Hi,
I am new to the forum so hello to all.
I am studying a BTEC in EE and going onto a HND hopefully next year. I am struggling a bit with reducing boolean algebra. I seem to be able to look at the same problem 3 times and come up with three different answers.
Can someone give me some hints. If some is NOT for example NOT A I write it as A'
I have this problem.... ABC + A'B'C' + ABC' + A'C'
this is one way I have "solved" it.
ABC + A'B'C' + ABC' + A'C'
ABC + A'B'C' + ABC' + A' . 1 (last C' cancelled against first C, turns into 1 (A+A' = 1))
AB + A'B'C' + C' + A' . 1 (cancelled first AB against second AB in third group)
AB + A'B'C' + A' . 1 (cancelled second C' as same as C' in second group)
AB + B'C' + A' . 1 (A' in second group removed as same as last A')
AB + B'C' + A' (A' . 1 = A')
Any help would be greatly appreciated!
Thanks!
I am new to the forum so hello to all.
I am studying a BTEC in EE and going onto a HND hopefully next year. I am struggling a bit with reducing boolean algebra. I seem to be able to look at the same problem 3 times and come up with three different answers.
Can someone give me some hints. If some is NOT for example NOT A I write it as A'
I have this problem.... ABC + A'B'C' + ABC' + A'C'
this is one way I have "solved" it.
ABC + A'B'C' + ABC' + A'C'
ABC + A'B'C' + ABC' + A' . 1 (last C' cancelled against first C, turns into 1 (A+A' = 1))
AB + A'B'C' + C' + A' . 1 (cancelled first AB against second AB in third group)
AB + A'B'C' + A' . 1 (cancelled second C' as same as C' in second group)
AB + B'C' + A' . 1 (A' in second group removed as same as last A')
AB + B'C' + A' (A' . 1 = A')
Any help would be greatly appreciated!
Thanks!