 We've introduced signals, communication signals. We've done some analysis, represented them in the time and the frequency domain. We've skipped some topic about analog and digital data. We may return to that later. We'll definitely at some stage, I don't know when, but then we went on to last week transmission impairments and there were three main impairments, attenuation. We transmit a signal with some strength and the characteristics of the medium that it passes via reduces the signal strength across distance. So the further the signal travels, the weaker it gets. That's called attenuation. There's noise when there's other transmissions. So I'm transmitting. If there's no other noise, the receivers can receive. But as others generate signals, from the perspective of receiving my signal, those other sources create noise and it makes it harder to receive the data. So noise is another impairment. The larger the noise, the harder to receive the data. And another one that we didn't mention, but also was related to those is distortion. There were a few points on that, but we will not cover distortion. Let's look at a signal and see the impact of noise on the signal. We finished last week with an example, this one where we went through, we sent some data and the signal that represented that data. We added in some noise, but this is what's received. The receiver receives the transmitted signal or the attenuated transmitted signal plus the noise. So you think they're additive and therefore the receiver has a signal and has to interpret what does this signal mean? So it maps the signal back to the data. Let's look at that again, but with a different example. And it's the one, did anyone bring it? The sheet from last week, the gray ones on the back of that, we'll use some of them to give some examples. Actually, I didn't bring any spares, so hopefully people bring theirs. Just look at your own, I'll show you, I've got one. Some plots like this, it doesn't come out so well when we print it. But hopefully after we go through it, it will make sense to see it on hand out. We want to look at if we transmit a signal and if there's some noise, well, what's the effect of that noise and the fact that it can cause errors? And similar to the previous example, but with some different impacts of noise. And I'm going to cover two cases. One with a very simple signal, this sine wave, okay? And we're going to use this same concept, let's say a high component. Transmit high if we want to send a bit one, transmit low if we want to send a bit zero. So this is a very simple sequence of bits, 1, 0, 1, 0, 1, 0 and so on. So this is the transmitted signal at the top. It's a sine wave, it's 1,000 hertz, so the period is 1 millisecond. Those details are not so important yet. We transmit this signal and I've introduced some attenuation. So we transmit the signal at some strength. The signal travels across a link and it gets weaker. And in this example, I'm going to say that the received signal, if we ignore noise, the received signal is half the strength of the transmitted signal. So if we see in our plot of the transmitted signal, it goes from minus 1 up to plus 1, that's the range, the received signal would be minus 0.5 up to plus 0.5. Just in this example, I'm saying how much signal strength do we lose across distance? I'm saying we lose half of it. In a later topic, we'll do some calculations to determine in practice how much do we lose related to distance and other factors. But for this example, I said the situation signal is half the strength of the transmitted one. But in addition to the attenuation, there's also noise. And the middle plot shows the noise. Noise typically appears random. There's some fluctuations due to the environment that has no structure. In this case, what I've done is generated some small random variations to represent the noise in this case, with one exception at this time instant or this time of about half a millisecond. Half a millisecond ain't noise. Just throw that half a millisecond and then back to just small random variations. I just created this just for this example. So from the perspective of the receiver, what the receiver receives is the transmitted signal, but attenuated, so shrunk in height, plus the noise. And you make that out here that you see the receive signal is following the sine wave, approximately. It doesn't go from minus 1 to plus 1. It's going from about minus 0.5 to plus 0.5. That's due to the attenuation. Plus, it's varying over time. That's due to adding in the noise. So we could say, and I'll try and draw what that wrong picture. I'll find that picture so we can draw on it. Same one. What do we notice? The system we can think is we've got a transmitted signal. Tx is short for transmitter or transmit. It, I don't know how to draw it, we transmit the signal. It gets weaker in strength. So if we draw the transmitted signal over distance, it gets weaker and weaker and weaker. That's the attenuation over distance. So it starts out strong and as it travels across the link, it gets weaker. In our example, it gets half as weak as the transmitted signal. Plus, there's noise added in. So we think that there's also noise. And the received signal is the addition of the noise and the received or the attenuated transmitted signal. That's the model of the transmission system. I think we drew a picture of that last week as well. Now, what we want to do is work out whether the receiver receives the data correctly. And in this simple example, we're assuming this is a bit 1, bit 0, 1, 0. That's the transmitted data. So what the receiver does is they want to record the received signal and see if it represents a bit 1 or a 0. How do they do that? How do they know if it's a 1 or a 0? You look at the received signal, how do you know if it represents bit 1 or bit 0? What's the scheme that we could use? Anyone? Sorry. Amplitude. So if we look at the amplitude of the received signal, what about the amplitude? If the amplitude is positive, if it's above 0, that represents a bit 1. If it's below 0, it represents a bit 0. There's one scheme, and that's what we'll use in this case, because the idea that transmitter will transmit a positive signal. It's above 0 when we have a bit 1 and below 0, negative when we have a bit 0. Now, how much above? Well, we're going to say as long as it's positive, it's a bit 1. If it's negative, it's a bit 0. What we could do is let's say look at, not just at one time instant, look at the average over a period of time over how long? Well, how long is each bit in this case? If this was a 1 and a 0, we can think that the bit 1 starts, sorry, that's not there, starts being transmitted here, finishes here, and the bit 0 starts here, finishes here. That is, at time 0.001 second, we transmit a signal to represent the bit 1. We finish that at time 0.015 and then start transmitting the signal to represent the bit 0 and so on. So the duration for each bit is half a millisecond. Follow those numbers. Two bits in, here we go from one millisecond to two milliseconds, two bits, so one bit in half a millisecond. So what the receiver can do at that same time instant, it's about here and here, let's say measure the signal strength during that time. Maybe take the average and if the average is positive, assume we received a bit 1. And then during this next period, measure the signal strength. If the average is positive, assume we got a bit 1. If it's negative, assume we got a bit 0. And I think you see the pattern in this period, the signal is above 0 mostly. That is, the average value would be positive, so that would be a bit 1 received. Here, the values are all below the line. Negative, 0 above. And you see it will follow. And I think we'll be successful if I think you'll see the pattern until we get to this highlighted part. And that's the grayed-out part there. So the received signal was not exactly the same as the transmitted one. It's attenuated, that's the first thing. It's much smaller in height, plus it has some random variations due to the noise. But on average, in this case, it's above 0. It's positive, so we still get a bit 1. Everything's okay until we get to here. Our noise has increased due to some electrical disturbance, for example. Just for a short time, the noise jumped. See how that impacts upon our received signal. What you do, the receivers, they look at the average of these values over time. It's hard to see if you look close. I think you'll find that most... Well, there's more time when the signal is above the line than below. Therefore, the average would be positive. If it's positive, what bit do we get? So bit 1 is received here. And I think at the same time, that should have been a bit 0. So that's our bit error occurring. Due to the noise, we've got a bit error. And we've seen this before. It's not so hard to follow this one. Let's try and put a few numbers to it because of what I want to show. The signal, in particular, if we change the number of components in the transmitted signal, we can start to reduce the chance of bit errors. We've made the statement before, the more components in a signal, the more accurate that signal is, the less chance of errors. How many components in this transmitted signal? If we think of our sine equation, what is it? There's just a sine 2 pi ft. It's just a simple sine. It's not 2 added together. We don't have those 2 humps at the top. So this is a simple signal with just one component. And we've got a bit error down here. Let's look at some of those values just during that period of the error. What can we put? Focusing just on this part. What's the average value? Anyone want to have a guess? If we take all of these points along here, what's the average value be? Point 5. Is it going to be larger than minus 1? No. If you look at this line, you'd see the first point is 0. Then minus 0 point something goes along up to minus 1 and then back down to 0. If we add up all those numbers and find the average, it's going to be somewhere between 0 and minus 1. And in fact, you can find it. And I think it's about minus 0.63. If you add it up, all those values, it's definitely not minus 1 because some of the values are less than minus 1. Actually greater than, sorry, the negative is confusing. It's to do with the integral of a sine wave to find the area there. Minus 0.63. So let's say that's the average value during this time. The average value of the signal being transmitted. What about the noise? During this period, what's the average value of the noise? Approximately. I don't know. Maybe it varies about this line. It goes up above it, sometimes below it. So if we took the average of those values, it would be something around here. And if we map that back, it's a little bit less than 0.5. Let's say it's 0.45. I'm just approximating here. The numbers are not, the exact values are not so important. It's less than 0.5, but it varies a bit. Let's say the average is 0.45. So the average transmitted signal is minus 0.63. The average noise is 0.45. What's the average value of the attenuated signal? Ignoring noise. The attenuated signal that we would receive. It's not shown there. While you're thinking about it, I'll write down here. So I think we transmit a signal, the average value is minus 0.63. The noise was 0.45. But we say when we transmit the signal, it gets weaker across distance. How much weaker in our example? Half. So if we transmitted at minus 0.63, we'd receive half of that. Minus 0.315. Let's just keep it at 2 decimal points. Minus 0.31. Half of minus 0.63. So we transmit a signal at this strength. Minus 0.63. It attenuates. It would be received at minus 0.31. But there's also noise. So we add in the noise of 0.45. So the ultimate received signal is the attenuated signal plus the noise. Add those two numbers together. What do you get? As an average value. It was at 0.14. This value is the average... They're all average values. This is the transmitted signal. This is the attenuated signal. That is transmitted divided by 2. This is the noise, the average value of the noise, just for that period that we're focusing on. And this is the received signal. Attenuated plus noise. Minus 0.31 plus 0.45. Positive or negative the received signal. It's positive. Which means we receive bit 1. Positive plus VE is short for positive. That means bit 1 is received. And that was our bit error, remember? I'll go back up. We transmit at some level. It's attenuated, adding the noise. And the average received value here, if we found the average, is about 0.14. It's above 0. Therefore we assume it was a bit 1. So just try to put some numbers to that concept of attenuated signal and then determines what is received. Any questions so far? That's the easy part. Now we'll move on to a different signal and see how the signal can improve the chance of avoiding errors. This was just make note, our signal, our transmitted signal, had one component. That was just a sine wave. Let's try again with a different signal. And it's in the top left plot that you have in front of you. Here everything is the same except of our transmitted signal. Instead of sending a sine wave we send a square wave. How many components in our signal to create a perfect square wave? How many components do we need? If we add sine, the solids together. Remember you added two together, you get two humps at the top. When three, it was a few more humps and we kept adding signs together. How many do we need to get a perfect square wave? An infinite number of components. So these two cases at the extremes. What if we have just one component versus what if we have an infinite number? It's actually not a perfect square wave but it looks close to one. This has an infinite number of components. Everything else is the same. The noise is the same. The attenuation is the same. So focusing on the time period where we had an error before everything's looking fine until this period here with this spike of noise. What's the average value here? The average value transmitted. If you look at, this is instantaneous, goes down to minus one, minus one, minus one. It's always minus one. Therefore the average is minus one. The noise is the same as the previous example. Nothing's changed there. So the average we said was 0.45. The transmitted signal is attenuated. It's reduced by half. So it drops down to minus 0.5. Yeah, 0.5. Let's write it. We transmit the average level of minus one. It's attenuated. It reduces in strength as it travels across distance. Therefore the attenuated one would be half. The noise is the same as the previous example. 0.45. And the received signal is therefore the attenuated plus the noise, which is minus 0.05. Negative or positive? Okay, what bit? It implies bit zero is received. We've got exactly the same noise, same attenuation, but two different signals, one with one component, one with an infinite number of components. With the infinite number of components, we don't get a bit error. With one component, we do get a bit error. This example is trying to show the fact that the more components you have in the signal, the more accurate it is, the less chance of bit errors. If you look here, if you look at the average value, you'll see it's just below zero. Minus 0.05. These numbers are not so important, the exact values. They're just trying to illustrate that, okay, adding in the noise does it bring us above zero or below zero? Why? Why using the square wave? We don't get a bit error, but we did with the sine wave. Well, the square wave is always minus one. So whereas the sine wave only summits, or only one time point is at minus one, the rest is slightly higher. So there's more chance that the noise will cause the transmitted signal to change from, in our case, from negative to positive. We transmitted a negative signal. We add in the noise. If the received signal is negative, we're okay. But if the received signal is positive, we get a bit error. In the sine wave, the received signal was positive, causing a bit error. But in this case, it's negative because adding in the noise doesn't impact on the entire transmitted signal as much as on the sine wave. How much noise on average do we need to cause a bit error for the square wave? If we transmit the square wave, what would the noise need to be to cause a bit error? Greater than point five. That is, the attenuated signal is minus zero point five. If the noise was greater than point five, when we add them together, we would swap the sine. It would go from minus to positive, negative to positive, causing a bit error. So the noise would need to be a particular strength to cause a bit error in this case. If it's below that strength, we're okay. Whereas with the sine wave, how much noise was needed to cause a bit error on average? I'll show you the numbers just to remind you where they are. On average, how much noise is needed to cause a bit error with the sine wave? It needs to be greater than what? I think greater than half of this minus, so point three one five. We had a greater than point three one five, point four five caused a bit error. If it was zero point one, it would not cause a bit error. So given the same conditions, two different signals give different quality of the received signal and different chance of errors. So this is just a simplified example trying to show the point. More components in our transmitted signal, the more accurate, the less chance of errors of that transmitted signal. Don't worry too much about calculating those numbers. That's not the point here. They're just there to illustrate, to accompany the picture. Any questions before we move on to another part of signaling? Any questions towards the back? Okay. In the question, I'll go back to the square wave. In the square wave, on average, what does the noise need to be? What's the average value of the noise that will cause a bit error? Zero point five. That is, if we're at minus one and we add in zero point five, sorry, minus one attenuated brings us to minus zero point five. And then add in noise of point five we're actually at zero. So if it was greater than that, it would bring us to positive. So the amount of noise needed to cause a bit error that when we add it to the received or the attenuated signal, it changes our sign from positive to negative or if we're going in another example from negative to positive. Because the way that the receiver worked was if my signal is positive, I assume I received a one. If my signal is negative, I assume I received a zero. So if we transmit a one, transmit positive, but the noise brings it down to negative, it will be a bit error. So the question I may have said the wrong thing before, it's not point o five, it's point five, point five. The amount of noise is in fact half of the attenuated signal or possible range of attenuated signal. Related to that, if the attenuated signal range from, let's say, minus two to plus two, maybe a square wave that went from instead of minus point five to plus point five, but the height was from minus two to plus two, how much noise to cause an error? What's the magnitude of the noise that would cause an error in this case? Two. If the, let's say, the attenuated signal received was minus two and we add in a noise greater than two, two point one, it would go from minus two to plus point one, error. Or if the signal received was plus two and we added in a noise with a magnitude of minus two point one, it would bring us from plus down to negative. So the noise needs to be greater than half of the range in this case. We may see that again when we look at, in a moment, another signal. Let's go back to one of our earlier simple signals and look at how we can increase our data rate and we'll come back to an example of noise which you have on the printouts on the bottom two pictures. Here's a, now let's for now assume no noise. Here's a transmitted signal or let's say a received signal. You receive this signal. It had two signed components added together using our normal pattern. What's the data received? What do you think it could be? Try and think what the data received could be. If you receive this signal, what would you interpret the data to be? There are different ways but I think following what we've done in the lectures up until now, we've said if the signal is plus one, represents a bit one. If it's low, negative one in this case or close to, bit zero. So a way to interpret this when it's received is to think again, measure the signal for a period of time. Over this period of time, measure the signal strength and look at the average and if you look at the average over this time it's I think about one. It goes above one here and here but there's below and here so average those values and you get one here plus one. And during this period it's an average value of minus one. So we could interpret that this received signal means data one zero next bit. This is an easy one and so on. This is just alternating ones and zeros. The simplest sequence what's the data rate? What's the data rate of this signal? Well in our one second look at the time frame down the bottom in one second, four bits. Four bits per second. We calculated this in a previous lecture this one. So we've seen this one before. Using the same scheme before we move on what's the duration of one bit? How long is one bit with respect to our signal? We have four bits in one second. Each bit is 0.25 seconds. 0.25 0.25 0.25 0.25 So each bit is represented by a signal for 0.25 seconds. We sometimes call this a signal element. You may see the terminology come up in the lecture notes later but this sending up on average plus one is sometimes called a signal element. And the duration in this example with duration a quarter of a second we'll see that that definition of signal element later but just since we're here one signal element another signal element in one second, four signal elements. Same scheme, different received signal. What data did you receive? A signal element lasts for a quarter of a second. So in the first quarter of a second what do we get? On average it's one. Bit one. One. Not so hard. One, let's write it all. Same conditions as before. The signal element is a quarter of a second. The data rate is four bits per second and all we're doing is when we have a bit to transmit when I have a bit one to transmit I send the signal so it's positive. If I have a bit zero then I really swap it so it's negative. I change the amplitude to be negative. So we can generate any sequence or we can generate a signal that represents any sequence of bits. How do we... Now the bandwidth that we have available is fixed. We cannot change it. The bandwidth in this one we've calculated before the frequency was what? We had a frequency of two hertz of our signal. I think we had a bandwidth of four hertz. It went from two hertz up to six hertz in this signal. The bandwidth of four hertz. Let's say we cannot change that. We're fixed with the same bandwidth. We cannot change the signal equation. That is the number of components. We still have two components. We can change some parameters but the number of components we cannot change. We cannot change the for the number of components. How can we increase the data rate? How can we send faster? I want to send faster than four bits per second. The frequency we've calculated before the frequency was two hertz. Remember the bandwidth with our signal equation with two components range from F to 3F. The bandwidth was 3F-F or 2F or in our case four hertz. That was the original signal. If you increase the frequency to four hertz you will increase the data rate. But the bandwidth increases as well. It would be 12-4 8 hertz. So yes, if you increase the frequency in this case you will increase the data rate but your bandwidth also increases and I'm saying you're not allowed to increase the bandwidth. Therefore you cannot change the frequency. How else can we increase the data rate? Can I change the bandwidth or frequency? Can I change the number of components? We've only got two. I still want to send faster. Any ideas? Let's try. Here's a different signal and I'm using a different scheme. What's different about this one? The amplitude. The shape is the same these two humps because it's two sine waves added together the frequency is the same the scale is different from the previous picture but in one second you would see there are two repetitions. It's still two hertz frequency but what have I changed? I've changed the amplitude. Sometimes it's the same as before plus one to minus one but I've also introduced another amplitude it's actually plus 0.33 or one-third to minus 0.33. So we changed the amplitude in some cases and we can use this to allow a signal element to represent more than one bit at a time the scheme that we'll use or I'll use in this example is let's say I transmit I'll show you the scheme and then it'll make sense I hope the transmitter may have any sequence of bits if there are two bits 00 then the scheme I'm going to use is to transmit a signal with an amplitude of minus one if it's 01 minus 0.33 10 plus 0.33 11 plus one that's the scheme that the transmitter uses we've got a sequence of bits to send if the first two bits are 00 then it will send a signal with an amplitude like this one it goes up to plus one but if the next two bits are 10 then the amplitude will be plus 0.33 Given that the scheme just remember that 00 minus 1 01 minus 0.33 what's the bits received in this case what bits, what sequence of bits do you receive here we have a receive signal which is plus one plus one corresponded to what 11 here we have minus one minus one corresponded to what 00 plus 0.33 10 if you cannot remember see 00 minus one 11 plus one 01 minus 0.33 plus one 10 plus 0.33 here we have 0.33 so it must be 01 minus 0.33 01 again plus one 11 down the bottom it's 0011 we'll come back to the scheme how many bits did I receive there are 16 there 16 bits in 2 seconds 8 bits per second between 01 there are 8 bits received so my data rate is now 8 bits per second in the previous case we had a data rate of 4 bits per second but we haven't changed the frequency or bandwidth the frequency if you look at the signal here's one repetition in half a second there are 2 repetitions in 1 second that is still a 2 hertz signal or better a signal element still lasts for a quarter of a second same as before that is this signal element the duration is a quarter of a second so within 1 second there are 4 signal elements but the way that we gain the data rate is that for each signal element we send 2 bits and the way that we can achieve that is by having 4 different values of the signal element 4 different magnitudes or 4 different levels instead of 2 levels plus 1 and minus 1 we now have 4 levels so each level can represent 2 bits so the number of levels that we have available can impact upon the data rate if you increase the number of levels you increase the data rate that you can achieve what data is received same scheme as before what did you receive remember the scheme was if the first bit is a 1 it is positive and if the first bit is a 0 it is negative so 0 0 0 1 were negative something 1 1 1 0 were positive something and 0 0 negative 1 1 1 positive 1 so in fact plus 0.33 alright was 1 0 1 1 1 1 again 0 1 0 0 so a different sequence of bits any sequence of bits we can cover using these 4 different levels questions before we see the impact of noise on these signals so this is a new concept we are no longer assuming 2 different levels something that we haven't done in the rest of the course questions is the question related to this one or I'll get to your quiz question later then any questions about the lecture or about this part but we'll come back to that so we've got a new concept we're not restricted to having just 2 levels to represent our data we can have as many as we like in theory data rate further here we're getting 8 bits per second in the previous case we had 4 bits per second I want faster than 8 bits per second what do I do be specific what do we do I want faster than 8 bits per second but again I don't want to change the bandwidth more more what think of these as the levels of the signal okay different values of the magnitude we'll go back to the previous one where we wrote it down in this scheme there are 4 different magnitudes each magnitude or each level represents 2 bits if I receive at one of these 4 I can map it back to the 2 bits in the previous scheme we had 2 levels positive and negative each level represented a single bit so how do I make it faster again more levels how many more what's the next step up this one's 4 levels we should go up to 8 levels where each level would represent 3 bits and with 8 levels each level representing 3 bits we'd have 3 bits per signal element here we have 2 bits per signal element 2 bits 2 bits if we had 3 bits per level we'd have 3 bits per signal element or in one second we'd have 3 by 4 12 bits per second go up to 32 64 levels 10 bits per signal element so the number of levels that we use to iterate what's the problem the bandwidth isn't changing by increasing the number of levels harder to analyse for you as a student maybe but more specific maybe you're right more chance of getting errors the more levels we have the more chance of getting errors and I would not spend too much time on it but the last 2 pictures on the print out now we have an infinite number of components different example but here we have 2 levels here we have 4 levels remember we said the noise the noise cause a bit error if the noise swaps the sign when we have 2 levels for example if I transmit a plus 1 and the noise brings the received signal down to a negative value then we'll get a bit error so the noise magnitude needs to be large enough to bring our transmitted signal more than half way down that is the received signal must be closer to the other value which in this case is a magnitude of 1 with 4 levels if I transmitted it here at 0.33 1 third and to get a bit error then the noise must be a magnitude that will take me to another level to maybe plus 1 or even minus 0.33 the difference is what 2 thirds 0.66 whereas here the difference between levels is 2 so the same amount of noise when we have more levels has a more chance of causing bit errors and that's the problem with having more levels it increases our data rate but in the presence of noise we get more errors it's not so clear on that picture I'll just leave it as that explanation that an error will occur if the signal swaps to a different level or is nearer to a different level and with the case of 4 levels if it's 0.33 and the noise is let's say minus 0.4 it will be nearer to minus 0.33 and the receiver will think it's a different sequence of bits with the same amount of noise more chance of bit errors when you have more levels but the more levels you have the higher the data rate so we have a trade off there with all other conditions the same like the bandwidth if we increase the number of levels and assuming we cannot control the noise the noise is always present then if we increase the number of levels then it increases the chance of errors now it may be that the noise is so small that we can have a high number of levels but eventually we'll reach some limit we cannot have an infinite number of levels because eventually the noise will be large enough to cause too many errors we'll see an example shortly of what was a typical value for an old system so we've seen in today's example we've seen in the previous lectures increasing the bandwidth increasing the data rate there's some relationship between bandwidth and data rate today we've seen that accuracy of the signal the number of components in the signal the more accurate the signal the less chance of errors and we've seen a new thing we can have more than just two levels and the more levels the higher the data rate except if we have noise the more levels the more errors go back to our lecture notes and the last 10-15 minutes we'll look at an equation that relates some of these factors together so that you don't have to go through all these calculations someone's done it for you and come up with an equation a relationship between the factors we've talked about during this topic it's called channel capacity think of channel as the link we have a link between transmitter and receiver it's often called the communications channel capacity is what what's capacity mean the maximum we can achieve the maximum we can fit in so in our case channel capacity what's the maximum data rate we can achieve across our link given a link with some characteristics how fast can I send bits per second that's what we're talking about maximum data rate at which data can be transmitted over a given communication channel or link and people have done analysis to relate data rate we'll see in our equations data rate is represented as C capacity in bits per second bandwidth B and hertz plus other factors like noise errors number of levels I think we'll see in the next one we're looking at two theoretical models two equations that people have developed that relate these factors the Nyquist capacity and Shannon capacity developed by Mr. Nyquist and Mr. Shannon so they come up with models that give us how fast we can send data on a link but under different conditions today we'll just do Nyquist capacity here it is C equals 2B log base 2 of M that's the equation last one to remember C is the capacity or if that's confusing to you think maximum data rate we can achieve data rate measured in bits per second B is the bandwidth of our link or a channel so how much bandwidth do we have available M is the number of levels that we have in our transmitted signal in our previous examples we had one with two levels and then we moved up to four levels so M was two and then we changed to M being four so the capacity is two times the bandwidth times log base 2 of the number of levels this assumes that there's no noise never true in practice there's always noise present but this is a simple model if noise was not present there was no noise then this would give us a relationship between bandwidth, levels and data rate let's do a few simple calculations just to see it in use and I think it's in the slides that the example I'm going to go through so let's say we have a dial-up modems a run dial-up internet access you had your modem a dial-up modem and that connected to the telephone line and that telephone line from your home went to some telephone exchange so our modem let's say this is our transmitter and the telephone exchange nearby is the receiver we want to know how fast we can send data via a dial-up internet access a telephone line the one that you have coming into your home not the mobile phone a telephone line uses analog signals and for talking for voice it has a bandwidth available about 3,000 Hz and the example we use 3,100 Hz the bandwidth we have available and in fact what your modem did your dial-up modem not ADSL or cable modem but the old dial-up modems they would take your data from your computer and convert it into some analog representation of that data and send it instead of sending a voice if you ever used a dial-up modem you would have known that you couldn't make a voice call while you're using the internet it was one or the other because the voice channel was used to send data and the voice channel had a bandwidth of 3,100 Hz so let's calculate how fast we can send data according to Nyquist capacity equation and let's say the signal transmitted is the simple one where we have high and low that is two levels m equal to 2 capacity 2 times bandwidth times by log in base 2 of m just plug in the numbers bandwidth in Hz base 2 of 2 remember logarithms if you don't you'll go home and practice tonight in preparation for tomorrow's lecture log base 2 of 2 is 1 so we get 2 times 3,100 times 1 or 6,200 bits per second be careful the units of bandwidth is in cycles per second but in each cycle how many bits can we send is determined by the number of levels so the resulting answer is in bits per second in our telephone line using a dial-up modem we can send at 6.2 bits per second did anyone actually use a dial-up modem anyone remember one the noise at least anyone remember the speed of your modem 56 kilobits per second was the typical speed of modems 56 kilobits per second our question says the maximum data rate we can achieve is 6.2 kilobits per second but you said how many kilobits per second how do you get 56 the bandwidth is the same the modem was still using the voice channel we can't change the bandwidth with a dial-up internet access so how do we increase the capacity m must go up the number of levels let's try what about m of 3 does that make sense m typically will be a power of 2 we will not calculate but m of 3 means 3 levels plus 1 0 and minus 1 but what bits would they represent so maybe don't write this one down because it's wrong but if we said 3 levels say 3 magnitudes plus 1 0 and minus 1 well we could have plus 1 represent bit 1 represent bit 0 well we don't need the third level in that case it's a waste what about plus 1 represents maybe 1 1 minus 1 represents 0 0 0 represents what I don't know 1 0 but what if we want to transmit 0 1 which level do we use so that's why we can't use 3 levels to transmit any sequence of bits if we had 3 levels we would never be able to send the bits 0 1 so we need a power of 2 in terms of the number of levels so that one we will not try try m equal to 4 what's the capacity same bandwidth times log base 2 of 4 base 2 of 4 is 2 so it's double before 6200 times 2 12400 bits per second so right we increased but we want around 56 kilobits per second how many levels try and find it if you want a capacity of 56 kilobits per second how many levels do you need in the transmitted signal try and calculate and to make what it would be so again the bandwidth in this example is fixed if we want a data rate of 56 kilobits per second I'll give you a hint it's not exactly 56 I think it's something like 55 800 maybe 600 800 not kilobits if we want to achieve 56 kilobits per second to be precise 55800 bits per second that's what your typical dial up modem could achieve and we have 3100 hertz what should m be again Nyquist tells us the capacity is 2 times the bandwidth log base 2 of m so what do we have B 55800 we know B 3100 we don't know m so to find m what do we get log base 2 of m should equal I hope I got this number right by just rearranging log base 2 of m should equal 55800 divided by 3100 calculate a time what's that 9 I think it's exactly 9 or very very close to 9 55000 divided by 6000 about equals 9 log base 2 of m equals 9 therefore m equals 2 to the power of 9 512 log base 2 of 512 equals 9 because 2 to the power of 9 equals 512 if we have 512 levels remember our first signal had 2 then we had an example with 4 imagine 512 different levels all within the same small space then using the bandwidth of 3100 hertz we could achieve the data rate of about 56 kilobits per second so that's what the modem did with a signal it mapped it to one of 512 levels how can we go faster why couldn't you buy a modem with 1 megabit per second so the modems were 56 kilobits per second well if we went up to the next one up would be 1024 which 1024 would be 10 and it would be something like 64 kilobits per second to get up to a reasonable speed say a megabit per second the number of levels you'd need would be just too many because although it's not captured in this equation the more levels the more chance of errors that summarise on Nyquist the trade-offs we see from the equation are increasing the bandwidth B increases the data rate we've said this multiple times before the other thing we've seen today increasing the number of signal levels M also increases the data rate that's a good thing that's what Nyquist equation tells us but the Nyquist capacity equation assumes there's no noise but when there is noise we have other problems increasing the number of signal levels M makes it harder for the receiver to understand what it receives there are more errors it's not shown in the equation but it's true in practice as M goes up the number of errors go up so that means there's limit or practical limit in technologies to what it's a good value for M high for high data rate low for low errors tomorrow we'll see Shannon capacity equation which takes into account noise and gives us a different relationship