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Elec 236 Logic CircuitsPowerPoint Presentation

Elec 236 Logic Circuits

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Elec 236 Logic Circuits

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Elec 236 Logic Circuits

Images from Chapter 3

Digital Systems 10th Ed. by Tocci

Prof. Tim Johnson

- 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.

- 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.

- 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.

- 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

- (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!

- 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

- 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

- 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.

- 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

BX + B

becomes

B(X + 1)

by distribution theorem

B∙1

by Basic Theorem #6

B

by Basic Theorem #2

- 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).

- Take for example, X + XY = X + Y
BU + ACBU

rearranged:

BU + BUAC

This reduces to:

BU + AC

This does not work with BU +ABCU because

BU ≠ B∙U (DeMorgan rule applies here)