Hello, I am trying to grasp this topic.
I have the following expression (see attached image).
I need to
i) simplify using deMorgans theorem
ii) then show how the expression could be implemented using only 2 input NOR gates.

I "think" I have the deMorgans simplification correct, but I am not sure how to do part ii.
Any help would be much appreciated.

 

write the equation so it has two OR gates, then negate the entire equation

 

thanks, but that is the problem.
I am not sure how to rewrite my deMorgan simplification as two NOR gates.

Can anyone please help with this ?

 

double negate the entire equation so you can deMorgans again to switch the gates

 

You used DeMorgan's theorem incorrectly. When you use DeMorgan's theorem, you split up the "not" bar and change the operation. For example:

(A+B)' = A'*B'

The ' symbolizes a "Not" bar. As you can see, the entire first part (A+B) is inverted, so you have a bar above all of it. You split the bar so that each part has the bar over it, and then you change the operation from OR to AND (+ to *).

EDIT: Sorry, you are correct. I misread the expression

 

You actually only need a single 2-input NOR gate.

Your simplification is wrong.

This is what you started with

J = [(G'*H')' + (H*I')]'

Applying DeMorgan's gets you

J = (G'*H')*(H*I')'

Applying DeMorgan's to the second term gets you

J = (G'*H')*(H'+I)

But at this point you just removed the parens around the second factor. You have to distribute one over the other.

Now take a step back and look at your original expression

J = [(G'*H')' + (H*I')]'

This is ALREADY a NOR gate!

J = NOR[(G'*H')', (H*I')]

So focus on the guts of this expression and in three easy steps it will reduce to the OR of just two terms.

 

You actually only need a single 2-input NOR gate.

Your simplification is wrong.

This is what you started with

J = [(G'*H')' + (H*I')]'

That is not correct. The bar goes over the entire G*H expression. It is NOT G'*H', it is (G*H)'.

The OP's simplification is done correctly.

 

Oops. That's what happens when a nearly blind man that sees double and triple images has to switch between windows and remember what lines he saw where.

It's a shame, too, because my incorrect version simplifies so beautifully!

But the OP's simplification is still wrong. The mistake I made does not affect the need to distribute one factor over the other.

So hopefully this is the correct starting point.

J = [(G*H)' + (H*I')]'

This is still a NOR gate at the outermost level.

Applying DeMorgan's

J = (G*H)*(H*I')'

Applying DeMorgan's to the second term

J = (G*H)*(H'+I)

The OP has this

J = G*H*H'+I

Which is wrong, since AND takes precedence over OR and this would simplify to J=I

Instead, what you have is

J = (G*H)*(H'+I)

J = (G*H)*H'+(G*H)*I

J = G*H*I

 

Applying DeMorgan's to the second term

J = (G*H)*(H'+I)

The OP has this

J = G*H*H'+I

Oops, you're right. Good catch!

 

A quick count, assuming I am correct that J=GHI, is that it will take six NOR2 gates.

 

Oops, you're right. Good catch!

Bwtween us we'll get it right!