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Combinational Logic + Boolean equation help - Page 2

Blogs > Raithed
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Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 04:01 GMT
#21
I think you're mixing that with something else fof, I didn't mean if they're the equivalence of anything but such as that:

bubble NAND gate....

[image loading]


NAND gates are basically (AB)' from the beginning, from that diagram alone, but if the input already gives A' and B' the output is already to be (AB)' wouldn't the NOTs cancel? Because the diagrams you provided have bubbles from input to bubbles of output of equivalent gates.
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
Last Edited: 2008-04-21 04:14:55
April 21 2008 04:10 GMT
#22
Thats not a NAND gate though. What you do is imagine the output bubble is far away. Then when it changes from that pic you have 2 bubbles on the output which cancel.

edit: might be the same thing your saying, I'm getting confused over this whole thread.
Do you really want chat rooms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 04:20 GMT
#23
This is a NAND gate: http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/nand.html
How is it what I drew not a NAND gate? A NAND gate as a bubble at the output, but I just added two bubbles on the input?
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 04:27 GMT
#24
[image loading]
Do you really want chat rooms?
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 04:29 GMT
#25
Thats how you solve it using "bubble logic". You can also do it algebraically with boolean algebra.
Do you really want chat rooms?
berated-
Profile Blog Joined February 2007
United States1134 Posts
April 21 2008 04:50 GMT
#26
thats some pretty drawing, boolean algebra is so much easier though
azndsh
Profile Blog Joined August 2006
United States4447 Posts
April 21 2008 05:28 GMT
#27
A' NAND B' = (A' AND B')' = A OR B
deMorgan's Law
sigma_x
Profile Joined March 2008
Australia285 Posts
April 21 2008 06:15 GMT
#28
"break the line change the sign" as they say. btw:
BC+B=BC+B*1 (identity law)
BC+B*1=B(C+1) (distributive law)

p.s. For those using examples from the real numbers to show this, why are you trying to explain properties of boolean algebra/rings using field axioms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 06:19 GMT
#29
Alright, and guys, how does...

X = (A’ + B’) + (B+C)
X = 1?

I get (A’ + B’) = (AB)' + (B+C) right?
What is the next step?
fight_or_flight
Profile Blog Joined June 2007
United States3988 Posts
April 21 2008 06:22 GMT
#30
X = (A’ + B’) + (B+C)

X = A’ + (B’ + B) + C

X = A’ + 1 + C

X = 1
Do you really want chat rooms?
Raithed
Profile Blog Joined May 2007
China7078 Posts
April 21 2008 06:29 GMT
#31
Thank you fof, I'd like to thank everyone for staying up with me while I continue to do this silly stuff. It's 2AM EST. I appreciate it, thanks again fof.
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