1 00:00:00,012 --> 00:00:04,555 In this video, we're going to talk about the Wireline Communication Channel. 2 00:00:04,555 --> 00:00:09,211 What that means is that the transmitter and the receiver, are connected by 3 00:00:09,211 --> 00:00:12,921 literally a wire. It also could turn out it means it could 4 00:00:12,921 --> 00:00:16,306 be optic fiber. It just depends on which situation. 5 00:00:16,306 --> 00:00:20,932 This has, these kind of channels have very different characteristics than the 6 00:00:20,932 --> 00:00:24,182 wireless channels, discussed in the previous video. 7 00:00:24,182 --> 00:00:29,410 And lets see how that word. The wireline channel I have a dedicated 8 00:00:29,410 --> 00:00:35,206 piece of wire connecting them, that's the idea, could be more than one transmitter 9 00:00:35,206 --> 00:00:41,486 and one receiver on a given piece of wire. It turns out to be the way ethernet works. 10 00:00:41,487 --> 00:00:47,525 We're not going to talk much about that but basically, that is also a wireline 11 00:00:47,525 --> 00:00:51,361 channel. In designing a wireline channel Maxwell's 12 00:00:51,361 --> 00:00:56,653 equation come into the act again because you have to minimize what's called 13 00:00:56,653 --> 00:01:00,480 crosstalk. So, let me take a simple example which is 14 00:01:00,480 --> 00:01:05,948 known as the two, two-wire pair. You need two wires because lets assume a 15 00:01:05,948 --> 00:01:11,800 sending station is a voltage source and you can't apply a voltage except if you 16 00:01:11,800 --> 00:01:17,900 have two places to apply that voltage. So, you need a, that's why you have a two 17 00:01:17,900 --> 00:01:21,739 wire pair. And let's assume the receiving system 18 00:01:21,739 --> 00:01:25,413 looks like a simple load resistor, for example. 19 00:01:25,413 --> 00:01:30,756 Well, suppose there is another two-wire pair sitting over here. 20 00:01:30,756 --> 00:01:35,756 Well it turns out both of these pairs look like antennas. 21 00:01:35,756 --> 00:01:40,081 What is sent by one will be picked up by the other. 22 00:01:40,081 --> 00:01:46,231 That's called crosstalk, and in general this is a very bad situation. 23 00:01:46,231 --> 00:01:52,334 It's like you picked up your telephone and all of the telephone calls, currently 24 00:01:52,334 --> 00:01:56,471 active in your neighborhood, you could hear them all. 25 00:01:56,471 --> 00:02:00,046 Particularly, that's not probably what you want. 26 00:02:00,046 --> 00:02:05,211 What it turns out a simple change in that two-wire system can mitigate the 27 00:02:05,211 --> 00:02:08,796 crosstalk, and that is by twisting the pairs together. 28 00:02:08,796 --> 00:02:15,286 So you take the two wires and nearly twist them, it's called a twisted pair and 29 00:02:15,287 --> 00:02:21,747 greatly reduces the crosstalk. It turns out if you solve masses equations 30 00:02:21,747 --> 00:02:29,163 for this much more complicated geometry, you can see that the crosstalk is greatly 31 00:02:29,163 --> 00:02:34,462 minimized. An even better way of minimizing crosstalk 32 00:02:34,462 --> 00:02:42,263 is to use what's known as a coaxial cable. And here I show a photograph of a piece of 33 00:02:42,263 --> 00:02:47,288 coaxial cable. You might recognize this as an TV cable 34 00:02:47,288 --> 00:02:53,444 that you use to connect to especially cable TV, but it's also used for 35 00:02:53,444 --> 00:02:58,825 electronic cable suite, cabling televisions together. 36 00:02:58,825 --> 00:03:05,539 Here I show a cross section of what it looks like, and here's a real live cross 37 00:03:05,539 --> 00:03:12,711 section of my piece of coaxial cable. So, the black part here is insulation to 38 00:03:12,711 --> 00:03:20,416 keep other wires from touching it. And inside you see this copper, the bright 39 00:03:20,416 --> 00:03:26,817 orange just gives away copper. So, that's one of the conductors. 40 00:03:26,817 --> 00:03:32,233 In between this central conductor, here's the other one. 41 00:03:32,233 --> 00:03:38,273 In between those two is the, what's here white, that's the dielectric. 42 00:03:38,273 --> 00:03:45,363 This case that I'm showing you is Teflon. It's a dielectric, especially designed for 43 00:03:45,363 --> 00:03:50,120 the piece of coax. And so a voltage is applied between the 44 00:03:50,120 --> 00:03:56,766 outer conductor and the inner conductor. And that's how the signals are sent. 45 00:03:56,766 --> 00:04:03,272 And it turns out if you solve Maxwell's equations for this geometry you'll see 46 00:04:03,272 --> 00:04:10,172 that it basically does not radiate at all and leads to a very good approximation so, 47 00:04:10,172 --> 00:04:13,860 multiple. Coaxial cables can carry separate 48 00:04:13,860 --> 00:04:19,153 communication signals without interfering with each other and that's good for 49 00:04:19,153 --> 00:04:23,277 wireline channels. Now, we need to explore the properties of 50 00:04:23,277 --> 00:04:26,451 the wireline channel, it's very interesting. 51 00:04:26,451 --> 00:04:32,199 And again going back to Maxwell's equations, you can show that a, this 52 00:04:32,199 --> 00:04:38,931 circuit model I show here this well describes the Maxwell's equations for the 53 00:04:38,931 --> 00:04:43,661 coaxial cable and for the twisted pair, for that matter. 54 00:04:43,661 --> 00:04:51,785 So here's the idea, this is the model for a very small section of coaxial cable of 55 00:04:51,785 --> 00:04:57,759 length delta x. Each section of coaxial cable is described 56 00:04:57,759 --> 00:05:05,559 by a resistor, in series with an inductor, and then there's a parallel chain of 57 00:05:05,559 --> 00:05:14,924 another resistor and a capacitor. And we notice the tildas over the RLGC are 58 00:05:14,924 --> 00:05:26,180 tilda is resistance per unit length, so that's why the resistance here is r tilda 59 00:05:26,180 --> 00:05:34,152 times delta x same for the inductor. This resistor that's in parallel it turns 60 00:05:34,152 --> 00:05:39,192 out its conductance is going to be a lot more useful than our former formula so I 61 00:05:39,192 --> 00:05:44,887 showed as conductance per unit length. This conductance per unit length here is 62 00:05:44,887 --> 00:05:49,746 different from this resistance. They're not the same at all. 63 00:05:49,746 --> 00:05:56,946 So what we need to do is figure out how the complex amplitude of the current at 64 00:05:56,946 --> 00:06:03,888 position x is related to the complex amplitude of the voltage at position x. 65 00:06:03,888 --> 00:06:11,766 So I'm assuming that the voltage, let's say, at position x is equal to the real 66 00:06:11,766 --> 00:06:21,255 part of some complex amplitude. Times a complex, complex exponential. 67 00:06:21,255 --> 00:06:28,014 So this should be, voltage is a function of x int. 68 00:06:28,014 --> 00:06:33,542 So, right away assuming sinusoidal source and everything we know about impedance 69 00:06:33,542 --> 00:06:38,766 will help us figure out the relationship between the voltage and the current. 70 00:06:38,766 --> 00:06:44,668 So, well what I'm going to do first is go back to fundamentals remember, Kirchoff's 71 00:06:44,668 --> 00:06:48,031 Current Law, KCL? Well, we need it right now. 72 00:06:48,031 --> 00:06:56,262 So I want to write KCL at that node, so the current coming into that node is I of 73 00:06:56,262 --> 00:07:01,583 x minus delta x. The current going south is given by the 74 00:07:01,583 --> 00:07:08,964 sum of those two currents, and that's given by the voltage across them divided 75 00:07:08,965 --> 00:07:16,568 times their conductance and emittances with the delta x, and that's equal to the 76 00:07:16,568 --> 00:07:23,003 current that's leaving in the I of x, so that's KCL at that node. 77 00:07:23,003 --> 00:07:28,985 We're going to write KVL along the loop and we'll pick that loop. 78 00:07:28,985 --> 00:07:37,529 So, this voltage is equal to this voltage, which is given by this plus the voltage it 79 00:07:37,529 --> 00:07:45,770 crosses parallel combination, which is just the voltage at there, so I think you 80 00:07:45,770 --> 00:07:51,138 see where this is headed. And now, I'm going to rearrange the 81 00:07:51,138 --> 00:07:58,107 equation let delta x go to 0 and arrive at they said are different two equations, one 82 00:07:58,107 --> 00:08:02,111 for the KCL equation, one for the KVL equation. 83 00:08:02,111 --> 00:08:07,646 Well I want to get one relationship here, so what I'm going to do is take a 84 00:08:07,646 --> 00:08:14,242 derivative of this equation which gives me a second derivative here, for the left 85 00:08:14,242 --> 00:08:19,500 side and on this side that gives me a derivative with respect to x. 86 00:08:19,500 --> 00:08:25,776 And now I can substitute in that equation down below in a while arrived at the 87 00:08:25,776 --> 00:08:32,448 simpler equation for the complex amplitude of the voltage at a given location x. 88 00:08:32,448 --> 00:08:38,088 And it turns out to be a very simple differential equation as we're going to 89 00:08:38,088 --> 00:08:43,634 see in a second where the second derivative of the voltage is related to 90 00:08:43,634 --> 00:08:47,708 the voltage times a frequency dependent constant. 91 00:08:47,708 --> 00:08:52,563 It does not end on x. That's why I considered a constant, but it 92 00:08:52,563 --> 00:08:57,054 is frequency dependent. We'll see how that works out in just a 93 00:08:57,054 --> 00:08:59,982 second. So, turns out the solution of this 94 00:08:59,982 --> 00:09:05,792 equation says, that voltage propagates just like in as you might expect because 95 00:09:05,792 --> 00:09:11,547 it's governed by Maxwell's equations. So, what's going on here is I'm going to 96 00:09:11,547 --> 00:09:17,471 call this gamma squared. And if you substitute this assumed 97 00:09:17,472 --> 00:09:23,876 equation into this equation. I'm pretty sure, I'm sure you can see, 98 00:09:23,876 --> 00:09:31,288 that when I take a second derivative of the voltage, that brings down gamma twice, 99 00:09:31,288 --> 00:09:35,928 giving me a gamma squared, and I get two solutions. 100 00:09:35,928 --> 00:09:42,754 The reason for the V plus and V minus we'll get to in a second, but you can have 101 00:09:42,754 --> 00:09:49,904 either a plus or a minus in the signs on those exponents and that solves this 102 00:09:49,904 --> 00:09:55,061 equation. So that's the differential equation. 103 00:09:55,061 --> 00:10:04,419 We see it's a solution looks like this and gamma is this rather complicated function. 104 00:10:04,420 --> 00:10:09,836 It turns out simplifying this is really not worth it. 105 00:10:09,836 --> 00:10:12,553 It doesn't really simplify very well at all. 106 00:10:12,553 --> 00:10:17,245 You can certainly multiply these two complex numbers together, but taking the 107 00:10:17,245 --> 00:10:22,499 square root means converting to polar form and trying to simplify and it just doesn't 108 00:10:22,499 --> 00:10:26,137 simplify. I suggest you try it for fun if you have a 109 00:10:26,137 --> 00:10:30,086 very strange sense of fun. It's not very nice at all. 110 00:10:30,086 --> 00:10:35,576 That's suffice to say we know it's a complex number so it has a real part and 111 00:10:35,576 --> 00:10:41,292 it has an imaginary part and those are the critical things we're going to need. 112 00:10:41,292 --> 00:10:48,690 Now we're going to assume that we always pick and you can prove that this is true 113 00:10:48,690 --> 00:10:56,867 that the a variable is positive. So, if we go back up to our formulas and, 114 00:10:56,867 --> 00:11:02,829 and stick in the solution here for what gamma is. 115 00:11:02,829 --> 00:11:09,822 If you had a positive, this says for x greater than zero that this will decay, 116 00:11:09,822 --> 00:11:14,904 all right, even minus ax. If you assume that for x positive you had 117 00:11:14,904 --> 00:11:18,638 the other term here, that would mean it would blow up. 118 00:11:18,638 --> 00:11:24,043 It would get infinitely big, that's a nonphysical answer so we throw it away. 119 00:11:24,043 --> 00:11:29,975 It doesn't, it mathematically fits, but it doesn't fit the physics of the situation. 120 00:11:29,976 --> 00:11:35,479 Power has to be conserved[INAUDIBLE]. So, that's why the plus and minus. 121 00:11:35,479 --> 00:11:41,465 This is the signal you get when the positive x axis from where a source might 122 00:11:41,465 --> 00:11:47,445 be let's say, the supplying a voltage and for the negative side you get that for 123 00:11:47,445 --> 00:11:51,946 approach. All right, so to put it all together, a 124 00:11:51,946 --> 00:11:59,812 voltage for a given point on the coaxial cable and in turn, notice this is a lower 125 00:11:59,812 --> 00:12:05,294 case v so I'm losing my real part. This is the critical part. 126 00:12:05,294 --> 00:12:13,160 I'm about to show you that when you have a solution that depends on 2 pi ft minus bx 127 00:12:13,160 --> 00:12:20,977 that corresponds to a propagating signal. And that's really important so let's 128 00:12:20,978 --> 00:12:28,323 proceed to show you that if this formula does means that voltage propagates. 129 00:12:28,323 --> 00:12:36,707 What I am going to look at is this exponent up here and I am going to think 130 00:12:36,707 --> 00:12:43,673 about it, evaluate it at t time t2 and let's say location x2. 131 00:12:43,673 --> 00:12:52,049 I am going to add and subtract t1 here and what you find is that this expression 132 00:12:52,049 --> 00:13:00,318 equals the same quantity at a different time and a different location x1. 133 00:13:00,318 --> 00:13:06,689 The separation between x1 and x2 is given by this quantity. 134 00:13:06,689 --> 00:13:13,882 So, lets plot what's happening here. Lets say that's x and that's t. 135 00:13:13,882 --> 00:13:23,946 We write x1 and here is t1 of course, if we go about to x2, we're at time t2 and 136 00:13:23,946 --> 00:13:33,950 signal here at x1, t1 is the same as the signal at a different location and a 137 00:13:33,950 --> 00:13:39,943 different time. That means the signal propagates. 138 00:13:39,943 --> 00:13:46,076 It's moving from x1, t1 location to the x2, t2 location. 139 00:13:46,076 --> 00:13:54,728 So any time you see a signal that depends only on time and space like this it means 140 00:13:54,728 --> 00:14:02,412 its a propagating signal. So, let's examine this more carefully. 141 00:14:02,412 --> 00:14:09,202 How fast is it propagating? You look at the units inside here, you see 142 00:14:09,202 --> 00:14:15,698 that's distance and that's time. So, that means this quantity has to have 143 00:14:15,698 --> 00:14:20,980 units of speed of velocity. So, the speed of propagation just by unit 144 00:14:20,980 --> 00:14:26,154 analysis is related to b. And once we have, when you know that, we 145 00:14:26,154 --> 00:14:33,154 now can calculate the wavelength and the wavelength is given from these classic 146 00:14:33,154 --> 00:14:37,369 formulas which applies to all propagating waves. 147 00:14:37,369 --> 00:14:44,701 That, the wavelength times the frequency is just the speed of propagation so we 148 00:14:44,701 --> 00:14:49,658 find that the wavelength is 2 pi over b. Well what's b? 149 00:14:49,658 --> 00:14:56,012 And if you recall, b is this equals this imaginary part of the, of the square root 150 00:14:56,012 --> 00:15:02,544 of gamma squared is equal to the imaginary part of gamma and it's a complicated 151 00:15:02,544 --> 00:15:06,436 expression. It turns out if you analytically try to 152 00:15:06,436 --> 00:15:10,406 find that square root, it's really quite difficult. 153 00:15:10,406 --> 00:15:17,552 And what I want to do is point out that it really simplifies though, if we consider a 154 00:15:17,552 --> 00:15:21,881 special case and that is when frequency gets big. 155 00:15:21,881 --> 00:15:28,976 When frequency is big, you can drop G tilde with a respect to that term. 156 00:15:28,976 --> 00:15:34,326 You can drop R tilde with respect to that term. 157 00:15:34,326 --> 00:15:41,401 And so what you wind up with is j2 pi f c tilde times j 2pi f L tilde. 158 00:15:41,401 --> 00:15:49,151 And that equals minus 4 pi squared f squared L tilde c tilde and we now have to 159 00:15:49,151 --> 00:15:55,840 take the square root. And the square root of this because of the 160 00:15:55,840 --> 00:16:03,329 minus sign is pure imaginary and as j 2 pi f square root L tilde C tilde. 161 00:16:03,329 --> 00:16:12,511 So the imaginary part is just j is 2 pi f square root of L tilde C tilde and the 2 162 00:16:12,511 --> 00:16:17,708 pi f is cancel. And lo behold, we to find out the speed of 163 00:16:17,708 --> 00:16:24,066 propagation is equal, is depend only on the characteristics of the cable. 164 00:16:24,066 --> 00:16:29,444 And is equal to that quantity depends only on the inductance per unit length and on 165 00:16:29,444 --> 00:16:32,554 the capacitance per unit length of the cable. 166 00:16:32,555 --> 00:16:40,478 Most well designed coaxial cables the speed of propagation is a that of third to 167 00:16:40,478 --> 00:16:47,888 have of the speed of lighting through space so its not as certainly not through 168 00:16:47,888 --> 00:16:54,550 space but its not far away from it just a small factor difference. 169 00:16:54,550 --> 00:17:02,627 Well, that's the propagating part. How about the attenuating part? 170 00:17:02,627 --> 00:17:07,437 What is a? Well, we know that a is given by the real 171 00:17:07,437 --> 00:17:14,782 part of, of gamma and It's a little bit trickier to show that the real part is not 172 00:17:14,782 --> 00:17:18,323 0. We've already shown that as frequency gets 173 00:17:18,323 --> 00:17:21,841 big, it looks like the imaginary part dominates. 174 00:17:21,841 --> 00:17:26,967 And that's true, because it turns out that imaginary part is proportional to 175 00:17:26,967 --> 00:17:30,525 frequency. But there is a small real part which is 176 00:17:30,525 --> 00:17:35,657 does not depend on frequency. It's a constant as frequency gets big and 177 00:17:35,657 --> 00:17:40,183 is easily seen. Once you do the analysis which is in the 178 00:17:40,183 --> 00:17:47,191 notes that a becomes the this quantity depends on the resistance per unit length 179 00:17:47,191 --> 00:17:51,142 and the conductance per unit length of the cable. 180 00:17:51,142 --> 00:17:57,880 And Z nought not is a quantity we're going to talk about in a later slide later part 181 00:17:57,880 --> 00:18:02,503 of this lecture. And this is square root of L tilde over C 182 00:18:02,503 --> 00:18:06,452 tilde. And this quantity shows once you figure 183 00:18:06,452 --> 00:18:12,280 out what it is for standard pieces of coaxial cable, this can be large. 184 00:18:12,281 --> 00:18:19,769 This can be 100 db per kilometer if you're not too careful at high frequencies. 185 00:18:19,769 --> 00:18:26,594 It turns out, this attenuation factor is smaller at low frequencies but if you 186 00:18:26,594 --> 00:18:33,734 operate the cable at low frequencies, the speed of propagation will now depend on 187 00:18:33,734 --> 00:18:37,270 frequency. And you'll get different delays for 188 00:18:37,270 --> 00:18:41,778 different frequencies. And we've already seen that, that, that 189 00:18:41,778 --> 00:18:47,906 won't be a very good situation. So wireline channels have the problem that 190 00:18:47,906 --> 00:18:52,871 the signal will attenuate, can attenuate very rapidly. 191 00:18:52,871 --> 00:18:59,741 Now it turns out these equations also apply to optical fiber and you'll find the 192 00:18:59,741 --> 00:19:06,709 attenuation factor is much smaller and you can get 1 db per kilometer or even less 193 00:19:06,709 --> 00:19:13,976 for very well designed optic cable, but for coaxial cable, it's a different story. 194 00:19:13,976 --> 00:19:20,660 Now, I just want to point out that of course the current propagates also, so 195 00:19:20,660 --> 00:19:27,334 we've been talking about the voltage here. And remember one of the differential 196 00:19:27,334 --> 00:19:31,972 equations was that for relationship between voltage and current. 197 00:19:31,972 --> 00:19:37,623 So I can just take the derivative of my expression and I'll solve for the current. 198 00:19:37,623 --> 00:19:43,569 And what you get for an answer, after all is said and done, plugging in for gamma, 199 00:19:43,569 --> 00:19:47,961 is that the current is given by a very similar expression. 200 00:19:47,961 --> 00:19:52,106 The important parts are the, the exponents here. 201 00:19:52,106 --> 00:19:58,910 So we get the same kind of thing in that 1, the plus side goes for positive x and 202 00:19:58,910 --> 00:20:05,178 the negative root goes for the negative x. Similar kind of things happen. 203 00:20:05,179 --> 00:20:12,202 The current also does not diminish, so this goes to one, and a high frequency 204 00:20:12,202 --> 00:20:18,607 limit and this corresponds to a propagation, so both current and voltage 205 00:20:18,607 --> 00:20:25,500 propagate, which you'd not expect. What was really interesting is this little 206 00:20:25,500 --> 00:20:29,338 result. If we're going to talk about a little bit 207 00:20:29,338 --> 00:20:32,992 later. Just like the complex amplitude of the 208 00:20:32,992 --> 00:20:37,532 current voltage. Looks like the, the impedance which is 209 00:20:37,532 --> 00:20:43,053 that ratio depends on frequency but we have a little surprise for you. 210 00:20:43,054 --> 00:20:49,577 So, let's summarize. The voltage on a piece of coax at any 211 00:20:49,577 --> 00:20:56,434 given point in space. When you apply a sinusoidal source, is 212 00:20:56,434 --> 00:21:03,481 given by this expression. And if things are designed right, this 213 00:21:03,481 --> 00:21:08,083 place goes to 1. So and that always occurs at high 214 00:21:08,083 --> 00:21:12,862 frequencies. So at sufficiently high frequencies 215 00:21:12,862 --> 00:21:20,632 determined by the characteristics of the coaxial cable signals propagate without 216 00:21:20,632 --> 00:21:28,069 loss at a given speed which is a pretty it's a fraction of the speed of light in, 217 00:21:28,069 --> 00:21:34,256 through space. Now let's worry about that ratio energy of 218 00:21:34,256 --> 00:21:40,431 the impedance. So this ratio which we just saw, is called 219 00:21:40,431 --> 00:21:48,687 the characteristic impedance of the cable. In it earliest frequency dependent and 220 00:21:48,687 --> 00:21:56,552 also depends on the component values that well describe each of coaxial cable of 221 00:21:56,552 --> 00:22:01,493 interest. Well what happens is frequency gets big? 222 00:22:01,493 --> 00:22:08,400 Well this frequency gets big, we can drop this term and drop that term. 223 00:22:08,400 --> 00:22:13,821 Ratio does not depend on frequency at all, becomes a constant. 224 00:22:13,821 --> 00:22:20,931 And I think this is one of the most amazing results about coaxial cable and 225 00:22:20,931 --> 00:22:25,074 twist of pair. Was that, at high frequencies, the 226 00:22:25,074 --> 00:22:30,258 characteristic impedance. The, a piece of coax or twisted pair of 227 00:22:30,258 --> 00:22:35,766 high frequencies looks like a resistor. Notice there's no j's in there. 228 00:22:35,766 --> 00:22:40,288 It looks like a resistor. I find it really, kind of, cool that 229 00:22:40,288 --> 00:22:45,806 regarding that resistor depends on the inductance per unit length and the 230 00:22:45,806 --> 00:22:51,340 capacitance per unit length. That's the way it all works out for this 231 00:22:51,340 --> 00:22:56,983 model is that at high frequencies the piece of coax, if you had to use a 232 00:22:56,983 --> 00:23:03,616 theravadon equivalent, you look down that piece of coax, at high frequencies it 233 00:23:03,616 --> 00:23:08,713 looks like a resistor. Well, this has very important consequences 234 00:23:08,713 --> 00:23:14,758 in actually[UNKNOWN] wireline channels. Now, there's a detail that I should have 235 00:23:14,758 --> 00:23:20,554 emphasized but didn't, is that the circuit model and the KCL and KVL equations were 236 00:23:20,554 --> 00:23:24,499 used to derive them assume the cable is very, very long. 237 00:23:24,499 --> 00:23:31,878 In other words that KCL equations apply at every point in space. 238 00:23:31,878 --> 00:23:37,091 That was very important. So, suppose you have a physical situation 239 00:23:37,091 --> 00:23:42,950 where you have some length of coax and you just cut the end off and you leave it 240 00:23:42,950 --> 00:23:47,577 open. Well at this point that current has to be 241 00:23:47,577 --> 00:23:51,790 0 because it can't flow between the 2 conductors. 242 00:23:51,790 --> 00:23:58,164 That's what the cutting it means. What happens now, suppose the voltage, the 243 00:23:58,164 --> 00:24:02,796 source from the transmitter sends this little blue pulse. 244 00:24:02,796 --> 00:24:09,489 And we know it's going to propagate, but look what happens because of the improper, 245 00:24:09,489 --> 00:24:15,330 and just cut it, it bounces back. And it bounces back theoretically with no 246 00:24:15,330 --> 00:24:19,462 attenuation. This greatly effects the transmission 247 00:24:19,462 --> 00:24:24,234 properties of the coax. You get the same thing back again, and 248 00:24:24,234 --> 00:24:29,240 guess what? If the impedance looking back that way is 249 00:24:29,240 --> 00:24:34,070 not right, it's going to bounce again and again and again. 250 00:24:34,070 --> 00:24:39,752 It's going to go on forever. Now let's consider the situation where I 251 00:24:39,752 --> 00:24:44,916 attach a resistor between the inner and outer conductors. 252 00:24:44,917 --> 00:24:52,021 Where it has a resistance that's equal to the characteristic impedance of the coax. 253 00:24:52,021 --> 00:24:58,026 As I said, that is the evident equivalent for an infinite piece of coax. 254 00:24:58,026 --> 00:25:03,991 So it looks like the cable goes on forever, just because you attach this 255 00:25:03,991 --> 00:25:08,324 resistor. Now, that means that when you send that 256 00:25:08,324 --> 00:25:14,867 pulse down the cable, it disappears. It gets absorbed, if you will, by the 257 00:25:14,867 --> 00:25:19,422 impedance. So it is very important to terminate the 258 00:25:19,422 --> 00:25:24,225 cable correctly. You'll find this in all well designed 259 00:25:24,225 --> 00:25:31,229 systems that is at make sure that both the load and that our termination resistor, 260 00:25:31,229 --> 00:25:37,821 the value equal to Z nought is used with nought, you're going to get reflections 261 00:25:37,821 --> 00:25:43,564 and it's not going to work at all. It turns out, for most coaxial cables, at 262 00:25:43,564 --> 00:25:49,328 least that I know of, of course this is in impedance in the range of 50 to 75 ohms. 263 00:25:49,328 --> 00:25:54,716 It's not totally big and just need to look up the specifications for the coax you use 264 00:25:54,716 --> 00:25:57,568 . I've seen some coaxial cables where the 265 00:25:57,568 --> 00:26:02,112 coaxial impedance is actually labeled on the surface of the coax. 266 00:26:02,112 --> 00:26:05,585 So it's pretty easy to find out what's going on. 267 00:26:05,585 --> 00:26:10,436 So let's summarize the characteristics of the wireline channels. 268 00:26:10,437 --> 00:26:19,380 They are especially designed to minimize interference using twisted pair of coaxial 269 00:26:19,380 --> 00:26:24,915 cable, no cross talk. So you have privacy in a wired 270 00:26:24,915 --> 00:26:30,070 communication system. Moreover, the bad news is, they have to 271 00:26:30,070 --> 00:26:32,939 be. Receiver and transmitter must be connected 272 00:26:32,939 --> 00:26:36,017 to the PC cable. Like I said, you could have multiple 273 00:26:36,017 --> 00:26:40,660 receivers in the, the transmitters. But they have to be connected together. 274 00:26:40,660 --> 00:26:45,577 There's no flexibility in moving around. That it's sort of tethered to the cable. 275 00:26:45,577 --> 00:26:51,254 If you design it well and pick your frequencies right, you have very little 276 00:26:51,254 --> 00:26:55,347 attenuation as long as you stay at high frequencies. 277 00:26:55,347 --> 00:27:01,359 If you don't, you can live that way, you'll discover though, that the amplitude 278 00:27:01,359 --> 00:27:07,453 will decrease exponentially with distance. And it depends on that factor a. 279 00:27:07,454 --> 00:27:12,485 How far you can go. Early, cables for the telegraph system 280 00:27:12,485 --> 00:27:16,746 back in the 19th century. Worked at low frequencies. 281 00:27:16,746 --> 00:27:22,070 And, they discovered that they couldn't transmit very far at all. 282 00:27:22,070 --> 00:27:27,175 And your only solution. Since this is before the electronics era. 283 00:27:27,175 --> 00:27:34,132 As they just applied a very big voltage to the telegraph wires in order to get the 284 00:27:34,132 --> 00:27:39,702 signals across, interesting. Let's also now compare wireless and 285 00:27:39,702 --> 00:27:43,760 wireline. Wireless has the great advantage let's you 286 00:27:43,760 --> 00:27:50,024 connect receiver and transmitter At will almost as long as they're within a proper 287 00:27:50,024 --> 00:27:53,802 distance of each other. There's no real limitations, no 288 00:27:53,802 --> 00:27:58,109 connections have to be made, just have to know, understand each others 289 00:27:58,109 --> 00:28:01,799 characteristics. The marked downside of a wireline is that 290 00:28:01,799 --> 00:28:05,736 it does require this direct connection. You're tethered. 291 00:28:05,736 --> 00:28:09,317 You can't move around. However, you have no noise, no 292 00:28:09,317 --> 00:28:13,157 interference. Wireless is definitely prone to noise and 293 00:28:13,157 --> 00:28:18,420 interference because of the linearity. Maxwell's equations, you, you have an 294 00:28:18,420 --> 00:28:22,169 antenna, you receive everybody else's transmissions. 295 00:28:22,169 --> 00:28:27,924 We haven't talked about noise yet. What noise also comes in and in a you get 296 00:28:27,924 --> 00:28:34,691 that too whereas in the wireline case transmitter and receiver is basically the 297 00:28:34,691 --> 00:28:38,490 reason of interference if there is no noise. 298 00:28:38,490 --> 00:28:43,956 We're going to talk about how we design now a communication system using these 299 00:28:43,956 --> 00:28:44,681 results.