21-04-2014
Philadelphia University
Faculty of Information Technology
Department of Computer Science
Computer Logic Design
By
Dareen Hamoudeh
Dareen Hamoudeh
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Canonical Forms
(Standard Forms of Expression)
Minterms and Maxterms
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Why STANDARD FORMS?
• All Boolean expressions, regardless of their form, can
be converted into either of two standard forms: the
sum-of-products form or the product-of-sums form.
• Standardization makes the evaluation, simplification,
and implementation of Boolean expressions much
more systematic and easier.
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Examples
 Product Terms : Terms that are ANDed together.
• XYZ
• (A+B)(C+D)(A+D)
 Sum Terms: Terms that are ORed together.
• X+Y+Z
• XYZ + VX
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Minterms
• So, if we have two literals (x,y) with AND operation
then we have 4 combinations:
X’Y’ X’Y
XY’ XY
Each combination is called minterm.
• Now,For 3 variables we have 23= 8 minterms:
X’Y’Z’ X’Y’Z …
XYZ
• In general, if a function has n variables there are 2n
minterms.
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Minterms
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Minterms
• If we have the function
When does it =1 ?
After implementing its truth table:
We notice that f1 =1 in the terms m1, m4, m7 … so:
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Minterms
•
•
•
•
Also called sum of products (SOP) or Standard Product expression contains:
– Only OR (sum) operations at the “outermost” level.
– Each term that is summed must be a product of literals:
f(x,y,z) = y’ + x’yz’ + xz
Every function can be written as a sum of minterms, which is a special kind of sum
of products form.
The sum of minterms form for any function is unique.
If you have a truth table for a function, you can write a sum of minterms
expression just by picking out the rows of the table where the function output is 1.
And can be expressed in the following notation:
f = x’y’z + xy’z’ + xyz = m1 + m4 + m7
= f(x,y,z) = m(1,4,7)
= f(x,y,z) = (1,4,7)
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sum of products (SOP)
•
Now, if we have the following truth table, can you extract the sum of
minterms for the function F2.
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sum of products (SOP)
The advantage is that any sum of products expression can
be implemented using a two-level circuit
• literals and their complements at the “0th” level
• AND gates at the first level
• a single OR gate at the second level
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sum of products (SOP)
“complement”
f = x’y’z’ + x’y’z + x’yz’ + x’yz + xyz’
= m0 + m1 + m2 + m3 + m6
= m(0,1,2,3,6)
f’ = xy’z’ + xy’z + xyz
= m4 + m5 + m7
= m(4,5,7)
f’ contains all the minterms not in f
x
y
z
f(x,y,z)
f’(x,y,z)
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
1
0
0
0
0
0
1
1
0
1
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Maxterms
• if we have two literals (x,y) with OR operation then
we have 4 combinations:
X’+Y’ X’+Y
X+Y’ X+Y
Each combination is called maxterm.
• Now,For 3 variables we have 23= 8 minterms:
X’+Y’+Z’ X’+Y’+Z …
X+Y+Z
• In general, if a function has n variables there are 2n
maxterms.
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Maxterms
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Maxterms
• If we have the function
When does it =0 ?
We notice that f1 =0 in the terms M0, M2, M3, M5,M6… so:
F1= (X+Y+Z)(X+Y’+Z)(X+Y’+Z’)(X’+Y+Z’)(X’+Y’+Z) = M0 . M2 . M3 . M5 . M6
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Maxterms
•
•
•
•
Also called Product of sums (POS) or Standard Sums expression contains:
– Only AND(product) operations at the “outermost” level.
– Each term must be a Sum of literals:
f(x,y,z) = y’ (x’ + y + z’) (x + z)
Every function can be written as a product of maxterms, which is a special kind of
products of sums form.
The product of maxterms form for any function is unique.
If you have a truth table for a function, you can write a product of maxterms
expression by picking out the rows of the table where the function output is 0.
And can be expressed in the following notation:
F = (X+Y+Z)(X+Y’+Z)(X+Y’+Z’)(X’+Y+Z’)(X’+Y’+Z) = M0 . M2 . M3 . M5 . M6
= F(x,y,z) = ΠM(0,2,3,5,6)
= F(x,y,z) = Π(0,2,3,5,6)
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Product of sums (POS)
•
Now, if we have the following truth table, can you extract the Product of
maxterms for the function F2.
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Product of sums (POS)
“complement”
f = (x’ + y + z)(x’ + y + z’)(x’ + y’ + z’)
= M4 M5 M7
= ΠM(4,5,7)
f’ = (x + y + z)(x + y + z’)(x + y’ + z)
(x + y’ + z’)(x’ + y’ + z)
= M0 M1 M2 M3 M6
= ΠM(0,1,2,3,6)
f’ contains all the maxterms not in f
x
y
z
f(x,y,z)
f’(x,y,z)
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
1
1
1
1
0
0
1
0
0
0
0
0
1
1
0
1
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Minterms and maxterms are related
•
Any minterm mi is the complement of the corresponding maxterm Mi
•
For example, m4’ = M4 because (xy’z’)’ = x’ + y + z
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Converting between standard forms
•
We can convert a sum of minterms to a product of maxterms
From before
and
complementing
so
f
f’
= m(0,1,2,3,6)
= m(4,5,7)
= m4 + m5 + m7
(f’)’ = (m4 + m5 + m7)’
f = m4’ m5’ m7’
[ DeMorgan’s law ]
= M4 M5 M7
[ By the previous page ]
= ΠM(4,5,7)
•
In general, just replace the minterms with maxterms, using maxterm numbers that
don’t appear in the sum of minterms:
f = m(0,1,2,3,6)
= ΠM(4,5,7)
•
The same thing works for converting from a product of maxterms to a sum of
minterms.
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Product of Sums
• If f(x, y, z) = sum of minterms (0, 1, 4, 5), represent f as a
product of maxterms
–
–
–
–
A: product of maxterms(2, 3)
B: product of maxterms(2, 3, 6, 7)
C: product of maxterms(0, 1, 4, 5)
D: product of maxterms(5, 6, 7)
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Sum Of Minterms
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Example 1
• Express the function F(A,B,C) = AB+A’C in
minterm notation
• First term AB misses the variable C, so AND term
with (C+C’):
AB= AB(C+C’) = ABC + ABC’
• Second term A’C misses the variable B so, AND
term with (B+B’):
A’C= A’C(B+B’)= A’BC + A’B’C
Now we combine the terms:
F(A,B,C)=ABC + ABC’ + A’BC + A’B’C
= m7 + m6 + m3 + m1
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Example 2
• Express E = Y’ + X’Z’ in minterm notation.
E = (X+X’)Y’ + X’Z’(Y+Y’)
= XY’(Z+Z’) + X’Y’(Z+Z’)+X’YZ’+X’Y’Z’
= XY’Z+XY’Z’+X’Y’Z+X’Y’Z’+X’YZ’
= m5 + m4 + m1 + m0 + m2
= m5 + m4 + m2 + m1 + m0
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Product of maxterms
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Example 1
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Example 1 cont.
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Note
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