Equations, Truth Tables and Circuits.

• These are equivalent.
• Starting with $f(a,b) = ab + a\overline{b}$ can you
  1. Create the circuit?
  2. Create the truth table?
• The circuit
  • And has higher precedence than or.
  • 
• Truth Table

  • $A$	$B$	$AB$	$\overline{B}$	$A\overline{B}$	$ AB + A\overline{B}$
    0	0	0	1	0	0
    0	1	0	0	0	0
    1	0	0	1	1	1
    1	1	1	0	0	1

• Wow, this looks like just A.
  • On page B-696 they give us several laws of boolean algebra.

  • $ f(A,B)$	$= AB + A\overline{B} $
    	$= A(B + \overline{B}) $
    	$= A(1) $
    	$ = A $

• We need to be careful of DeMorgan's law.
  • $\overline{AB} = \overline{A} + \overline{B}$
  • $\overline{A + B} = \overline{A} \cdot \overline{B}$
  • Pick one, prove it by truth table, prove it by circuit.
• Can you translate this circuit into a Truth Table and a Boolean Expression?
  • 
  • Answer
    • $f(C,D) = \overline{C}D + C\overline{D}$

    • $C$	$D$	$ \overline{C}$	$\overline{C}D$	$\overline{D}$	$C\overline{D}$	$\overline{C}D + C\overline{D}$
      0	0	1	0	1	0	0
      0	1	1	1	0	0	1
      1	0	0	0	1	1	1
      1	1	0	0	0	0	0

  • By the way, this is exclusive or (xor) $C \oplus D$
• Can you translate this truth table into the other two?

  E	F	f(E,F)
  0	0	1
  0	1	0
  1	0	0
  1	1	1

• Answer

  • E	F	f(E,F)	Minterm	Include?
    0	0	1	$\overline{E}\overline{F}$	Yes
    0	1	0	$\overline{E}F$	No
    1	0	0	$E\overline{F}$	No
    1	1	1	$EF$	Yes

  • $f(E,F) = \overline{E}\cdot\overline{F} + EF$
  • 
• If you wish, the circuits are in equations.v
• More than two variables. Do the same thing.

  • G	H	I	f(G,H,I)
    0	0	0	0
    0	0	1	1 ($\overline{G}\overline{H}I$)
    0	1	0	0
    0	1	1	0
    1	0	0	1 ($G\overline{H}\overline{I}$)
    1	0	1	1 ($E\overline{H}I$)
    1	1	0	0
    1	1	1	1 ($GHI$)