab\overline{ab}=\overline a+\overline b,\qquad \overline{a+b}=\overline a\,\overline b

Moving an output inversioninversion The NOT operation exchanges 0 and 1. An overline applies NOT to the entire expression covered by the line. to both inputs also changes AND to OR, or OR to AND. Merely moving the circles preserves neither the symbol nor the function. At inputs 01, NAND gives 1 but an AND of the complemented inputs gives 0.

Move the output inversion
  1. NAND: Y = NOT (a AND b)
  2. Invert each input: Produce NOT a and NOT b
  3. Combine with OR: Y = (NOT a) OR (NOT b)

Active-low describes assertion

An active-lowactive-low A convention in which a signal is asserted at 0. The suffix _n often marks this convention; the specification defines the behavior. request is asserted at zero. The suffix _n is a naming convention, not an automatic circuit operation. If either of two active-low requests must trigger an active-high alarm, then

alarm=req0_n+req1_n=req0_nreq1_n\mathrm{alarm}=\overline{\mathrm{req0\_n}}+\overline{\mathrm{req1\_n}}=\overline{\mathrm{req0\_n}\,\mathrm{req1\_n}}

Each short overline complements one named signal. The final long overline complements their AND as a whole, so a NAND implements this function. Inputs 11 mean no request and produce alarm=0; inputs 01 assert the first request and produce alarm=1. Keep a signal’s numeric value, asserted meaning and inversion symbols distinct.

Tự thử

Which gate gives go=1 only when both active-low permits permit0_n and permit1_n are asserted: NAND or NOR?

Đọc giải thích

NOR. Both permits are asserted at 00, so go=NOT(permit0_n OR permit1_n). The phrase “both asserted” must be interpreted using the signal polarity.

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