Complement of a Boolean input variable in a circuit?

[Coder]

I am having trouble understanding that how this complement of input variable is working in the circuit below.

 

Please help.

 

 

The Process:--

 

 

 

 

 

Our motive is to draw a circuit diagram with only a NOR gate of the Boolean expression :--

 

 

Step 1:--

 

 

 

 

Step 2 :--

 

 

 

Step 3 :--

 

 

 

 

 

 

 

Now from step 2 to step 3 we have to remove the inverters and complement the input variables as marked in the image.

 

So, my question is how this complementation is working? In some places it remains the same (in below 'D' & 'A') and in some places its complements.

 

 

Any help will be appreciated.

[KipIngram]

So you're trying to see how to get from step 2 to step 3?

A one-input NOR gate is just an inverter; if you have the inverse of the input to that gate already available, then you can just remove the gate and use it. That's the gist of how they got from 2 to 3.

I think there's an error, though - I think the D in diagram 2 should have become /D in diagram 3.

Were these diagrams given, as part of the problem, or did you produce them?

In the final diagram two things have happened: 1) one-input NORs of some signal X have been dropped because /X is also available directly, and 2) the two one-input NORs on either side of the wire marked with the arrow A+BD have been dropped, since inverse of inverse is no change.

Edited June 4, 2017 by KipIngram

[Coder]

So you're trying to see how to get from step 2 to step 3?

A one-input NOR gate is just an inverter; if you have the inverse of the input to that gate already available, then you can just remove the gate and use it. That's the gist of how they got from 2 to 3.

I think there's an error, though - I think the D in diagram 2 should have become /D in diagram 3.

Were these diagrams given, as part of the problem, or did you produce them?

In the final diagram two things have happened: 1) one-input NORs of some signal X have been dropped because /X is also available directly, and 2) the two one-input NORs on either side of the wire marked with the arrow A+BD have been dropped, since inverse of inverse is no change.

 

This question is an example in a pdf I have.

 

Yes I also think that the D must be inverted to /D.

 

Now I noticed that A is inverted 2 times so it remains same.

 

And thanks for helping once again.

[KipIngram]

Sure - I don't really get to work in this area these days, in spite of it being my core training. So these opportunities are like sips of fine wine.

[Strange]

DeMorgan's Theorem: http://hyperphysics.phy-astr.gsu.edu/hbase/Electronic/DeMorgan.html