By Richard Sandige, Michael Sandige

This article is meant for an introductory electronic layout direction for college students on the freshman point; it is also meant for an introductory machine layout path with meeting language programming for college kids on the sophomore point. this article makes use of a spiral educating method by means of introducing a layout challenge after which, within the similar bankruptcy or a later bankruptcy, both (1) reemphasizing an identical innovations while a unique layout is gifted, or (2) operating an analogous challenge utilizing a unique method. this is often performed to extend the possibility of retention.

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The NAND gate is referred to as a functionally complete gate, because a set of just NAND gates can be used to design any logic circuit. 15 shows NAND gate equivalent circuits for the NOT, AND, and OR gates. 15 NAND gate NOT, AND, and NAND gate OR gates A equivalent circuits for the NOT, AND, and OR gates equivalent circuits A A B A∙B A B A+B ≡ A ≡ A B A A∙B A∙B = A∙B A A B B -A ∙B = A + B ≡ The NOR gate is also a functionally complete gate, because a set of just NOR gates can be used to design any logic circuit.

Observe that the label xor_1 must be used in the architecture declaration because it is the label that is used in entity declaration. • In the architecture declaration, the Boolean expression X xor Y is assigned to the signal Fxor via the signal assignment symbol (,5)—that is, Fxor ,5 X xor Y. The Boolean function is Fxor ,5 X xor Y placed in the architecture declaration between begin and end Boolean_function. 4 a comment may be placed in VHDL code by using two hyphens in series, that is, “--”.

Y. Y because the signals X Y 5 11 occur at the same time or have an intersection. Y each provide a 1 at the same time or at a different time, they have a union. Because FXOR 5 0 for these expressions, the complement FXOR must be used when combining the Boolean expressions to form the Boolean function, or FX OR 5 X # Y 1 X # Y . Complementing both sides of the Boolean function FX OR 5 X # Y 1 X # Y results in the Boolean function FX OR 5 X # Y 1 X # Y . Apply DeMorgan’s theorem to the Boolean expression, and we obtain FX OR 5 1 X 1 Y 2 # 1 X 1 Y 2 .

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