Post on 02-Jan-2016
description
Conversion from one number base to another
Binary arithmetic
Equation simplification
DeMorgan’s Laws
Conversion to/from SOP/POS
Reading equations from Truth Tables
Boolean expression to Karnaugh Map
Minimization using Karnaugh Maps
Minterm and Maxterm Equations
Minimization using don’t cares
Logic to Boolean Expression conversion
Word problems
Determining how many gates and inputs a boolean expression has
Determining Prime Implicants and Essential Prime Implicants
Logical completeness
Review for Exam 1
Conversion from one number base to another
356.8910 to Hexadecimal (2 digits)
Conversion from one number base to another
Binary arithmetic
23 6 | 141 -12 21 -18 3
Equation simplification
Simplify and convert to SOP(A’ + B + C’)(A’ + C’ + D)(B’ + D’)
Y = (AB’ + (AB + B)) B + A
Equation simplification
(X + Y)(X + Z) = (X + YZ)
X + XY = X
X + X’Y = X + Y
X + XY = X
DeMorgan’s Laws
G = {[(R + S + T)’ PT(R + S)’]’T}’
DeMorgan’s Laws
G = {[(R + S + T)’ PT(R + S)’]’T}’ = [(R + S + T)’ PT(R + S)’] + T’ = [ R’S’T’ PT(R’S’)] + T’ = R’S’T’PTR’S’ + T’ = R’S’P(T’T) + T’ = T’
Conversion to/from SOP/POS
(X + YZ) = (X + Y)(X + Z)
Reading equations from Truth Tables
A B C D F
0 0 0 0 1
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1
0 1 0 1 0
0 1 1 0 1
0 1 1 1 1
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 1
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0
Reading equations from Truth Tables
A B C D F
0 0 0 0 1 A’B’C’D’
0 0 0 1 0
0 0 1 0 0
0 0 1 1 0
0 1 0 0 1 A’BC’D’
0 1 0 1 0
0 1 1 0 1 A’BCD’
0 1 1 1 1 A’BCD
1 0 0 0 0
1 0 0 1 0
1 0 1 0 0
1 0 1 1 0
1 1 0 0 1 ABC’D’
1 1 0 1 0
1 1 1 0 1 ABCD’
1 1 1 1 0
Boolean expression to Karnaugh Map
AB
CD 00 01 11 10
00
01
11
10
AB + C’D + A’B’C + ABCD + AB’C
Boolean expression to Karnaugh Map
AB
CD 00 01 11 10
00 1
01 1 1 1 1
11 1 1 1
10 1 1 1
AB + C’D + A’B’C + ABCD + AB’C
Minimization using Karnaugh Maps
AB
CD 00 01 11 10
00 1
01 1 1 1 1
11 1 1 1
10 1 1 1
AB + C’D + A’B’C + ABCD + AB’C
AB + C’D + B’C
Minterm and Maxterm Equations
F(ABCD) = m (0,2,4,7,9,12,14,15)
AB
CD 00 01 11 10
00
01
11
10
BC’D’ + BCD + ABC + A’B’D’ + AB’C’D
Minterm and Maxterm Equations
F(ABCD) = m (0,2,4,7,9,12,14,15)
AB
CD 00 01 11 10
00 1 1 1
01 1
11 1 1
10 1 1
BC’D’ + BCD + ABC + A’B’D’ + AB’C’D
Minimization using don’t cares
AB
CD 00 01 11 10
00
01
11
10
A’B’ + AD
F(ABCD) = m (0,1,2,11,13) + d (3,9,12,15)
Minimization using don’t cares
AB
CD 00 01 11 10
00 1 x
01 1 1 x
11 x x 1
10 1
A’B’ + AD
F(ABCD) = m (0,1,2,11,13) + d (3,9,12,15)
Logic to Boolean Expression conversion
Logic to Boolean Expression conversion
F = (XY + W)Z + V
F = (B+C)A + BC
Word problems
Determining how many gates and inputs a boolean expression has
levelsgatesinputstransistorsinputs/gate max
levelsgatesinputstransistorsinputs/gate max
F = (XY + W)Z + V
Z = A’B’C’ + ABC + BCD +B’C’D’
Determining how many gates and inputs a boolean expression has
4 levels4 gates8 inputs16 transistors2 inputs/gate max
2 levels5 gates16 inputs32 transistors4 inputs/gate max
F = (XY + W)Z + V
Z = A’B’C’ + ABC + BCD +B’C’D’
Determining Prime Implicants and Essential Prime Implicants
AB
CD 00 01 11 10
00 1 1 1
01 1 1 1 x
11 x x 1
10 1
Determining Prime Implicants and Essential Prime Implicants
AB
CD 00 01 11 10
00 1 1 1
01 1 1 1 x
11 x x 1
10 1
6 prime implicants
3 essential prime implicants
Logical completeness
Inverter
AND gate
OR gate
Logical completeness
Inverter
Inverter AND gate
NAND
NAND gateInverter
InverterOR gate