Logic Gates with XOR gate need to verify solution

[Femme_physics]

Homework Statement

Basically and most importantly... I wanted to see if I got my F function correct. That's at clause 2.

At clause 3 I'm told

a = b = 1
c = d = 0

and to calculate

At clause 4 I'm asked if the logic value of P is 1, if the logic value of Q matters...

The Attempt at a Solution

http://img528.imageshack.us/img528/4162/annnnswer.jpg

 

[I like Serena]

What did you get for "p"?

 

[Femme_physics]

Hey ILS

What did you get for "p"?

To what question? Only in the last question they refer to "P", and I wrote my answer in this scanned paper.

 

[I like Serena]

Hey ILS

Hi :shy:

To what question? Only in the last question they refer to "P", and I wrote my answer in this scanned paper.

In your diagram a point is marked with the letter "p".
The corresponding part in your formula seems a little off.

 

[Femme_physics]

In your diagram a point is marked with the letter "p".
The corresponding part in your formula seems a little off.

It's just a point in the diagram, it has no relevance to the formula

 

[I like Serena]

It's just a point in the diagram, it has no relevance to the formula

Your formula contains $\overline{b}cd + \overline{d}$.
This corresponds to the result of the OR gate "P" on the signals at p and at q, which is $p+q$.

This means that effectively you have $p=\overline{b}cd$ and $q=\overline{d}$ in your formula.
The expression for q is correct, but the expression for p is not.

 

[lectrohed]

N=Logic NOR

 

[Femme_physics]

Your formula contains $\overline{b}cd + \overline{d}$.
This corresponds to the result of the OR gate "P" on the signals at p and at q, which is $p+q$.

This means that effectively you have $p=\overline{b}cd$ and $q=\overline{d}$ in your formula.
The expression for q is correct, but the expression for p is not.

Sorry I must've missed this post.

Is there any other posts I've missed for over a week?

To reply: I don't think the actual signals in q and p matter at all. Only if they are one or zero. if we know one of them is one, then the other's irrelevant. Doesn't it make sense?

 

[NascentOxygen]

In part (2) you have not recognized that Z is an inverting gate, it's NAND.

 

[I like Serena]

Is there any other posts I've missed for over a week?

You've missed:
https://www.physicsforums.com/showthread.php?p=3827895#post3827895
https://www.physicsforums.com/showthread.php?p=3835966#post3835966
https://www.physicsforums.com/showpost.php?p=3793606&postcount=34

 

[earlofwessex]

I don't think the actual signals in q and p matter at all. Only if they are one or zero. if we know one of them is one, then the other's irrelevant. Doesn't it make sense?

but we don't know what they are, as they are dependent on the input, so we can't write either one off as irrelevant. whether or not you care about what the signal is, in order to state that the output of P = /bcd + /d, as you have done in your answer to clause 2, you need to be sure that those equations are correct.
(which they are not, and that has caused you to answer part 3 incorrectly)

(note P the gate ≠ p the signal)

edit: just realized you may be talking about clause 4, in which case you are correct, however I Like Serena was pointing out a mistake in clase 2.
the labels p and q are the signal inputs to the gate P.

 

[Femme_physics]

edit: just realized you may be talking about clause 4, in which case you are correct, however I Like Serena was pointing out a mistake in clase 2.
the labels p and q are the signal inputs to the gate P.

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!

NOW I understand :)

So let me get back to the first reply :P

What did you get for "p"?

That would be
http://img442.imageshack.us/img442/9803/peepee.jpg Ahh...I see :) I forgot a big tag over everything. *smacks forehead*

And now to correct the next question based on that:

http://img855.imageshack.us/img855/1918/abc800.jpg

You've missed:
https://www.physicsforums.com/showthread.php?p=3827895#post3827895
https://www.physicsforums.com/showthread.php?p=3835966#post3835966
https://www.physicsforums.com/showpost.php?p=3793606&postcount=34

Thanks, I'll get on it :)

 

[I like Serena]

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!

NOW I understand :)

So let me get back to the first reply :P

Ahh...I see :) I forgot a big tag over everything. *smacks forehead*

And now to correct the next question based on that:

All good now.

Thanks, I'll get on it :)

You still seem to have missed the last one!

 

[Femme_physics]

All good now.

You rock!

You still seem to have missed the last one!

Heh, on yea...that.. as soon as I fix my hair :) Thank you

 

[space_buster]

In X we get b' = 0
In Y we get a.b = 1.1 =1
In Z we get (b'.(c.d))'=b+(c.d)'=b+c'+d'=1+1+1=1 DeMorgan's Law (a.b)'=a'+b'
In N we get (d.d)'=d'=1
In P we get b+c'+d'+d'=b+c'+d'=1+1+1=1
In R we get 1+1 , now since inputs are even in number , XORing gives result 0 , i-e f=0
Hope it helps you ...

 

[space_buster]

Note here you mentioned N here to be NAND , its NOR but for this case the answere will come the same ...
In N we get (d+d)'=d'=1