 These are the hints from, where are they? This is the front page of last year's, no, of this year's exam, 2013. Tuesday the 6th of August, so check the time and room for you. This exam has 18 pages. You write your answers on the question page, question sheet. Closed book exam, no dictionary. Calculator is allowed and encouraged. Bring a calculator, it will make your life easier. There'll be some questions you will not be able to answer probably without a calculator. Like, the log of 10 to the power of 0.2, you cannot easily calculate that accurately without a calculator. Most things there are standard. Don't use your phone as a calculator. Some non-standard things. If I present a sequence of bits like these eight bits here and I talk about the first bit, then I mean the leftmost bit. So the first bit of this sequence is zero, the last bit or the eighth bit is one. So just read from left to right. To avoid any confusion in terms of significant bits. Speed of light, three by 10 to the eight meters per second. If nothing's given about the speed of transmission, assume that value. Free space path loss model is given. So then we have four equations given, which you don't have to remember, but you need to know how to use them. So these are given at the front of the exam. Things that are not given, because that's all that's given. What is, what's the equation for calculating transmission delay, propagation delay? Transmission delay, data size divided by data rate. Propagation delay, distance divided by speed. They're not given, you need to know them. Logarithms, not logarithms, decibels. DB equals 10 log base 10 of some power ratio. That's not given. So you need to know how to convert between, say, DBM and milliwatts in the opposite direction. These four are given, and we'll see in some of the questions in last year, you need to know what the values mean. What is M? B, bandwidth, C, capacity or maximum data rate. M, number of levels. So we'll see in some questions that once you work out the values of these parameters, then it's easy to solve. You find that for the unknown. But knowing how to apply these equations is the challenge. You need to know that, for example, in the free space path loss model, the gains are expressed in the absolute value. If I say the gain of the antenna is six DBI, you cannot plug six into this equation. It will not work. It will not give you the right answer. You need to convert that six DBI back to an absolute gain of 10 to the power of 0.6 and then plug that value in. So these take absolute values, not decibel-scaled values. Same with signal to noise ratio. We'll see that that takes absolute signal divided by the absolute noise level. There are nine questions, 100 marks. First question is a fill in the blanks. We'll see last year, it's similar to that. The next eight questions each have multiple parts and calculations, draw pictures, some concepts, advantages, disadvantages, and so on. Everything covered up until signal encoding techniques. That is, including signal encoding techniques. Always show your calculations. If you calculate on a separate piece of paper or on your calculator and get the answer of 23 milliwatts, and the correct answer is 50 milliwatts, then you get zero marks. Unless you show some calculations that show that you're following the correct steps and then you'll get some partial marks. So if you show your calculations, even if your final answer is wrong, you still may get a significant number of marks in that case. For example, if you just make a simple mistake in the last step and instead of tizing by two, you divide by two, then you'll get most of your marks. So show your calculations. In some cases, if you have a long answer, some people have an answer and they make a mistake and they remove some and start writing again, if it becomes messy on the page, then clearly mark your final answer, okay? That is, if you start writing in three different columns on the one page, then sometimes it's hard for me to know which one's your final answer. Most times it's obvious. On that, always give units when appropriate. If you don't give units or you give incorrect units, then you'll lose marks. And sometimes I can be, you know, you may get zero marks. If it's a simple question, what is the wavelength? And you don't give a unit of meters, then in some cases I've said, okay, zero, even if the value is correct. So units are important. Prefixes, you may choose which prefix. If the answer is 10 milliwatts, then an answer of 0.01 watts is also correct, okay? Or if you express it in DBM, it can also be correct. So the prefix, whether it's milliwatts, megawatts, kilowatts, watts, it doesn't matter. That is, as long as it's correct in the end. But the units are important. Try to use sensible prefixes. You must know what they mean. You must know what the M milliwatts here. You must know that lowercase b is bits, uppercase b is bytes. You must know those units used. There's nothing, no penalty if you give a wrong answer. So attempt all questions, especially the fill in the blanks. If you cannot, you don't know the answer, then you can guess something, okay? You will not lose marks if it's wrong. Use past quizzes, your quizzes that you've done this, this semester so far and the practice questions online plus past exams to practice for the upcoming exam. Any questions on the structure or the format or the content of your midterm exam in about two weeks? Yes, this file is already on the website so you can view this already. What we're going to go through now is last year's exam and that's also on the course website. You'll find it somewhere. And the last year's exam is in your handouts. If you go to the end of your handouts, you'll find last year's exam. So I'm just going to go through, unless you have any specific topics you want me to address, I'll go through last year's exam and we'll calculate and work out some answers. Last year's exam, you'll see a similar front page to this year. First question last year was a set of 15 mock, fill in the blank type questions and same structure this year, okay? There'll be different number of questions possibly and different questions, but I give you a set of possible answers in this table here. So a list of terms of names and then a question and you need with a blank space to complete the statement, you need to fill in that blank space with one of these possible answers. What's the answer? The application layer protocol used by web browsers to download web pages is HTTP, okay? Easy, and most of these should be easy for most students because you select from this set. Now, importantly, select only from this set of answers. Don't make up other words, okay? The answers are here, okay? So use these only, give only one answer. Even if there are two correct answers, give just one, okay? If you give two answers, I don't know, here you've got, right, HTTP and also wavelength, okay? You write two words here. HTTP is correct, wavelength is incorrect, you get zero marks, okay? You can't just list all possible answers and expect me to mark it correct if just one of them is correct. So just give one answer and students do that. They give multiple answers with the hope that one of them is correct, but no, one answer, if one's wrong, they're all wrong. And you may reuse answers. There may be multiple questions with an answer of HTTP, that's okay, just because this one's HTTP, doesn't mean the others are not. So fill in the blanks there. I will not go through them now, you can look through and practice them. Even in, this is 2012, in previous years, previous to that, there may be even some example, fill in the blanks. All of them are online, as are the answers. We don't have time to go through all these questions. Some will calculate, or I'll give you a chance to calculate. Let's just browse through them first and classify them. This one's about capacity, Nyquist Shannon capacity. Come back to them. This one is about, question three is about overhead. That just, question three is about overhead and throughput, which is some things that we covered early on in the course when we looked at performance of delay, throughput, efficiency. There may be some other things there too. Question four is about wireless systems, antenna, antenna gain, free space path loss. I'll write just path loss, that's the main question. Using the equations of free space path loss. Question five is about signal equations. I'll come back to them. I just wanna select some of them. So our topic on data transmission, we looked at the basic signal equations and bandwidth, frequency, data rate, some plots to look at the accuracy of different signals. All right, that's a long one. Question six is about delay. Transmission and propagation delay and other delay components. Let's call it delay. What's seven? Seven is about, in this case, digital or signal encoding techniques. In this one, digital data as digital signals. That's the set of questions in last year's exam. There may be more, or in fact, there are more this year and they are different. Which one should we go through first? We don't have time to answer all of these questions. Any preferences? Which one's the hardest? Capacity, throughput, delay, path loss, signal equations, any preferences? Are they all easy? Which one do you think is hardest? In fact, you have the exam in front of you. Is there any specific questions you want me to start with as opposed to just go in order? If we go in order, we may miss some towards the end. Any suggestions? Question four, path loss. Okay, let's start there. We will not go through all parts of each question again. We just won't be enough time. Question four, have a read through the question description and then try and answer part A yourself. So have a couple of minutes to read through, try and work out what the question is asking and then the first part, part A, should be quite simple to answer. Try that. And again, you should have this exam in the back of your handouts. Try and answer part A and B. They should be the easy ones. Part C together. What's the answer of part A? In DBM, we have, let's explain the scenario first. If you read through the description, we see that there's a wireless network, a wireless LAN. There's an access point like the one on the wall and then there are some users with tablets. So they are connecting to the access point. And we know characteristics of those devices. For example, they both transmit a signal with a power of 0.1 watts. A receive power threshold means this is the lowest power which they can receive a signal and understand a signal. It's the threshold. If we receive a signal with a power level lower than the threshold, my device cannot understand it. If the signal is higher than the threshold, then it can understand it. That's the meaning of this. The received power threshold, sometimes called the receiver sensitivity, is some level, some characteristic of the hardware such that if I receive a signal weaker than the threshold, then my hardware cannot understand or receive that signal. So think of the minimum power that I can successfully receive. So it's some receive power level, some frequency of the signal I'm sending. I know a characteristic of my access point antenna. It has a gain of six DBI. I did some experiments and I measured in an open field, an open environment, the distance between an access point and my tablet. And I measured that the maximum distance I can be apart which we can still communicate is 169 meters. So if I go 170 meters, the signal is too weak. The signal received would be lower than the threshold and they would not be able to communicate. 168 meters, fine. 169, okay, 170 meters, too far apart so they cannot communicate. That experiment assumed free space path loss to make things simpler. Now, before we get into the details, the first one, what is the transmit power of the access point measured in DBM? So the transmit power is given as 0.1 watts. What is the value in DBM? Well, what is DBM? One way, okay, how many milliwatts? 0.1 watts is how many milliwatts? 100, which is how many DBM? And here you need to remember your equation for DB, the general equation. The DB value equals 10 log in base 10 of our power level or our ratio. So if we have 100 milliwatts, reference to one milliwatts, that's what the DBM, the M means. M in DBM means one milliwatts. So 100 milliwatts relative to one milliwatts, log of 100 is two times by 10 is 20, 20 DBM. Everyone got that? So know how to convert between watts, DB, M, DB, W, milliwatts and so on. So know the general or remember the general equation when we're converting, it's actually P1 divided by P2. Some value in DB is 10 log base 10 of one power divided by another power. DBM, that other power P2 for DBM, P2 equals one milliwatts for DBW P2 equals one watt. That's the definition. So really you just need to remember this. If you have some power level, 100 milliwatts in DBM, it's 100 milliwatts divided by one milliwatts, which is 100 log, which is two times 10 is 20 DBM. And you should be able to go the opposite direction as well, given 20 DBM, how many watts? The next question is, what is the wavelength of our signal? What's the answer? How do we calculate wavelength? The wavelength is the speed of light C divided by the frequency of our signal. And in the question, the frequency is 2.4 gigahertz. In our case, well, it's our frequency 2.4 gigahertz and with your calculator, 0.125 meters is the answer there. So we have our wavelength of our signal. We know the transmit power. In fact, we have it in this case in three different values, 0.1 watts or 100 milliwatts or 20 DBM, whichever one we want to use, they're all the same. That are the easy parts. Part C asks, what is the gain of the tablet antenna? How are you going to calculate that? You know something about the setup, you have an access point, say on the wall, someone has a tablet and we know that if we're 169 meters away from each other, we can communicate, but any further, we cannot. Let's try and draw that and keep track of the values that we have. So for example, if we have our access point on the wall and we have our tablet, then over a distance of 169 meters, we can communicate. Let's say the access point transmits with some power PT and the power is given as 0.1 watts or 20 DBM. This maximum distance of 169 meters tells us that when we transmit at some power PT, 0.1 watts, we receive at a power which is just strong enough for the receiver to understand because if we go any further, we'll receive at a power which is too weak to understand. So we receive at a power level PR which is equal to our threshold, which is given in the question to be 2.77 by 10 to the minus six milliwatts. So what it's saying is if we separate by 169 meters and one of them transmits with a transmit power of 0.1 watts, then the other one, if we're in free space, we'll receive with a power of 2.77 by 10 to the minus six milliwatts. Because if we go to 170 meters, the received power will be lower, which will be below our threshold. If we were closer, the received power would be higher than the threshold, which is good. But if we go further apart, it would be lower than the threshold, which means we would not be able to receive. So in fact, if we separate by 169 meters and transmit a 0.1 watt, then we receive at 2.7 by 10 to the minus six milliwatts. Let's record those values. So over distance of 169 meters, transmit power is, what, 100 milliwatts. Receive power of 2.77 milliwatts by 10 to the minus six milliwatts. Over distance of 169 meters, we have a antenna at the access point with a gain equal to six DBI, says in the question. We have a signal with a wavelength of, we just calculated, 0.125 meters. Is that correct? Yeah, 0.125 meters. And in free space, we can relate all these factors together. What the question asks is, what is the gain of the tablet antenna? What is GR in this case? We know PR, PT, D, GT, Lambda. We don't know GR, the gain of the receive or the other device antenna, but we also know the free space path loss model, that equation that relates all these factors together. If you go to the front of the exam up the top, we have the equation that connects these factors together. We know PT, 100 milliwatts. We know PR, 2.77 by 10 to the minus six milliwatts. We know Lambda, D is 169 meters. We know GT is six DBI. We need to find GR. Since we have this equation, it's just a matter of rearranging the equation to find GR. With the one thing to carefully consider, the gains in this equation must be in their absolute value, not in their decibel value. So we need to do a conversion before we use this equation because our gain of six DBI needs to be converted. Six DBI in absolute value that is 10 to the power of 0.6. There are no units on the game. It's a factor ratio. Because from this equation, six DBI equals 10 log base 10 of some value. Log in base 10 of 10 to the 0.6 is 0.6, times by 10 is six. So now we have GT, 10 to the power of 0.6, PR, PT, Lambda, D. Plug them into your equation, find GR. And you have the answer. I'm not gonna do the calculation. You can check that. Just need to calculate it to calculate. So the hardest thing here is to first determine that this question is asking something about power lost in a wireless system. Come back down. We know it's a wireless system. We know we have one device transmitting to another. We know the characteristics of those devices. Gives us the hint here. We perform an experiment in the conditions of free space path loss. So we can use the free space path loss model here to relate all the factors together. And maybe the hardest step is to realize, well, what is the receive power? Where if we're at the maximum distance of 169 meters, the receive power will be exactly the threshold. And therefore we know PR. Then it's a matter of determining the wavelength, Lambda, and converting our gain into an absolute value, 60 BI to 10 to the power of 0.6. Then you can use a provided equation to get the answer. And I've done it before of two or three DBI. If you plug in those values, you'll get GR equal to two approximately. And you don't need three DBI, but just for further info, that's the same as about three DBI. 10 log of two times 10 is three. You said that was the hard question. Is it easier now? Any problems? So recognize to use the free space path loss model and then find the parameters. And given the known values, there should be one unknown and find that unknown value. The answers to the exams are also available online. So you have the exam questions in front of you. The answer sheet is also on the website. You can check later the detailed values and the exact calculations. There's another part, and we won't have time to go through it. It takes some time. It's maybe harder again. At least it looks hard. The free space path loss model is one model to determining how much power do we lose between transmitter and receiver? There are other mathematical models. So we don't have to use the free space path loss model. There are others. This question gives you one other. It needs a detailed description of how it works, but it gives you an equation for saying how much power is lost between transmitter and receiver. You need to, in this question, use that to find some or to answer some questions. A good practice question for you to go through in your own time. Let's try another one. Any preferences? Your experts of path loss now. Which other question looks hard? Have the lecture notes. At the end of the lecture notes, there's the exam. Go to the end and you'll find the exam. Not this year's exam, last year's exam. Question seven, okay. Jump through to question seven. I think we can go through parts of it reasonably quickly. What's the data? Quickly solve this one. Here's a digital signal. The technique is called non-return to zero invert on ones. NRZI. What is the data? Well, you need to know what the scheme NRZI does. And the I part is the hint. I means invert or inverted. Means we change the level whenever we have a bit one descend. When we have a bit zero, we don't change the level. We invert on ones. So if the blue line is the signal, what's the first bit? It's a one because at the start of that signal period, we invert the level. We change the level from low to high. So the first bit must be a one. So this bit corresponds to one. What's the second bit? Zero, we don't invert. That is, we're here at high. The next bit, well, we maintain the level which means it must be a bit zero here. The next bit, well, we invert which means we have to a bit one. Invert again, one, one, one. And then we maintain the level for the last two, zero, zero. So to solve this, you need to know what does, how does NRZI work? And you need to remember that. I will not give that to you. Easy way to remember, invert on ones. NRZI, invert on ones. The other scheme is NRZL, NRZ level. In that case, it's simply high for bit zero, low for bit one. That was the easy part. Now draw the signal for Manchester encoding. So the question gives you the definition of Manchester encoding. You don't need to remember that. You need to remember NRZ level and invert on ones. But the other schemes, I will give you the definition if needed. But you need to know how to apply them or to do the mapping from data to signal. To save some time, we see the answer here. With Manchester encoding, when we have a bit zero to send, in the middle of the interval, we'll make a transition from high to low. So the first bit is in fact a bit zero. So here's our bit interval. In the middle of that interval, we transition from high to low. And the second bit is a bit zero. So again, in the middle of the interval, we must transition from high to low. Since we were at low before, it means at the start of the interval, we must go up to high so that we can go from high to low in the middle. So with zero, it's always, if we look at this part, high then low. If you look zero, zero. And one is the opposite from low to high. One, zero, one, one, one, zero. So you just need to make sure in terms of the signal that if we are low and the next bit is a bit zero, then we need to make a transition up to high. Here we were low after the bit zero. The next bit is a one. We don't need to change the level at the start only in the middle. So always a transition in the middle where they're Manchester encoding, high to low, bit zero, low to high, bit one. Sometimes a transition at the start of the interval, only when needed. There's a part C, good one to have a read through in your own time. I will not go through all parts. Of course, you may get questions about different schemes. We've gone through some in the lecture. There may be new schemes, but I will not require you to remember them. I'll just ask you, well, here's the definition of Manchester encoding. Now solve some problem. Same with the other schemes. Remember NRZI, NRZL. Well, a little bit of that one done. Any other preferences? Three. Okay. Actually, that's the answers. Let's go back to the questions. Have a read through. We have two computers connected by one cable and they're using some specific protocols to communicate. Now first, remember back to our five layer stack. You need to memorize those five layers from the top application, transport, network, data link, physical layers. They're the general or they're the names of those layers. Here, in each of those layers, we use a specific protocol. In this question, it tells us what protocols are being used. It says, we're transferring a large file from one computer to another using a protocol called TFTP. We haven't mentioned this one before, so it's something new. It's called the trivial file transfer protocol. It's for transferring files, but that doesn't matter. So the application uses TFTP. So that's the application layer protocol which uses UDP as a transport protocol. So at the transport layer, we have UDP. What's the protocol of this set? So if this is application, UDP is transport, which one's network layer? We've got one, two, three, four, five. It almost gives you the answer there. If TFTP is application, UDP is transport. The question says that. Which one's network layer? If you read through, the last two are physical and data-link. So it leaves just one remainder, remaining, IP. Our first question is, draw a protocol stack labeling each layer. Well, let's draw it in detail. And so, there are the five layers that you need to remember. From the question, what protocols are used in each layer? At the application layer, we use TFTP. That's for transferring the file. The question says that TFTP uses as a transport protocol UDP. So the protocol here is UDP. The question also says, physical layer, here's the hints, physical data-link. Is IEEE 802.3 Ethernet? In fact, they're the same name. So I'll say at the data-link layer, in fact, the data-link layer and the physical layer use the same standard for their protocols, although they're different protocols. So what's left? The network layer from our question, sorry, IP, the internet protocol. So that's a detailed protocol stack for one of the computers. That is, my application has a 100 megabyte file. We use TFTP to transfer that file to the other computer. But what happens is that TFTP adds some extra data, splits the file into parts, sends it to UDP, which adds some header, sends it to IP, which adds some more header, and so to the bottom two layers and then transmit that across the link to the receiving computer. Recall that the protocols commonly add some header information. For example, for addresses, for sequence numbers, to keep track of things so that the protocol works correctly. Coming back to our question, how much overhead is in each packet sent by the source computer? Well, let's work out how big is each packet sent by the source computer. We start with a 100 megabyte file. What, at the start, the top layer we use is TFTP. It says the maximum allowed data size is 512 bytes. So we've got 100 megabytes, but TFTP must break it into smaller chunks, which are no more than 512 bytes long. So we can do the calculation later of how many chunks, but we know that we split that 100 megabyte file into chunks of 512 bytes long. So we have 512 bytes of data, and we add a four byte header. Then we send that to the next layer, which is in fact UDP. That allows up to 65,000 bytes. Well, in fact, we're gonna have much less than that, 512 plus four, adds a header, sends it to the next layer, which is IP, adds a header, and sends it to the subsequent layers. Let's keep track of how much header is added to that original amount of data. So TFTP adds four bytes. Let's write it here. I'll just write, these are the number of bytes added in terms of header to every packet sent. UDP adds eight bytes, IP 20 bytes, the data link layer adds a header of 14 bytes, and a trailer of four bytes. So a total of extra over header adds is 18 bytes. The header is at the front of the data, and the trailer is at the end, but they're both counted as overhead. And the physical layer adds four bits. All right, we're counting bytes here, let's say half a byte. So when we have our original piece of data, which we'll see is 512 bytes long, we add all of these headers and trailers, and then we send that entire packet across the network. So if we add them up, what do we get? 50.5 bytes, 50 and a half bytes. So we started with 100 megabyte file, but our application limits, well, specifically TFTP limits to send just 50 or 512 bytes at a time. So in fact, we have to break that 100 megabytes into 512 byte chunks. How many chunks? This is not solving part B, but it's actually leading us to part C. So we'll do it while we're here. If I have 100 megabytes, 100 million bytes, and divide it into 512 byte chunks, I have 195 312 packets of 512 bytes, 312, and the last packet would have the leftover. The maximum size allowed is 512, but if you work out 195 312 times 512, it's slightly less than 100 megabytes. It's the wee one packet, which is 256 bytes. You can check that, and if you multiply 512 by 195 312, add 256 bytes and you get exactly 100 megabytes. So what we've done is we start with 100 megabytes. We have a limitation. We can't send any more than 512 bytes at a time. So we break it into 512 byte chunks, 195 312 chunks. Plus at the end, there's it's leftover 256 bytes. So that's the last chunk. And for each of those, we take the data and attach, if we sum up the other parts, 50.5 bytes of header. That is for every one of those packets or chunks, we attach 40, sorry, four plus eight plus 20, plus 18 plus a half a byte of overhead. 50.5 bytes is the last chunk. 50.5 bytes is 404 bits. What was part, what was the question? Because we've solved multiple questions here. How much overhead is in each packet sent by the source computer, answer is 404 bits. It's the extra information we send. So part B answer 404 bits or 50.5 bytes. How many packets must be sent by the source computer to deliver the entire file? If there are no errors, what's the answer? It's 195,313. Okay, it's 195,312 at this size, plus one at this smaller size. So how many packets in total? It's the sum of those two. And what's, what is the total number of bits transmitted across the link from source to destination? So how many bits in total? With every packet, we have 404 bits of overhead. For this, well, for the first 195,312 packets, we have 404 bits of overhead plus 512 bytes of data. And then for one packet, we have 404 bits of overhead plus 256 bytes of data. Add them all up and you get the answer to this part D, how many bits in total. Or another way to calculate that is to say, okay, the amount of bits in total, we have 100 megabytes of data plus we have 195,313 chunks of overhead. So 195,313 times by 404 bits is the overhead. And the data is 100 megabytes, the original data. And you add them together and that's the total bits sent. Total, we have, that's the number of packets times by the overhead bits plus, that's the overhead in total plus the original 100 megabytes. 100 by 10 to the six bytes. So times by eight to get bits. And if you add them up, you get the answer. Total bits sent is data plus all overhead. And I've done the calculation before and it equals, let's finish this question. Around 878 million bits is a total sent. And remember, our data was 100 megabytes or 800 million bits. We started with 800 million bits of data plus overhead. So now we can determine an efficiency of using this link because to deliver 800 million bits, we actually need to send 878 million bits. So the efficiency of using this link to transfer our original data is the ratio between the original data and the total. You can express it as a percentage. So 800 million divided by 878 million gives us an efficiency of 87.89%. Let's check, not 87, I need my calculator. 800 million over the total, 91%, 91.02%. Which is saying that 91.02% of the time we're sending real data, the other 9% of the time we're sending overhead because the overhead contributes to that 78 million bits. So that was simply 800 million divided by 878 million to give our efficiency expressed as a percentage. And I think from that you can answer all of those parts of that question. Any questions on that calculation? That one's maybe one of the more complicated ones. One more in the last five or 10 minutes. Which question? Two, five or six. Two is about Shannon and Nyquist capacity. Five is about delay calculation. No, five is signals and six is the delay calculation. Any preferences? Apart from go to lunch, what's your preference? Two, five or six, which is harder? Five, okay, good. Plot the signal. Here's a signal equation, draw a plot of that signal in the frequency domain. Recall in the frequency domain, we plot the frequency components versus their amplitude, their peak amplitudes. So first step, determine the frequencies of each of the four components. We have four components in this signal, easy. What are their frequencies? That's your first step. And that equation you need to remember or need to know to solve this or help you solve it. The general equation for our sinusoid, the peak amplitude sine two pi times the frequency times t, the time plus phase five. Well, look at the pattern or the each of those components individually. 10.5 sine 40,000 pi t. Phase is zero, we're gonna ignore that. Peak amplitude A is 10.5. F is what? What's the frequency of this first component? We have 40,000 pi t. The general form is two times pi times the frequency times t. What is F? 20,000. So the frequency of the first component is 20,000 hertz. Then do that for the last three components. Find their peak amplitude and their frequency. First one, frequency of 20,000 hertz. Second one will be 60,000. 100,000 hertz. 140,000 hertz. And the peak amplitudes are easy. They're just the multipliers in the front. Given those values, how do we create the plot? I have the answer here. Here's our plot. It's quite simple. At each of those frequencies, we draw an impulse with the height equal to the peak amplitude. So 20,000 hertz. This is measured in kilohertz, the scale here. 20,000 hertz, peak amplitude 10.5. 60,000 hertz, 3.5. 100 kilohertz, 2.1. 140 kilohertz, 1.5. So there's our answer. The plot of a signal in the frequency domain is the components and their amplitudes. What's the absolute bandwidth of this signal? Bandwidth, just look at the picture. Look at the width of the spectrum. It ranges from 20 up to 140 kilohertz. So the width is 120 kilohertz. The difference between the minimum component and the maximum. So bandwidth of this signal, 120,000 hertz. 120 kilohertz. What is the value of the frequency of this signal? If you check the frequencies of the components, we see that they are all integer multiples of one of those components. We have 20 kilohertz. We have 3 times 20 kilohertz. 5 times 20 kilohertz. And 7 times 20 kilohertz. In that case, it means the resulting signal has a frequency of 20 kilohertz. Because there are all integer multiples of one of them, that one of them is the fundamental frequency. The others are harmonics, harmonic frequencies. So the answer is the signal frequency is 20 kilohertz. And it's only true when the other components are integer multiples of that one. Last one. Here are eight plots. If we zoom out, there are eight plots given. Which one is the signal that we just analyzed? So now we have the signal in the frequency domain. Now we plot signals in the time domain. So time versus amplitude. There are eight different signals plotted here. Which one do you think is the one that we started with? Well, the hint, look at the frequency in the plots. You cannot see at this level of zoom. But with a frequency of our signal of 20 kilohertz, it means the period is 1 divided by 20 kilohertz. Which is, well, 50 microseconds. So then if you look in the scale here, you at least find the one which has a period of 50 microseconds. I can't remember which one it is. All right, then ours had four components. It's not this signal, because we know this is just a single sine wave. There's just one component here. So it's not these two. It's unlikely to be this one, because a perfect square wave would need more than four components. And I think then if you can remove the ones which have the wrong frequency, you can find out the answer is C. You have to check the details. With a period of 50 microseconds, some of them have the wrong period. They are not our signal. Some of them have the wrong shape. They are not our signal. And it leaves just one as the possible answer. And the answer is C, I think, in that case. So that's one where you need to work out and remove the ones which are not possible.