Chapter 3

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# Chapter 3 - PowerPoint PPT Presentation

Chapter 3. Logic Gates &amp; Boolean Algebra. Algebra Example. Simplify this expression:. ACD + ABCD. Factor out common variables:. CD(A + AB). Apply theorem 15a:. CD(A + B) or ACD + BCD. DeMorgan’s Theorems.

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Presentation Transcript

### Chapter 3

Logic Gates & Boolean Algebra

Algebra Example

Simplify this expression:

ACD + ABCD

Factor out common variables:

CD(A + AB)

Apply theorem 15a:

CD(A + B)

or

ACD + BCD

DeMorgan’s Theorems

Demorgan’s Theorems allow us to break up a “bar”which is over an entire expression:

(X + Y) = X ● Y

(X ● Y) = X + Y

DeMorgan’s Theorem Example

(A B + C)

Simplify this expression:

Apply Demorgan’s:

(A B) ●C

Apply Demorgan’s:

(A + B) ●C

(notice importance of keeping the parentheses!)

Cancel double bars:

(A + B) ●C

or

AC + BC

DeMorgan’s Theorem Example

(A + C) ( B + D)

Simplify this expression:

Apply Demorgan’s:

(A + C) + (B + D)

Apply Demorgan’s:

(A ●C) + (B ● D)

Cancel double bars:

AC + BD

DeMorgan’s Theorem Example

(A + BC) (D + EF)

Simplify this expression:

Apply Demorgan’s:

(A + BC) + (D + EF)

Apply Demorgan’s:

(A ● BC) + (D ● EF)

Apply Demorgan’s:

(A ● (B+C)) + (D ● (E+F))

Distribute:

AB + AC + DE + DF

Alternate Gate Representations

(X + Y) = X ● Y

X

Y

X + Y

X

Y

X

Y

X ● Y = X + Y

NAND Gates are Universal Gates

X

X ● X = X

X X

X

Y

XY

X ● Y

XY

X

Y

X

Y

X + Y

NAND Gates are Universal Gates

X

Y

X ● Y =

X + Y =

X + Y

X + Y

X

Y

X

Y

X ● Y

X

Y

Using NAND Gates to Simplify

2 CHIPS:

7408 = AND

7432 = OR