 So that what we're going to do in channel capacity is look at two Two formulas that tell us the relationship between the bandwidth of our signal and the data rate that we can achieve So they give us a good indicator if we know something about our communication signal We can work out or how fast can we send bits using that signal using that communications link the channel capacity But to do that we need Some a little bit of background knowledge that some of you may have some may not so let's go through for everyone We we need to talk about and introduce decibels so some of May have covered it in other topics. Let's try and make sure everyone understands what we mean by a decibel and The issues of gain and loss If if you have 1,000 bar for example and you Give that to your friend you borrow you lend the money to your friend 1,000 bar and in a week's time your friend returns and gives you 1,000 bar plus an extra 1,000 2,000 bar in total in return What's your profit? Absolute value absolute profit you start with a thousand you give it to your friend. He gives back 2,000 next week What's your profit a thousand bar you've gained 1,000 bar okay now from a Relative to the initial amount of money you had how much have you gained? Relative to the initial 1,000 bar you've you've gained 100% or think of it as a ratio or factor You've doubled your money Okay, you start with 1,000 You get 2,000 back so you've doubled your money So we can say The game in this case is a factor of two So the double times by two So we use And when we look at communication systems We look at the gain of systems and also will see the loss of systems for the same can be applied in many different systems So if I start with 1,000 and end up with 2,000 then we can say the game as a ratio or as a factor is two There are no units to this game. It's just a multiplier of increased by a factor of two What if I start 1,000 my friend gives back 250 250 bar I've got a gain of what? He's in trouble. Yes, or I'm in trouble. I've got lost money Okay, so the gain I've lost 750 bar. So then as a ratio relative to the initial amount The game we say is 0.25 one quarter. I start with 1,000. I End up with 0.25 So relative to the 1,000 I've got one quarter of it So we can say the gain in that system is one quarter zero point two five Another way to express this gain of zero point two five is what? Did I gain money? I lost money We can express that as a loss. What's the loss in that case? The loss as a ratio, how much did I lose in that case? I? Start with 1,000 but I end up with 250 bar How would you express the loss in that? Well the gain is point two five one quarter We can say That is a loss of a factor of four. I Start with 1,000 divide by four in this case. I end up with 250 so We can express these ratios as either a gain or a loss We're really gain a loss of the inverse of each other The loss is a negative of a gain when we deal with communication systems. We also need to look at gains and loss of systems We transmit With some power level We transmit a signal the signal propagates through our communication system and is received with some weaker lower power level We can say there's a loss of power in that case and we want to be able to express that Let's go through some examples and that this will lead us to decibels and the basic notation or the basic formula for decibels Now that was the example about money. Let's consider an example about say the amplifier for this audio system When I talk I produce a signal of some output strength It goes into the microphone and eventually into the amplifier in the cupboard which increases the signal strength So the signal that comes out of the speaker is Louder than what's coming out of my mouth. We have some gain in the system Let's express that and calculate as with some simple examples So We can draw this system We have some system We have some input and an output So here's our system the rectangle and the input. Let's express We have some input and we with communication systems. We often talk about power Measuring watts for example, so I'll say we have some power in P in And some power out I have a system which takes as an Some input power and if it's an amplifier it increases the input power and produces an output power and By looking at those two power levels. We can determine the gain or loss of that system So let's take an example. Let's say my input power is 1000 let's say that's to give it units watts W for watts 1000 watts in some amplifier and the output it produces is 2000 watts So zero What's the gain of my amplifier? What's the gain of the amplifier? I've started with a thousand in I End up this is a zero as what 2000 out the absolute increase is 1000 watts But relative to the input we say the output is two times larger So simply the gain of this system is two So we can express that let's say g equals to the gain is to there are no units for gain It's a factor or a ratio of multiplier output is two times larger than the input what if I started with my 1000 watts, but the output was 250 watts So I start with an input of 1000 watts and the output powers 250 watts What's the gain? What's the gain first? point two five You're ahead One quarter in this case so the inputs the same I start with 1000 watts The output is 250 watts the gain is one quarter point two five in this case It's not an amplifier or not a very good amplifier because it's decreased the output signal in this case It's produced a loss in fact so the loss in this case we could express as Four G for gain a Gain of point two five is the equivalent to the loss of a factor of four in this case four times Less than the input the output is start with one thousand divide by four and I get the output of 250 So we say the loss is a factor of four The gain is a factor of point two five So with any system when we have some input an output We can talk about its gain and loss whether it's money a financial system whether it's communication system Whether it's anything that has some input quantified input and output we can measure the gain and loss We usually at least in communication systems are talking about ratios or factors Two times larger four times smaller another example Our input is The input on the left side. Let's give a different numbers 100 milliwatts and the output we end up with One micro watt what's the gain of this system or maybe easier. What's the loss of this system? so the input power Now consider a communication system my wireless Transmitter on my laptop transmits a signal The strength of that signal that's coming out has a power level of 100 milliwatts that signal propagates through the air and goes up to the access point the receiver and We measure the received signal to be one that is a new one micro watt What's the in this case? It's a loss. We've gone from 100 milliwatts down to one micro watt What's the loss of this communication system? Try and calculate 100 milliwatts down to one micro watt. It will be 100,000 It's 100 one micro watt is 100,000 times smaller than 100 milliwatts So be careful your prefixes here 100 milliwatts is Millie to convert to micro we multiply by 1000 So 100,000 micro watts So the loss is a factor of 100,000. So I won't write the game, but the loss in this case 100,000 the output is 100,000 times smaller than the input Now in communication systems we often talk about gain and loss and Another way to express gain and loss is using decibels And the convenience of decibels is it We use a logarithmic scale and it reduces this usually reduces the size of the numbers so that they're more manageable And in fact we'll see later that we can do operations much easier So Let's convert this this loss into decibels and The general formula is And we'll see it you've got on some of the handouts in one of your initial handouts definitions and concepts we had some acronyms and Towards the bottom after units and prefixes, so this is after logarithms decibels and signal strength We'll only go through part of this But we see the general equation to convert some in this case a gain Into from the unit loss as a factor into decibels 10 times log base 10 of our gain and similar with a loss 10 times log base 10 of our loss Convert some number in the decibel form. I'll write that on the board So we can switch back So the absolute gain of a system is the power out divided by the power in The output divided by the input If my output was two thousand my input was one thousand my gain is two That's what we've calculated so for so far. The loss Is the opposite or in this case the input? divided by the output if I start with in one thousand Watts and I get an output of 250 watts and I say that's a loss of four a factor of four So that's the way to calculate the loss It's just the inverse of the game in the absolute values And that's what we did in this example. We said the input was 100 milliwatts. The output was one micro watt the input 100 milliwatts divided by one micro watt and the loss is 100,000 there are no units here. It's a factor or ratio To convert into decibels you've got it on Your handouts, but they get death decibels for gain we take 10 log base 10 of G in this case of the absolute value think of this as the absolute value. This is the value of decibels Where G is really p out divided by p in and the loss in decibels is the same equation 10 log base 10 of P in part of a p out In fact a general equation for decibels is simply 10 times long base 10 of some factor We're gonna apply it to any system So remember that if you know some absolute ratio Then you can convert it to decibels by taking the logarithm of that ratio and multiplying by 10 Let's do it on our example The loss the absolute value is 100,000 in decibels. That's equivalent to well. We take the log In base 10 of 100,000, which is you don't need a calculator there What's the log of 100,000? 100,000 now this is your high school mathematics And if you can't remember you go and take one of the less there's a lesson online You can do not many people have done it 10 to the power of five log base 10 of 10 to the power of five is simply five That's an easy one. So log of 100,000 in base 10 is five times by 10 We get 50 so we can say this loss of 100,000 Is the same as a loss of 50 DB 50 decibels They are the same. This is the absolute value. This is the value given in decibels. What about our gain? If we go back up to our first example, we went from 1,000 watts to 2,000 watts. We had a gain of two Again, we can convert that into decibels We take the logarithm of two and Multiply by 10 there you need your calculator. Okay, the logarithm of two so 10 times log of two Is about three the log of two is about zero point three zero point three oh one so ten times that is three that is our gain of two is the same as three DB three decibels Okay, so we can express our gain or loss as either an absolute value or a in decibels Gain of point zero point two five again We take the logarithm of point two five and multiply by ten So the log of zero point two five is minus zero point six Times by ten About minus six minus six point oh two So a gain of zero point two five is the same as about minus six DB a gain of minus six DB is a loss of How many DB and this is where using decibels becomes easy a gain of zero point two five is equivalent to a gain of minus six decibels a Gain of zero point two five is equivalent to a loss of Four a gain of zero point two five. We reduced by a factor of four It is the same as a loss of four. We've reduced by a factor of four So a gain of a quarter is the same as a loss of four and the expressed in decibels minus six DB The loss will be plus six DB You can do the calculation take four times by log in base ten of four and base ten of Four times by six is plus six DB And this is where decibel one reason where decibels become easier to work with measured in decibels The loss is the negative of a gain Measured measured in absolute values is the inverse But because we're taking the logarithms the properties of logarithms turns that into the negative value So a gain of minus six DB is the same as a loss of six DB a gain of one hundred DB Is the same as a loss of minus one hundred DB? So they're just the negative of each other in decibels We'll see later when we go through different communication systems many things are measured in decibels They're generally easier to work with especially when we have large values That is a loss of 100,000 Well, a large number is simply 50 DB when we have a loss of one billion Then we can easily calculate and express in decibels so what you need to know is given some ratio Again is just a ratio of output divided by input to convert into Equivalent for decibels log in base ten of that ratio multiply by ten Gain is output divided by in loss is the inverse In divided by out any questions about decibels so far easy Some of the simplest mathematics you'll see in your degree and the concepts are quite simple to be clear The loss and gain have no units. There's no units here. It's a ratio or a factor similar We say decibels is not a unit think of it. It's just indicating a different scale about loss or gain as We go through the course we'll see some variations of decibels DB watts DB milliwatts and so on but not yet given that Now that we know about DB Some simple or some common ones which are useful to remember doubling a gain of two is about 3db and if you double again a gain of four you add 3db And you get 6db so a gain of 12 db 3 plus 3 plus 3 plus 3 is equivalent to an absolute factor of 16 a gain of 16 doubling is Equivalent to 3db or about three decibels That's useful when you want to do quick analysis and comparisons of different values Okay, that's a bit about decibels. We're going to see decibels used in our capacity equations. That's why I introduced it But before we move on to there, let's look at some examples of signals So back to signals Remember we can represent a signal in the time domain or frequency domain Some signal we can break it into the sine components and find the frequencies of those components Here's part of our lecture yesterday recorded so to less than two minutes of audio Just a wave file. I opened up in some audio editor audacity And this shows the signal as a function of time so we can see the signal strength varying over time over that first One minute 45 Of course, it's not a simple sine wave Signals are more complex than that. We can zoom in and see some details and we can zoom right in to see That's our signal varying over time so the time scale now that this portion is about Five milliseconds so five milliseconds of my voice from the lecture yesterday is shown in this signal here It's varying over time in magnitude Again, it's not a perfect sine wave We've got signals are much more complex than the simple ones we've looked at in the lecture but in theory it can be broken down into a combination of sinusoid and We can identify the frequency components of this signal this software has some Capabilities to do that for us it can so currently shows the signal in the Time domain You can do some analysis and plot the spectrum which is really the signal in the frequency domain and I need to Select All of that analyze plot the spectrum and it shows me that same signal but in the frequency domain Where on this axis we have the frequencies and Here's the signal strength and one reason to introduce DB because it's commonly expressed in decibels But the main way to interpret this is that of that audio The strongest signals In these frequency ranges So from zero Hertz or tens of Hertz up to around one thousand Hertz When I'm talking the audio signals in the range from say one thousand up to two thousand not as strong as in the original frequency zero up to one thousand and Although the audio has Components ranging from five thousand up to also twenty thousand that are measured in this system They are very weak Note that this scale is using DB So from minus thirty DB to minus thirty six DB there's a difference of six DB or a factor of four six DB is equivalent to a factor of four which means the signal here Well actually here at minus thirty six is about four times weaker than at minus thirty So you need to be careful when this is a logarithmic scale So these signals are in fact all these components are very weak. The amplitude is very small This is showing that the main frequency components of the audio recorded has greater strength in this area around One hundred to one thousand Hertz That's this area. If we look at signals, we can see the main set of frequencies that they can contain Now we can move on and look at Some aspects of capacity given a signal How fast can we send data using that signal or in general that communications link What capacity in terms of sending bits? Do I have remember capacity is that the upper limit? So when we talk about channel capacity we mean the communications channel or the communications link How fast can we send bits through that communications link? That's what we care about We saw some examples yesterday in calculating bits per second given a particular signal this some people have generalized this and Come up with some equations or some theoretical models to calculate Capacity or bits per second data rate given the signal characteristics The maximum data rate at which data can be transmitted across a given channel or link and there are two Theoretical models that will go through Called Nyquist capacity and Shannon capacity named after two different people who worked on these models and they tell us What data rate we can achieve and in the equations will see the data rate expressed as C because it was called capacity The highest data rate in bits per second For a particular communications link that has some bandwidth be measured in Hertz and They may in some of the models will also consider noise and error rate. We'll see them as we come across them So what we care is given a communications channel. How fast can we send our bits? the first model is was by Nyquist or Nyquist did the main the major amount of work on this So Nyquist was a guy who did some analysis of communication systems and Come up with a fear theoretical model eventually an equation that says if we have a communications channel and there's no noise So in the perfect world, there's no noise in the perfect lecture, there'll be no noise There'll be no one talking. There'll be no air conditioning sound and so on We know in reality that there's always some background noise noise in the system, but if we assume that there was no noise Nyquist did some analysis and arrived at this formula that is The capacity of our communications channel in bits per second is Equal to two times the bandwidth of that channel times by log base two of M Where M is the number of levels of our signal So we know about if if we know the bandwidth of our channel B Say is one megahertz Then we need to know something about our signal This value M and if we can determine that we can determine the the Nyquist capacity how fast we can send bits per second So to understand this we need to understand what is M So we need a different example to show M the number of signal levels And then we'll come back to some examples of calculating Nyquist capacity Let's draw some pictures to see what we mean by the number of levels of our signal recall When we went through examples yesterday, we said with a sinusoid signal if it's high it represents one bit and low another bit Let's Remind you of that what we had yesterday was something if we Want to transmit the bits 0 1 1 0 what we could have done was sent a signal which looked like a Zero we could have sent a low portion one high the second one another high and Zero low portion So that was a simple scheme that we used or we assume where To send bits Depending on whether it's zero one We can use a different type of signal in this case either a low signal or a high signal So here we have just the simple portion of the sine wave our signal doesn't have to be a sing Single sine wave it can be a combination of sinusoid. So we could express that also say as a digital signal So another way to draw that or another signal that represents that same data is We could draw that as a square wave for example Zero we could draw a low and for one Hi So there's another signal that represents the same information low for bit zero high for bit one zero one one zero We would say this signal has two different levels high and low and The mapping was quite easy. We map one level to one bit But we can do more advanced than that we can have more than two levels in a signal and and map More than one bit to each level Let's see some examples If we want to transmit a longer sequence of bits if we want to First use just two levels We have a sequence of bits and I've made up a set of bits that we want to send Let's say just a random sequence zero one one one zero zero we want to send these ten bits and First we're going to use a signal which is as we've done before low for bit zero high for bit one So for bit zero the first bit we'd send a low signal and then high for the second bit bit one In fact, we have three ones one two three Then two zeros zero Then a bit one high again bit zero Then to finish two bit ones one one So there's an example of a signal to represent that sequence of bits where the mapping Was simple we said zero was low One was high Now let's do a different signal where we have four different levels and the mapping over the side is here We're going to have very low low high and very high We'll map zero zero to very low zero one to just low as one zero to high and One one to very high and I'll draw the signal of that same sequence of bits using this second scheme In this case we consider two bits at a time We consider the first two bits zero one and from our scheme it says we should have a low signal draw that here For zero one and then the next two bits one one very high. What do we see? First two bits zero one from our scheme. We need a low signal I've drawn at this level. It's negative The next two bits one one very high Here and then the next two bits zero zero very low So I need to come down and continue generating the signal zero zero one zero and one one So we've got a more advanced mapping from bits to signal levels with four different levels We map two bits to each level and we send the signal at a particular level for a fixed period of time So in the first case where I bat one bit to a level this was one bit two three four and so on in the second case the green case where I map two bits per level Two bits the level is low zero one the next two bits one one very high Next two bits zero zero very low here one zero one one using this scheme Under the same conditions Look at the time it takes to transmit this 10 bits in the first scheme If it's to the same scale that took to here with the green one we finish in half the time We have sent the same sequence of bits In half of much time in other words. We've sent at two times the data rate If we say that this took One second then that would be 10 bits per second and here we have 20 bits per second So by increasing the number of levels we get a higher data rate for our data transmission That's the main point that we're arriving at any questions so these schemes that I've designed here are Just examples Okay, they don't have to be the same as this but we have a mapping from bits to signal level And in this case every two bits map to one of four signal And we get if we want to send those 10 bits we get two different signals And the main point here is that When we use more levels We can send the bits faster Because every time we transmit a signal at one level we're sending two bits In this case one bit per signal period The number of levels we use in the signal Is the value m in the Nyquist capacity equation This value m So now if we know something about our communications channel in particular the bandwidth b And the signal that we're sending across it and in particular the number of levels that it has m Then Nyquist capacity equation tells us that we can calculate some theoretical data rate for that channel 2 times b times log base 2 of m the number of levels Let's go through some calculations and see that in in work consider a telephone system The modem That connects your computer to the telephone network Allows a bandwidth of 3,100 hertz What is the maximum data rate that we can achieve in this case maximum theoretical data rate? First let's make an assumption that We have a very basic signal with just two levels Find the maximum data rate in this case if the number of levels in our signal is two Use Nyquist capacity equation to find the The value of c the data rate We're not you're too far ahead. We're not on transmission media yet Go back keep going back Yeah We know the bandwidth of our channel and to get started Let's assume that the number of levels for the signal that we transmit is just low and high two levels Find the data rate for that channel the bandwidth in this case the example was 3,100 hertz for our channel And the number of levels is two so m Is two so using the Nyquist equation we can find the capacity or the data rate c Is two Times the bandwidth times log base two Of the number of levels m Log base two of two is simply one so it's two times 3,100 6,200 and be careful here the units of bandwidth is in hertz The units of capacity is in bits per second So the answer is 6,200 bits per second So that's a simple application of the Nyquist capacity equation Two times the bandwidth times log base two of the number of levels of the signal that we use That was assuming we had a signal with just two levels What if we use a different type of signal which now supports four levels? In the same channel, what do we get? What if the number of levels goes from two up to four? Everything's the same except m is now four Bandwidth is still the same We can calculate the capacity again And that's an easy one because log base Two of four is simply two So we get twice as much 12,400 bits per second bits per second the the The bandwidth was in hertz The data rate is in bits per second if the bandwidth is in megahertz Then with times by 10 to the power of six mega 10 to the power of six and get megabits per second And we see generally increasing the number of levels used in our signal will increase our data rate According to the Nyquist capacity equation That's better The more levels the better Everyone anyone remember the old home modems the dial-up modems Anyone else dial-up modem The one that made the the quacking sound like a duck When it Connects Remember never used one dial-up modem Before your adsl modem. Yeah, some people have used them Do you remember how fast your dial-up modem was? A common one was 28 kilobits per second 56 was about the maximum that we got to 56 kilobits per second for the dial-up modem The dial-up modem uses the telephone network The bandwidth supported across the telephone network is what we have in this question 3100 hertz That modem took your bits created a signal with some number of levels And transmitted them across the telephone network How many levels did that 56 kilobit per second modem use? Try and determine How many levels did your modem use? So The old dial-up modems the maximum data rate that we can achieve was about 56 kilobits per second Maybe slightly different So That means the data rate The capacity We achieved was about 56 000 bits per second And that was using the same bandwidth channel 3100 hertz How many levels did the modem use when it generated a signal? What is m? What answer? Approximate So the closest power of 2 So if you use the Nyquist capacity equation, you know vc You find m Too many, I think Oh Where was the decimal point? You know too many then 512 sounds right 5500 5500 Oh, I saw the yellow one 55800 All right If you do the calculation If m is 512 If m is 512 log Of 512 in base 2 is 9 Log of 512 is 9 So our equation becomes 2 times b Times 9 log base 2 of 512 is 9 Which is 9 times 6200 Which is 54 55.6 Kilobits per second So with these This bandwidth this number of levels you get approximately 56 kilobits per second That is the old dial-up modems Used a bandwidth of 3100 hertz and generators a signal They took the digital data from your computer Generators signal with 512 different levels Not four But if we now have 512 levels And varied the number the level depending upon the input bits How do we go faster? Okay, we want to use the same equipment We've got the same telephone network We've only got 3100 hertz bandwidth How can we go faster? More levels Just increase the levels And in practice it was usually a power of two So let's go up to 1024 levels 2048 levels Keep going up to a million levels And in theory the data rate will go up In practice The more levels we have The more chance of errors due to noise And if you go back to the assumption about the Nyquist capacity equation There's the assumption that there's no noise So in theory just increase M and you get a higher data rate In practice there is noise And when there is noise the more levels you have The more errors you'll get Which is a bad thing So you cannot just keep increasing the number of levels or M But it does show us some trade-offs Increasing the bandwidth increases the data rate Increasing the number of signal levels increases the data rate That's what this equation shows us But in practice We cannot have an infinite number of signal levels There are some limits The more signal levels The harder it is to implement a receiving rate And implement a receiver to receive that signal and make sense of it And therefore the more chance of errors that occur So usually there's some practical limit of M And in the modems it was about 512 That limited our data rate to about 56 kilobits per second Any questions before we look at the last equation About number of levels How that impacts upon our signal and our data rate Okay Sure Not sure about which Log Use your calculator Yeah okay So two questions Not sure about log Okay Be sure about log Use your calculator The calculator logarithms and so on Fine you can have it a calculator in the exam But there's a lesson on the website on logarithms Goes back to some very basics of properties of logarithms And note sometimes we have a different base Okay So Nyquist capacity uses base two So the the opposite operation is the exponential Two to the power of something Two to the power of four The log of base two to the power of four Is four The log of the base in in base two of two to the power of 20 is 20 So be careful of the base If I don't give a base Assume it's 10 So check your knowledge on logarithms The second question about okay What if the bandwidth was megahertz For example the bandwidth of one megahertz Let's say number of levels was four Our equation would give us Two b log base two of m Which is two times the bandwidth One what's one mega Mega is a multiplier of 10 to the power of six One times by 10 to the power of six Times by log base two of four Log base two of four is two times two is four times one is four It was four by 10 to the power of six bits per second So the the prefix mega giga kilo or no prefix Is just a multiplier When in a short Short cut if we have megahertz here Then the answer we could express as four mega bits per second Or four by 10 to the power of six bits per second Last or the second and last equation Nyquist assume that there's no noise And come up with this model If we know something about the channel If we assume there's no noise Then we can determine its capacity In reality in every communications channel there is noise Shannon another guy come up with a different model That took into account noise When we have noise We start to get errors And to avoid errors we If we have a higher data rate we get more errors To avoid that then in the presence of noise our Data rate or capacity is generally lower And if there's no noise So Shannon came up with this Different model relating bandwidth The to capacity or data rate C And he takes into account the noise And the signal strength or the signal power So we transmit a signal If we receive it with some power level And there's also noise in the system Which has a power level To note it as n or the noise power Then we can say the ratio between the receive signal And the noise is the signal to noise ratio s and r If we know that for a communications channel Then using Shannon capacity equation We can determine the data rate in that channel So we now have three things Receive signal strength Signal power in this equation The receive noise So transmissions from others If we know that Then we can determine the signal to noise ratio s and r And if we know the bandwidth of the channel Then we can calculate the capacity With this equation The signal to noise ratio is a factor here An absolute factor Sometimes it's expressed in dB decibels Because again it's just a ratio of two power levels Here's an example We have a channel that uses a spectrum between Three and four megahertz The channel we've measured the signal to noise ratio That is The ratio between the receive signal strength And the receive noise Is measured to be 24 dB Before answering the second part First answer the question Given that channel and the signal to noise ratio What's the capacity? What is the capacity according to Shannon's equation? If we know the spectrum and the signal to noise ratio First what's the bandwidth? Spectrum ranges from three to four megahertz The bandwidth is one megahertz Let's record what we know So with the spectrum three to four We get a bandwidth of one megahertz And we said the signal to noise ratio was 24 dB Now be careful Shannon's equation SNR is not measured in dB This value we cannot put the dB value in here We need to convert it back to the absolute value Using our general decibel equation So let's try and do that So 24 dB equals 10 times log base 10 of the absolute value of SNR Note I distinguish between the absolute value of SNR And SNR measured in dB So the subscript dB here To use Shannon capacity formula We need this value We know in dB we need this value So you need to solve this equation for SNR So bring 10 across to the other side Divide by 10 We get 2.4 equals log base 10 of SNR So what's SNR? 10 to the power of 2.4 Divide by 10 We get 2.4 equals log base 10 of SNR So SNR is 10 to the power of 2.4 And you need to calculate it for that one Anyone tell me the answer 200 and 251 of out So we just have to be careful here The signal to noise ratio of 251 means The received signal is 251 times greater than the noise Signal to noise ratio So signal 251 times greater than noise But of course we can also express that in dB 10 times log base 10 of 251 is 24 dB So two different expressions of the same value in this case The Shannon capacity equation uses the value 251 So we now get capacity is bandwidth log base 2 of 1 plus SNR And our bandwidth is 1 megahertz 1 by 10 to the power of 6 Times log of base 2 of 1 plus 251 which is 252 SNR was 251 You'll use your calculator and you'll get approximately 8 megabits per second Log base 2 of 252 is about 8 A little bit less than 8 Times by 1 million gives us 8 million 8 million bits per second or 8 megabits per second So in this case We know the channel bandwidth B was 1 megahertz We know the signal to noise ratio in decibels We needed to convert that to the absolute value And we got the value 251 for SNR 251 plus 1 252 log in base 2 is about 8 Times by a million capacity or data rate of this channel Is 8 megabits per second So we have two different models of how to determine the capacity of a communications channel They make different assumptions One is assume no noise One assumes noise If you assume no noise you just need to know the bandwidth and number of levels M But if there is actually noise Then you need to be able to measure that Or more precisely measure the signal to noise ratio And we don't have a way to measure it in this case In this class or a way to calculate it We would be measured in practice So it would be given for example in a question But carefully SNR the absolute value was used in the Shannon capacity equation Not the dB value So we determined the capacity of the channel What was our question A channel uses a spectrum of 3 to 4 megahertz Giving us a bandwidth of 1 megahertz SNR is 24 decibels We converted that to an absolute value of 251 And we determined the capacity in that case is 8 megabits per second The question is how many signal levels are required to achieve 8 megabits per second Anyone have another answer? How many signal levels are needed to achieve 8 megabits per second 16 is the correct answer Our capacity is 8 megabits per second Our bandwidth B We know to be 1 megahertz Now if we apply Nyquist equation Remember Nyquist says C equals 2B log base 2 of M The question is how many signal levels? What's the value of M? So C is 8 megabits per second 8 by 10 to the power of 6 Times 2 times the bandwidth of 1 by 10 to the power of 6 Times log base 2 of M It's an M Therefore you can find M It would be 16 Because 8 divided by 2 is 4 Log base 2 of 16 equals 4 So in that case we combined or used both of those capacity equations to determine the answer So let's summarize about the two capacity equations Nyquist came up with one equation assuming no noise So a perfect environment From that we can determine the capacity or data rate Shannon had a different model that took into account noise If we know the noise In particular the signal to noise ratio And there's another way to determine the capacity They are both theoretical models In real life we can never achieve the capacity because there are other impairments But they give us an upper limit So we can approximate what we can achieve We may not get to the 8 megabits per second But we should be able to get close under good conditions So it gives us an upper limit as to what we can achieve with a particular channel And some of the trade-offs we see Increasing bandwidth increases our data rate B goes up, C goes up Increasing the signal power also increases the data rate Transmit with a higher power, receive with a higher power The SNR goes up if everything else is fixed If SNR goes up the capacity or data rate goes up Increasing the noise, the noise power, SNR goes down and the capacity goes down More noise, lower data rate Now something the equations don't show but happens in practice Generally the more bandwidth, the larger the bandwidth, the more noise is possible So another trade-off And if you transmitted a high power, use high signal power We said high data rate but it can cause more interference Again having negative effects on other transmissions So again in practice it's not as simple as this equation But at least gives us an approximation of what we can achieve in some channel There are other practical factors that impact upon the data rate And we've finished this topic on data transmission Next week we'll look quickly about different types of media Different types of cable technologies and wireless technologies And then return and look at this in more depth Some specific schemes to map bits to signals Data to signals And we'll look at a variety of real schemes Not just some made up ones I've given him Any questions on capacity? Very useful Go and read about Shannon Claude Shannon, the guy who created this equation He did many things other than this equation Regarding communications, regarding digital circuits And regarding cryptography So many things this guy come up with Which are very important in communications, in security And in digital systems