(i) No. of inputs and outputs
Number of inputs = 4
Number of outputs = 4
(ii)Assigning letter symbols
Symbols of inputs – A, B, C, D
Symbols of outputs – W, X, Y, Z
(iii)Truth table
INPUT |
OUTPUT |
||||||
A |
B |
C |
D |
W |
X |
Y |
Z |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
0 |
0 |
1 |
0 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
1 |
0 |
0 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
0 |
1 |
0 |
0 |
0 |
0 |
(iv)Boolean equation
Z
All the entries of Z are compliment of the entries of D.So,
Z=D’
Y
All the entries of Y and C are same in the truth table.So,
Y=C
X
X = BC’+ B’C = B(XOR)C
W
From the truth table,
W = A’B'C’D’ + A’B'C’D = A’B'C’(D’+D) = A’B'C’
(v)Logic Diagram
thanks a lot
Thanks alots
Yoy are So AMAZING!
How it is possible…Please tell…
9′s complement of BCD 0 is 9 i.e. 0000(0) becomes 1001(9) and so on. I hope this clears your doubt.
wounderfull thanks
Why u have directly written the ans for w means why it is not solved by using kmap
Hi Kirti,
If you observe the truth table, there are only two instances where W is true. So, solving it using boolean algebra is much simpler. However, solving it using a K-Map would obtain the same result.
Super. Thank u.
nice thank