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I'm having some trouble understanding how I can convert a boolean expression to a NOR-gate only expression.

What I'm working with looks like \$T = A B' C + A' B C' + A B\$. How would I go into implementing it with NOR gates?

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    \$\begingroup\$ You might want to start by minimizing the logic expression. It's not a requirement. But it often helps solve the problem faster, if you do that first. Then, ask yourself, "What is the template for a NOR gate?" There is some imagination involved. But it's not hard to just start with that template and work towards it. Can you show any thoughts or work? \$\endgroup\$
    – jonk
    Commented Jan 5, 2019 at 23:07

5 Answers 5

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NAND and NOR gates are universal. So one way to solve this problem is first reduce the logic using K-maps or whatever, then draw it out with AND, OR, and NOT gates. Then use bubble pushing identity techniques to convert the gates to the desired type.

schematic

simulate this circuit – Schematic created using CircuitLab

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I'll provide some direction to head. Since your question isn't so much about simplification, I'll do the simplification without any further ado. Then I'll provide a direction to head from that point but I'll leave some work you'll need to perform yourself.

Simplify

Let's simplify the expression:

$$\begin{align*} T &= A\,\overline{B}\,C + \overline{A}\,B\,\overline{C} + A\,B\\ &= A\,\overline{B}\,C + \overline{A}\,B\,\overline{C} + A\,B\,\overline{C} + A\,B\,C\\ &= A\left(\overline{B}\,C + B\,\overline{C} + B\,C\right) + \overline{A}\,B\,\overline{C}\\ &= A\,\overline{\overline{B}\,\overline{C}} + \overline{A}\,B\,\overline{C}\\ &= A\,B + A\,C + \overline{A}\,B\,\overline{C}\\ &= A\,C + B\left(A + \overline{A}\,\overline{C}\right)\\ &= A\,C + B\left(A + \overline{C}\right)\\ &= A\,C + A\,B + B\,\overline{C} \end{align*}$$

I'm not going to explain all the details above. It isn't the only, nor even the best approach to simplification. It's just one I decided to write out, off-hand. Still, you should try to take the time to understand each transition on your own. (Or not. It's up to you.)

From there, you can see that if \$A\$ and \$B\$ are both true, the expression \$T=A\,C+B\,\overline{C}\$ already captures the term, \$A\,B\$, as one or the other is picked up regardless of \$C\$.

So the simplified version is:

$$T=A\,C+B\,\overline{C}$$

Applying the NOR gate template

The basic model of a NOR gate, as I'm sure you know, is \$T=\overline{R+S}\$. That's the template. The question is, how do you take an arbitrary expression and make it conform?

I'll start you out:

$$\begin{align*} T&=A\,C+B\,\overline{C}\\\\ &=\overline{\overline{A\,C+B\,\overline{C}}}\\\\ &=\overline{\overline{A\,C}\cdot\overline{B\,\overline{C}}}\\\\ &=\overline{\left(\overline{A}+\overline{C}\right)\cdot\left(\overline{B}+C\right)}\\\\ &=\overline{\overline{A}\,\overline{B}+\overline{A}\,C+\overline{B}\,\overline{C}}\\\\ &=\overline{\overline{A}\left(\overline{B}+C\right)+\overline{B}\,\overline{C}} \end{align*}$$

(If it's not immediately clear, take note that in the first transition I used a double negation. The reason is that I know the basic NOR template requires a negation of the entire underlying expression and that a very obvious and simple way to achieve that is to double-negate the entire expression. This guarantees a negation that I can keep at the top, while applying the remaining negation to the underlying expression.)

At this point, you can see that if \$T=\overline{R+S}\$ then \$R=\overline{A}\left(\overline{B}+C\right)\$ and \$S=\overline{B}\,\overline{C}\$. So you are closer to an answer, now. But you have two new situations to resolve.

I'll solve \$S\$:

$$\begin{align*} S&=\overline{B}\,\overline{C}\\\\ &=\overline{\overline{\overline{B}\,\overline{C}}}\\\\ &=\overline{B+C}\\\\ \end{align*}$$

And that easily fits the model of a NOR with no additional work.

You get to resolve \$R\$, now. Once you walk yourself through that process, you should find a requirement for exactly five NOR gates. I've only shown you the first two of them. You get to find the other three.

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  • \$\begingroup\$ The original equation simplifies further than you suggest... to AC+BC’ \$\endgroup\$
    – vicatcu
    Commented Jan 6, 2019 at 0:54
  • \$\begingroup\$ @vicatcu If you READ the text I wrote, you will see that expression!!! And you will see me USE IT, too. Take a look, once again. You're not telling me anything new. \$\endgroup\$
    – jonk
    Commented Jan 6, 2019 at 0:58
  • \$\begingroup\$ yes, I see that now, meant no offense just missed it the first read through \$\endgroup\$
    – vicatcu
    Commented Jan 6, 2019 at 1:16
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Your Boolean equation has the NOT, AND, and OR operators in it. DeMorgan's Law says that you can perform an AND function with a NOR gate or an OR function with a NAND gate. If you tie the two inputs of a NOR gate together, what kind of function does that give you?

Since this looks like homework, I'll let you fill in the details.

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I know this question is old, but maybe this tip will help students in the future, as it has helped me.

When we're looking to represent a logic expression in terms of only NAND or NOR gates, it's useful to use the POS (maxterm) and SOP (minterm) forms depending on which gate you want it in.

For instance, if we want to "phrase" a logic expression in terms of only NAND gates, we can use a K-Map to minimize the expression into SOP form, and then apply DeMorgan's Theorem twice to transform the expression into one that is in terms of only NAND gates.

The same applies when we consider NOR gates. Simplify your expression into POS form, then apply DeMorgan's theorem to phrase it in terms of only NOR gates.

I've found that even for rather complex expressions this trick works beautifully and really makes the work easier, especially if you're tasked with doing the heft of the work using logic expressions (as opposed to graphically).

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I'll give you a hint. If you missed DeMorgan's Theorem in lecture, you need to look it up. Briefly, the following two gates are identical

schematic

simulate this circuit – Schematic created using CircuitLab

The rest is up to you.

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