REVISION AND PRACTICE

Week 3 Quick Revision

On this page 3 headings

Start here Ten minutes before the exam? Read this page. It has every Week 3 idea in one place, plus practice questions in the exam’s style.

The whole week in 8 lines

SOP → circuit: NOTs → one AND per term → one OR. This is two-level logic.
Canonical ≠ minimal. $\Sigma m \left(3 , 4 , 5 , 6 , 7\right) =$ $A + B C$ .
POS uses the 0-rows. Each gives a maxterm: $0 \to A , 1 \to \overline{A}$ , joined by OR. Then AND them all.
Σ ↔︎ Π → use the missing numbers. F ↔︎ $\overline{F}$ → also the missing numbers. Both → the same numbers.
NAND and NOR are universal. Each one alone can build anything.
Gate counts: NAND: NOT 1, AND 2, OR 3, XOR 4, XNOR 5. NOR: NOT 1, OR 2, AND 3, XNOR 4, XOR 5.
AND– $\operatorname{OR} = \operatorname{NAND}$ –NAND, and OR– $\operatorname{AND} = \operatorname{NOR}$ –NOR (same shape, swap the gates).
Timing diagram: slice wherever an input changes → truth table → simplify. “Any one flip changes the output” = XOR.

Mistake	Correct way
Using the SOP literal rule for maxterms	Maxterm: $0 \to A , 1 \to \overline{A}$
Treating a NOT as free in a NAND-only circuit	Every NOT costs 1 NAND
Counting gates before simplifying	Simplify first. The answer may be just 1
Counting terms instead of literals	“ $a b + \overline{a} \overline{c}$ ” has 4 literals
Forgetting rows go to 15 for 4 variables	Rows = 0 to 2ⁿ − 1

$F \left(A , B , C\right) = \Sigma m \left(0 , 1 , 5\right)$ . Write F as $\Pi M$ .

\Pi M \left(2 , 3 , 4 , 6 , 7\right)

\left(A + \overline{B} + \overline{C}\right) , M_{3}

4 (it’s XNOR).

$F = \left(\overline{X} + \overline{Y}\right) \left(Z + W\right)$ . Minimum 2-input NAND gates?

G1 = NAND(X,Y) gives $\overline{X} + \overline{Y}$ for free; G2 = NAND(Z,G1); G3 = NAND(W,G1); $F = \operatorname{NAND}$ (G2,G3).

Minimum literals in $a b + \overline{a} \overline{b} \overline{c} + \overline{a} b \overline{c}$ ?

4 (it simplifies to

a b + \overline{a} \overline{c}

A timing diagram shows $Y = 1$ whenever an odd number of A, B, C are 1. Minimum 2-input XOR gates?

2 \left(Y = A \oplus B \oplus C\right)

True or false: an AND–OR circuit becomes a NOR–NOR circuit when you swap every gate for NOR.

False. AND–OR becomes NAND–NAND. It’s OR–AND that becomes NOR–NOR.

It usually needs fewer gates (cheaper and smaller), even though it is a bit slower.

\Sigma m \left(2 , 5\right)

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