How to construct a truth table with 1 and 0

[Her-0]

I have written the function for the circuit as --> F= (W'.Z)(X.Y')+(Y'+X)+(Z+W)

I am very confused on how to set up the truth table can someone help me?

 

[berkeman]

I have written the function for the circuit as --> F= (W'.Z)(X.Y')+(Y'+X)+(Z+W)

I am very confused on how to set up the truth table can someone help me?

You appear to have 4 inputs and one output. Make your table with 5 columns, with the 1st 4 columns counting binary, and the 5th column with the result for each combination of inputs

0000 F(0000)
0001 F(0001)
etc...

 

[Her-0]

You appear to have 4 inputs and one output. Make your table with 5 columns, with the 1st 4 columns counting binary, and the 5th column with the result for each combination of inputs

0000 F(0000)
0001 F(0001)
etc...

So what you are saying is set up 5 columns W,X,Y,Z than the last as F but how would I come about calculating the result that's what I am confused at.

 

[NoMoreExams]

Assume you had just 2 variables: X and Y, there would be 4 combinations for them:
(0,1), (1,0), (1,1), (0,0)

Let's assume your function was X and Y, so figure out what would T and T = ? T and F = ? , etc.

 

[Her-0]

Assume you had just 2 variables: X and Y, there would be 4 combinations for them:
(0,1), (1,0), (1,1), (0,0)

Let's assume your function was X and Y, so figure out what would T and T = ? T and F = ? , etc.

Okay, but than what about if its an and/or the values would differ
For and:
X Y
T T T
T F F
F T F
F F F

For or:
X Y
T T T
T F T
F T T
F F F

 

[NoMoreExams]

Yes so you would have shown that X and Y is not logically equivalent to X or Y which you should've known.

 

[Defennder]

I'll work with 1's and 0's. It's easier to read the symbols. As for evaluating the logic expression, you have to work with certain results from Boolean algebra.

Take this as an example:
Suppose the configuration from a particular line in your truth table is 1101 which is an ordered input for WXYZ. Then after substituting the Boolean values for W,X,Y,Z the Boolean expression becomes (0*1)(1*1)+(1+1)+(1+1) where * denotes AND and + denotes OR.

Evaluate this in order of parantheses:
(0*1) = 0, (1*1)=1, 1+1=1, 1+1=1

so it now becomes: 0*1+1+1 = 0+1+1 = 1.

So for that particular input configuration F=1. Now apply this method to all the other possible inputs for WXYZ and you have the truth table.