디지털 시스템: 이진수 표현의 이해

### 이진수 시스템: 소개
* 🔢 "전압 측면에서 논리 1과 논리 0의 의미"와 **임의의 논리 게이트에 대한 논리 1과 0의 특정 전압 범위**에 대해 논의했습니다.
* "나머지 과정에서는 이 전압 범위를 **0으로, 이 전압 범위를 1로 취급**하고 다양한 형태로 1과 0에 대해 이야기할 것입니다."
* 🎚 "이번 주에는 **논리 1과 0의 수학적 프레임워크와 그 표현**에 대해 알아볼 것입니다."
* ⚖ "저울의 1kg 추를 사용하여 각 위치가 **2의 거듭제곱에 해당하는 정수**를 표현하는 방법에 대해 논의했습니다."
* 📊 "이진수 시스템은 0 또는 1 값을 가질 수 있는 **2진수 값인 비트**를 사용합니다."
* 🔟 "십진수 체계에서 숫자는 **10진법으로 표시**되며, 각 자릿수는 0에서 9까지의 값을 가질 수 있습니다."
* ✨ "수학적으로 **십진수와 이진수 체계는 모두 숫자의 동등한 표현**"이므로 일관된 산술 연산을 가능하게 합니다.

디지털 시스템에 대한 강좌에서는 전압 범위를 통해 논리 1과 0의 물리적 의미와 수학적 프레임워크를 설명합니다.

[Music] [Music] Hello all and welcome back to this course on digital systems. So in the last week, we looked at the meaning of a logic one and a logic zero quite elaborately, right? In the sense, we looked at what does the logic one mean in terms of voltages and what does a logic zero mean in terms of a voltage. Again, so we discussed that there is a certain range, you know, from Z to VIL. So for example, logic Z, any voltage, right, to a logic gate, if is greater than zero or less than or equal to VIL. Okay, or logic 1, the input must be greater than VI and must be less than or equal to the supply voltage, right? So this is for any arbitrary logic gate. UI used the inverter in the last case to take you through the transfer characteristics and so on, right? So U, so having established the physical meaning of these logic levels, we can now simply treat and this voltage range as zero, this voltage range as one. So in the rest

of the course, we will only be talking of ones and zeros, right, in various forms, uh, in terms of the truth table, in terms of mathematical operations, everything. We will be talking of ones and zeros but at a voltage level. This is what we will eventually mean, okay. So, in this week, I plan to get get into the mathematical framework. So, the module one of the course, which is basically the mathematical framework, all right, uh. So, the logic 1 and zero U also have a very specific reason why that works mathematically and this week we will discuss that in great detail. So, again, to Let me refresh your memory with respect to what we had discussed in the electronic system, think electronic systems thinking course, right, uh, and this had to do with the representation of an integer. So, from the ADC, we got an integer and we wanted to represent them as ones and zeros because ultimately the computer is going to Now understand ones and zeros through these voltages through these voltage ranges, right, and therefore any integer that is

given to us, we must be able to convert it to a binary representation. Okay, and it was in this context that I had discussed that suppose you have a weighing scale like this. Not a weighing scale, a weighing, um, you know, balance, right? So I have something like this, right, and there is a central point at which you can hold it. Okay, and now I'm going to place an object here, okay, which is anything between 1 kg, right, to 15 kg, but only in integer steps. So it could be 1, 2, 3, so on up to 15. Okay, now you do not know what the weight of this object is, and what you are only given is, let's say, four different weights, each of 1 kg. Okay, this is 1 kg, 1 kg, and I have four such weights with me. Now I have to figure out what is the exact weight of this unknown red object that is placed on this side using four such 1 kg weights. So we had discussed that the one way to do this is

you know, first of all, if you are given, um, you know, not just 1 kg weights, but you were given 1 kg, 2 kg, and so on, then you had to go in powers of two in order to be able to measure every integer between 1 and 15, right? Of course, zero is very trivially achievable, right? Now, in order to use only 1 kg weights, we said we have to construct a very special balance, which is like this, right? I have here, okay, so effectively, if this is distance X, right, then this is also distance X, right, on which I can place a 1 kg. Then at 2x, I have another balance, right, and let me move that, um, here, this thing here, okay, and I have then, of course, so this is another X, okay, then I have another at 4X, okay, I need to have a balance like this, which is going to sit here, okay, and from the center, this balance is going to be 4X, okay, then I have another thing all the way quite far away, which is going to

be at a distance of 8X. Okay, the last one is not to scale, so I'm saying this is another 4X. Okay, in terms of distance. So, what happens is if I place a 1 kg block here, the effective weight that is going to see with respect, uh, the effective, uh, torque, if you, if you wish, right, or the moment, right, in physics parlance, that would, it would experience the weighing balance would experience on the right hand side, is going to be 2X into 1 kg. Now, on the other hand, if I place the 1 kg here, right, and I'll just take this off, then it would experience a moment of 4X into 1 kg, right. Likewise, on the last thing, if I place it, it would experience a moment of 8 into 1 kg. Okay, so now in order to identify what this unknown weight is, what would you do? We would start with, let's say, the largest possible weight that I can measure with just 1 kg. I place it on the last balance.

so let me call uh let me assign some numbers here. Okay, I'm going to call this 0, 1, 2, and 3. Okay, on the right hand side. And because it's on the right hand side, I'll call it R0, R1, R2, and R3. So I place one of these one kg blocks on R3, and I see if it is heavier or lighter. Now, of course, it could be equal, right? If it is equal, then I will just say the weight of the unknown object red object is just 1 kg into 8. So it's an 8 kg block. Okay, but the block that I have placed on the left hand side, the red block could be heavier, right? Now, if it is heavier, then what I need to do is I need to go ahead and now place another block here. Okay, so the total weight that is appearing because of the moments on the right hand side is basically two. It is 4X into 1 kg, okay, plus 8X.

into another 1 kg. The net weight because of these two blocks is going to be like this. Effectively, it will be 4 + 8 = 12 kg. Even though I'm placing only a 1 kg block on each of them right now. You look at the weighing balance again. Suppose it is equal. Then the weight is equal to 12. Now suppose it were not equal. Okay, then what do I do? Then I need to say, let's say this time after I placed the 4, I mean the one 1 kg on R2. Let us say that the right hand side was heavier. Okay, in which case, obviously I cannot be placing the block 1 kg block on R2. Therefore, I will take this off and then place it on R1. Okay, now if it is placed on R1, then the net weight that I'm going to see because of one 1 kg on R3 and 1 kg on R1 is going to be 2x. Right, this thing will be 2x into 1 kg plus 8x into 1 kg.

so effectively, it is like 8 + 2. So, it is 10 kg as opposed to 12 kg. Right. Now, let's say that the red block this time happened to be in the earlier case. The red block was lighter. Let's say this time it is heavier. It is heavier, right? So, what we know till now, right? What are the, we did three weight measurements. Okay, so I will write that down here. Step one, okay, we did three weight measurements. Step one, we compared the weight against 8 kg. Right, and the weight was greater than 8 kg. So, this much we know, right? In step two, we said the weight was, we added one more to R2, right? And therefore, the effective moment that we saw was 8 + 4, 12. But we then saw that the right hand side was heavier. That means the weight is less than 12 kg. Okay, in step three, what we saw is by placing the 1 kg on R1, right? What we are able to see here is that

the left-hand side, the red block is now heavier than this. Okay, so what we infer now is W is greater than 10 kg. Okay, actually, you look at step two and three, you will, you will actually know what the integer weight should be. It is greater than 10, but it is less than or equal to 12, right? And therefore, it has to be 11. So how do you do that? We know that the left-hand side is now heavier, so I will place an additional block of 1 kg on r0. Okay, so because of this, the net weight in step three, okay, step three, the net weight is 8 + 2 + 1. There's nothing on r two, okay, this one is because of step two, and this is step one. Is that no, this was step two, right? This was um step three, and this step four, okay, so what we are saying now is step 4, the weight will therefore be exactly equal to 11 kg. Okay, so what we have effectively been able to

do right is represent this information of the weight of the unknown red object as 4 1 kg weight s with a certain weighted addition. Okay, so what we see here is in each of you know the balances on the right hand side you can either place a 1 kg block or there is nothing there. This corresponds to the one and zero that we know. If the weight is there, it means it's one. If the weight is not there, it is zero, right? So therefore what we are now effectively we have done right is we took the unknown object W okay, which was an integer and we converted this okay in blue to let me just go back r0, R1, R2, R3. Okay, I'm going to write it in um in right to left for a certain reason. Okay, r0, R1, R2, sorry, and R3. Okay, so r0 is basically saying look that if that object so each R for example r i could be either one of zero or 1 okay so the effective weight w okay in

red I can represent it as R 3 into 8 + R 2 into 4 + R 1 into 2 + r 0 into 1. Okay, and that's exactly coming out of the fact of the way this balance has been constructed, right? R 0 is at distance X, exactly the same distance as the uh weight that is being placed on the left-hand side. R 1 is being placed at a distance 2x. R 2 is being placed at a distance 4X. R 3 at a distance uh 8X, right? So, therefore, what we can do now is to to rewrite this as R 3 into 2^ 3 + R 2 into 2^ 2 + R 1 into 2^ 1 + r 0 into 2^ 0. Okay, now each of these RIS effectively can take only one of two values, zero or one, either the weight is there or it is not there, and therefore these are essentially binary valued numbers. Okay, this is a binary number, and a binary number can take only one of two values, zero or one, right? U

and because they are binary numbers, they are also called a bit. Okay, so in fact, because they are called bits, these RS are actually represented by, represented by, let's say B 0. Okay, B 0 2^ 0. Okay, just to say that it's a bit B1 into 2^ 1 + B2 into 2^ 2 + B3 into 2 power 3. Okay, so I can write this as summation b k 2^ k, k = 0 to 3, right? So effectively any integer, right, if you follow this algorithm of placing the weight, you know, on the right hand side, on the right extreme, and then if it is heavier, you follow a certain thing, remove the weight, if it is lighter, then you add another weight, and so on, right? If you continue doing this, you can represent any integer on the left hand side of the balance with a four binary bits on the right hand side, which basically are, you know, these one 1 kg weights, a combination of these 1 kg.

weights, right? So, uh, this is exactly what we had, I had mentioned in my initial introductory lecture as the binary number system. So, if you look at this, right, this is nothing but the binary number system, okay, where each bit can take either zero or one, and you're doing a weighted addition in terms of the powers of two, okay? So, compare this to the decimal system, decimal number system, okay, where the number can be represented in the base 10, okay? So, any number, if you take this W, the same W that we are talking about, it can be represented in terms of, let's say, let's say that the weight was 12 kg, okay? What does 12 mean? It is actually 1 into 10^1 + 2 into 10^0, okay? Now, if I had more than 15 kg, let's say I had a three-digit, I could have up to 120, 125 kgs, okay? What does 125 mean? It is 1 into 10^2 100 + 2 into 10^1 + 5 into 10^

0. Okay, so this one is getting represented as K equals to 0 to uh let's say 3 or 2. Okay, um some number a into 10 power K where a is 0, 1 all the way to 9 any integer between 0 and 9. Okay, so in the decimal number system, the moment you exceed 9 you go to the number 10, right? You will add an extra digit. All right, on the other hand side in the binary number system because you are dealing only with zero and one you go from count from zero, you reach one. The moment you reach one, the next number has to be 1, 1, and 0, right? That's the equivalent of the number two, right? So mathematically, both these number systems are equivalent and you can work with them as the as identical representations of the same number, right? Maybe not identical, they are different representations, but they are, yeah, maximum I can say is they are equivalent representations and therefore I can do addition with it. I can do multiplication with it. I can do whatever I want with it.

and the answers will come out to be consistent in that particular number system. [Music] Right.

W2_L1_Binary Number system

IIT Madras - BS in Electronic Systems

디지털 시스템의 논리 수준과 수학적 프레임워크 이해하기

논리 1과 0은 특정 전압 범위에 의해 정의되며, 이제 이 과정에서는 이러한 범위를 모든 논의와 작업의 기초로 취급하는 데 중점을 둘 것입니다.

다음 모듈에서는 정수를 이진 표현으로 변환하는 데 중점을 두고 논리 1과 0의 수학적 프레임워크를 살펴볼 것입니다.

전자 시스템 사고 과정에서 정수를 2진수로 표현하는 개념을 설명하기 위해 저울과 표준 추를 사용하여 미지의 물체의 무게를 측정하는 과정을 사용했습니다.

1kg 블록을 사용하여 알 수 없는 무게 측정하기

1kg 블록만 사용하여 미지의 무게를 측정하려면 2의 거듭제곱에 따라 1kg 블록을 배치하는 거리를 늘려 특수 저울을 만들 수 있습니다.

저울에 1kg 블록을 서로 다른 거리에 배치하여 유효 무게와 모멘트를 계산하여 미지의 무게를 식별할 수 있습니다.

이 과정에는 저울의 오른쪽 모멘트를 비교하고 위치를 조정하여 미지의 물체의 무게를 결정하는 과정이 포함됩니다.

빨간색 블록의 무게는 저울의 다른 쪽에서 일련의 무게 측정과 조정을 통해 11kg으로 결정됩니다.

저울의 추를 체계적으로 추가하고 제거하여 빨간색 블록의 무게가 10kg보다 크지만 12kg 이하라는 결론을 내립니다.

이 과정에는 추를 8kg과 비교하고, 저울을 조정하고, 어느 쪽이 더 무거운지에 따라 추론하는 과정이 포함됩니다.

가중치를 이진 표현으로 변환하기

미지의 빨간색 물체의 무게는 이진 시스템을 사용하여 표현되었으며, 각 R(R3, R2, R1, R0)은 2의 거듭제곱(2^3, 2^2, 2^1, 2^0)에 해당합니다.

저울의 무게추 배치는 이진 표현 패턴에 따라 거리(X, 2X, 4X, 8X)에 대응합니다.

이 이진 표현은 각 R 값이 이진수와 유사하게 0 또는 1만 될 수 있다는 개념을 기반으로 합니다.

비트로 표현되는 이진수는 2의 거듭제곱을 사용하는 가중치 더하기 시스템을 따릅니다.

이 시스템을 사용하면 이진 비트의 조합을 사용하여 모든 정수를 표현할 수 있습니다.

이와 대조적으로 십진수 시스템은 10의 거듭제곱을 기본으로 하여 유사한 가중 덧셈 프로세스에서 숫자를 표현합니다.

십진수와 이진수 체계는 모두 숫자를 나타내지만 기호와 계산 방식이 다릅니다.

십진수 체계에서는 숫자가 9부터 한 자리씩 증가하는 반면, 이진수 체계에서는 1부터 한 자리씩 증가합니다.

이러한 차이점에도 불구하고 둘 다 일관된 결과로 수학적 연산에 사용할 수 있습니다.