Martin Posted November 14, 2010 Share Posted November 14, 2010 Now I know this may be lost on some off you, but I also know (hope) others understand it (because I don't) Now a little key . = AND+ = ORBOLD = NOT e.g D = NOT D Original expression(A.B) + (A.C) + (A.D) DeMorganised expression_________________(A+B) . (A+C) . (A+D) If anyone can tell me whether this is correct or not I'd me very appreciative! I'll even give you an interweb high five! Link to comment Share on other sites More sharing options...
meekstaaa Posted November 14, 2010 Share Posted November 14, 2010 Yeah thats correct.... Link to comment Share on other sites More sharing options...
sidrick Posted November 14, 2010 Share Posted November 14, 2010 Lost of me, but I'm sure google would know the answer. Link to comment Share on other sites More sharing options...
oxheylady Posted November 14, 2010 Share Posted November 14, 2010 Not sure if its right been a while since i done all this,but I did find this if its any help http://www.allaboutcircuits.com/vol_4/chpt_7/8.html Link to comment Share on other sites More sharing options...
Martin Posted November 14, 2010 Author Share Posted November 14, 2010 Thanks I;ve looked through Google and it's just confusing me more Link to comment Share on other sites More sharing options...
oxheylady Posted November 14, 2010 Share Posted November 14, 2010 Same as lolI put the link up thinking you might understand it Link to comment Share on other sites More sharing options...
eetaylog Posted November 14, 2010 Share Posted November 14, 2010 Yep that is 100% correct. The way i was always taught at uni was that if you break the bar, you change the logic type (AND <-> OR gate). By removing the bar above a NOT gate, you also change it to NOT NOT (ie, the logic 1 or 0 doesnt change). You can check your answer using a Karnaugh map as well if you want, but i wont go into that, youll have to google it! Link to comment Share on other sites More sharing options...
Craig855S Posted November 14, 2010 Share Posted November 14, 2010 I've done a bit on Digital circuits and Logic but never heard of this. I dont really need to understand it in depth though, just how to read the gates on circuit diagrams Link to comment Share on other sites More sharing options...
Martin Posted November 14, 2010 Author Share Posted November 14, 2010 Thanks very much! I got the original expression from my karnaugh map and had to do DeMorgans on it. It's hard because the woman teaching it isn't sure if she's doing it right so she'll tell you one way, next lessons she'll tell you it was wrong so teaches you her new way then next lesson she says go back to method1 Link to comment Share on other sites More sharing options...
caled Posted November 14, 2010 Share Posted November 14, 2010 Thanks very much! I got the original expression from my karnaugh map and had to do DeMorgans on it. It's hard because the woman teaching it isn't sure if she's doing it right so she'll tell you one way, next lessons she'll tell you it was wrong so teaches you her new way then next lesson she says go back to method1 And how much are you paying for this course? Link to comment Share on other sites More sharing options...
eetaylog Posted November 14, 2010 Share Posted November 14, 2010 Looking at it again, i think the expression can be simplified further to: A.(B+C+D) By the way, wheres my high 5? Link to comment Share on other sites More sharing options...
Martin Posted November 17, 2010 Author Share Posted November 17, 2010 http://classicfun.ws/wp-content/uploads/internet-high-five-place-hand-here-right-480x444.jpg sorry it's late, heating packed in at my house so had to move out on Sunday. Just back! Link to comment Share on other sites More sharing options...
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