[Music] [Music] So we looked at, you know, what happens when there are ones okay in the output of the truth table. Now, is it possible to look at the zeros and actually arrive at a similar expression, right? So let us look at that also. So let me take, maybe, let me take the same example that I started with. I'm going to start with, um, I'm going to look at A2, A1, A0. I'm going to look at Y0. Okay, 0 0 0 0 0 1 0 1 0 1, right? So what is Y0? We said it was just the inversion of A0. So it was going to be like this: 1 0 1 0 1 0. And of course, we looked at the min terms. But now, can I implement a Boolean logic or the Boolean function? Can I express it in terms of the zeros as well? Okay, so now let us look at what happens here. You see this, there are four

zeros. Okay, now if I were to implement, you know, the Boolean function using these zeros, I first need to understand what I did when I implemented it using the ones, okay? So let us look at the meaning of these Min terms. The Min terms was basically a product, right? It basically said A2 bar A1 bar a not bar, okay? And then this was A2 bar A1 and a bar, okay, something like that. And then we wrote it as a sum, right? Now, what does the and condition mean, right? The product, what does that mean, right? It basically says that the product can be one only if each term in that product, in each term of that Min term, happens to be a one because it's an and gate that we are implementing there, right? So, therefore, if I want to deal with the Min terms, which is a product, I am I have to deal with an and gate and where basically I'm saying every term has to be one in order to be in order for the minterm to be one, right? And the

next thing that we did was to do sum of products. We took those products and put them through an OR gate. So, what does that mean? It means that the final sum could be one if any one of those minterms happen to be one. And that's exactly what we want, right? If any of these combinations appear at the inputs, we want the output to be one. For why not to be one, that's exactly the definition of the truth table. Now, if I want to do the same thing with zeros, what should I do? Okay, so the zero is, you know, slightly, you know, it's the dual of the one, right? So in the minterms, we were able to consider the product. In the max term, I'll have to consider the sum, not the product. Okay, for example, I will have to do this A2 bar. No, no, no. What is it? What is the value of the input that will result in a zero? Okay, so therefore, that has to be A2.

or A1 or a not bar. Okay, some uh let me write out, you know, the Min terms for the other cases also. So this is um A2, A1 bar, a not bar. The last one is A2, A1, a. Okay, now we got this thing here. Similarly for the max term, right, for the next zero we finished this. Now we looking at this, it has to be A2 or A1 bar or a not bar. Okay, and then we are saying this has to be A2 bar or A1 or a not bar. Finally, we are saying it has to be A2 bar, uh, sorry, this is wrong. This has to be A2, A1, a not bar, right? So the last Max term will be A2 bar plus A1 bar plus a, yes, a not bar. Okay, so in some sense we now have the cases, the input combinations where we know the output has to be a one. Okay, so we wrote one y not sop as this expression, A2 bar, A1 bar, a bar plus A2 bar, A1, a bar plus A2, A1 bar, a.

bar plus A2 2 A1 a not bar right uh this is what we got because uh so how do we how do I verify this it is A2 A1 bar A2 bar A1 yeah so this is right I am now going to write something known as the product of some expression right that is I can write the same y not okay as a product of sums each of these is a sum each of this is a sum all right it's an orgate and if I put all of them through an and gate what does it mean again we have to see what this means because of the sum we know that the input combination if it is that it will come out to be a zero now when when will the uh you know when will the output we want the output put to be a zero if any of these combinations occur right that's the definition again of the truth table or the functionality that we want to implement and therefore if I take these sum X terms and put them through an and

gate any one term being zero will actually result in the output being zero right so therefore I can write the same y not as a product of sums which is A2 + A1 + a bar into and get A2 + A1 bar + a bar into A2 bar + A1 + a bar A2 bar + A1 bar + a bar. These two representations are the same you have to get the same a not bar as the answer after you simplify this whole thing here right so um in order to do that you know you'll have to do a lot of uh Boolean simplification around here and all that but effectively the same expression output for y not can be it can it can be represented in two different ways either the sum of product or the product of sums so I leave this as an exercise do the Boolean simplification and show that you are able to get this so where will you where would one want to use the sum of product or product of sums right so let us look

at a very simple example of the and nand gate um Y and Y and okay so this is 0 0 0 1 1 0 1 1 okay. The and gate is like this. The nand is like this. Okay, now if I were to go ahead and do this, you know, using uh the sum of products in both cases, the and gate would give me the answer straight away because there is only one this a into B. Okay, on the other hand, nand gate, if I were to do it using the sum of product terms, we know that we had to start with three minan terms, right? M0, M1, M2, add them up, do Boolean simplification, and then show that is that is the same as a into B whole bar. Now what you can do is you can simply look at this zero. Okay, I will Express this in the product of some terms. So here I will do minterm and for that it is simply going to be AB. On the other hand, for the nand

I will do Max term, where I will simply look at the zero and what is the combination of the inputs when written as a sum that will give me a zero. Well, that has to be A bar + B bar, okay? And therefore, I can say Y nand is also equal to A bar + B bar. This is nothing but A and B whole bar by De Morgan's laws. So, in some sense, what we are saying is the two representations, sum of products and product of sums, are the same. All you have to do is, if you have more number of ones and very few zeros, sparsely populated zeros in the truth table, it may be better to deal with the product of sum implementation. On the other hand, if you have sparsely populated ones in the truth table, it may be better to deal with the sum of product implementations, right? Of course, if you embark on doing Boolean simplification to both of them, you will end up with the same answer, irrespective of what canonical representation you use, right? So, in summary, you know, you can

now, take a black box. Get the truth table and get the Min terms or the max terms. Simplify the whole thing and, you know, implement the Boolean expression with as few Gates as possible, right? And the gates are AND, OR, and INVERT, right? Inverter. So, with this, you can implement any arbitrary Boolean function that you have. That is given to you, right? And who gives it to you? That usually comes from the spec of a product you want to do this particular task. And therefore, from that, you can get the truth table. And then you proceed from there. So, in the next lecture, what we will do is we will look at some other techniques of, you know, instead of doing this Boolean simplification by hand and by expressions like this, we will look at alternate Boolean simplification techniques, which is known as the Karnaugh map technique. Thank you.

W4_L7_Canonical Products of Sum representation

Scripts