 So we know something about antennas, the real antennas, we can measure their gain relative to an isotropic antenna. Come back to our simple model of any wireless system. We transmit some signal. We start with an electrical signal at the source device, generates an electrical signal at some power level. It goes into the transmit antenna and we will think that that transmit antenna has some gain. It amplifies the signal. So some signal comes out and as that signal propagates between the transmitter and receiver, it loses power. It attenuates. Then it's received by the receive antenna and we'll think that this receive antenna has some gain as well. It receives some input radio waves and the shape and design of the antenna determines how much that is amplified by as it's received and then passed to the actual receiving device, the electrical signal. So a simple model and we'll draw it in a moment is we start with some transmit power here at the transmitter. We send it to the antenna, the antenna acts as an amplifier. It increases that signal. As the signal comes out and propagates through space, through the air, it loses strength. There's some loss and then it's received and the receive antenna effectively amplifies that received radio wave and produces an output electrical signal, which we will measure as the received power. So we'll try and look at the relationship between if we start with some transmit power. We have two particular antennas which have some gains and if we know how much power we lose across some distance, then we can work out the received power. So that's what we'll focus on today. If we draw that, I think we transmit with some power PT. There's a transmit antenna. T means transmit, P the power level. The transmit antenna produces some gain. We often will denote as GT, the gain of the transmit antenna. The signal propagates and then goes to the receive antenna. And we receive a power with strength PR and the receive antenna has a gain GR. That's my simplified picture of the slide. We start with a power level. The first antenna increases it. The signal propagates to the receiver which will increase it and we'll get the resulting receive power. The other point is that between those two antennas the signal gets weaker. It attenuates and the attenuation will denote as some loss. And we'll introduce and talk about the path loss. If you think there's a path between transmit antenna and receive antenna, the signal comes out of the transmit antenna with some strength and it attenuates. It gets weaker and weaker and weaker. By how much? We'll denote as the loss across that path. L path or I'll just later write it as LP. So we want to look at these five values and see how they're related. We can draw that slightly different. We can say we start with a transmit power. Then we have some component. We've drawn pictures like this before which has some gain. The gain of the transmitter or transmitting antenna. Then as the signal comes out of that antenna there's some loss. I'll denote as LP instead of writing path LP. The loss across the path that is between the two antennas. And then it arrives at the receive antenna and that has a gain of GR. And then we receive the receive power. So this is a simple model of what happens with our power level. We start with the power level. There's some gain due to the transmit antenna. There's some loss due to the fact that signals attenuate across distance. And then there's some gain at the receive antenna. And the resulting power level is determined by those four values. Now in real life if you think we have a transmitter and then an antenna there's actually some wiring between the transmitting device and the antenna. Maybe even a cable. There may be loss there but usually it's very small. So we'll keep it simple and think. We start with the transmit power, a gain, a loss, a gain and then a receive power. Let's put some numbers to it and see the relationship between those. T for transmit, R for receive, G for gain, L for loss. P, lowercase P here as a subscript is for path loss. The loss across the air between those two devices. Let's just put some numbers. I'll make some up. Let's say we transmit at some power level of 10 watts. And that we know the gain of the transmit antenna. We bought the antenna. We know it's a factor of five. What that means is that this antenna relative to an isotropic antenna produces a signal at the same distance away which is five times stronger. A gain of a right times five meaning it's five times stronger than our ideal isotropic antenna. So we think we're going to amplify the input signal by a factor of five. And let's give the others some number. Let's say the loss is a factor of 25. And the gain of the receive antenna is a factor of three. Then what's the receive power? What is the receive power in this case? We just have a starting power and a set of gains and losses. Anyone with an answer? Quick answer? Today we have many calculations. So sometimes we'll go through quite quick especially the larger numbers. In this case we start with 10 watts. A gain of times five means the output here will be 50. Five times larger than the input. So the output is 50. Then we have a loss by a factor of 25. What does that mean? We have an input of 50. The output will be 25 times smaller. Maybe a right that is divide by 25. A loss of a factor of 25 means it will divide the input by 25. So we start with 10. After the first gain we have 50. After the loss we end up with two. And then after the gain of the receive antenna we end up with six. So the receive power is six watts. That's a simple model of any wireless communication system. If we know the gains of the antennas and we know the amount of power which is lost between the antennas given the transmit power we can find the receive power. And you see how do we calculate PR? In this case generally PR was 10 watts times by five the factor of five for the gain divide by 25 for the loss and the last part times by three for the receive antenna. Gains we multiply loss we divide when we deal with factors. This is not about DB. It's not decibels. Not yet. You may see in general we could say that PR is the transmit power Pt multiplied by the gain of the transmit antenna times multiply the gain of the receive antenna divided by the path loss and we'll use this general equation and I'll show it to you in a moment. You've got it on your lecture slides or somewhere in your lecture notes. Where is it in the lecture notes? It's actually in one of the extra handouts. I hope you have it. Just go through a few more pages. Let's hope keep going. After your antennas. Yes, path loss model. You have this document somewhere towards the end of those the current lecture slides a path loss model. So this describes some of the things that we're going to write down in a moment. This is a more general description without the numbers but we just arrived at this equation. Power received is the transmit power times the gain of the transmit antenna times the gain of the receive antenna divided by the path loss. That is in your lecture notes. You have this print out there already. Another example. Let's say we have keep the number simple that we know the transmit power is 10 watts. We know the gain of the transmit antenna is a factor of eight. We don't know the path loss. We know we have a gain at the receive antenna of a factor of two. And we know the receive power is four watts. Start with 10 watts. We know the characteristics of both of our antennas. We also know the receive power. What's the path loss? Use the same equation to find the path loss. Anyone have an answer? Which is 40. 10 and you can check. It's just a rearrangement of this equation. Here we know PR four watts. We know PT 10 watts. We know GT is times eight. GR times two. What's LP? What's the path loss? Rearrange and you find it's 40. Just check. 10 watts times eight brings us up to 80. A loss of a factor of 40 brings us down to two. We start at 80. A loss or a factor of 40 we divide by 40 down to two. A gain of two brings us up to four. So we receive it four watts. So we can do any rearrangement with this equation. If we know four of the values we can find the fifth. And in fact this one is common because in practice let's say you want to set up a wireless link between two points or maybe you want to know how far you can separate your two devices. You normally know the transmit power of your device because it's a characteristic of the device that you buy. It's usually limited. And you usually know the gain of your antenna. Again you buy an antenna and the gain is specified. So you know the transmit power and the gain. You also know the gain of the receive antenna and another characteristic of most devices is the minimum power which it can receive. Sometimes called the receive power threshold or the receive sensitivity. What's the smallest power which we can receive successfully? So if we know that then we often want to find well what path loss is acceptable. And we'll come back to how we use that with a few more examples. Any questions on our path loss model so far? Let's do one more example. Let's say we transmit with 100 milliwatts. The gain of our transmit antenna is a factor of 2. The path loss is a factor of 10,000. And the receive antenna is a factor of 4. What's the receive power? Maybe you need a calculator. At least know how to determine it. 0.08 watt. Not watts, milliwatts. 0.08 milliwatts. 100 times 2 is 200 milliwatts. Divided by 10,000 is what? 200 divided by 10,000 is 0.005. 0.02 times 4 is 0.08. You can do the math better than me. I've done it before though. It's 0.08 milliwatts. Note that we start with milliwatts. These have no dimensions. There's no units here. It's a factor. Times by 2 divided by 10,000. Times by 4. So we end up with milliwatts. Now, this is all using just factors. But we know we can express gains and losses in decibels. Let's convert. Convert these values to their dB equivalents. Remember the general equation? 10 times log base 10 of the ratio or the factor. So, for example, with an antenna gain, a factor of times 2 is 10 times log base 10 of 2. Gives us the gain in dB and dbi, which is about 3. So times 2, you can use your calculator. It's about 3 dbi. Again, just take this value of 2 log base 10 and you get 0.3 approximately. Times by 10, you gives us 3 dbi. Remember the gain dbi means decibels relative to an isotropic antenna, our ideal antenna, i for isotropic. The path loss. Forget about the divide. Just a number of 10,000. A factor of 10,000 is what? Log base 10 of 10,000 is 4 times by 10 gives us 40 db. 4, you can calculate, is 6 dbi. And also, we can express the power levels in dB. In this case, dBm. Decibels relative to 1 milliwatt. So 100 milliwatts divided by 1 milliwatt is 100. Log base 10 of 100 is 2 times by 10 gives us 20 dBm. And can someone do it on the calculator for me? 0.08 milliwatts. Log base 10 of 0.08 times by 10. Approximately. All we're doing is logarithm of the number times by 10. 0.08 milliwatts converted to dBm is approximately minus 10.969. Approximately minus 11. It's about minus 11 dBm. So decibels is just a different way to express those power or those factors. We transmit at 100 milliwatts. A gain of 2, a loss of 10,000, a gain of 4 gives us 0.08 milliwatts. We just multiply by the gain divide by the loss. But in dB, we transmit at 20 dBm. A gain of 3 dbi. A loss of 40 dB. A gain of 6 dbi gives us a received power of minus 11 dBm. Anyone see the relationship between the green numbers? How do we get PR there? A gain we add. A loss we subtract. A gain we add. Gives us our minus 11. 20 dBm plus 3 is 23. Minus 40 is minus 17. Plus 6 gives us minus 11. So you can choose either. The absolute factors in this case where you multiply for gain divide by loss. Or the dB versions where you add for gain and subtract for loss. Why? Why does multiplication become add and division become subtraction? You know based on your basic logarithms. You can see in one of these slides where properties of logarithms. Logarithm of A times B was a log of A plus the log of B. And this, I will not go through it now, but this document gives that proof why when we use the absolute values we multiply and divide and why when we convert to dB it becomes plus and subtract. You can go through and check and confirm that. Because if you take the log of both sides and multiply by 10 it can simplify to be addition and subtraction of those values in dB. And that highlights one of the benefits of using dB. If you know the values in dB determining the received power in this case is very easy. Just add and subtract. And for most people adding and subtracting is easier than multiplying and dividing. Especially when you have very large numbers here. If this division is by 4 billion then it's often easier to deal in subtracting some small decibel value. Because with logarithm it becomes much smaller. So in practice it's usually convenient to deal with dB. In general we get the power received in, I write it here, dB. If it's expressed in decibels, dB milliwatts, dB watts whichever is equal to the transmit power in our example, I write dBm. It's a sub-subscript so it's getting small. Plus the gain of the transmitter in dBi plus the gain of the receiver in dBi minus the path loss in dB. The top one is when we use the absolute factors we multiply and divide and the bottom green one is when we use the decibel versions of those values. We add and subtract. You can use either. So in an exam question if I ask you, given this system find the received power if some of the values are in dB and some are in the absolute values you'll need to convert one to another. They all need to be using the same. Either dB or not. And then solve however you like, whichever one's easier. Generally using dB will be easier to calculate. But if all the values are in the absolute factors then it's okay just to use them. Another example staying in dB we know actually before we write that we'll do a realistic example. I go and buy a wireless router. It's a TP-Link wireless router. Just like the one on the wall but much newer. And we go and look at the specs for this wireless router and we zoom in. Many details. The ones that we're interested in, the wireless aspects. First start here. Transmit power. The transmit power of this device. When you buy the device the maximum transmit power is 23 dBm when we're using the frequency of around 5 GHz. So that's the characteristic of the device. You buy that you can't go higher. Usually you can go lower. Let's say this is our transmit power. The other aspect of the device is that if we can transmit at 23 dBm the other one is the receiver. What is the capability of the receiver at receiving signals? And that's here the reception sensitivity. There are many different values because this device supports both frequencies of Wi-Fi. 2.4 GHz, the very common one plus the less common 5 GHz channels. So let's just focus on 5 GHz because it's nicer in our example. We transmit at 23 dBm the 5 GHz receiver. 11A is an old technology. 11AC is one of the newer ones. That says that this device when it receives a signal think of PR in our equation if the signal power is above minus 71 dBm it will successfully receive. But if the signal power that it receives is below minus 71 dBm like minus 72, minus 73 then it cannot successfully receive. So this is called the reception sensitivity or sometimes the receive power threshold. This is the smallest power which this device can receive that it can make sense of. So that's an important characteristic because if we transmit at this power and we know we must receive a power above this this is the minimum power we can receive below and our devices cannot communicate then that tells us something about the path loss that is acceptable. So let's take these two values. This is the transmit power, 23 dBm our receive power is minus 71 dBm. What about the antenna gains? Unfortunately it's not on this spec but they're just the standard dipole antennas like the one on the router on the wall and the gains are about 2 dBi maybe a little bit larger but the gains of these antennas are about 2 dBi. So let's say we have two of these routers and they want to communicate with each other. We know the transmit power. We know the receive power. We know the gain of the transmit antenna, 2 dBi the receive antenna, 2 dBi, they're the same. We can work out the path loss which is acceptable. So transmit power, 23 dBm a gain of our antenna of 2 dBi the receive antenna is the same and the receive power was, what was it, minus 71. We know those characteristics of the two devices we want to allow to communicate. What's the path loss? You just rearrange this equation and the path loss will be the transmit power 23 plus the gain of the transmitter 2 brings us to 25 plus another 2, 27 minus minus 71 27 plus 71 which will be 98 dB path loss in dB will be 23 plus 2 plus 2 minus minus 71 and just check in here, 23 plus 2 is 25 minus 98 plus 2 should equal minus 71. Any questions so far? What we can use this for is we have our two wireless routers. One thing we want to know is how far apart can we separate them such that they can still communicate? I put one on in one location and I want to put the other one as far as possible such that they can still talk to each other. Well, we know now that with these devices the path loss between the two devices must be less than or equal to 98 dB. If the path loss was 100 dB if it was 100 dB we'd get 23 plus 2 is 25 minus 100 minus 75 plus 2 gives us minus 73. If the path loss was 100 the received power would be minus 73 dBm but the specification of the device says it must be above minus 71. Remember minus 73 is below minus 71. So if the path loss was 100 we would not be able to receive. Therefore this is the largest path loss that we can allow such that two devices can communicate. That's sometimes confusing all the negative numbers and so on. I don't know if this will capture. This is our receive threshold or sensitivity and in our case it was minus 71 dBm. That's a characteristic of my receiver and what it means is if I receive a signal which has a higher power level let's say minus 65 if my receiver receives a signal with a power level of minus 65 dBm it'll be successfully received. But if it receives a signal with a power level of say minus 73 dBm below the threshold unsuccessful. It would be too weak for my receiver to make sense of it. That's how we interpret the threshold which is a characteristic of the device. So in our case if we we have our 23 dBm transmit power plus the gain of 2 minus a path loss of let's say if the path loss was 92 plus 2 dB what do we get? 23 plus 2 is 25 minus 92 is minus 67 plus 2 gives us minus 65. So if the path loss was 92 we would successfully receive. But if the path loss was something else like minus 100 or 100 so we subtract 100 here we'd get minus 73. 23 is a transmit power the gains of each antenna 2 and 2 minus the path loss becomes what? 27 minus 100 minus 73. So the maximum path loss that we can accept is 98 dB. Anything higher than 98 means we won't be able to receive. The negative numbers can be confusing sometimes. Why do we have a negative number? What does it mean? Remember in decibels it means it's less than one. Actually less than 10 as a factor. Logarithm of a number less than one will give us a negative value. Any questions before we move on or use this path loss value? So far in our lectures I think I see people writing very small on election notes. There's not much space. The next version I'll try and give a bit more space. Some people write on post-it notes. Maybe next time bring a larger piece of paper than a post-it note. But in the next version of the slides we'll have some after the midterm some more slides. Don't worry. I'll add some more space. So now we know with my two wireless routers the path loss must be greater that must be less than 98 dB less than or equal to. The next thing is can we relate path loss with distance? And the answer is yes. The amount of power we lose depends upon the distance and some other factors. And it depends upon where you are. Whether you're inside a building in a city outside the path loss will differ. But there's a simple model which is very common which assumes that we've got no other obstructions and I think it's on your lecture slides but it's also in this handout that you have an equation for the path loss in what's called free space. Imagine there are no obstructions. There are no atmospheric effects. You're out in space and the signal propagates perfectly. Then the path loss is measured as 4 times pi times the distance all squared divided by the wavelength squared. Not in dB, this is in as a factor. So I'll just copy that equation down 4 pi d squared on lambda squared. 4 pi d all squared on lambda squared. We know in our case the path loss needs to be, has a limit of 98 dB. LP here is not measured in dB. It's in its absolute factor. So we need to convert 98 dB back to the actual factor. We'll do it in a moment. But let's first, what's our wavelength? Remember wavelength is the speed of light divided by the frequency. Speed of light is 3 by 10 to the power of 8 meters per second. The frequency, I don't know if I wrote it down but these devices that we chose 5 gigahertz. So the frequency of the signal in this example is 5 gigahertz. And you do that calculation, you get 0.06 meters, wavelength is in meters. 6 centimeters. So, we know lambda. We don't know d. We want to find the distance d in meters between our two devices such that the path loss will be 98 dB. I'd like to know how far I can separate them such that they can still communicate. I know the path loss in dB is 98 dB. I need to convert that back to the absolute factor. Let's do that now. We know the path loss in dB, we calculated to be 98 dB which is the same as an absolute factor, 10 to the power of 9.8. Remember the inverse of calculating our decibels, 10 log base 10 of the factor. So the inverse, we take our decibels divided by 10 and that's the exponent with the base of 10. So these are the same numbers, one in dB, one as the absolute factor. So, given our path loss equation in free space meaning in imperfect conditions, we have 10 to the power of 9.8 equals 4 times pi, that's a pi, times the distance squared divided by lambda squared. We can rearrange and find d. We have an equation with one variable, so some rearrangement and you'll find d. I will not go through the calculations. I've done it before. It's 379 meters. Therefore, d, the distance is about 379 meters. What that tells us is that with our two wireless routers using the specs that we found on the website if we separate them by a distance of 379 meters and we're operating with no other obstructions around in free space that is, there's no trees in between. In fact, there are no atmospheric effects. It's like in a vacuum. So in perfect conditions if we separate them by 379 meters when we transmit a signal with the antennas that we chose the received signal will just will be the same as the received power threshold. So we say that that will be successfully received. If we bring them a little bit closer, 378 meters the received signal will be slightly stronger than threshold, even better. And the closer we bring them the lower the path loss as the distance gets smaller, the path loss gets smaller as the path loss gets smaller then the received power will be higher and we'll be able to receive successfully. If the distance was 380 meters the path loss will be slightly larger than 98 dB. If D goes up, LP goes up the path loss goes up and we said that the limit was the path loss of 98 dB. If it was 99 dB then our received power would be below the threshold and we would not receive. So it now gives us a relationship if we know transmit and receive powers we know the antennas we can use this path loss equation, this one to work out, well what distance can we separate those devices such that they can communicate? And that's very valuable for building a wireless system because you can now predict well where do you locate your transmitter and receiver such that they'll be able to communicate successfully. Any questions on our example? Should be many questions because there are a few hard things there and a few assumptions that we may have glossed over. Okay, exam in what, a bit over a week? Maybe two weeks, is it? Two weeks. Can calculate the path loss, calculate the distance between two Wi-Fi routers? It's a very common exam question. Here's the scenario for example find the distance between the devices or some other rearrangement of that. Find the minimum received power with this given distance. One last thing to continue this example or actually let's go back to our slides and see some of the things we've skipped. Let's go to the slides that show that equation but in a slightly different form. We have to go through here called the free space loss model. It's slightly different from what we wrote down but you'll see it expresses the same values. It's a mathematical model to say how much power is lost between two points in free space. So assuming there's no obstructions between our transmitter and receiver, perfect conditions, this is the model that we get. How does that relate to what we've seen in our example? It's in fact a combination of two. This equation P R equals P T G T G R on L P. This general equation and if you replace L P in the case of free space path loss, L P is this equation. If you substitute that in you get this with a little bit of rearrangement. Transmit power, receive power, distance, gain of transmit antenna, gain of receive antenna and wavelength lambda. You're lucky. I will give you this equation or a similar one in the exam. So I check the previous exams and things like Shannon capacity, Nyquist capacity equations are given in the exam and so is the free space path loss model. It may not look exactly like this. It may be rearranged. For example, we can rearrange that to be what do we have? We had P R, the first equation P T G T G R divided by L P, the path loss which is 4 pi D squared on lambda squared which becomes, sorry, P T G T G R lambda squared divided by 4 pi D squared and the one on the slide is just a rearrangement of that. How do we increase the receive power to make P R higher? Let's say we know the threshold. We know the minimum P R. We need to get the receive power above some threshold which is characteristic of the equipment. Well, we've got different options. Increase the transmit power. Larger transmit power, larger receive power. Increase the antenna gains either the transmitter, receiver or both. Increase the wavelength. Remember, wavelength is C divided by F. Increasing the wavelength is the same as saying decreasing the frequency. Lower frequency, wavelength goes up. P R goes up. Or how do we get the receive power to go up? Decrease the distance. Bring the devices closer together and the receive power will go up. This model is called the free space path loss model. There are other mathematical models to represent more realistic conditions. This one's the easiest to deal with. But if you're sending a signal between the access point on the wall and devices inside this building, well, the signal as it goes through the walls and the doors and bounces off all these different obstacles, the path loss may be much larger than in free space. So there are other models that people have developed to try and represent what happens in real life like inside cities, inside suburban areas, for TV broadcast stations. So there's a TV station which has a transmit antenna and it transmits to TVs in homes. There are mathematical models to work out how much path loss occurs between the station and your home and indoor models. So there are mathematic models to try and determine how many, depending upon how many floors you have in the building because the ceiling, the floor creates an obstruction and the shape and the structure of the building, there are models to work out how much path loss across a particular distance. We will not see them, but just be aware there are other than the free space path loss. This one is the easiest one to deal with, easiest one to perform calculations with. Let's go back to our example. Recap. We arrived at the general model. The received power is the transmit power times the gain times the, of both antennas divided by the path loss. Path loss is how much power we lose between transmit antenna and receive antenna. This applies in any system. We can convert that to the DB form and the nice thing about DB is that multiplication becomes addition and division becomes subtraction. Let me use the DB form. You can use either. In our examples, we found some Wi-Fi devices, transmit power, particular antenna gains. Where do we find the antenna gain from? What's the antenna gain depend upon? We did this last week. We looked at, if we transmitted with an isotropic antenna, say one meter, and then used our directional, our real antenna one meter, the antenna gain is the ratio between those two values. How much stronger is our antenna compared to isotropic? That we saw in our example is a characteristic of the antenna. You can actually calculate that in some cases and jump back to the slides, go back a few slides, we've skipped over this one. You can calculate the antenna gain given some knowledge about the size of the antenna, the frequencies it's transmitting. So G of any antenna can be calculated with this equation. We'll see an example in a moment, but let's explain. The gain of antenna is 4 times pi times the effective area of that antenna divided by the wavelength squared. Wavelength depends upon the frequency we're sending. So if we know the frequency, we can find the wavelength. 4 times pi is just constants. Effective area. So it's related to the physical size of the antenna. The larger the antenna, the larger the effective area, and the larger the gain. But it differs depending upon the design of the antenna. The dipole antennas on the access point on the wall are a different shape than say the parabolic dish antennas you may have for satellite TV. And they'll have different effective areas. We will not try to calculate the effective area. In an exam question or a quiz, I would say, here it is, this is the effective area or assume it's calculated based upon this. And a common one, a parabolic dish antenna. If you look at those dish antennas face on, what's the shape? Imagine, you know, the satellite dish antennas. If you look at it direct in front of it, what's the shape that you see? A circle, okay? It's a parabolic dish, but if you look on front on it, it's a circle. So we often approximate the area of that dish is just the area of the circle. So if we know the dish is say one meter in diameter, the radius is half a meter, the area of the circle is pi r squared. That's the physical area or approximately the physical area of the dish. The effective area is usually a smaller value than the physical area. And it depends upon the design of the antenna actually, the materials and how it's designed. And usually that would be specified say, the effective area is half of the physical area. Again, I would have to tell you that in an exam. I don't expect you to remember. It's not always the same. Some antennas, it's a half, sometimes it's larger, sometimes smaller. Let's do an example. Given that, given our free space path loss model, free space loss model there. Let's do this example. We have a wireless system. Transmitter and receiver, they're using two antennas of the same shape, same design. They're both parabolic dish antennas. Diameter of one meter. Frequency five gigahertz. That's our signal frequency. We transmit at one watt. We transmit across a distance of 1,000 meters. So the two devices are separated by one kilometer. We want to know what is the required receive power threshold or the receiver sensitivity. What's the minimum receive power that the device should have? And the assumption that you'll need to know in this case, let's assume the effective area of a dish antenna is 0.5 times the area of the circle. So if we know the diameter, we know the radius, we can find the area of the circle and then say the effective area is half of that, half of the area of the circle. It's not written here, but let's assume it. Find the receive power, PR. Try it for a few minutes. Try to write down the values with respect to the things that were the PR, PT and so on and see what you need to calculate. While you're calculating, I will try and draw the system but give you time. Try and work out the antenna gains because you'll need the antenna gains to use in the free space path loss model. And then try and work out PR. And in this case, the transmit and receive antenna are exactly the same size. So in fact, you only need to calculate the gain for one of them, then you know the gain for the other. So as a hint, the first step, calculate the gain of our antenna and the assumption that we're making is that the effective area of our antenna is a half, 0.5 times the area of circle. If we look at the antenna face on, it appears as a circle. The effective area of our antenna, Ae, is half of that area of the circle. This one you don't need to do in dB. You could convert everything to decibels but there's no need to. We'll need to find the gain of an antenna. We have the equation. The gain of an antenna is 4 times pi times the effective area divided by the wavelength squared. So find Ae and lambda. What's the area of our circle? Our dish is one meter diameter. So pi r squared, it's not one meter, it's half a meter squared. There's no connection between this 0.5 and this radius of 0.5. It's just a bad coincidence in this example. There's no connection there. If our diameter was 10 meters, then the radius would be 5 meters but still 0.5 out the front. I will not calculate this yet. You can use your calculator and find the value. I will not do it yet. We can just to save some time and space but now we can find the effective area. Half of the area of the circle of the antenna where our antenna has a diameter of one meter. Effective area is meters squared. Area is meters squared. 0.39 something, okay? Good. The other thing we need is the wavelength. C divided by F. Speed of light divided by 5 gigahertz. We did it before. 0.06. So now we plug those values into our gain equation. The gain of the transmit antenna, 4 pi times the effective area divided by the wavelength squared. Someone will calculate that. Plug in all of the numbers in your calculator. Anyone? 1, 3, 6, 9 points something. You may have approximated it at some point, did you? When I did it, I got 1, 3, 7, 1. Be careful. It doesn't matter that both numbers would be right in the exam but be careful. If you calculate AE and then you say it's 0.0036 when it's really 0.00359 something and then plug it in here, it's best to calculate just at the end. Otherwise, all the inaccuracies of approximating earlier on start to have an effect as they go through. But in the exam, not a problem. If it's close, it's correct. But use your calculator and... So instead of calculating AE and writing it down, I just plug all these values into this equation and get, again, it's an approximation. You had 1, 3, 6, 9. I had 1, 3, 7, 1. So let's try 1, 3, 7, 0. It won't make much difference in the end. What are the units? There are no units. The gain in this equation is not in DBI. It's the absolute factor. So it says my antenna at the same distance away compared to an isotropic antenna is 1,370 times stronger. I think it has an amplifying effect of 1,370 times. What's the gain of the receive antenna? We calculate the same. We're lucky in this question. The size of the receive antenna is the same as the transmit antenna. Therefore, the gain will be the same. It doesn't have to be the same. The antennas can be different sizes. They may have different gains. But in this question, they're the same. We're almost there. We have this free space path loss model. We know GT. We know GR. We know Lambda. We know PT. It's 1 watt. And we know D, 1,000 meters. Rearrange, find PR. Rearranging 4 pi D all squared, all squared. In fact, we're finished. Well, you can do the final step of plugging in the numbers. You know all those values. Just make sure you use the right units. PT, 1 watt, so 1, GT, 1,370, GR, 1,370. Lambda, 0.06. Lambda squared. D is 1,000 meters. Distance in meters, not kilometers. So it's not 1, it's 1,000. Plug it in and you get some number as output. About, again, I think it's 0.0428 or something. 42 microwatts, around 43 microwatts is the received value. So we transmit, starting at a power level of 1 watt, using our two antennas, there's some amplification. Across 1,000 meters, the signal gets weaker and weaker and weaker such that finally we receive with a received power of 42 microwatts. Start at 1 watt. Finish at 0.042 milliwatts, 42, 43 microwatts. Any questions? It's not as hard as it seems now that you know the equations. And in fact, you'll see that a lot of the questions involve some rearrangement of this. Given the transmit power, the received power and the gains find the distance. Or what transmit power do you need to cover this distance? So in exams and in quizzes, you'll see, and in fact in real life, you'll see that it's about rearranging this to find the value that you're after. You must know the other values, of course. We will not do it, but you could check. You could have did it all in dB. It could have converted the gain from 1370 to, I think, 31.4. But in this case, there was no benefit of converting to dB because our equation, most of our values are in the absolute factors. So we've covered a number of things today. The relationship between transmit and receive power in a wireless system. We've covered antenna gain can be calculated if we know the characteristic, the effective area of the antenna. We can determine the antenna gain. And now we see this model for how much power is lost between transmit and receiver. It allows us to work out the distance between two devices. Other questions may be, you know the transmit power, you know the receive power, you know how far you want to transmit, choose an antenna. And that's a very practical thing you may want to consider when building, say, a wireless link. Choose the antennas, which have enough gain so that you can communicate across a particular distance. Any questions before we just spend five or so minutes seeing what we've skipped in these slides? The last five minutes are not as important as what we've covered up until now. Can calculate okay? Given the equation, you don't have to remember the equations. You'll be given them in the exam. But know how to use them. Know that when you have GT here, it's not in DBI. Most of the equations we give use the absolute factors, not in DBI. So you need to be careful. Can we move on? Questions? So we did this one just to remind you that the antennas don't have to be the same. In this case, they were, but they don't have to be. So let's go back, see what we've skipped. Last week we spoke about antennas. We described the concept of antenna gain. We've just introduced the slides that I skip over but don't talk about like this one, extra detail that are not relevant for us at this stage. This is just how a parabolic antenna works. For any antenna, if we know the effective area, we can find the gain. What does it say? Higher area, which usually means larger the antenna, larger the gain. Remember last week we had pictures of the dish antennas. You can have one for your satellite TV, it's about half a metre. But the ones that communicate up to space and pick up receptions from space are maybe 300 metres across. Much larger antenna, much higher gain. Smaller wavelength, smaller, a larger gain. This one we skipped, but let's quickly mention. How does a signal propagate? And there are three main models. If the signal is less than around 2 MHz, when we transmit a signal from some point, it roughly follows the curvature of the earth. We're not going to explain why. You need to look at the physics of the atmosphere and why a signal would do that in the first two especially. They're a bit strange, but that's what happens when your signal is a low frequency, less than 2 MHz, it will basically go around the earth. When it's between 2 and 30 MHz, the signal will bounce off the earth and the ionosphere. The particles in the ionosphere are such that if we transmit the signal up, it will bounce back down off the earth and keep going. As a result, we can transmit it again around the earth. And the other one, and this is called sky wave propagation, the first one's ground wave propagation, the last one's line of sight propagation. Basically, the signal goes straight. And because of the curvature of the earth, if our antennas are around too far, the signal will not be able to communicate. The two devices will not be able to communicate. The signal cannot go through the earth. We mainly deal with line of sight communications, with communication systems. Some use the lower frequencies, especially some radio systems. So if you want to communicate across a very long distance without using satellite communications, then you need to use these lower frequencies. We're not going to ask questions about this in the exam, but it's a nice thing to know if you want to study, especially shortwave radio or radio systems that allow long distance communications. And this is the last thing for today. And again, it's worth looking at in your own time. Remember ELF, VLF, HF, and all those frequency ranges? This talks about, for those different frequency ranges, some of the characteristics of how the signal propagates. And the main point that you need to pick up is that depending upon the frequency we use, a signal may be obstructed in different manners. The best example is when you use infrared or similar, the signal will not go through the wall. The wall obstructs the signal. When you use Wi-Fi, the signal will go through the wall. It will be obstructed, but if the signal is strong enough, part of it still gets through the wall and gets to the other side. So the different frequency signals are obstructed by different particles inside the wall. Some signals, for example, using satellite communications, are attenuated by water. So when it rains, the signal gets weaker and you may notice with satellite TV sometimes and the old systems that when it rained heavily, the signal could degrade and the reception, what you see, may be poor quality. So this just talks about with different frequency signals, there are different characteristics of how that propagates. And the result is depending upon your application, you should choose a frequency that suits your needs. If you don't care about going through a wall, then infrared is okay. So remote control of TVs, for example, makes sense, but if you want to communicate through different obstacles, then you need to consider the frequency for the communication system. So it gives some examples of each different systems and we will stop there. All we've got remaining here, we've talked about free space path loss, the others we were not. That's the main one. And I'll let you have a look at some examples of wireless media, satellite, terrestrial, broadcast radio. But that's about it for the examples. You know examples. We're not going to ask about those individual examples. Just be aware of many different types of wireless systems. In an exam, I'm not going to ask about these examples, mainly about those calculations that we've gone through. So we'll stop there. Tomorrow we'll say a little bit about signal encoding and that will finish us up until the midterm exam.