 We introduce that decibels is used to express a ratio between two levels. We're dealing with power levels and A ratio for example between two power levels is just one power level divided by the other If we want to express that ratio in decibels, then we take the logarithm in base 10 and then multiply by 10 so if one power level is ten times or is One power level is a hundred times greater than another than that ratio is a hundred to one Or one is a hundred times bigger so that the ratio or the factor is one hundred there There are no units there But we can also express that in db. We have a hundred log of a hundred is two times by ten We get twenty db so we can say one power level is twenty db greater than the other and we spoke about We introduce the concept of gain and then finish with loss. So let's go back to our example from last lecture We had a simple example of an amplifier take some signal input and produces some signal out and we can measure the power levels of those signals and Would say the input signal has power P in and the output signal has power P out an Amplifier we know should amplify the signal the output should be larger than the input but we can measure that and we can Specify a characteristic of that amplifier being by how much larger the gain So what we did is we gave some values if the input was one watt and the output was let's say a more nicer example if the output was two watts Then the output is two times larger than the input. So we say the gain of that amplifiers to a factor of two There are no units. It's not two watts two seconds. It's two two times but we can also express that factor as In a logarithmic scale that is in decibels We take two we log by ten and log of two is about zero point three Zero point three times ten gives us about three So a factor of two is the same as 3db three decibels What's the units here 3db? There are no units don't be confused db is not a unit It's just really think of specifying a scale Convert to with no units into another quantity That quantity doesn't have units as well So even though we write the letters db after three, it's not like three seconds three meters three watts Where seconds meters and watts are units. They're measuring some quantity Db is not a unit. It's just indicating the scale. We're using So we can convert to db and we saw a very nice one to remember We double something a factor of two Approximately 3db not exactly. We're about 3db You have something You reduce it by a factor of two then you reduce it by a factor of about 3db So 3db is nice to remember because we often double or half something and The properties or logarithms is such that if we have a factor of two and we double it again to a factor of four Then that's the same as taking our 3db and adding another 3db Giving a 6db. So a gain of a factor of four is the same as about six decibels A gain of 10 is 10db and we saw some other values. So just some example values a nice thing about using decibels is even when we have very large Factors here in decibels. It's only a small number a million is 60db a billion is 90db So we're dealing with smaller numbers and Then finally we mentioned at the end loss Gain is when we increase something losses when we decrease that the inverse of each other So if we had a Trans a transmit power of one watt and a receive power of 0.1 watts We start with one we end up with a smaller value Well the same way as before the gain is 0.1 Output divided by input The input is zero point one time. Sorry the output is zero point one times the input So the gain is zero point one the log of zero point one Minus one times by ten gives us minus 10db. So we say the gain in this case is minus 10db But we can also think of that same quantity as a loss We started with one watt. We ended up with zero point one watts We have reduced by a factor of ten So we can say in this same case the loss is a factor of ten The output is ten times smaller than the input Or the output is zero point one times larger than the input. It's the same a loss of ten The output is ten times smaller than the input take the logarithm of ten. We get one times by ten. We get ten db These four values a gain of point one a loss of ten or a gain of minus ten db a loss of ten db are all the same They all express the same quantity Just using different approaches When we're using that the the factor the gain a loss of One divided by each other the inverse one divided by zero point one gives us the loss of ten One divided by the loss gives us the gain When we're using decibels the properties of logarithms instead of divide and Multiply we get subtract and add additions. So a Gain of minus ten db the equivalent loss is plus ten db a Gain of minus 50 db the equivalent loss is plus 50 db so they're the inverse in that they're additive in the inverse in this case and Another benefit of using db is that when we do operations of Sequential operations then we do addition and subtraction and Addition and subtraction is easier than multiplying and dividing when we use the factors So let's continue with a different Maybe example that will again use db and introduce some new concepts But before we do that any questions on the example from yesterday You should be able to convert a Factor or a ratio to db just take the logarithm in base 10 multiply by 10 and convert backwards Given 20 db convert back to 100 What do you do 20 db take 20 divide by 10 gives us two and Ten to the power of two gives us 100. So remember how to do both ways Let's consider another example. Let's consider our audio system that we use in this lecture room so when I speak I generate a signal and Eventually it gets to your ears But in fact if we think about the audio system that goes through multiple components when I speak The audio signal comes out of our mouth that goes to the microphone The microphone sends a signal down this wire and in my pocket. There's this wireless transmitter Okay, it goes to this box This box sends a signal wirelessly to if you look on the desk There's a box here with some antennas that receives the signal That sends a signal to an amplifier in the cabinet here the amplifier Amplifier amplifies the signal Increases it. We have a gain sends a signal up to these speakers on the walls And then it goes down through the air to your ears So we'll draw that and we'll put some numbers to that as to some examples of what is the gain and loss of each component and look at the receive power I'll try and draw that and to represent what we have So we have the transmitter We start with me standing here and I speak and it goes to the microphone. We'll have to squeeze it in Here's the microphone and The microphone is connected to this transmitter in my pocket And then it goes wirelessly to some box on the desk That's the wireless receiver on the desk And that has a cable to an amplifier and that amplifier has cables to the speakers and Then here you are With your ears that's going to receive that signal. So we have different components in this communication system Let me make up some numbers to Indicate the loss and gain of some of those components and when I say make up I'm going to just choose some numbers that Easy for me to calculate. They may not be so realistic. Okay, we don't Not too concerned about being realistic here. I just want to show some example calculations when I talk I create a Sound out of my mouth and then we can think that sound has some signal strength some strength How do we measure sound? Usually we think about sound pressure There's some pressure But let's to make it easier to cheat a little bit and say that the the sound that comes out of my mouth has some signal strength measured in watts some transmit power So I'll make up a number and say we transmit it about What When I transmit With some power PT Let's say it's 10 micro watts That's what we start with coming out of my mouth Now that's not realistic again. That is we cheated it should be a measure of pressure or microwatts per area when the Audio comes out of my mouth that goes to the microphone And then the few centimeters and then it travels along this cable to the transmitter in my pocket so from this point to this point if we measured the power at the wireless transmitter here would it be greater than Equal to or less than 10 microwatts Between my mouth where it starts at 10 microwatts goes to the microphone and then to the device in my pocket The power level at the device in my pocket will be will it be greater than equal to or less than 10 microwatts Hands up for greater than Hands up for equal to the same as Hands up for I don't know Hands up for less than All right, let's try a different one. Everyone put your hand up. See if your arms are working today Everyone okay now put your hand down if you think is put your hand up Put your hand down if you think it's less than That is the receive power here is less than the transmitter power Hand down put your hand down. It's less than Why well we know of an impairment called attenuation when I transmit a signal When it travels some distance that signal attenuates always So I transmit a signal if you measure it coming out of our mouth to be 10 microwatts When it hits the microphone, it's going to be less than 10 microwatts because over that five centimeters The signal has got smaller. It's attenuated and As that signal travels from the microphone through the cable to the transmitter. It also attenuates it gets smaller Okay, so whenever a signal travels some distance it gets weaker So it's always going to be less than what it started out as and Unless there's some component in there that amplifies Well, let's this is not amplifying at this stage. So let's assume it Just loses power. So we'll denote Between my mouth and the wireless transmitter, I'll give it a value by how much does it get weaker by let's say So we'll actually have a loss between here and here and I'm going to make I'm going to make up a value Not so realistic but easy to calculate and we say the loss is a factor of true Instead of me writing a loss of two, I'll write divide by two With a loss we start with a power The receive power will be less than by how much by a factor of two. So effectively we divide by two in other words if I transmit at 10 microwatts if the loss is a factor of two then at this point I'd receive at five microwatts It's reduced by a factor of two Let's consider the other components Between the wireless transmitter and the wireless receiver on the desk. There's also some loss The signal needs to travel through the air as it propagates through some distance. It gets weaker and Let's say I just put some nice numbers in here. It loses by a factor of 10 Across this link here will be 10 times smaller than it was at this point from the wireless transmitter to the Amplifier, there's some cabling and there's a small loss across that cable And I'll put it to be a loss of 1.6 Be 1.6 times smaller across this link The amplifier is our component that introduces a gain It amplifies It will take an input signal whatever comes in and the output should be should be bigger So I'll say that the gain of the amplifier and I'll make one up is 1,000 and I write that as multiply What that means is this amplifier takes a signal with some power in P in P out would be 1,000 times larger than P in If the input is one watt the output is a thousand watts if the input is one Millie watt the output is 1,000 milliwatts. That's all the gain of 1,000 we multiply by 1,000 It sends a signal to the speakers on the walls and let's say the the loss across the cable is also 1.6 There's not no real no reality in these numbers here I've just chosen them so that they'll be easy to calculate and then From the speakers to your ears. Let's say there's also a loss as the signal gets weaker of a factor of 10 what we want to know if that's the system What power do your ears receive the signal at if I transmit at 10 microwatts? What do you receive? Calculate it There's nothing to do with DB yet. This is just a calculation of gains and losses How would you calculate it? Well? Start with 10 microwatts divide by 2 the first loss we reduce it by a factor of 2 It'll be down to 5 then divide by 10 because that five microwatts is then reduced by a factor of 10 The next number divide by 1.6 then times by 1,000 and then continue and what's Received at your ears is the value that you get it when you calculate we start with 10 we divide by 2 Then the next stage we divide by 10 we reduce by another factor of 10 and Then a factor of 1.6 Then we times by 1,000 increase Divide by 1.6 Then to your ears a factor of 10 and that should give us the receive power 19.53 Microwatts the units we were using there as Microwatts we started with 10 we end up with 19.53 or about 20 about 20 microwatts This is a you or a mu the Greek letter mu. It's not an M. Just be careful with my writing micro Micro is 10 to the power of minus 6 Now let's do that again, but use decibels Exact same calculation, but with a different scale We said the first component had a loss of a factor of 2 convert that to db 2 How many db is 2? we 3 right that's a good one to remember half We reduced by 3 db Why is it 3 or about 3 log of 2 is 0.3 times by 10 gives us about 3 You can use your calculator, but Let's approximate and say that this component 2 is equivalent to 3 db and instead of me writing a loss Similar to where I wrote divide well, we can think where we lose 3 db It's actually a gain of minus 3 db a factor of 10 convert to db log of 10 is 1 times by 10 Gives us simply 10 db 1.6 the reason I chose 1.6 is because it converts to about 2 db log of 1.6 Times by 10 is about 2 so 2 db For the next value. We're going to lose 2 db our Amplification by a factor of 1,000 how many db calculate log of a thousand 3 We're doing log base 10 here 3 times 10 30 db Not negative. It's plus. That's a our gain. So we go up in that case Divide by 1.6 again is another minus 2 db and Divide by 10 another minus 10 db. So we start with 10 micro watts and those components of loss and gain we can measure them as 3 db or a Gain of minus 3 db or a loss of 3 db, whichever way you want to look at it a Gain of minus 10 db a gain of minus 2 db a gain of 30 db a negative gain is a loss a Loss of 2 db a loss of 10 db and we would what we could do is And I will not do it. We'll do another step first what we could do is say okay the total system Gain or loss is just the summation of those components Minus 3 minus 10 so we have minus 13 minus 15 Plus 30 brings us up to 15 db Minus 2 is down to 13 db Minus 10 is 3 db The total gain in this system is the summation of the component gains Which is 3 db if you add up those numbers you get 3 So does that make sense? Well we transmit at 10 micro watts The total gain is 3 db 3 db is equivalent to a factor of 2 We start with 10 micro watts. We gain by a factor of 2. We should end up with 20 micro watts. All right, that's what we calculated before so Sometimes dealing in db is easier because instead of dividing and multiplying on my calculator I can add and subtract in my head and find this is 3 db Which is equivalent to a factor of 2 If we consider all of those components if we add them all up we end up with a Total gain or a system gain of 3 db or a factor of 2 We convert back that becomes a factor of 2 So if our system gain is 2 then we start with 10 then we must end up with 20 and that makes sense with our previous Calculation that's what we got So it's just a different approach to do the same problem any questions on that Conversion to db don't get confused between gain and loss Okay, be aware when we're talking about one or the other now the next part is the new part Which is confusing for some people so any questions on db so far no questions good Remember we're using db to indicate a Ratio between two power levels one divided by another But just express it on a logarithmic scale It turns out we can also use this decibel or logarithmic scale to express a ratio of say our transmit power 10 microwatts Relative to some reference point so we'll introduce a new concept and let's say We want to compare two power levels, but we'll fix one of them We'll fix one to a reference point It's always the same and that will allow us to compare the other to that reference point So let's do that by our example then and come back to signif explain the significance of it Let's say we have a reference point of 1 watt we want to compare power levels to 1 watt remember the gain is One power level divided by another Let's fix This power level to be one watt always and That will allow us to compare some other power level to one watt So if our reference is one watt, let's consider our transmit power PT and Let's consider on a logarithmic scale What do we got we have ten log base ten of The actual power level ten microwatts Relative to our reference point of one watt we had a transmit power of ten microwatts We can also write that using DB on a logarithmic scale, and this is how we do it We compare that ten microwatts to our reference point of one watt Simply ten micro de watts divided by one watt. What's the answer of this? Here's when you need to remember your prefixes Micro is ten to the minus six Ten times ten to the minus six is ten to the minus five Ten to the power of minus five watts divided by one watt The watts cancel out One we divide by one we end up with ten to the power of minus five log in base ten of ten to the power of minus five is minus five times by ten Minus five times ten minus fifty, so you can check that Log in base ten of ten to the minus five is simply minus five Times by ten we get minus fifty. Here's the new thing DB. We've converted to DB That's the ten log part Instead of writing ten minus fifty DB with a reference of one watt We write W here DBW meaning our transmit power of ten microwatts is minus fifty DB Greater than one watt the W refers to one watt here, so We often use one watt as a reference point to express in a single power level But sometimes we use a different reference point Another one that we will commonly see is one milliwatt So let's do the same again same transmit power, but compare it to one milliwatt That is how does ten microwatts compare to one milliwatt and on our logarithmic scale ten log base ten What was our power level? Same again ten microwatts Divided by our reference point, which is now one milliwatt ten micro We said was ten to the power of minus five watts We need to have the same prefix to be able to do the division. Let's convert it just to simply watts One milliwatt is ten to the power of minus three watts Ten to the power of minus five divided by ten to the power of minus three. What do we get? Ten to the power of minus two Log of that gives us minus two times by ten minus twenty DB Reference point is no longer a watt the reference point is a milliwatt. So instead of writing DBW we write DBMW everyone following So this is not simply decibels. It's decibels Reference to a particular value and Instead of me saying always reference point one watt or reference point one milliwatt. We write it here minus 50 DB referred to one watt or in this example Minus 20 DB reference to one milliwatt. So we write DBMW here, but here's the trick We're lazy sometimes instead of writing DBMW We leave off the W and you'll see it just DBM. So that's the confusing part for some people. We should write DBMW, but many people will not write that W And it looks like DB meters, but it's not that M there refers to milliwatts and that's the common Notation that you'll see in practice. What does all this mean? We started with a transmit power We said it was 10 microwatts. We could also write that as minus 50 DBW Decibels minus 50 decibels relative to one watt or it's the same as minus 20 decibels Relative to one milliwatt M for milliwatt They're the same numbers same values just on different scales and It turns out in communication systems often DBW and DBM are used Because in many communication systems the power levels are either very big or very very small so in Wi-Fi the received power Maybe nanowatts a very small value Converting it to DBM makes it a nice value like minus 70 or something So often if you buy a wireless transmitter or receiver in The specs they'll specify the transmit and receive power and they'll be expressed in DBM or sometimes DBW Not in watts milliwatts and microwatts So it's important to know that any questions on converting one power level To DBW or to DBM then convert 20 microwatts. What was the received power in our case? With the first question the received power we've determined was 20 microwatts Convert that to DBW and DBM that is do the conversion, but PR equals 20 microwatts Remember from our original calculation. We determined PR is 20 microwatts What is it in DBW? DBM you may need your calculator to find the DBW equivalent divide it by one watt and Then logarithm times by 10 Here's a chance to use your phone instead of playing games and browsing the web in the lecture then use your phone to as a calculator we have 20 micro divided by one and Then log and times by 10 anyone have an answer 20 micro was logarithm of 20 by 10 to the minus 6 and In in exams I always allow you to use your calculator But no phones of course, so make sure you do have a calculator By the time the mid-term exam comes Anyone get an answer? We're going to have 20 micro is that right 20 by 10 to the power of minus 6 is the 20 microwatts and We need to take the logarithm of that because when we divide by one watt it becomes that value and then times by 10 about minus 47 DBW and in DBM we have 20 microwatts Divider by one milli watt Take the logarithm times by 10 What do you get? You'll calculate and you'll get about minus 17 DBM. Why do I know that? Look at the difference between DBW and DBM We have the same power level to calculate DBW. It's derived by one watt To calculate DBM it's divided by one milli watt There's a 1000 times difference there from one watt down to one milli watt So a factor of 1000 log of 1000 is 3 times by 10 is 30 DB difference So there's always 30 DB difference between DBW and DBM To convert you can add 30 DB Minus 50 plus 30 is minus 20 minus 47 plus 30's minus 17 Now you can calculate the the manual way or you can remember the difference between the two The other thing that we see We went from 10 microwatts up to 20 microwatts. We doubled We have an increase of a factor of two or an increase of 3 DB Minus 50 DBW plus 3 DB minus 47 Minus 20 DBM plus 3 DB minus 17 DBM So multiplying by two is the same as adding 3 DB The last thing in this example coming back to our Total gain Remember the components we had minus 3 DB minus 10 and so on let's write them again We start with Let's do it in DBM. We started with a transmit power of minus 20 DBM that was the starting point then we had a loss of 3 DB Then the next component was a loss of 10 DB then a loss of 2 DB then we had a gain of 30 DB Then a loss of 2 DB. So this is just from the top example and Then a loss of 10 DB What do we end up with add them up? So these are from the the losses and gains from our audio system But we started with a transmit power of minus 20 DBM What do we end up with a received power of? minus 20 and Here's the trick which confuses people remember DB is not a unit Decibels is not a unit. It's not like meters or seconds. So we can add these up so Don't be confused by seeing DB here and DBM here. Do we need to convert them? No We can add them up as is So we get minus 20 minus 3 so minus 23 minus 33 minus 35 Negative 17 is the answer. You're too fast for me. Good Minus 35 plus 30 we get up to minus five Minus seven minus 17, correct minus 17 watt not watts minus 17 DBM Relative to one milliwatt. So the answer of our received power is minus 17 DBM Well, that's what we expected. We said minus 17 DBM is the same as 20 microwatts when we calculated the very first way We ended up with 20 microwatts. So it's the same just a different approach The point is that when we know the component gains and losses in DB and If we know the transmit power in DBM or DBW It's very easy to find the received power. Just add and subtract As opposed to multiplying and dividing and many communication systems are analyzed in that way minus 17 DBM Is the same as our first answer 20 microwatts