Digital Electronics - Boolean Algebra and Logic Simplification - Discussion
Discussion Forum : Boolean Algebra and Logic Simplification - General Questions (Q.No. 17)
17.
How many gates would be required to implement the following Boolean expression after simplification? XY + X(X + Z) + Y(X + Z)
Discussion:
13 comments Page 1 of 2.
Rushikesh said:
8 years ago
Given:
XY+X(X+Z)+Y(X+Z)
= XY+XX+XZ+YX+YZ .......(By Using Distributed Law)
= XY+X+XZ+YX+YZ .......(Idempotent we substitute here :(XX=X) )
= X+XY+YX+XZ+YZ .......(Rearrange the equation for simple simplify )
= X+XY+XZ+YZ .....(By Using Commutative Law : ( XY=YX) due to we substitute here
XY n again its common so again use here Idempotent)
= X+XZ+YZ .......(By Using Absorption Law : X+XY=X )
= X+YZ .......( By Absorption Law)
X+YZ.
XY+X(X+Z)+Y(X+Z)
= XY+XX+XZ+YX+YZ .......(By Using Distributed Law)
= XY+X+XZ+YX+YZ .......(Idempotent we substitute here :(XX=X) )
= X+XY+YX+XZ+YZ .......(Rearrange the equation for simple simplify )
= X+XY+XZ+YZ .....(By Using Commutative Law : ( XY=YX) due to we substitute here
XY n again its common so again use here Idempotent)
= X+XZ+YZ .......(By Using Absorption Law : X+XY=X )
= X+YZ .......( By Absorption Law)
X+YZ.
(4)
EBoyones said:
1 decade ago
XY + X (X+Z) + Y (X+Z) = XY + XX + XZ + XY + YZ.
By close inspection there are two XY we retain just one of this form and XX = X.
= X + XY + XZ + YZ.
= X (1 + Y + Z) + YZ.
The form: (1 + Y + Z) = 1 for what ever value of Y and Z.
= X + YZ.
This Boolean expression can be built using only 2 gates which is an and gate and or gate respectively.
By close inspection there are two XY we retain just one of this form and XX = X.
= X + XY + XZ + YZ.
= X (1 + Y + Z) + YZ.
The form: (1 + Y + Z) = 1 for what ever value of Y and Z.
= X + YZ.
This Boolean expression can be built using only 2 gates which is an and gate and or gate respectively.
Otman said:
1 decade ago
Solution Please note that ----> XX = X; 1+X = 1; 1+Y = 1; 1+Z=1.
XY+X(X + Z)+Y(X + Z).
XY+XX+XZ+YX+YZ.
XY+X+XZ+YX+YZ.
X(Y+1+Z+Y+YZ).
X(1+Z+Y+YZ).
X(1+Y+YZ).
X(1+YZ).
= X+YZ.
Thanks.
XY+X(X + Z)+Y(X + Z).
XY+XX+XZ+YX+YZ.
XY+X+XZ+YX+YZ.
X(Y+1+Z+Y+YZ).
X(1+Z+Y+YZ).
X(1+Y+YZ).
X(1+YZ).
= X+YZ.
Thanks.
(2)
Ballabh said:
1 decade ago
xy require one gate that is AND gate
x(x+z)........req.two gate that is OR GATE ,OTHER IS AND GATE\
NOW y(x+z)....req............one OR,other is oR gate
total gates r 5
x(x+z)........req.two gate that is OR GATE ,OTHER IS AND GATE\
NOW y(x+z)....req............one OR,other is oR gate
total gates r 5
Richal said:
1 decade ago
You have missed 1 step...
its xy+xx+xz+yx+yz
xy+(x+xz)+yx+yz ....using AB+A=A and AB=BA
xy+xy+x+yz ........using A+A=A and AB+A=A
x+yz.......
its xy+xx+xz+yx+yz
xy+(x+xz)+yx+yz ....using AB+A=A and AB=BA
xy+xy+x+yz ........using A+A=A and AB+A=A
x+yz.......
DINESH KUMAR JADON said:
1 decade ago
xy+x(x+z)+y(x+z).
=xy+xx+xz+yx+yz.
=xy+x+xz+yx+yz.
=x+xz+yx+yz.
=x+yx+yz.
=x+yz.
One AND Gate for yz and one OR Gate for x+yz.
Total 2 gates required.
=xy+xx+xz+yx+yz.
=xy+x+xz+yx+yz.
=x+xz+yx+yz.
=x+yx+yz.
=x+yz.
One AND Gate for yz and one OR Gate for x+yz.
Total 2 gates required.
(1)
Kornelis said:
10 years ago
xy+x(x+z)+y(x+z) <=>
xy+xx+xz+yx+yz < => use aa = a, ab = ba && a+a = a.
xy+x+yz use <=> ab+a = a.
x+yz.
xy+xx+xz+yx+yz < => use aa = a, ab = ba && a+a = a.
xy+x+yz use <=> ab+a = a.
x+yz.
Maher Yassin said:
1 decade ago
xy+x(x+z)+y(x+z)
xy+xx+xz+yx+yz (A.A)=A
(xy+x)+xz+yz (AB+A)=A
(x+xz) +yz (AB+A)=A
x+yz
xy+xx+xz+yx+yz (A.A)=A
(xy+x)+xz+yz (AB+A)=A
(x+xz) +yz (AB+A)=A
x+yz
(1)
RIZIKI GRAPHICS said:
4 years ago
= XY+X+XZ+XY+YZ,
= XY+XY+X+XZ+YZ,
= XY+X+XZ+YZ,
= X(Y+1)+XZ+YZ,
= X+XZ+YZ,
= X(1+Z)+YZ,
= X(1)+YZ,
= X+YZ.
Answer is X+YZ.
= XY+XY+X+XZ+YZ,
= XY+X+XZ+YZ,
= X(Y+1)+XZ+YZ,
= X+XZ+YZ,
= X(1+Z)+YZ,
= X(1)+YZ,
= X+YZ.
Answer is X+YZ.
(12)
Rajeshwari said:
1 decade ago
On solving,
XY + X(X + Z) + Y(X + Z).
We get,
x+yz.
This can be implemented with AND and OR gate.
XY + X(X + Z) + Y(X + Z).
We get,
x+yz.
This can be implemented with AND and OR gate.
(1)
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