Babylon University
The college of science for girls
Logic Design
Lecture 3
Boolean Function Representation
The Boolean functions can be represented algebraically using binary variables (x,y,z,…) and a logic operators (OR,AND,NOT).the output of the Boolean function is either o or 1 depending on the values of the binary variables. For example the function g=xyz`
The Boolean function is equal to 1 if x=1 and y=1 and z`=1(z=0);other wise g=0.
A Boolean function may also be represented in a Truth Table .The number of rows in the Table (combinations) is where n is the number of binary variables in the function . the value of the function is either 0 or 1 as shown in the following table:
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X
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Y
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Z
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g
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0
0
0
0
1
1
1
1
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0
0
1
1
0
0
1
1
|
0
1
0
1
0
1
0
1
|
0
0
0
0
0
0
1
0
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Table (1)
There are two methods to Transform a given Truth Table into an algebraic expression Boolean function to manipulate, minimize, and convert to a logic diagram of logic gates these methods are:-
1-Canonical form.
2-Standard form.
3-Canonical form.
A Boolean function may be expressed algebraically from a given Truth Table in two ways. These are:-
a)Sum of minterms:
A binary variables may appears either in its normal for (x) or in its completed form (x`).if we have 2 binary variables x,y then there are four possible combinations when these variable combined with AND these are :-x`y`,x`y,xy`,xy .each of these terms is called a minterm. In similar Manner , n variables can be combined to form (2 ) minterms. Each minterm is obtained from an AND term of n variables ,with each variable being primed if the corresponding bit of the binary number is a 0 and unprimed if a 1 . A symbol of each minterm is (M ) as shown in Table (2).
Ahmed M. Shhaab
b)product of Maxterms:
The second method to express a Boolean function is the product of maxterms .in this method the binary variables are combined with OR with each variable being primed if the corresponding bit is 1 and unprimed if 0; with each term called maxterm.
There for the maxterm is obtained from an OR term of the binary variables and symbol is (M ) as shown in Table(2).
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X
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Y
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Minterms
Term Designation
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Maxterms
Terms Designation
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0
0
1
1
|
0
1
0
1
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X`.Y` M
X`.Y M
X.Y` M
X.Y M
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X`+Y` M
X+Y` M
X`+Y M
X+Y M
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Table (2)
Example:-
Convert the Truth Table given below into Boolean function and with the two types of Canonical form .
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X
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Y
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Z
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F1
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F2
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0
0
0
0
1
1
1
1
|
0
0
1
1
0
0
1
1
|
0
1
0
1
0
1
0
1
|
0
1
0
0
1
0
0
1
|
0
0
0
1
0
1
1
1
|
The algebraic expression for the first function (f1) using the sum of minterms method is:
F1=X`Y`Z+XY`Z`+XYZ
=M +M +M F1(x,y,z)=?(1,4,7)
and for the second function (2):
F2=X`YZ+XY`Z+XYZ`XYZ
=M +M +M +M F2(x,y,z)=?(3,5,6,7)
Using the product of maxterms method the two function can be expressed as:-
F1=(X`+Y`+Z`).(X+Y`+Z)+(X+Y`+Z`).(X`+Y+Z`).(X`+Y`+Z)
=M +M +M +M +M
Ahmed M. Shhaab
And the second function
F2=(X+Y+Z).(X+Y+Z`).(X+Y`+Z).(X`+Y+Z).
=M +M +M +M
And this can be written as:
F2(X,Y,Z)=? (0,1,2,4)
1)Simply F1 (Sum of minterms):
F1=x`yz`+x`yz+xyz`
=yz`(x`+x)+x`yz
=yz` . 1 +x`yz [(x`+x=1)]
=yz`+ x`yz
=y (z`+x`z)
=y[(z`+x`)(z`+z)]
=y[(z`+x`) . 1] [(z+z`=1)]
=y (z`+x`)
=yz`+yx`
2)Simply F1 (product of maxterms)
F1=(x+y+z).(x+y+z`).(x`+y+z).(x`+y+z`).(x`+y`+z`)
=[(x+y)+zz`].[(x`+y)+zz`].(x+y`+z`)
=[(x+y)+ 0 ].[(x`+y)+ 0 ].(x`+y`+z`?)
=(x+y) . (x`+y) . (x`+y`+z`)
=(y+xx`) . (x`+y`+z`)
=y (x`+y`+z`)
=yx`+yy`+yz`
=x`y|+yz`
=y(x`+z`)
Ahmed M. Shhaab
Conversion between Canonical Forms:-
To convert from one canonical from to another, interchange the symbols ?&? and list those NO. missing from the original form.
E.X:- f(x.y,z)=?(0,2,4.5)
f(x,y,z)=?(1,3,6,7)
2)Standard Form :
Another way to express Boolean function is in standard form.
In this configuration the terms that form the function may contain one ,two or any other number of letters.
There are two types of standard forms:-
a)Sum of Products
b)Product of Sums
The Boolean function expressed in these Two forms is the minimized (simplified) version of the Boolean function in sum of minterms and product of maxterms respectively
F1=y`+xy+x`yz` Sum of Product (S O P)
F2=x.(y`+z).(x`+y+z`+w) Product of Sums (P O S)
Note:-
sometimes A Boolean function may be expressed in a non standard form as follows:
F=(AB+CD).(A`B`+C`D`)
It is neither in sum of product not in product of sum. It can be changed to a standard form by using the distributive law to remove the parenthesis and as follows:-
F=ABA`B`+ABC`D`+CDA`B`+CDC`D`
= 0 +ABC`D`+ CDA`B`+ 0
=ABC`D`+A`B`CD Sum of Product
Example:-
Express the Boolean function F=A+B`C` in a Sum of minterms then write the corresponding Truth Table
Solution:- the function has three variables A,B,and C the first term A is missing two variables ; therefore:-
A=A(B+B`)=AB+AB`
Ahmed M. Shhaab
A=AB(C+C`)+A`B`(C+C`)
=ABC+ABC`+AB`C+AB`C`
Combining all terms , we have:-
F=A+B`C`
=ABC+ABC`+AB`C+AB`C`+AB`C+A`B`C
=ABC+ABC`+AB`C+AB`C`+A`B`C
=M+M+M+M+M
F(A,B,C)= ?(1,4,5,6,7).
And the corresponding Truth Table is:-
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F
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Z
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Y
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X
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0
1
0
1
0
0
1
1
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0
1
0
1
0
1
0
1
|
0
0
1
1
0
0
1
1
|
0
0
0
0
1
1
1
1
|
Ahmed M. Shhaab
المادة المعروضة اعلاه هي مدخل الى المحاضرة المرفوعة بواسطة استاذ(ة) المادة . وقد تبدو لك غير متكاملة . حيث يضع استاذ المادة في بعض الاحيان فقط الجزء الاول من المحاضرة من اجل الاطلاع على ما ستقوم بتحميله لاحقا . في نظام التعليم الالكتروني نوفر هذه الخدمة لكي نبقيك على اطلاع حول محتوى الملف الذي ستقوم بتحميله .