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I have 4 inputs; (A, B, C, D) and 3 outputs; (X,Y,Z). 1)X is true when the input is less than 0111. 2)Y is true when the input is greater than 0111. 3)Z is true when the input is 0111.

Can someone help me out with the Boolean Expression for X? I have already obtained the expressions for Y and Z which are as follows:

Y = A
    _
Z = A . (B . C . D)
Clifford
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user3526197
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1 Answers1

1

X is true when neither Y or Z are true:

    _   _
X = Y + Z

or

    _____
X = Y . Z

The expansion of which can be simplified, hint:

_   _   _
A + A = A

From first principles, any expression can be obtained from the truth table by OR'ing the true AND expression for each row that has a true result (then simplifying where possible); for example:

A B C   X
---------   _   _   _
0 0 0   1 = A . B . C
0 0 1   0
0 1 0   0
0 1 1   0
1 0 0   0
1 0 1   0
1 1 0   0
1 1 1   1 = A . B . C

     _   _   _
X = (A . B . C) + (A . B . C)

alternatively:
     _________
X = (A + B + C) + (A . B . C)

For large truth tables, this can become cumbersome (which is why my example has only three variables), in these cases a Karnaugh Map could be used instead.

Clifford
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  • answer should be 0 when A B and C are turned on at the same time. It should be true for all the combinations of inputs even when all are off but it should be false when all 3 are turned on. – user3526197 Sep 28 '14 at 10:43
  • @user3526197 : and...? What is your point? You have 4 inputs not three. My three variable example is an illustration of a *general technique* for solving *similar problems*, not an answer to *your problem* - that is given by `X = NOT(Y AND Z)`. I am leaving some work for you to do. – Clifford Sep 28 '14 at 11:01
  • I managed to solve the problem. I was just thinking of it all wrong in the first place. Thanks for your help – user3526197 Sep 28 '14 at 11:46