MHB Analysis. Fill in the F column values in the truth table for the circuit.

[shamieh]

Fill in the F column values in the truth table for the circuit.
Need someone to check my work.

View attachment 1444

My Answer:

• x y z | f | x! and y! | x! XOR z|
• 0 0 0| 0 | 1 | 1
• 0 0 1| 1 | 1 | 0
• 0 1 0| 1 | 0 | 1
• 0 1 1| 0 | 0 | 0
• 1 0 0| 0 | 0 | 0
• 1 0 1| 1 | 0 | 1
• 1 1 0| 1 | 1 | 0
• 1 1 1| 0 | 1 | 1

 

[Ackbach]

So, let's define the $\uparrow$ symbol as the Sheffer stroke, which corresponds to the NAND logic gate you have there. Then $f=(x \uparrow y) \oplus (x \oplus z)$. The truth table is then
$$
\begin{array}{c|c|c|c|c|c}
x &y &z &x \uparrow y &x \oplus z &(x \uparrow y) \oplus (x \oplus z) \\ \hline
0 &0 &0 &1 &0 &1 \\
0 &0 &1 &1 &1 &0 \\
0 &1 &0 &1 &0 &1 \\
0 &1 &1 &1 &1 &0 \\
1 &0 &0 &1 &1 &0 \\
1 &0 &1 &1 &0 &1 \\
1 &1 &0 &0 &1 &1 \\
1 &1 &1 &0 &0 &0
\end{array}
$$

 

[shamieh]

Okay, I see where I went wrong. But you're saying x NAND y in the first row is 0 NAND 0 which is really 1 AND 1 = 1. I get that. Then you are saying in the 2nd row the same thing 0 NAND 0 = 1 because it's really 1 AND 1. Then in the 3rd row you are saying 0 NAND 1 which is really 1 AND 0 = 1. But how does that = 1? 1 AND 0 = 0. It only equals 1 if both are 1.

 

[Ackbach]

Okay, I see where I went wrong. But you're saying x NAND y in the first row is 0 NAND 0 which is really 1 AND 1 = 1.

Actually not. The NAND operation means "not both". If I say $x \uparrow y$, or $x$ NAND $y$, that is equivalent to $\overline{xy}$. By DeMorgan, $ \overline{xy}= \bar{x}+ \bar{y}$, not $ \underbrace{\overline{xy}= \bar{x} \bar{y}}_{\text{Wrong!}}$.

I get that. Then you are saying in the 2nd row the same thing 0 NAND 0 = 1 because it's really 1 AND 1. Then in the 3rd row you are saying 0 NAND 1 which is really 1 AND 0 = 1. But how does that = 1? 1 AND 0 = 0. It only equals 1 if both are 1.

Again, this reasoning is flawed. If you need to, calculate NAND's like this: to compute $x \uparrow y$, first compute $xy$, and then negate the result. You cannot compute NAND's by negating the $x$ and $y$ first, and then AND'ing the results.

 

[shamieh]

Ahh! I see! So DeMorgans law changes the ANDing of xy to OR, while negating the terms as well. x! + y! ok. Going to re-work it and make sure I get the correct solution. I'll be back (Wait)

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Awsome, thank you for the detailed explanations. (Sun)