Qfyilyi

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?

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.

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

^{simulate this circuit – Schematic created using CircuitLab}

The rest is up to you.

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:

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

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 $T=A,C+B,overline{C}$ is the simplified version.

The above isn't the only, nor even the best approach to simplification. It's just one I decided to write out, off-hand.

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:

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

At this point, you can see that if $T=overline{R+S}$ then $R=overline{A}left(overline{B}+Cright)$ 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.

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.

^{simulate this circuit – Schematic created using CircuitLab}