Hello everyone !

So i have to turn this circuit that is made of nand gates into the equivalent of this but this time only with NOR gates. This is the circuit :




First i tryed to turn the circuit into some boolen algebra statement but nothing happned, actually it became harder =/ i am strungling trying to solve this loads hours ....help me please...

 

Take a look at DeMorgan's Theorem.

hgmjr

 

Take a look at DeMorgan's Theorem.

hgmjr

Well ok De morgan's theorem, but i need an expression 1st for the circuit, there is where i am strungling , i am getting confused all the time....

This is stupid but anyway... my conclusion is: (((((pp)'p)')'(rr)')')'

this is not right, right?

 

Well ok De morgan's theorem, but i need an expression 1st for the circuit, there is where i am strungling , i am getting confused all the time....

This is stupid but anyway... my conclusion is: (((((pp)'p)')'(rr)')')'

this is not right, right?

First observation is that there is no q in your boolean expression.

Have you noticed that four of the nand gates are configured as simple inverters?

hgmjr

 

First observation is that there is no q in your boolean expression.

Have you noticed that four of the nand gates are configured as simple inverters?

hgmjr

indeed theres no q, it was meant to be like this (((((pp)'q)')'(rr)')')'

but still not sure if this is .

 

HINT: If you feed the same signal into both inputs of a NAND gate, all that is doing is inverting the input.

hgmjr

 

HINT: If you feed the same signal into both inputs of a NAND gate, all that is doing is inverting the input.

hgmjr

yeah i know that, thats why so many '''''' xD

ill try again and post

 

Using your boolean notation, I get simply:

(p'q)r'

because (pp)' = p'

hgmjr

 

i tryed again in a bit different way,

isnt it ( (p' NAND q)' NAND (r') )' ?

 

i tryed again in a bit different way,

isnt it ( (p' NAND q)' NAND (r') )' ?

That looks correct although it would be clearer if some simplification were performed.

hgmjr

 

hmmmm i tried to simplify it a bit, not sure again if its right

( (p' NAND q)' NAND (r') )' =

((p' q) nand (r') )' =

(((p' q) (r') )' )' =

((p+q') + r)' =

(p'q) r'

i believe theres a mistake somewhere

 

If you take a look at post #8 in this thread, you will see that I got the same simplified answer you have in post #11.

I guess your next step is to implement this boolean expression using NOR gates. Right?

hgmjr

 

If you take a look at post #8 in this thread, you will see that I got the same simplified answer you have in post #11.

I guess your next step is to implement this boolean expression using NOR gates. Right?

hgmjr

yeah as long as this result is ok for the given circuit, ah btw forgot to mention that i hd already proved that p'q = p nor (p nor q) , the problem is the r' now, should i call for example p'q=d and say r'd = r nor ( r nor d) ?

Thanks in adance

 

My thought was that if you needed an inversion, you would feed the same signal into both inputs of a NOR gate just as was done in the NAND gate version of the circuit.

hgmjr

 

My thought was that if you needed an inversion, you would feed the same signal into both inputs of a NOR gate just as was done in the NAND gate version of the circuit.

hgmjr

indeed i could feed a nor gate with 2x r so the result is r' , but still theres a multiply , how can do that as a nor gate . i mean --> (p'q) multiply r'

 

You do realize that there is not an actual multiply operation involved. That is simply the indication of an AND function.

hgmjr

 

See if the information at this website helps you out.

hgmjr

 

You do realize that there is not an actual multiply operation involved. That is simply the indication of an AND function.

hgmjr

ehm lol yeah, i think i finally figured out, ill post the scematic as soon as possible

 

nor version of this (p'q) r' is :





hopefully

EDIT: i think that the second NOR (where r entrance is ) is not needed tho.

 

Your diagram has a couple of errors. Take another stab.

hgmjr