SPEAKER 1: air or the vacuum, and the wireless communication can be actually considered as the link between the port input and the port output.
Now, the only thing that I want you to know here, you see the dBm. The lower script m stands for one milliwatt. As you know from the physics classes that you have already taken in high school or after that, the power is measured in watt, the energy is measured in joules. And power is the rate of energy transfer. Any word including rate means with respect to time. Therefore, joule divided by second will be equal to watt.
Now our measurement always here for the power is in watt. Now, if I'd like to give the dBm, I have just a reference point. My power in will be always one milliwatt, that's where the m stands for. One milliwatt is thousandth of the watt, that means 1 divided by 1,000 or 10 exponent minus 3.
Therefore, here for example, if I have dBm power of an antenna, here there is an example I am giving you now here. What is the dBm power of a wireless antenna having a 200 milliwatt power transmission?
200 milliwatt divided by 1 milliwatt is 200. Therefore, let us work out the 200. How come I got this 23? Let me show you how I could get that one. Let me just get it here for you.
Now, I will just leave the 10 for now. Let me work out the log 200, and it will be equal, as you can see to log 2. Why 2? Because this is one of the two things I want to remember, log 2 is 0.3, almost, it is there. And log 3 is a 0. 477. Log 2, log 3, and we know log 1 is 0. Log 10 is 1. I needed only log 2, which is 0.3 and log 3, which is 0.477, OK.
Now, let me continue working out this. It is 2 multiplied by 100, and that will be equal to log 2 plus, as we know, because I am multiplying, log 100. And the answer will be obvious, log 2 is 0.3, roughly speaking. You can just make it more precise by using the calculator. Engineers are dealing with estimations.
That is 2, log 100 is 2. And the answer will be 2.3. But don't forget, I have to multiply that one. Therefore hence, 10 log 200, the log 200, I have it. Log 200 will be equal to 10 multiplied by 2.3, and that is equal to 23 dBm, OK? That will be equal to 23 dBm.
Here, that is the answer, actually, for the previous question as you saw it here. I wrote it directly, assuming that you will have. While you are practicing, maybe you will get a very nice way of creating things very quickly.
Now let me calculate the gain in decibel if the enclosed gain is 2. Now, we are given an enclosed gain as 2. Now this is a reverse way of doing it.
Now, I have 10 log 2. That, again, is 2. That means I am amplifying the signal, the gain is twice. I put 1, I get 2. I put 3, I got 6. I put 10, I got 20.
Why I'm insisting on the 2, because it is actually doubling the power of the signal. It is very important. The reverse of it, losing half of the signal. Therefore, either multiplying the signal by 2, gaining 100% more power or dividing it by 2, losing 50% of my signal. These two trivial boundaries are very important, as we will see now. Therefore, log 2 is 0.3 multiplied by 10 is three decibel.
Now, keep in your mind, this is according to most of the textbook, one of the benchmarks. Now, they tell you there are rules of the tenth and the rules of the three. Oh wonderful, where from I got that one? From this. Root of 10, when you multiply it by 10, usually, you are doing what? You are adding 10 decibels, as you can see.
Multiply any number by 10, in the multiplication, you're going to add 10 log 10, which is 10. Therefore, if you have a value of 10 log gain, it's equal to something. 10 log 10 of that gain will be equal to that something plus 10.
Now, I multiply by 2. Anything, think you multiply it by 2, you add to previous value 3 decibel. Multiply by another 2, you are there another 3 decibel, so on so forth.
Therefore, let me just do this thing here now with you. If we know that log 2, I just put the signal now, 2 means I am multiplying the input by 2, gaining. I input 1, I got 2. Then OK, I am adding 3 decibel.
What if I have 4 here? What did I do for this 2? I multiplied by 2. That means I am adding another 3 decibel. And the answer will be 6 decibel.
What if I have 8? I can see that if it is log 4, it is 6 decibel. Now look 4 multiplied by 2 is 8. Therefore, I'm going to add to the 6 decibel another 3 decibel. That will make it 9 decibel. And so on and so forth.
Now as you can see me here, I did the reverse. I am losing half of the signal. The power out is half of the power in. That means, what do I expect? Now it is the negative 3 decibel over there. You see now I'm multiplying by 2, I'm gaining, I am boosting the signal twice. I am gaining 3 decibel. I am losing half of the signal. I am changing the 3 decibel to minus 3 decibel.
We call this half power point. Very, very critical point in that.
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