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Unformatted text preview: x, x · x = x
(P2) Identity: x + 0 = x, x · 1 = x (P3) Domination: x + 1 = 1, x · 0 = 0 (P4) Complement: x + x = 1, x · x = 0
¯
¯ (P5) Involution: x = x.
¯ Proofs by perfect induction January 24, 2013 Lecture 3 AWB More properties (P6) Commutativity: x + y = y + x, x · y = y · x (P7a) Associativity: x + (y + z ) = (x + y ) + z , (P7b) Associativity: x · (y · z ) = (x · y ) · z (P8) Covering x + x · y = x, x · (x + y ) = x (P9a) Distributivity: x · (y + z ) = (x · y ) + (x · z ), (P9b) Distributivity: x + (y · z ) = (x + y ) · (x + z ) (10a) Combining: x·y+x·y =x
¯ (10b) Combining: 22 (x + y ) · (x + y ) = x
¯ January 24, 2013 Lecture 3 AWB Distributivity  by perfect induction 23 Prove property (P9b): x + (y · z ) = (x + y ) · (x + z )
x y z L = x + (y · z ) R = (x + y ) · (x + z ) 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 1 January 24, 2013 Lecture 3 L=R AWB Distributivity  from properties 24 (P9b) also follows from (P1)(P9a)
R = (x + y ) · (x + z ) = (x + y ) · x + (x + y ) · z
(P 9 a ) = x · (x + y ) + z · (x + y )
(P 6)) = x · x + x · y+z · x+ z · y
(P 9 a ) = (P 9 a ) x +x · y + x · z + y · z = x + y · z
(P 1) January 24, 2013 (P 6) (P 6) Lecture 3 (P 6) (P 6)+(P 8) AWB Completeness of NAND and NOR
NOT A A AND B A OR B A NAND A NOT( A NAND B ) (NOT A) NAND (NOT B) A NOR A (NOT A) NOR (NOT B) 25 NOT (A NOR B) We can built circuits from NAND or NOR gates
only!
January 24, 2013 Lecture 3 AWB De Morgan laws
¯¯
A+B =A·B
A
B 26 ¯¯
A·B =A+B
A Y B A Y A
Y B B A
B January 24, 2013 Y A Y B Lecture 3 Y AWB De Morgan laws 27 In general:
¯¯
¯
• A1 + A2 + · · · + An = A1 · A2 · · · An
.
.
. ¯
¯
¯
A1 · A2 · · · An = A1 + A2 + · · · An
.
.
. January 24, 2013 .
.
. Lecture 3 AWB Bubble pushing 28 ’Bubble’ represents negation.
When ’bubble’ on the output is pushed back to the
inputs
• OR chnages to AND, AND changes to OR
• bubbles appear on inputs
• two bubbles in sequence same as no bubble January 24, 2013 Lecture 3 AWB Bubble pushing A·B+C ·D 29 (A · B ) + (C · D ) A·B·C ·D A A A B B B Y Y Y C C C D D D January 24, 2013 Lecture 3 AWB Bubble pushing 30 A A B B C C D E
F January 24, 2013 D Y Y E
F Lecture 3 AWB Systematic design 31 Terminology:
• Complement: negated variable
• Literal: variable or its complement
• Product term: logical AND of literals
• Sum term: logical OR of literals
• Minterm: product term where all input variables
appear exactly once
• Maxterm: sum term where all input variables
appear exactly once
January 24, 2013 Lecture 3 AWB Systematic design  minterms 32 minterm
m0 = a¯c
¯b ¯ a b c s = F (a, b, c) 0 0 0 0 m1 = a¯
¯bc 0 0 1 1 m2 = abc
¯¯ 0 1 0 1 Sum minterms m3 = abc
¯
m4 = a¯c
b¯ 0 1 1 0 for which 1 0 0 1 F (a, b, c) = 1 m5 = a¯
bc 1 0 1 0 m6 = abc
¯ 1 1 0 0 m7 = abc 1 1 1 1 Canonical SOP form: F (a, b, c) = m1 + m2 + m4 + m7 January 24, 2013 Lecture 3 AWB Twolevel logic
F (a, b, c) = 33 m1 + m2 + m4 + m7 =
{ 1 ,2 ,4 ,7 } a¯ + abc + a¯c + abc
¯bc ¯ ¯
b¯ = Twolevel logic (AND array followed by OR gate):
a b c m1
m2
s
m4
m7 January 24, 2013 Lecture 3 AWB FA  bit of carry
minterm
m0 = a¯c
¯b¯ a b cin cout = F (a, b, cout ) 0 0 0 0 m1 = a¯
¯bc 0 0 1 0 m2 = abc
¯¯ 0 1 0 0 m3 = abc
¯
m4 = a¯c
b¯ 0 1 1 1 1 0 0 0 m5 = a¯
bc 1 0 1 1 m6 = abc
¯ 1 1 0 1 m7 = abc 1 1 1 34 1 F (a, b, cin ) = m3 + m5 + m6 + m7 January 24, 2013 Lecture 3 AWB Systematic design  FA
F (a, b, cin ) = 35 m3 + m5 + m6 + m7 =
{ 3 ,5 ,6 ,7 } abc + a¯ + abc + abc
¯
bc
¯ =
a b cin m3
m5
cout
m6
m7 January 24, 2013 Lecture 3 AWB Systematic design  simpliﬁcation 36 F (a, b, c) = abc + a¯ + abc + abc
¯
bc
¯
bc
¯
= abc + abc + a¯ + abc + abc
¯
= (¯ + a)bc + a¯ + abc + abc + abc
a
bc
¯
= bc + (¯ + b)ac + abc + abc
b
¯
= bc + ac + ab(¯ + c)
c
= bc + ac + ab
Is there any other way? January 24, 2013 Lecture 3 AWB Next .... 37 More Boolean algebra
Karnaugh maps
Read Sections 2.12.5 January 24, 2013 Lecture 3 AWB...
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This note was uploaded on 09/26/2013 for the course ECE 2300 taught by Professor Long during the Fall '08 term at Cornell University (Engineering School).
 Fall '08
 LONG

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