 These hints are on the course website so you can access these so let's briefly go through them This is the front page of this year's mid-term exam. Okay? It's a three-hour exam most things You should be aware of in terms of the structure and format of the exam but some instructions, okay? It's 18 pages long It's a closed book exam No dictionary a calculator is allowed and encouraged that is bring a calculator just a normal standard calculator not your mobile phone So even if you some questions if you can calculate in your head Still use your calculator to be sure what the answer is and save yourself time Okay, so use your calculator These things are normal Okay, these last three specific to the exam when I talk about a sequence of bits I Talk about it in the order of left to right that is If I talk about the first bit of some sequence is the leftmost bit the last bit is the rightmost bit Okay, the way that you read it left to right some people Have asked questions about that before so in this example sequence of eight bits the first bit is zero The last bit is bit one is value one Assume the speed of transmission is three by ten to the eight meters per second unless I say otherwise you can assume that throughout the exam And then I give you four equations, which are the main ones that we've covered this This semester so far the free space propagation path loss model So relating all of these factors that indicate how much power do we lose across a link? Of course, you need to know how to use this equation If I give you a question with gain in DBI You need to realize that you cannot simply insert the DBI value into this equation These values of G are the absolute gains, so you'll need to convert that DBI value Back to the absolute value Okay, so be careful and be aware of what values each of these variables are They're not the DB values. They're not the decibel values Lambda of course you need to know how to determine lambda the wavelength is the speed of light divided by the frequency of the signal The equation for the gain of antenna so Depends upon the effective area and again in in react in real systems The effective area is dependent upon the size of the antenna But we don't have an easy way to calculate the effective area of some antennas So I may have to give you information like okay, I assume the effective area is 10 square centimeters Then determine the gain of that antenna and then the two capacity equations are given so Nyquist capacity and you need to understand What does M mean? B and C and similar CB and SNR. What do they mean? So I give you those equations. You need to know how to use them There are no other equations provided that we've gone through I think that the main ones we've seen in this course For example, how to determine decibels DB That's not provided. You need to know that So this is for the front page of This is given on the front page of this year's midterm exam for those arriving late then some hints So again, no need to copy these down. You can download this file on the website and read it yourself There are nine questions 100 marks in total Question one is a fill in the bank fill in the blanks type question We'll see an example of it from last year's exam. It's similar different types of different questions you need to put in the the most correct word into some statement and Two to nine are some general questions with calculations and multiple parts. They're not just a single part We cover everything up until and including signal encoding techniques So last week we finished on PCM pulse code modulation. That will be covered Some some guidelines or some hints all right show your calculations Don't just write your final answer so don't do the calculations on your on your Calculator on another piece of paper and just write the final answer because if it's wrong you'll get zero marks But if you show your calculations and your final answer is wrong, but some steps are correct You'll get some partial marks If it's a long calculation, then you do many different variations. That is you you write all over the page Make sure it's clear to a reader that Where is your final answer? What is your correct answer? So in most cases not a problem, but some people write in many different columns on the same page So just make sure it's logical to follow and it's clear to me. What your answer is Some students give multiple answers to one question multiple different answers in That case unless all of them are correct. I'll mark you wrong Okay, so some people okay, what is this value in what is the value of the power? they'll give three different answers and One may be correct if the other two are wrong, then I mark the entire question wrong give just one answer to the question All right for practice for preparation then use the the quizzes the online quizzes past exam So today we'll go through last year's exam at least some selected questions from that to give you an indicator of the types of questions Always include your units when appropriate when needed Okay, you'll lose marks if you don't include units the prefixes You can choose okay, whether it's micro watts milli watts mega watts As long as the number is correct with the right prefix that as long then I'll mark you correct But the units watts Must be included or the appropriate unit and Understand what the units and prefixes mean. What's the difference between the lowercase m milli watts lowercase b uppercase b bits bytes Okay, I will not answer questions from you. What does mb mean in the exam? No penalty for wrong answers. So attempt all questions. Okay, so Especially the fill in the blanks don't leave one of them blank if you have no idea then guess one Okay You've got a small chance or try and give give an intelligent guess but cut it down and and you you may be successful So that the hints from for this year's midterm exam again on the website you can Download and read through that and your own time Any questions about the midterm? So everything's covered Everything that we've gone through in lectures at least everyone's prepared Well, you have two weeks. So you have a bit of time to prepare for the midterm So what we'll do if there are no questions about this is go back to last year's midterm Just scroll through and and do some of the questions as we go as we we discuss them and You have last year's midterm in your handouts. I hope Go to the go to the end of your handouts and see if you have the Last pages you have last year's midterm at the end of your handouts. So that's what we'll go through now So you have it on your handouts. I'll show it on the screen Not much to see at the start. I will not go through all the questions because we will not have time I don't think but we'll go through some of them or I'll ask you to answer some of them The first one and similar for this year is this filling the blanks type question. I give you a set of Possible answers a set of terms or words and Then a set of questions or statements And you need to select the most appropriate Answer and fill it in here write it in here Okay, so the application layer protocol used by web browsers to download web pages is What's the answer? HTTP okay, we've mentioned that for web browsing several times earlier lectures So what you'd need to do is the answer must come from this set Other other answers will not be marked. Okay, so it must come from this set So simply you'd write HTTP here. You don't have to expand it just HTTP write it here So we'll not go through all of these now that's some easy ones you can practice Again wrong answers No penalty, so attempt everyone You may use these terms more than once So there may be two or more questions which have the answer of HTTP Just because we use HTTP here. It doesn't mean it's not a correct answer in another one and There may be although there may be multiple correct answers Only give one answer, okay Even if there are two two answers in here that correct just give one and That will be marked correct. Okay, just give one of the answer one answer in each question If again if you give two and one of them is correct one's incorrect then I'll mark you wrong So just give one answer for each question Any questions about these types of filling the blank questions? I can't remember how many we have this this year, but maybe 10 to 20 questions here There were 15 or so questions. They don't take long, but think carefully about them when you answer Some are not obvious some are Generally about concepts that we've covered in that in the courses Let's now just go through in this exam I'm not sure the total the number of questions that will go through some of those questions to give you a chance to Recall for what we've covered in previous Earlier parts of the semester Try this question an Encoding scheme maps 10 bits of digital data into one signal element Part a in a noise-free environment with a bandwidth of 20 megahertz What's the maximum theoretical data rate possible? You should be able to calculate that in a couple of minutes so try and do it for the next few minutes and At least try and connect or what what topic are we talking about? What information do I need to use from this question and then how do I calculate the answer? So in your handouts you have this exam Right at the end of the handouts. It's also on the website if you need Where do you get started on this? first Main question of the exam. Can you solve it as a guide? Maybe try and From the question identify the important information the numbers And then try and work out. Well, what what's the question asking for? Gives us something about 10 bits of digital data one signal element BAM width 20 megahertz It's asking for data rate And there's some other information there, so you need to try and work out. Okay now. I need to determine the data rate Maximum data rate possible. I know the bandwidth. I know something about the data and the signal elements What do we use to determine the data rate? The maximum possible data rate. I see some people have the right equation Another term we use to talk about the maximum theoretical data rate is the capacity If we have some link We want to send bits across it how many bits per second. Well, that's the data rate What's the maximum we can achieve? Well, that's the capacity of that link That's if you can solve that you can solve the solve the question No So first you need to connect. Okay. It's asking for something about capacity and It's saying in a noise-free environment. Well We know two equations that relate BAM width to capacity Nyquist capacity Shannon capacity. They're at the front of the exam What's the difference between them? Well Nyquist capacity assumes You're operating in a noise-free environment. There's no noise So this suggests that to solve this we need to use the Nyquist capacity equation which is Given at the start of the exam the capacity See or the maximum possible data rate equals two times the bandwidth log base two of M Okay, you know the bandwidth B is given in the question 20 megahertz. So, you know B you're trying to find C Then the last part is well, what is M? So now you need to know well in this equation, although I give it to you. What does M mean? And how does that relate to our question? We know B 20 megahertz We want to find C, but we need to know M What's the definition of M? What's the definition of M? What how do you describe M? Levels the number of levels Yes, so M in our equation, sorry not here M is the number of levels so Think of All right our sine wave We can have a high and a low two levels, but in fact more complex signals We can break into okay a high amplitude a medium amplitude or high medium negative and then a Large negative amplitude that is we can break it into four different heights or eight different heights so the number of levels of our signal because all of our our digital data maps to one of those levels So yes M is the number of levels Well, how do we determine M ten bits of digital data into one signal element when we have Say two levels high and low The very basic scheme we map one bit to each level let's say That that is two levels M equals two one bit per Level or per signal element so the question says okay now we have ten bits in one signal element So we have some level Which corresponds to ten bits and then we have another level that corresponds to another ten bits And we have more levels each corresponding to a different sequence of ten bits So ten bits per level ten bits per signal element So how many levels are there well ten to the power of To two to the power of ten in that case a thousand and twenty four Because if you list them all out then you'll get up to level One thousand and twenty four with ten bits ten bit numbers The number of possible values we have is two to the power of ten or a thousand and twenty four So in this case each level corresponds to ten bits That is ten bits of digital data Maps into one level or the term here we use one signal element Whatever that level is it doesn't have to be high or low as some amplitude some phase some Frequency so now we know M two to the power of ten and now you simply solve for Solve for C What do we get to we have our bandwidth? We know M Is two to the power of ten or a thousand and twenty four? Because there are ten bits per signal element, which means one signal element represents His represent represents ten bits So we need a thousand and twenty four different signal elements to represent any of those combination of bits And then we just use the Nyquist capacity equation C equals two times the bandwidth log base two of M log base two of a thousand and twenty four is You don't need your calculator there log base two of two to the power of ten is of course ten So we get ten times two times B 400 2 times 20 is 40 times 10 400 Here bandwidth is in megahertz Capacity will be in mega bits per second. So same prefix But the units are not hurt, but bits per second. So I answer 400 megabits per second Any questions so we'll go through some in more detail some will skip over or go quickly Any questions about this first one Easy the mathematics is easy the hard part is connecting the question to the the right equation to the right concept again Bring your calculator to the exam. So you have no reason to make simple mistakes in calculations And you can be fast. You don't have to use your head to calculate all of the answers For this exam you can find online. So there's also an answer sheet online There's no need to copy down or what I have here So you'll find the answers online Okay, so here's a variation if the level of noise was measured at twenty six point eight seven seven DBM received signal strength thirty three DBW With a channel of bandwidth 30 megahertz what's again the maximum theoretical data rate What's the capacity of our channel of our link and here? well, you need to connect the question which gives us asking for capacity gives us bandwidth Then it gives us something about noise and Receive signal strength Remember Nyquist Equation assumes no noise Shannon capacity Takes into account the signal to noise ratio So here you need to use Shannon capacity equation So look at the equation and then determine the the parameters in that equation So quickly try and solve that one and then we'll Move on and of course Shannon capacity equations given at the front Our maximum data rate or our capacity C is what we want to solve We have the bandwidth B. It's given The B times log base two of one plus SNR. You need to now know what SNR is signal to noise ratio So it's signal Received divided by noise. It's a ratio of the signal received to the the noise received So I've noted the bandwidth that was given as the the question says the signal received So I don't know that is s 33 DBW and N the noise received and 26 point eight seven seven DBM and Shannon capacity equation be log base two of one plus SNR Almost there Except you need to realize that well, what is SNR signal to noise ratio is easy. It's signal divided by noise so We need to know the signal and the noise value, but the problem is that in the question or In the Shannon capacity equation the values of the signal and noise are in the absolute values not in their DB values So even though SNR is simply s divided by n signal divided by noise We need to make sure the values of SNN are in the Using the right scale not using DB. So in fact, we need to convert s in this case, let's say to watts and Convert n into the same units with the same prefix and Then we can plug them into the equation so How many watts is the received signal? What is s? 33 DBW again or now you need to know that the DB equation 10 the DB value is 10 log in base 10 of some power level Let's write it here 33 DBW equals 10 log Base 10 of Some power level P. I'll say What is P? What is the original power level? So you just need to solve for P there, right? Which is in fact what I'll write here is the answer 33 divided by 10 is 3.3 and Solve for the the original power level is 10 to the power of 3.3 so convert our Decibel value into the original absolute value Using this equation so divide by 10 we get 3.3 log base 10 of something equals 3.3 That's something equals 10 to the power of 3.3 watts DB watts the resulting value is in watts and Do the same for the noise in this case? using the same approach you'll find the noise of 26.877 dBm is 10 to the power of 2.6 877 What's the unit? milliwatts dBm the m stands for milliwatts not meters decibels Relative to a milliwatts So the units here is milliwatts How many watts now if we're going to calculate signal to noise ratio S divided by n they need to have the same prefix So we have 10 to the power of 3.3 watts, whatever that value is and 10 to the power of 2.6 877 milliwatts You can use your calculator to solve and then convert the the value of From milliwatts into watts have a calculator 10 to the power of for example 2.6 877 is 487 milliwatts so then How many watts? Divide by 1,000 So the noise level in watts 0.487 1 9 834 and And then we can solve SNR which is Signal divided by noise which was our let's do it again 0.487 Is our noise value and our signal is 10 to the power of 3.3 So 10 to the power of 3.3 the signal divided by the noise about 4095 so We know the bandwidth in the equation was 30 megahertz We have now the signal to noise ratio s divided by n 4095 plus 1 4096 log of 4096 in base 2 you'll use your calculator or you'll remember Is 12 is 12 so log of 4096 is 12 so it becomes 12 times the bandwidth 12 times 30 megahertz gives us 360 Megabits per second should be your final answer in this case again Use your calculator to solve these problems again the difficult part is using the equation and Using it correctly by making sure that the values you plug into the equation are in the correct scale the correct units Convert from decibels to the absolute value for example. Let's move on We've gone through that one in detail. There's a third part you'll try and solve Let's have a look at some of the other questions We'll come back to this one because it takes a while to calculate. It's a question about Okay, without reading there's something about the protocol stack draw a protocol stack labelling each layer for one of the computers So go back to some of our earlier lectures that talked about the five layer protocol stack application transport network data link physical layer and and Related that question you can draw a picture of that stack a protocol stack is a protocol architecture that the a Picture with those five layers for example If we have time at the end, we'll come back to the calculation of overhead in this one Let's go through some others So questions about overhead bits transmitted and throughput question for is about If you read through the details about free space path loss How much power do we lose in a particular scenario we have a wireless LAN access point The workers in the company have some tablets with wireless LAN capabilities They connect to the access points So we have some access point and we have the users with their tablets connecting Some characteristics of both the access point and tablets transmit power Receive power threshold frequency and antenna gain of 6 dbi You do some experiments and measure the maximum distance between Access point and tablet that they work that they communicate is 169 meters All right, the first one. What is the transmit power of the access point measured in dbm? All right, we did a conversion in the opposite direction in the previous question of Converting db or dbm to absolute value this part a requires you to convert point one watts into dbm easy one to do What is the wavelength of the transmitted signal? How do you calculate the wavelength? wavelength what's the equation for wavelength lambda equals You need to remember it's not given in anything lambda equals Speed of light divided by the frequency. Okay, so that's all you need here What is the wavelength lambda equals that see the speed of light which is given on the front page 3 by 10 to the 8 meters per second? divided by the frequency 2.4 by 10 to the power of 9 hertz Okay, calculate the wavelength What is the gain of the tablet antenna? This is the hard part if we look at the question And if you look at the free space path loss model, we'll see that there's a transmit power Pt There's a receive power the power that we can receive the signal PR Frequency and therefore wavelength lambda we know We know something about the antenna gain 6 dbi We know that or the antenna of the access point We know the distance 169 meters if we assume free space path loss What is the gain of the tablet antenna? so Let's write down the our equation for free space path loss Again, it's the front of the exam This equation is given. So if we look at the question, we know Pt transmit power We know PR receive power We know lambda the wavelength we calculated that The question gives us the distance between the two devices at 169 meters. So we know D 4 pi D all squared One of the antennas is the antenna of the access point. The other is the antenna of the tablet They say GT for the access point the transmitting device Was given a 6 dbi so we can determine GT the question asks us. What's the gain of the tablet? say gr So we have an equation. We know all variables except one gr Then it's just a matter of rearranging to solve for gr Let's list the values. I will not solve for it go through the detailed steps, but we can list the values transmit power Pt from the question was point one watts or say 0.1 Watts Was given to be 2.77 by 10 to the minus 6 milliwatts We need to make sure both of those trans those powers in use the same prefix watts both use watts or both use milliwatts The D the distance D was 169 meters Lambda We should have calculated in the previous part But you check and find it's 0.1 to 5 meters and The gain of one of our transmitters Say gain of the transmit antenna was 6 dbi and Remember in our free space path loss equation the gain is not in decibels, but in the absolute value So convert that in that case it becomes 10 to the power of 0.6 Convert that the DB value Into the absolute value remember our general equation 10 log base 10 of some power level equals the DB value so going backwards 6 divided by 10.6 and 10 to the power of 0.6 is our absolute value so if you now look at your Equation for free space path loss you have these five variables The ones missing is g gr the gain of the antenna at the tablet So you just solve for gr plug those values actually Maybe convert our transmit power to milliwatts. So everything's in the same prefix 0.1 watt is 100 milliwatts Since the two powers are now using the same prefix both milliwatts we can plug them into the equation So we have 100 milliwatts 2.77 10 to the minus 6 milliwatts 169 meters 0.1 25 meters gain 10.10 to the power of 0.6 Take those values put them into this equation and There'll be something missing gr, and you'll rearrange and that's what you'll get gr Again, the mathematics is easy. It's how to use the equation in the right way. That's the difficult part and You'll find gr is about 2 is the absolute value, which is if you solve the equation, which is about 3 dbi Let's see what other questions we have If you don't catch the answers I draw here again. They're on on you can find the answer sheet on the website Let's look at some other questions. There's another part to that one. You'll have a look So we can get through it one Part of each question Here's a quick one Here's a signal Draw a plot of that signal in the frequency domain Try that so you need to remember the basic form of our signal equation and note that this signal has four components Determine the frequency of each of the components the peak amplitude of those components and Then plot that in this in the frequency domain So determine the component the frequency of each of those four components. I will Find the answer What's the frequency of the first component first component? 20,000 20,000 Hertz For this you just and again it wasn't given at the front of the exam But you'll need to remember the general equation for our signal or a sinusoid You need to remember that If you know that Then you know that okay these four components we can determine the value of F the frequency of each component The 2 pi f t we have 40,000 pi t therefore f the frequency of that component must be 20,000 20,000 times 2 is 40,000 times pi times t So frequency of the first component is 20,000 Hertz 20 kilo Hertz Same approach the second component will be 60,000 100,000 and 140,000 Hertz So you determine the frequencies of those four components the value of f in this equation Also the peak amplitudes the value of a 10.5 3.5 2.1 1.5 that's easy and then the plot in the frequency domain Shows impulses or spikes at those four frequencies with a height This out of the peak amplitude What does that look like? I have a plot here The four components in in kilo Hertz here 20 kilo Hertz 60 kilo Hertz 100 kilo Hertz 140 kilo Hertz and the height 10.5 3.5 2.1 1.5. So we have our signal plotted in the frequency domain the part B question says what's the absolute bandwidth and You can determine that quickly by looking at the plot The absolute bandwidth remember the difference between the maximum and the minimum frequencies the width Between 20 and 140 So simply 120 kilo Hertz is the absolute bandwidth. So what is the value of the absolute bandwidth? 120,000 Hertz with between Visually between these two points Part C. What is the value of the frequency of s one of two T that is out our signal? Well, we need to note that in this case that all of the components are Frequencies which are multiple integer multiples of 20 kilo Hertz 20 kilo Hertz 3 times 20 kilo Hertz 5 times 20 kilo Hertz 7 times 20 kilo Hertz all of the components are multiples of 20 kilo Hertz and Hence the resulting frequency of the end of the addition of those components is simply 20 kilo Hertz So when all of the components are integer multiple or the component frequencies are integer multiples of one of the others Then that other one is the signal frequency Also called the fundamental frequency So we have a fundamental frequency of 20 kilo Hertz and the other three components are harmonics So the answer of path C is 20 kilo Hertz There's a few more parts to that one again worth you reading through. So let's skip to a different question to see Get a wide coverage of the topics. Let's go to a question, which is not in the last year's exam but is in I think a quiz one of that practice quizzes So you don't have this in front of you You're making a voice call on your computer using some software like Skype the software uses a PCM encoder and Has 64 code numbers or 64 levels if remember PCM we take some analog input and Sample it and map it to one of the many levels in this case. There are 64 possible levels The software takes that analog input and Produces digital data and sends that across the internet and it sends that at a rate of 80 kilobits per second or generates at a rate of 80 kilobits per second What's the sampling period? Try and solve that one remember with PCM. We're mapping analog data to digital output What we do is we take the analog input say this blue signal here and we take samples at some regular interval we take samples of the analog input and Each of those sampled values. So the Magnitude here maps to some level. So we divide this space into a set of level levels So this sample maps to this third level this sample to this sixth level and so on So in this question your voice is the analog input and Your your software on your computer is using PCM to sample your voice so your microphone records your voice the software samples it at some interval and The number of levels used is 64 in this case. So there are 64 levels on On the vertical axis there We take samples we produce a number of bits in each sample and Then those bits are sent in this case the rate at which the bits are sent is 80,000 bits per second Determine the sampling period How many bits per sample in this case? Six bits per sample why six why six why to the power of six? 64 64 levels Our question says you're right our 64 code numbers or 64 levels means each level Over our diagram. We take a sample at this point maps to save this level three. We've got 64 different levels How many bits do we need to represent? Values zero to 63 that is 64 different values six bits because two to the power of six is 64 that means if the Decimal value was three then we need to use a six-bit value This is binary three six bits long because it goes up to the maximum of 63 and To represent the value decimal value 63 If we started at zero 64 levels zero to 63 we need six bits to represent 63 six ones in that case so every sample produces six bits of data and We're producing data at a rate of 80,000 bits per second So how many samples per second in that case? Trying to determine the sampling frequency the samples per second. We have 64 levels therefore we determine six bits Six bits are needed to represent any of those levels so six bits are created per sample But the question says we're generating 80,000 bits per second So now well how many samples per second? six bits per sample 80,000 bits per second samples per second 80,000 divided by six Which is We have 80,000 bits If we want to work the samples per second You're you're correct for the final answer. Let's take a long approach. We'll see why in a moment. We have 13,333 samples per second each sample consists of six bits giving us our 80,000 bits per second Then the question was how many What's the period the sampling period the time between samples? So samples per second. We take the inverse and we get the the number of seconds So the inverse of 13,333 gives us You're right 775 microseconds if there's 13,333 samples per second there are One sample every 75 microseconds, so that's our answer in the end 75 microseconds is a sampling period the time between each sample Which is in our diagram this duration? 75 microseconds 75 microseconds so every 75 microseconds we generate six bits Let's extend this question not in the quiz So we'd say our sampling frequency in this case is 13,333 samples per second Let's say that our voice contained frequencies from zero up to four well, that's a Three kilohertz so our application our software on our computer was taking the analog input of the voice and Producing bits as an output. Let's say that voice contained frequencies from zero to three kilohertz from zero to three kilohertz is the original analog data What's the ideal sampling frequency at this case? So if we know something about the analog input It has a bandwidth of three kilohertz the maximum frequency component is three kilohertz Then what's the ideal sampling frequency in this question that the sampling frequency was 13,333 samples per second or 13,333 Hertz But what's the ideal sampling frequency if this was the input and this was the last thing that we covered really in the previous lecture Maybe we didn't cover it in enough detail Anyone want want to attempt? What's the answer? well in the I Don't have them easy But in the when we looked at PCM at the end there was something called the sampling theorem It's actually the Nyquist Shannon sampling theorem a theorem that says that if our input data has a maximum frequency component of three kilohertz Then the ideal sampling rate a sufficient sampling rate is double that six kilohertz and The description it was more general than that but in practice if we have an input with a range of frequency components and the maximum is b say then the Ideal sampling frequency is two times that so in this case the ideal frequency would be six kilohertz Ideal in that We don't need to go larger than that less than Would give us poor quality Greater than six kilohertz would not increase the quality so Go back to your slides on the sampling theorem, and that's a useful equation Two times the maximum frequency component is the ideal sampling rate sampling frequency if it wasn't voice that I was sending but music and The music range from 10 Hertz up to 15 kilohertz The maximum frequency component is 15 kilohertz therefore the The ideal sampling frequency is 30 kilohertz two times the maximum the ideal sampling frequency or rate In this case would be two times the maximum which will be six kilohertz Two times three kilohertz. It's not two times the bandwidth. It's two times the maximum frequency component Let's go back to our exam last five minutes At least see some of the questions that are remaining Question about signals which ones are better So of the following plots there are some questions of Is a better than B or in what? Characteristics is a better than B. Okay, so think about signal accuracy and The bandwidth consumed and the data rate Again, the one that's closer It turns out in this question the one which is closer to the square wave is more accurate and less chance of errors But we'll see consumes more bandwidth if everything else is the same So there's a trade-off So there's some questions about the signals. What is an advantage of one compared to another and different signals compared Here's a question with a network with three devices and two links a Via link one to be B via link two to C First two paths asked for the transmission delay and the propagation delay. So again, there's no equations given You need to remember transmission delay is The data size divided by the data rate Propagation delay is the distance in meters divided by the speed Say three by ten to the eight meters per second. So transmission delay for a to be data size 125 bytes from a to be Link one rate of 10 megabits per second. So be 125 bytes Divided by 10 megabits per second. So make sure you convert bytes to bits in that case Then what is the propagation delay from B to C? distance is 30 kilometers Our speed of light is assumed three by ten to the eight meters per second So we can calculate the propagation delay 30,000 meters divided by three by ten to the eight meters per second Now the easy parts the hardest part In fact, it's not much harder but Now consider the entire network the two links we send data from a Through to C via B and then from C back to a so the data across both of those links and This requires you to determine the total delay and it is quite easy because all you do is you determine the the delays of each component and Then add all of them up So you determine the transmission and propagation delays queuing delays processing delays Calculate them all in both directions add them up and you get the final answer Remember delay is additive Which means if we have multiple components we add them up to get the total delay almost out of time so let's Just show you the last questions Question seven What's the data this one says we're using non return to zero I invert on ones What's the first bit in this case? Remember and RZ I invert on month ones whenever we have a bit one we invert the signal we change the the level So in the first case we change the level from low to high therefore it must be bit one Then we maintain the level so it must be zero. It's not a one. So one zero then we change one We change one one one zero zero So determine the data in that case In in this year's exam and it was the same as last year you need to remember NRZ I and NRZ L The two non return to zero techniques. I would not describe what they are But the other techniques I will give a description if I ask a question about it. So like this Manchester encoding is described as zero Transition high to low in the middle one transition from low to high in the middle That's the definition of Manchester encoding Given that For this data draw the signal So I don't ask you to remember Manchester encoding Just how to apply it if the description is given and same for the other encoding techniques the other digital data to digital signal techniques and There's another part, but that gets us to the end of last year's exam. We've covered some of the questions You will go and do the others over the next two weeks in practice for Your mid-term exam