Karnaugh Map for Active Low Output Y*

[JustinCk]

Hi All,

I am given a canonical expression with active low output Y*

Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)

I would like to use a karnaugh map to obtain the Product of Sum.

However, I am confused because of the terms "active low".

Is my truth table correct?

A B C D Y*
0 0 0 0 1
0 0 0 1 0
0 0 1 0 1
0 0 1 1 0
0 1 0 0 0
0 1 0 1 1
0 1 1 0 0
0 1 1 1 1
1 0 0 0 1
1 0 0 1 0
1 0 1 0 1
1 0 1 1 0
1 1 0 0 1
1 1 0 1 1
1 1 1 0 0
1 1 1 1 1

Thank You!

 

[nlantz]

Active low means the same thing as logical NOT. It is the same thing as if you put a line over Y.

However, for the case:
Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)
your answer is opposite because they are asking you do define Y*

If the question asked for Y then you would have to invert your output.

 

[JustinCk]

Active low means the same thing as logical NOT. It is the same thing as if you put a line over Y.

However, for the case:
Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)
your answer is opposite because they are asking you do define Y*

If the question asked for Y then you would have to invert your output.

Hi,

Thanks for your reply.
just to confirmed soo my answer should be:
A B C D Y*
0 0 0 0 0
0 0 0 1 1
0 0 1 0 0
0 0 1 1 1
0 1 0 0 1
0 1 0 1 0
0 1 1 0 1
0 1 1 1 0
1 0 0 0 0
1 0 0 1 1
1 0 1 0 0
1 0 1 1 1
1 1 0 0 0
1 1 0 1 0
1 1 1 0 1
1 1 1 1 0

All my output should be inverted. Am I correct?

 

[nlantz]

Oops, sorry I was looking at POS instead of SOP. (It's been a few years) I think you were right the first time.

Y*(A, B, C, D) = πM(1, 3, 4, 6, 9, 11, 14)

ABCDY* Y
0 0 0 0 0 1
0 0 0 1 1 0
0 0 1 0 0 1
0 0 1 1 1 0
0 1 0 0 1 0
0 1 0 1 0 1
0 1 1 0 1 0
0 1 1 1 0 1
1 0 0 0 0 1
1 0 0 1 1 0
1 0 1 0 0 1
1 0 1 1 1 0
1 1 0 0 0 1
1 1 0 1 0 1
1 1 1 0 1 0
1 1 1 1 0 1

Does the term "Active Low" make sense though? You are going to see that a lot. In this case you can ignore it because your truth table has a star on it. But in general its important.