 This definitions and concepts gives a set of common acronyms you'll see across this course and a few others to look up. If you can't remember them during the lecture, we scroll through. There's some common units, seconds, meters, bits, hertz, radians, watts and decibels. Remember them, what they're used for, some explanation, some common prefixes, and tarot through to picot. Remember them, we've been using them as we go. We probably won't use tarot too much or picot, but you still see them in practice. You see terabytes, of course, when you buy a hard disk. Picot, we do not see so much, but nice to know those prefixes. But now we want to look at decibels, dB. And on the way, look at logarithms. Because in fact, when we calculate a Shannon and Nyquist capacity, we calculate a log-base-2 of something. How do you calculate log-base-2 of 100 on your calculator? Pull out your calculator, your phone, and calculate log-base-2 of 100. You're allowed a calculator in the exam, but at this point you can try it on your phone. Calculate log-base-2 of 100 without using the internet. Don't look it up on Google or Wolfram or wherever. How do you calculate log-base-2 of some number? Anyone? Log-base-2 of 100? Right, you need to do a few steps to solve this. Your calculator doesn't have log-base-2. It's probably like this one. I have log, which is actually log-base-10. And even ln, your calculator may have that, which is log-base-e 2.718. We don't use that so much. What about log-base-2? Log of 100 is 2. This is base-10. 10 to the power of 2 is 100. Log of 1000 is 3. The log function on your calculator is usually in base-10, but we want to do it in base-2. A nice property of logarithms is that the log of a number in some base, we can calculate using a different base. Let's see how we do it. The log function in my calculator is base-10. Log in base-10 of 100 divided by log in base-10 of 2, the base that we want to use. I want to find log in base-2 of 100. The way to do it is take log in base-10 of 100 and divide that by log in base-10 of 2. It gives us 6.64. Another example to convince you, log in base-10 of 16 divided by log in base-10 of 2 is 4. Log in base-2 of 16 we know is 4. So this is how you do it in your exam. If you want to use your calculator, you need to use log base-10 because that's all your calculator has. You do a log base-10 of the number divided by log base-10 of the base that you want to normally. That property is the one at the bottom here. Here are some properties of logarithms. Log of any base m of x equals log of some other base n of x divided by log of that other base n of m. Remember that one. Very useful. The other properties there are nice to know as well. If you can't remember your logarithms, then just have a refresher reading through some of them. They become useful. We've seen an example of a ratio, signal to noise ratio. A ratio has a power level divided by another. A ratio has no units. It's dimensionless. There's no dimensions for it. Ratios we use in communication systems a lot, signal to noise ratio. We talk about gains and losses of system. We'll see them in a moment. Another way to express a ratio is in decibels. We'll see the way to convert and we'll see that decibels are much more common in communication systems because they're often easier to deal with. There's some examples in the slides here in this handout, but we'll do some on the screen and let you calculate a few as we go. Let's consider an example. Let's say our first one was the signal to noise ratio that we used in the previous example. We had 251. That was the example we used, which meant that the signal power is 251 times larger than the noise power. That's what SNR means, ratio of signal power to noise power. Let's put some numbers to them. Let's say if the noise was 1 watt, then what was the signal? I'll just keep it short as s. 251 watts. If the noise was 2 watts, then the signal would be 2 times 251 watts. It's just a ratio. We don't know the exact values of these, but we know that signal is 251 times larger than noise. If we had this case, then SNR would have been calculated as 251 watts divided by 1 watt. 251. SNR is a ratio of two power levels. We can express that in decibels. To express it in decibels, we take the logarithm in base 10 of that ratio and multiply it by 10. The general formula for dB, decibels, is 10 times log base 10 of the ratio. In this case of our signal to noise ratio. You need your calculator there. The log base 10, which your calculator does have, is the log button of 251. Someone tell me the answer. About log base 10 of 251. Log of 251. 2.399 something. 2.399. But multiplied by 10 out the front, so it becomes 23.99, which is about 24. 24 decibels. And you see in your lecture notes, that's where the question was expressed in dB, but I gave you in the absolute value. What that means is a ratio of 251, something is 251 stronger than something else, means is the same as 24 dB, 24 decibels. So we use decibels to talk about ratios. Not just signal to noise ratios, ratios of any two things, and especially power levels. We'll give an example about an audio amplifier. It has an audio input in, and it amplifies the audio and gets some output. We can talk about the ratio of the input and output and express that as dB. So 251 is the same as 24 dB. Just different expressions of the same value. It turns out when we deal with communication systems, you'll see the ratio is expressed as decibels, rather than the absolute ratio, because it turns out they're usually easier to deal with in calculating values, and we'll see a few examples of that. Okay, take it slowly. What else can we do? We have values. Let's say we have the ratio, and we want to know the equivalent in dB, decibels. Ratio of one. That is, for example, if it's signal to noise ratio, ratio of one means the signal is one time stronger than the noise, which means the signal and the noise are the same level. So if we have a ratio of one, how many dB plug it into the equation, the general equation for decibels? 10 log base 10 of the ratio. The ratio of one is how many decibels? And this is the general equation. It's written a little bit different, but the 10 times log base 10 of a ratio between two power levels. In this question, I'm giving you the p-out divided by p-in. It's one. So it's 10 times log base 10 of one. One. What's log base 10 of one? Zero. Zero, so zero times 10, it's zero dB. The ratio of one is zero decibels. What about two? Something is two times stronger than something else. Try it on your calculator. Remember log base 10, that's easy on your calculator. It's about three. Not exact. Log base two is what? 0.3 something. Times by 10, it's about three dB. It's a nice one to remember, in fact. Let's do some easier ones, and we'll come back to that in a moment. What about 10? A ratio of 10. The signal is 10 times larger than the noise. Log base 10 of 10 is one. Times by 10 gives us 10 dB. 100, something is 100 times larger than another value. 20 dB. Log base 10 of 100 is two. Two times 10 is 20 dB. And this one, you'll see the pattern here. A thousand is 30 dB. So that is something we may see. As long as you know how to calculate or convert from the absolute ratio to decibels. Just take the log in base 10 of the ratio and in this case we get three. Multiply by 10 gives us 30 dB. Or convert backwards. What is 30 dB? Back into a normal ratio. Or you need to apply that equation in the opposite direction. We'll see some examples in a moment. Come back to two. So two is 3 dB. What about four? The ratio of four. Signal for example is four times stronger than the noise. Log base 10 of four. Get your calculators out. Check my calculations. If you don't have a calculator you can still solve this in your head. Log base 10 of four is about six. Log base 10 of four is 0.6. Times by 10 gives us about six. It's not exactly. What about eight? Try it. Log base 10 of eight. Multiply by 10. I think some people have found 9 dB. 16 will be 12 dB. Doubling a ratio of two is approximately 3 dB. So if you double again from two to four the beauty of using decibels is that we are not multiplying but we can add the values. Decibels, remember, are calculated using logarithms. And one property of logarithms if you log two numbers multiplied together it's the same as taking a log of one of those numbers and adding it with a log of the other. So it turns out with decibels instead of multiplying numbers. Two is four. Four times two is eight. Eight times two is 16. We know a factor of two is a 3 dB. So four, which is two times two is this double. So three dB plus three dB is 60 dB. Another factor of two, another doubling brings us to 9 dB and double again to 12 dB. That becomes useful for quick calculations. You've always got your calculated available to check. Let's just use for shortcuts. You don't need to remember these values. You can calculate. Let's try some other examples. I have an amplifier in the cabinet here that we have for the audio in this classroom. The amplifier, not amplitude, not amps but an amplifier. Take some audio input. The amplifier comes from... My voice goes to the microphone, goes to the wireless receiver, then sends an audio input into the amplifier. The amplifier amplifies that audio signal and produces some output. And I think you know amplification is you increase the output. So let's look at that. We have some power going in. We denote as P in, and some power comes out of the amplifier. We denote it as the power level coming out. We'd like to look sometimes as, well, by how much does this amplifier increase my input power? Let's say that the amplifier increases by a factor of 10. That is, if the input is 1 watt, the output will be 10 watts. If the input is 2 watts, the output will be 20 watts. This factor we can talk about is the gain of the amplifier. We have an input signal. It gains strength when it comes out. It increases strength. So the gain factor or just the gain of the amplifier. So, for example, if P in was 1 watt, with a gain, G equal to 10, G is the gain, then P out would be 10 watts. If P in was a different value, 2 watts, with the same gain, P out would be 20 watts. Simply calculated by multiplying the input by the gain. Input times gain is the output in this case. But often we want to deal with these gains and the powers using dB, using decibels. What is our gain in dB? Well, the ratio is 10. So the gain is log base 10 of 10 is 1, multiplied by 10 is 10 dB. So the gain factor of 10 is equal to 10 dB, 10 decibels. Don't be confused at different values. This is 10, this is 10 dB. So if the power input is 10 watts and the gain is let's try a different one. 20 dB not 10 dB what's the output? So be careful, I've changed the gain here. We no longer have the same amplifier. We bought a better one. What's the output now? Try, calculate what the output power is. Just so you understand the relationship between dB, the absolute value and what gain means. Gain of 20 dB different ways to calculate it. The best way, the full way is to convert this back to a ratio. 20 dB and in fact we did it before, didn't we? 20 dB is 100. 20 dB is 100. So the gain as a ratio is 100, meaning the output is 100 times larger than the input. So if the input is 10 watts the output would be 10 times 100, 1000 watts. In general, gain in dB is 10 times log base 10 of the output divided by the input. So we talk about the gain of some system. We think about, well, how much was output of that system, how much was input of that system, the ratio of those two and then convert to dB. Any questions before we move on to another example? We may return to this when we have time at the end and do a few modifications to a different system. Let's say now we have a we transmit a signal. We have a communication system. We transmit a signal. It goes across some distance. It attenuates and we receive a signal. Ignoring noise at this stage. We transmit some signal. There's no noise. That signal gets weaker across distance we've said. It attenuates. If we transmit a signal at a level of 20 mW and we receive at a level of 1 mW what's the gain of this system? And calculate it as a ratio and then calculate it in a dB form. So the gain as a ratio if we transmit 20 mW but receive 1 mW the answer you calculate gain by input. We start with 20 going in. 1 comes out. So the gain is 1 divided by 20. 1 mW out 20 mW in 1 divided by 20 or 0.05 That means if I transmit at 20 mW the gain we multiply by 0.05 to find out what our receive power is. 20 times 0.005 is 1. So it's a multiplier, the gain. Input times gain gives us output. Now express it in dB. The gain in decibels is equal to what? Well you have your ratio 0.05 log base 10 of that times by 10 what do you get? Someone tell me the answer. Minus 13 sounds right. Minus 13 dB. Now this is a bit confusing. Did we gain? Did we increase the power? No, it got smaller. What's the opposite of gain? Loss So we have a gain of 0.05 or minus 13 dB The other way we can look at it is that we have a loss in this case. Well what's the loss? Let's calculate it. The gain was the output divided by the input. The loss L is the input divided by the output. Our input was 20 mW The output was 1 mW So the loss is a ratio of 20 That means I transmit my signal at 20 mW The received signal is 20 times lower than the transmitted signal. That's what the loss means. It's a factor of 20 it decreases by. Again means I transmit the received signal is 0.05 times the transmitted signal. So loss and gain are inverse of each other. And in dB 20 a loss of 20 equals what? 13 dB Remember decibels we take the logarithm of a value. So in that ratio the inverse of 0.05 is divided by 0.05 which is 20. That makes sense. In dB we got logarithms. One of the properties of logarithms is that the log of a over b equals the log of a minus the log of b. So it turns out that if the gain is minus 13 dB the loss is just plus 13 dB. Let's move on to another concept and we'll see a few more examples. Everyone can calculate dB. Just remember 10 times log base 10 of the ratio What is the ratio? It depends upon what we're dealing with. Is it signal to noise ratio? Is it just input divided by output which is loss or output divided by input which is gain? Now another thing that we'll see in communication systems so we'll often see the gain of a system expressed in dB or the loss of a system expressed in dB but sometimes we'll also see the transmit and receive power expressed related to dB. So let's look at that. We received at a power level the receive power in this example 30RX was 1 mW The general form for decibels is 10 times log base 10 of one power level divided by another. That's the general form. One power divided by another. So with gain it's output divided by input with loss it's input divided by output with signal to noise ratio signal divided by noise So in general some power level divided by another So it's a ratio of two power levels who will sometimes express a single power level relative to some standard or some reference point. Let's say the reference point is and we'll refer to that in the above equation P2 is 1 mW No, don't copy that one. Let's start simpler or start harder but be more interesting. 1W So what we want to say is the receive power was 1 mW What is the receive power relative to some reference point in this case the reference point is 1W and express that in dB And so what we do P2 is 1W P1 is 1 mW So 10 log base 10 of 1 mW divided by 1W equals what equals 10 log base 10 1 mW divided by 1W is 0.001 mW So 1W is 1000 mW So we get 1 mW divided by 1000 mW We need to have the same units or the same prefix there 1 divided by 1000 is 0.001 log of 0.001 is minus 3 mW by 10 you get minus 30 dB W This is the new concept 1 mW relative to 1W that's where this W comes from is minus 30 dB larger This is a new concept It's not just a decibel which is a ratio between two powers We have a ratio between two powers where one of them is fixed and in this case it's fixed at 1W Therefore when we write the answer we write dB relative this is how you read it decibels relative to 1W that's what the W means So sometimes you'll see expressed say a receiver or a transmit power not as milliwatts but in this case as dBW dBW In other words 1 mW is the same as minus 30 dBW where W means the reference point is 1W Correct It's a reference point in that and usually there's only two or three that we'll see in practice in our course we'll just use two One reference point will be 1W Another reference point we'll commonly see is 1 mW but it needs to be defined and the way that we know which one's used is this letter here dBW means the reference point is 1W Let's try that one or a similar example Our transmit power in the previous example was 20 mW What is it? Expressed in dBW The transmit power PTX from a previous example we said was 20 mW Expressed in dBW what is the value? So it's the ratio of that 20 mW to the reference point of 1W gain 10 log base 10 of our power level 20 mW divided by the reference point 1W That's what the W means here the reference is 1W What do you get? There's 20 mW divided by 1,000 mW So 20 divided by 1,000 that's 10 of 0.02 That is 20 mW is 0.02 times 1W 0.02 times 1W is 20 mW Calculator What does it give us? Log of 0.02 times by 10 gives us minus 17 dB Watts So the answer 20 mW is the same as minus 17 dB relative to 1W 20 mW relative divided by 1W and then convert to dB So dB on its own expresses a ratio between two values dBW a power level relative to 1W Let's add more fun What is this value in dB mW decibels relative to milliwatts dBW is decibels relative to 1W dBMW is decibels relative to 1 mW Make sure you can read that dB mW Well the same approach 10 log base 10 of our power level 20 milliwatts divided by the reference point dBMW the reference point is 1 milliwatts becomes 10 log base 10 of 20 milliwatts divided by 1 milliwatts is 20 which is what 13 dB mW and we're lazy sometimes we don't write the W we just write dBM but it really means dBMW you'll see in many documents it's just dBM So a power level of 20 milliwatts is the same as a power level of minus 17 dB watts which is the same as a power level of 13 dB milliwatts and sometimes we remove the W here just write it as dBM many times in fact dBM refers to dBMW not dB meters we talk about power it's just an abbreviation to dBMW we'll see some examples over time and you'll start to see some patterns about the relationship between those values questions before we go through one last example put all this together so my transmit power is 13 dBMW which we just calculated I have a system I transmit at some power level there's a gain at this component and then some output and the second component in the system produces a loss L where the gain is let's say 20 dB and the loss is 30 dB and the result is the received power the Rx what is the received power so the scenario is I transmit a signal at some power level 13 dBM there's some component in our communication system that amplifies the signal by how much by a gain of 20 dB then the output of that goes into another component which causes a loss maybe the signal travels across some distance the loss of 30 dB what is the received power how strong is the power that's received well how would you solve this two ways the long way and the quick way how long do we have we can just fit in the long way and the quick way will be easy convert everything to our normal units that's how many milliwatts what's 13 dBM we calculated it before 20 milliwatts that's why I used that value we know it already you could calculate if it was a different value so the input is 20 milliwatts the gain is 20 dB 20 dB is a ratio of what what's the ratio 20 dB it's 100 isn't it that is log base 10 of 100 is 2 times by 10 gives us 20 so it means we have an input of 20 milliwatts we increase by a ratio of 100 so what's the output here the input is 20 milliwatts the output would be 20 times 100 2000 milliwatts so at this point we have 2000 milliwatts the loss of 30 dB what's 30 dB so let's say this is P Px that is the intermediate power level Px would be 20 milliwatts times by 100 2000 milliwatts the gain is a multiplier the loss of 30 dB is a factor of 1000 so PRX we have 2000 milliwatts going in a loss of 1000 we divide by 1000 remember it's a loss so it goes from 2000 in down to 2 coming out 2000 milliwatts as the input divided by the loss of 1000 means 2 milliwatts come out so in this system we had an input of 20 milliwatts a gain of 20 dB a loss of 30 dB and the result was that we receive at 2 milliwatts not so hard for this one that was the long way the short way when you deal with dB you can just add and subtract and I'll leave you to explore that but let's write it down the power received is the transmit power 13 dBm plus the gain of 20 dB minus the loss of 30 dB 30 equals 13 plus 20 is 33 minus 30 gives us 3 dBm that's an M M means milliwatts the receive power is 3 dBm if you want convert 3 dBm to milliwatts and you'll find it's 2 milliwatts get the same answer the power of using decibels is that we can do calculations with addition and subtraction and for simple for quick analysis it's much easier to look at decibels we add the gain, we subtract the loss as opposed to in the absolute ratios we apply by the gain and divide by the loss for most people addition and subtraction is easier than multiplication and division especially with large numbers decibels are commonly used in communication systems I will let you confirm yourself if this is the case why this is true and why we get the same number here we'll stop there and we'll continue with transmission media and look at how what media do we have available to send our signals across and then we'll start to see some use of dB in practice