Simplification of Boolean Expression

Status
Not open for further replies.
B

blade88

Junior Member level 2
Joined
Jul 29, 2014
Messages
22
Helped
1
Reputation
2
Reaction score
1
Trophy points
3
Activity points
185
Hello,

I am trying to simplify the following expression:

F=A'BC'D+A'BCD'+AB'C'D+AB'CD'+ABCD

However, the Karnaugh Map doesn't show any pairs of ones.

Are there any suggestions?

Thanks
 

not in terms of not-and-or logic. Maybe there is something using xor.

For FPGA applications, it is a 4LUT -- K-maps don't make sense for FPGA applications. (you should still know them for interviews)
 
Hi,

It's been a long time since I posted here.

I solved it:

F=A'BC'D+A'BCD'+AB'C'D+AB'CD'+ABCD=
=(A'B+AB')C'D+(A'B+AB')CD'+ABCD=
=(A'B+AB')(C'D+CD')+ABCD=
=(A XOR B)(C XOR D)+ABCD

Using Karnaugh Map, I should find F' using zeros in the table and I could find the same result.

Thank you
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…