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.
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)
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.......
Maulik said:
1 decade ago
Thank for answering.
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
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)
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)
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)
Nag said:
1 decade ago
@Otman your answer is correct but procedure is wrong.
1+something is 1.
1+something is 1.
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.
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.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers