Yes you got it right in scan0009.jpg (without the B' of course, for reasons I explained above). I should give credit where credit is due

Scan0010.jpg is not strictly incorrect, it is just that you have moved further from the original question.

Although it may not feel like it, you are getting it

Dave

 

Let me look at your circuit in scan0009.jpg...

Dave

 

Thanks alot Dave! This has to be turned in tomorrow night and I was about to throw all this in the trash and not turn anything in! I cannot believe how hard of a time I have picking this logic stuff up! So you are saying in scan 9 I need to totally leave the B out? I can see your point where the DD' would take care of it. Then that would change the entire circuit. At least if I have the first 2 done right in NAND gates then I will get half credit for the homework. I have 4 weeks left in this class, what do you charge to tutor.....lol. I'll buy your dinner or anything.
Thanks again

 

You've nearly got it. The B' part is not right, as described above. Furthermore you have not quite got the last NAND-gate right - it should NAND the whole function and hence you need a NOT-gate at the output (see the attachment).

Dave

 

FYi, there is another (more elegant but complex) solution:

((A'B) + (A'CD))'

Take common factors:

(A'(B + CD))'

Apply Demorgans AB' = A' + B':

A'' + (B + CD)'

Note A'' = A, so:

A + (B + CD)'

Apply DeMorgans on (B + CD)':

A + (B'(CD)')

Double NOT the function:

A + (B'(CD)')''

Apply DeMorgans to the whole expression (A + B)' = A'B':

(A'(B'(CD)')')'

The circuit for this is shown in the attachment.

Dave

 

Is this even any where near being right? I need to draw this circuit
A+(A+B')+(A'+C') using only NOR gates.
Thanks Dennis

 

Is this even any where near being right? I need to draw this circuit
A+(A+B')+(A'+C') using only NOR gates.
Thanks Dennis

Sadly not. The problem you have is that when 2 (or more inputs) go into a NAND or NOR gate the whole expression is ANDed or ORd (not the individual inputs, as you have done). For example:

Inputs: A and B' go into a NAND-gate

Output: (AB')'

Inputs: A and B' go into a NOR-gate

Output: (A + B')'

Clear? You will need to use a NOR-gate with the inputs tied together to remove the NOT across the whole expression.

Dave

 

The circuit for this is a lot simpler if you note the Associative Property of Addition (ref. http://www.allaboutcircuits.com/vol_4/chpt_7/4.html). It states that:

A + (B + C) = (A + B) + C

For your original expression:

A + (A + B') + (A' + C')

We can write it as:

(A + A) + B' + (A' + C')

A + A = A, therefore it simplifies to:

A + B' + (A' + C')

Using the Associative Property again, we can write this expression as:

A + (A' + C') + B' = (A + A') + C' + B'

A + A' = 0, therefore it simplifies to:

C' + B'

Dave

 

So if I understand this right you should always simplify a problem before trying to draw it out? I didn't realize you should simplify it first. I thought you would just DeMorgan it then draw out what you got. It sure would be alot easier trying to plot out C' +B' then A+(A+B)+(A'+C').
Thanks Dennis

 

Hopefully this diagrahm is a little closer.

 

Here's another.

 

So if I understand this right you should always simplify a problem before trying to draw it out? I didn't realize you should simplify it first. I thought you would just DeMorgan it then draw out what you got. It sure would be alot easier trying to plot out C' +B' then A+(A+B)+(A'+C').
Thanks Dennis

Yes, if possible simplify first then draw your circuit - you want the least gates as possible.

Hopefully this diagrahm is a little closer.

That is the correct circuit for A+(A+B)+(A'+C').

Dave

 

Here's another.

That is the correct circuit for (A+B')+(A+C'+D'). It may be able to be simplified, but the simplification is not obvious in terms of gate-reduction.

Looks like you're getting the hang of it now

Dave

 

Dave you definitely have secured your spot in Heaven........lol
Thanks Dennis

 

As long as you feel you learnt something from it, and that you get a good mark of course .

Best of luck with the rest of the course.

Dave