Boolean-Function
Question 1 |
XY’+Z’ is a minimal SoP expression which represents the function (X,Y,Z).
The expression XY’ + YZ’ + X’Y’Z’ can be reduced to XY’+Z’
XY’ + YZ’ + X’Y’Z
= Y’(X+X’Z’) + YZ
= Y’(X+Z’) + Y
= XY’ + Y’Z’ + YZ’
= XY’ + (Y’+Y)Z’
= XY’ + Z’.
The expression (X+Z’)(Y’+Z’) is a PoS expression which also represents the same function (X,Y,Z).
Question 2 |
f(w, 0, 0, z) = 1
f(1, x, 1, z) = x + z
f(w, 1, y, z) = wz + y
The number of literals in the minimal sum-of-products expression of f is __________.
6 |
f(w,0,0,z)= 1 If x=y=0, then the sum of the corresponding minterms be 1.
The minterms with literals x’ and y’ are wx’y’z(9), w’x’y’z(1), wx’y’z’(8), w’x’y’z’(0) .
If x=y=0, then we get wz+w’z+wz’+w’z’ = 1.
f(1,x,1,z)= x+z.
The minterms with variables w and y in true form and x or z or both in true form.
The corresponding minterms are wx’yz(11), wxyz’(14), wxyz(15)
If w=1 and y=1, then we get x’z+xz’+xz= x+z.
f(w,1,y,z)= wz+y
The corresponding minterms are w’xyz’(6), w’xyz(7), wxyz’(14), wxyz(15), wxy’z(13).
If x=1, then we get w’yz’ + w’yz+ wyz’ + wyz+ wy’z = y + wz
So, the function f(w,x,y,z)= Σ(0,1,6,7, 8,9, 11, 13, 14, 15,).
Therefore, the k-map will be:
Therefore, the minimal expression will be: X’Y’ + WZ + XY
Thus, the number of literals will be 6.
Question 3 |
Σ(1, 3, 5, 7, 9) | |
Σ(3, 5, 7, 9, 11) | |
Σ(3, 5, 9, 11, 13) | |
Σ(5, 7, 9,11,13) |
Question 4 |
n2 | |
2n2 | |
2n | |
2 to the power of 2n |
Number of variables= n
Number of input combinations is 2n.
Each “boolean” function has two possible outputs i.e 0 and 1.
Number of boolean functions possible is 22^n.
Formula: The number of m-ary functions possible with n k-ary variables is mk^n.
Question 5 |
n^2^n | |
2^n | |
2^2^n | |
n^2 |
Question 6 |
x.y’ | |
x.y’+z | |
x.z’ | |
none of the options |
Question 7 |
Identify the matching Boolean Expression.
X $ Y' | |
X' $ Y
| |
X' $ Y' | |
None of the above |
So,
X $ ¬Y= X+Y
¬X $ Y= X' + Y'
¬X $ ¬Y= X' + Y