Implement the following function with NOR gates:
F = A’B’C + ABC’
Answers
Answer:
). I’m going to use Option #1 because that’s the way I like it. If any students are reading this, I suggest that — after we’ve finished — you redo everything from this point on using the Option #2 Karnaugh map to ensure you really understand the process.
Looking at the Option #1 map, observe where we show “AB”. To the right are the four combinations of 0s and 1s that we can have associated with the AB inputs: “00”, “01”, “11”, and “10”. It’s very important to observe that these combinations are presented in such a way that they form a Gray code, which means only a single bit changes as we move from one value to an adjacent value. This is key to the way in which the Karnaugh Map works.
In the binary code, when we transition from 01 to 10, two bits change. By comparison, if we look at the corresponding transition in the Gray code (01 to 11), only a single bit changes.
Furthermore, consider the very last line of the binary code. If we were to “wrap around” and transition from this line to the first line (11 to 00), then — once again — two bits change. However, if we look at the corresponding lines in the Gray code (10 to 00), we see that — once again — only one of the bits changes.
OK, so now let’s populate our Karnaugh Map. We do this by adding 1s into each of the boxes corresponding to the product terms in our equation. Once again, we will do this step-by-step as illustrated below (the small numbers 1 through 6 in the circles correspond to the product terms in our original equation):
Another point to note here is that we don’t need the truth table in order to populate the Karnaugh map. All we have to do is stroll along our equation and, for each product term, we add a 1 into the corresponding Karnaugh map “box.”
The next stage is to use our Karnaugh map to minimize the logic. Looking at the final Karnaugh map (f) above, we immediately see that we can reduce this to only three terms. As usual, let’s do this one step at a time.
Observe the two 1s highlighted in the image below. We know that this means the output is 1 in both these cases. For each of these boxes, A = 0 and C = 1, so these values are important. However, B = 0 for one of these boxes and 1 for the other box. What this says is that so long as A = 0 and C = 1, we don’t care whether B is 0 or 1.
Next, let’s look at the second group of two 1s highlighted in the image below. For each of these boxes, A = 1 and C = 0, so these values are important. Once again, however, B = 0 for one of these boxes and 1 for the other box. What this says is that so long as A = 1 and C = 0, we don’t care whether B is 0 or 1.
Last, but not least, let’s look at the group of four 1s highlighted in the image below (one of the tricks about Karnaugh maps is that we can use the same 1s as part of multiple groups).
In this case, for some of these boxes A = 0, for others A = 1, which means we don’t care whether A is 0 or 1. Similarly, for some of these boxes C = 0, for others C = 1, which means we don’t care whether C is 0 or 1. In fact, the only input that is constant across all for boxes is B, which is always 1.
What all of this means is that we can use the results from our Karnaugh map minimization to write an optimized sum-of-products equation as follows:
From this, we can easily draw a corresponding gate-level schematic using NOT, AND, and OR gates as illustrated below:
Before you email me, I know that we’re not using the !B signal (I’m using the ‘!’ character here to indicate NOT(B) because I can’t draw a horizontal line over the letter in this text), but we will be using it in the not-so-distant future. Speaking of which, the future is closer than you might think. This is the point where we have to put our thinking caps on, because the task posed by the loathsome lecturer was to present the final circuit using only NAND gates or only NOR gates.
Implementing NOT gates using NANDs or NORs
Let’s start with the low-hanging fruit, which would be the three NOT gates in this example. First, let’s remind ourselves that the truth tables for the five common primitive gates are as illustrated below:
What this