XOR question

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Jhon Paul Jaspe

Joined May 13, 2017
23
C'DE'F + CD'EF'

Can this be XOR?

Since XOR -> A'B + AB' = AxorB

if i let A' = C'E' and B = DF

then it would be like A'B + AB' ?????
and it would be like
CE xor DF ????

is this correct?
 

Deleted member 115935

Joined Dec 31, 1969
0
you only have 4 variables, so 16 states,
draw out the truth table for both,

what do you get ?
 

Papabravo

Joined Feb 24, 2006
21,159
No. Only two combinations of the input variables produce a 1 for an output. It is a two-state decoder, and the states are 0101 and 1010. To be sure they are complements of each other.
 

WBahn

Joined Mar 31, 2012
29,979
C'DE'F + CD'EF'

Can this be XOR?

Since XOR -> A'B + AB' = AxorB

if i let A' = C'E' and B = DF

then it would be like A'B + AB' ?????
and it would be like
CE xor DF ????

is this correct?
So expand (CE) xor (DF) into it's full SOP (sum of products) form and see if it is the same thing. It's a good exercise for you to do.
 

Deleted member 115935

Joined Dec 31, 1969
0
No. Only two combinations of the input variables produce a 1 for an output. It is a two-state decoder, and the states are 0101 and 1010. To be sure they are complements of each other.

well. that gave them the answer...
 

WBahn

Joined Mar 31, 2012
29,979
well. that gave them the answer...
Not really. It's not obvious, particularly for someone new to digital logic, whether this is or is not the same as CE xor DF. After all, A xor B and AB' + A'B are each two-state decoders and they ARE equivalent.
 

Deleted member 115935

Joined Dec 31, 1969
0
Re how does a truth table help,

the question was, is C'DE'F + CD'EF' the same as ???

so suggestion was,

truth table both, there are only 16 states,
if the answer is the same, then they are equvilent,
 

Papabravo

Joined Feb 24, 2006
21,159
Re how does a truth table help,

the question was, is C'DE'F + CD'EF' the same as ???

so suggestion was,

truth table both, there are only 16 states,
if the answer is the same, then they are equvilent,
That's certainly what I was thinking, but I thought I'd let someone else have a crack at it.
 
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