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Young Won Lim9/18/13

Truth Table (2A)

Young Won Lim9/18/13

Copyright (c) 2011-2013 Young W. Lim.

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Truth Table (2A) 3 Young Won Lim9/18/13

Truth Table

x

y

z?

1

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

x y z F

F

I/O relationship

Truth Table (2A) 4 Young Won Lim9/18/13

Truth Table and minterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when x=0 and y=1 and z=1

the case when x=0 and y=1 and z=0

the case when x=0 and y=0 and z=0

the case when x=0 and y=0 and z=1

the case when x=1 and y=1 and z=1

the case when x=1 and y=1 and z=0

the case when x=1 and y=0 and z=0

the case when x=1 and y=0 and z=1

inputs

All possible combination of inputs

x y z

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

For the output of an and gate to be 1, all inputs must be 1

x=1y=1

z=1

x=1y=0

z=11

Truth Table (2A) 5 Young Won Lim9/18/13

Truth Table and minterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when the minterm

the case when the minterm

the case when the minterm

the case when the minterm

the case when the minterm

the case when the minterm

the case when the minterm

the case when the minterm

inputs

All possible combination of inputs

x y z

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

x y z = 1

3

2

0

1

7

6

4

5

index

m0 =

m1 =

m2 =

m3 =

m4 =

m5 =

m6 =

m7 =

m5minterm index 5

binary 101

x y z = 1

0

bar

x=1y=1

z=1

1

Truth Table (2A) 6 Young Won Lim9/18/13

Truth Table and MAXterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when x=0 and y=1 and z=1

the case when x=0 and y=1 and z=0

the case when x=0 and y=0 and z=0

the case when x=0 and y=0 and z=1

the case when x=1 and y=1 and z=1

the case when x=1 and y=1 and z=0

the case when x=1 and y=0 and z=0

the case when x=1 and y=0 and z=1

inputs

All possible combination of inputs

x y z

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

For the output of an or gate to be 0, all inputs must be 0

x=0y=0

z=0

x+ y+ z = 0

x=1y=0

z=10

Truth Table (2A) 7 Young Won Lim9/18/13

Truth Table and MAXterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

the case when the MAXterm

inputs

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

M 0 =

M 1 =

M 2 =

M 3 =

M 4 =

M 5 =

M 6 =

M 7 =

M 5minterm index 5

binary 101 1

bar

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+ z

x=0y=0

z=00

Truth Table (2A) 8 Young Won Lim9/18/13

1

0

0

0

xy

Maxterm and minterm Conditions

1 1

1 0

0 1

0 0

x y

1

1

1

0

x+y

1 1

1 0

0 1

0 0

x y

0

0

0

1

xy

0 0

0 1

1 0

1 1

x y

0

1

1

1

x+y

0 0

0 1

1 0

1 1

x y

1

01

01

1

1

1

0

0

0

0

Truth Table (2A) 9 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with minterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

F The output F becomes 1, for one of the three following cases

(the case when x=0 and y=1 and z=1)

(the case when x=0 and y=0 and z=1)

(the case when x=1 and y=0 and z=0)

or

or

x y z = 1

x y z = 1

x y z = 1

m1 =

m3 =

m4 =

Domain Range

m1

m3

m4

m0

m2

m5

m6

m7

1

0

Truth Table (2A) 10 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with minterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

F

F = m1 +m3 +m4

The output F becomes 1, m1=1either m3=1 m4=1or or

For the output of an or gate to be 1, at least one must be 1

m1 +m3 +m4=1 F = 1

m1

m3

m4

Fm0m2

m7

Fm5m6

Truth Table (2A) 11 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with minterms (3)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

F

F = m1 +m3 +m4

The output F becomes 1, m1=1either m3=1 m4=1or or

For the output of an or gate to be 1, at least one must be 1

m1 +m3 +m4=1 F = 1

The output F becomes 0, either

F = 0

m0=1 m2=1 m5=1 m6=1 m7=1or or or or

m0+m2+m5+m6+m7=1

F = m0+m2+m5+m6+m7

Truth Table (2A) 12 Young Won Lim9/18/13

(the case when x=0 and y=1 and z=0)

(the case when x=0 and y=0 and z=0)

(the case when x=1 and y=1 and z=1)

(the case when x=1 and y=1 and z=0)

(the case when x=1 and y=0 and z=1)

1

0

1

0

0

0

0

1

Boolean Function with Maxterms (1)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

FThe output F becomes 0, for one of the five following cases

or

or

Domain

RangeM 1

M 3

M 4

M 0

M 2

M 5

M 6

M 7

1

0

x+ y+z = 0

x+ y+z = 0

x+ y+ z = 0

x+ y+z = 0

x+ y+ z = 0

or

or

ororor

Truth Table (2A) 13 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with Maxterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

FThe output F becomes 0, M 0=0either M 2=0 M 5=0 M 6=0 M 7=0or or or or

For the output of an and gate to be 0, at least one input must be 0

F = 0M 0⋅M 2⋅M 5⋅M 6⋅M 7=0

F = M 0⋅M 2⋅M 5⋅M 6⋅M 7

M 0M 2

M 7

M 5M 6

M 1

M 3

M 4

F F

Truth Table (2A) 14 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with Maxterms (2)

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

FThe output F becomes 0, M 0=0either M 2=0 M 5=0 M 6=0 M 7=0or or or or

For the output of an and gate to be 0, at least one input must be 0

F = 0

The output F becomes 1, either

F = 1

M 0⋅M 2⋅M 5⋅M 6⋅M 7=0

F = M 0⋅M 2⋅M 5⋅M 6⋅M 7

M 1=0 M 3=0 M 4=0or or

M 1⋅M 3⋅M 4=0

F = M 1⋅M 3⋅M 4

Truth Table (2A) 15 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Complimentary Relations

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

inputs output

All possible combination of inputs

x y z

3

2

0

1

7

6

4

5

index

F

The output F becomes 0,

F (x , y , z) = M 0⋅M 2⋅M 5⋅M 6⋅M 7

M 0=0either M 2=0 M 5=0 M 6=0 M 7=0or or or or

For the output of an and gate to be 0, at least one input must be 0

F (x , y , z) = m1 + m3 + m4

m1=1either m3=1 m4=1or or

For the output of an or gate to be 1, at least one must be 1

The output F becomes 1,

F (x , y , z) = m0 + m2 + m5 + m6 + m7

F (x , y , z) = m0 + m2 + m5 + m6 + m7

= m0⋅m2⋅m5⋅m6⋅m7

mi = M i

M i = mi

Truth Table (2A) 16 Young Won Lim9/18/13

Boolean Function Summary

for the cases

The output F becomes 1,

1) when

M 1=02) when M 3=0 M 4=0or or

F=1

m1=1 m3=1 m4=1or or

F (x , y , z) = m1 + m3 + m4

F=1F (x , y , z) = M 1⋅M 3⋅M 4 (F=0)

for the cases

The output F becomes 0,

1) when

2) when

F=0

F=0

(F=1)

1

1

10 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

x y z

3

2

0

1

7

6

4

5

F

0

0

000

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

x y z

3

2

0

1

7

6

4

5

F

M 0=0 M 2=0 M 5=0 M 6=0 M 7=0or or or or

F (x , y , z) = M 0⋅M 2⋅M 5⋅M 6⋅M 7

m0=1 m2=1 m5=1 m6=1 m7=1or or or or

F (x , y , z) = m0+m2+m5+m6+m7

Truth Table (2A) 17 Young Won Lim9/18/13

Boolean Function Summary

F=1F (x , y , z) = m1 + m3 + m4

F=0

F=0

(F=1)

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

x y z

3

2

0

1

7

6

4

5

F

0

0

000

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

x y z

3

2

0

1

7

6

4

5

F

F (x , y , z) = M 0⋅M 2⋅M 5⋅M 6⋅M 7

F (x , y , z) = m0+m2+m5+m6+m7

1

1

1

0

0

000

1

1

1

F=1F (x , y , z) = M 1⋅M 3⋅M 4 (F=0)

Truth Table (2A) 18 Young Won Lim9/18/13

SOP and POS

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

AND

AND

AND

AND

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

OR

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

AND

AND

AND

AND

0 1 10 1 00 0 10 0 0

1 1 11 1 01 0 11 0 0

OR

1

0

1

0

SOP

POS

Truth Table (2A) 19 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with minterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F

1

11

11

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F1

11

1

1

1

F F

Truth Table (2A) 20 Young Won Lim9/18/13

1

0

1

0

0

0

0

1

Boolean Function with Maxterms

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F

0

00

01

0

1

0

0

0

0

1

0 1 1

0 1 0

0 0 1

0 0 0

1 1 1

1 1 0

1 0 1

1 0 0

x y z

3

2

0

1

7

6

4

5

F0

0

0

0

0

FF

0

Truth Table (2A) 21 Young Won Lim9/18/13

Truth Table

Young Won Lim9/18/13

References

[1] http://en.wikipedia.org/