Elec 236 Logic Circuits

Images from Chapter 3 Digital Systems 10th Ed. by Tocci

Prof. Tim Johnson

Basic Boolean Theorems

Communiative Theorems

• x + y = y + x• It doesn’t matter to the output where input x

and input y are connected. • Similarly…• X*Y = Y*X• The inputs to an AND gate don’t care which

one is hook up where, switching the connections will not affect the output.

Associative Theorems

• X + Y + Z = X + (Y + Z) = (X + Y) + Z• XYZ = X(YZ) = (XY)Z • In these cases, a three input OR/AND gate is

the same as 2 two-input OR/AND gates where X and Y or Y and Z are connect to one of the input of the next OR/AND gate. The parenthesis represent the grouping of two of the inputs feeding into the next level.

Device schematic for associative expressions

Distributive Theorems

• X(Y + Z) = XY +XZ• (W + X)(Y + Z) = WY + WZ + XY + XZ• In these cases, ordinary arithmetic rules apply

and can alter the gate structure from 1) a two-input OR gate feeding an AND gate INTO two AND gates feeding a two-input OR gate, 2) two 2-input OR gates feeding a 2-input AND gate INTO four 2-input AND gates feeding a 3-input OR gate.

Device schematic for distributive expressions

Special rules for a 2-input OR gate where one input is an AND gate• X + XY => X• An input OR’ed with itself AND another input

makes the other input unnecessary. • X + XY => X + Y• X + XY => X + Y• If one of the OR inputs is the INVERSE of one of

the AND inputs…you keep the single input by itself and drop the inverse input keeping the OR gate

Device schematic: 2-input OR gate with AND input

DeMorgan Theorems

• (X + Y) = X Y• (X Y) = X + Y • To implement these rules, looking at the left

expression: break the bar and change the sign.• This changes the output inversion to an

inversion of the inputs and changes the type of gate used!

Device schematics forDeMorgan Theorems

Double Negative

• A double negative is an input, output, or group that has two bars across the input, output or group. Examples

• X, X+Y, ABC, L+MN• Become• X, X+Y, ABC, L+MN• Ignore double bars of the same length over

the same inputs. You can delete both bars

Device schematic of double negative

Points to remember

• The letters used in these rules can represent groups:

• X 1 = X can also be written as AB 1 = AB∙ ∙• X + Y = Y + X ≈> DY + CE = CE + DY• F + ABC = ABC + F• C + LC = C + CL = CL + C• X(Y + Z) = XY +XZ ≈> DMA + CMA = (D + C)MA

More Points to Realize

• The Boolean rules apply across the equal sign meaning the change can go both ways:

• X(Y + Z) = XY + XZ means XY + XZ = X(Y + Z)• Think of the equal sign as meaning

CDE + ABC => CDE + CAB => C(DE + AB) or skipping the second step you can just take the common term out of CDE + ABC as you would in Algebra.

Moving on to more complex expressions

• BX + B is one of those 2-input OR gates using an AND gate as one of the inputs. Here’s why its correct: B(X+1) pulls the common term out. One (1) is a Boolean symbol that means always TRUE (or high). We have an elementary Boolean rule that deals with X+1 right? X+1 = 1, so we can substitute in a 1 for the X+1 giving us: B·1. Don’t we have a rule for X·1 ? X AND 1 is X. Thus BX + B => B

Writing a reduction solution to BX + B

BX + B becomes

B(X + 1) by distribution theorem

B 1 ∙by Basic Theorem #6

Bby Basic Theorem #2

Inverted terms

• The common term can be a NOT input, X• ABC + XYC => C(AB + XY) • Carefully observe that the above expression

does not meet all the criteria to apply the other rules for 2-input OR gates…dropping the other inputs to the AND gate. Those rules don’t apply here because the NOT input is not by itself (running solo).

More expressions with NOT terms

• Take for example, X + XY = X + YBU + ACBU

rearranged:BU + BUAC

This reduces to:BU + AC

This does not work with BU +ABCU because BU ≠ B U ∙ (DeMorgan rule applies here)