n binary variables can be combined to form 2n terms (AND terms),
called Minterms or standard
products. x’y’(00),x’y(01),xy’(10),xy(11)
In a similar fashion, n binary
variables can be combined to form 2n terms (OR terms), called Maxterms or standard sums.
Boolean
Functions expressed in Sum of MinTerms or Product of MaxTerms are said to be in
Canonical Form
* Note that each maxterm is the complement of its corresponding minterm
and vice versa.
1
Each
Minterm is obtained from an AND term. Variable primed if corresponding bit 0,
vice versa Each Maxterm is obtained from an OR term. Variable primed if
corresponding bit 1, vice versa
x
|
y
|
z
|
Minterms
|
Maxterms
|
||
0
|
0
|
0
|
x’y’z’
|
mo
|
x+y+z
|
Mo
|
0
|
0
|
1
|
x’y’z
|
m1
|
x+y+z’
|
M1
|
0
|
1
|
0
|
x’yz’
|
m2
|
x+y’+z
|
M2
|
0
|
1
|
1
|
x’yz
|
m3
|
x+y’+z’
|
M3
|
1
|
0
|
0
|
xy’z’
|
m4
|
x’+y+z
|
M4
|
1
|
0
|
1
|
xy’z
|
m5
|
x’+y+z’
|
M5
|
1
|
1
|
0
|
xyz’
|
m6
|
x’+y’+z
|
M6
|
1
|
1
|
1
|
xyz
|
m7
|
x’+y’+z’
|
M7
|
2
§ Given the truth table, express F1 in sum
of minterms
  x |
y
|
z
|
F1
|
F2
|
||
0
|
0
|
0
|
0
|
1
|
||
0
|
0
|
1
|
0
|
|||
1
|
||||||
0
|
1
|
0
|
0
|
1
|
||
0
|
1
|
1
|
0
|
1
|
||
1
|
0
|
0
|
1
|
0
|
||
1
|
0
|
1
|
0
|
0
|
||
1
|
1
|
0
|
0
|
0
|
||
1
|
1
|
1
|
1
|
0
|
||
F1 (x, y, z) = å(1,4,7) = m1 + m4 + m7
= (x' y' z) + (xy' z'
) + (xyz)
§ Find F2 3
§ Given the truth table, express F1 in
Product of Maxterms
x
|
y
|
z
|
F1
|
F2
|
||||
0
|
0
|
0
|
0
|
1
|
||||
0
|
0
|
1
|
1
|
0
|
||||
0
|
1
|
0
|
1
|
|||||
0
|
||||||||
0
|
1
|
1
|
0
|
1
|
||||
1
|
0
|
0
|
1
|
0
|
||||
1
|
0
|
1
|
1
|
0
|
||||
1
|
1
|
0
|
1
|
0
|
||||
1
|
1
|
1
|
1
|
0
|
||||
F1(x, y, z)
= Õ(0,2,3) = M 0
|
× M 2
|
× M3
|
||||||
= (x + y + z)(x
+ y'+z)(x + y'+z')
|
||||||||
§ Find F2 4
Express the Boolean function F = x + y' z in a sum
of minterms.
AIM: Convert it to 3 variable AND Terms and take their SUM
x = x( y + y' ) = xy + xy'
xy = xy(z + z' ) = xyz + xyz' xy' = xy' (z + z' ) = xy' z + xy' z'
y' z = y' z(x
+ x' ) = xy' z + x' y' z
Adding all terms and excluding
recurring terms:
F (x, y, z) = x' y'
z + xy' z'+xy' z + xyz'+xyz
F (x, y, z) = m1 + m4 + m5 + m6 + m7 = å(1,4,5,6,7)
5
Express the Boolean function F = xy + x' z in a
Product of Maxterms.
AIM:
Convert it to 3 variable OR Terms and take their Product
Tip: Postulate 4b (Distributed)
F = xy + x' z = (xy + x' )(xy + z) (xy
+ x' ) = (x + x' )( y + x' )
(xy + z) = (x + z)( y + z)
F =1.( y + x' )(x + z)( y + z)
|
All Term missing one variable X+X’=1
|
||||||
x'+ y + zz' = (x'+ y + z)(x'+ y + z' )
|
|||||||
x + z + yy' = (x + z + y)(x + z + y' )
|
|||||||
y + z + xx' = ( y + z + x)( y + z + x' )
|
|||||||
Remove any recurring terms
|
|||||||
F = (x + y + z)(x + y'+z)(x'+ y
|
+ z)(x'+ y + z' )
|
||||||
M0M2M4M5
|
= F(x,y,z)=Î (0,2,4,5)
|
6
|
|||||
•
Find Sum of Minterms and Product of MaxTerms of F= xy+x’z
(Same
as before)
– Truth Table
Find Sum of
Minterms & Product of Maxterms from truth table
– F(x,y,z)= ∑(1,3,6,7)
– F(x,y,z)= П (0,2,4,5)
7
• Complement
of a function expressed as sum of minterms equals sum of minterms missing from the
original function
• F(A,B,C,D) =
∑(1,4,5,6,7)= m1+m4+m5+ m6+m7
• Complement
is F’(A,B,C,D)= ∑(0,2,3) = m0+m2+m3
• Complement of F’ is F that
is different
F = (m0+m2+m3)’ = = m’0. m’2 . m’3 = = M0.M2.M3= П(0,2,3)
8
•
Sum of Minterms and Product of Maxterm may not be most
efficient way to draw a circuit as all variables must be used.
•
Standard Functions have one two or three terms and Two
Levels. (AIM: Minimum Gates)
•
SUM OF PRODUCT
– F= y’ + xy+ x’yz’
– Sum denotes
OR of these terms
– Two Levels
•
PRODUCT OF SUM
– F= x(y’+z)(x+’y+z’)
– Product
denotes AND of these terms
– Two Levels
9
SUM OF
PRODUCT F= y’ + xy+ x’yz’
Sum denotes OR of these terms Draw the circuit (2 Level)
x’ y z’
y’
x y
F
10
PRODUCT
OF SUM F= x(y’+z)(x’y+z’)
Product
denotes AND of these terms Draw the circuit (2 Level)
x’ y z’
x
y’
z
F
11
•
Boolean Expression in non standard form F=AB+C(D+E)
.
– Neither SOP or POS
– Implemented in 3 levels
A
|
||
B
|
F
|
|
C
|
||
D
|
||
E
|
– Convert to Standard Form by distributive law
– F=AB+C(D+E) = AB+CD+CE
– Draw 2 Level
Standard Sum of Product Circuit
12
•
F0=0 Constant 0 or Null
•
F1= x.yè AND (x and y)
•
F2 = xy’è x but not y
•
F3 = x
•
F4 = x’yè y but not x
•
F5 = y
•
F6 = xy’ + x’y èExclusive OR ( x or y but not both
•
F7= x+y èOR (x or y)
•
F8 = (x+y)’ è NOR (Not OR)
•
F9 = xy + x’y’è Equivalence (x equals y)
•
F10 = y’ èNot y
•
F11 = x+y’è If y then x
•
F12 = x’ (Not x)
•
F13 = x’ + y If x then y
•
F14 = (xy)’ è NAND
•
F15 = 1 Binary constant 1
13
14
15
2-input exclusive-OR (XOR) logic
gate:
F = X Ã…Y
x
|
F
|
|
y
|
||
X
|
Y
|
F
|
0
|
0
|
0
|
0
|
1
|
1
|
1
|
0
|
1
|
1
|
1
|
0
|
16
2-input exclusive-NOR logic gate:
F = ( X Ã…Y )'
x
|
F
|
|
y
|
||
X
|
Y
|
F
|
0
|
0
|
1
|
0
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
1
|
17
• AND F=xy 
• OR F=x+y 
• Inverter F=x’
• Buffer F=x
• NAND
F=(xy)’
• NOR F=(x+y)’
• Exclusive OR- XOR
• Exclusive
NOR
18
3-input exclusive-OR (XOR) logic
gate:
F = X Ã…Y Ã… Z
x
|
F
|
y
|
|
z
|
Implemented
by 3 Input Gates
x
y 
F
z
Implemented by 2 Input Gates
X
|
Y
|
Z
|
F
|
0
|
0
|
0
|
0
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
1
|
0
|
0
|
1
|
1
|
0
|
1
|
0
|
1
|
1
|
0
|
0
|
1
|
1
|
1
|
1
|
19
§ Using only OR and NOT gates, draw a
schematic for the
following function: F = xy + x' y'+ y' z
(F ' )'= ((xy + x' y'+ y' z)')'
= [( xy)'.(x' y' )'.(y' z)']'
= [( x'+ y' ).(x + y).( y
+ z' )]'
= (x'+ y' )'+(x + y)'+( y + z' )'
x
y
F
z 
20
20
§ Using ONLY NAND gates,
draw a schematic for the following function: F = (a.b)+(b.c)
a
b
F
c
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