1 00:00:03,520 --> 00:00:07,110 Welcome back to our continuing series on linear circuits. 2 00:00:07,110 --> 00:00:09,335 And today we will be talking about transfer functions. 3 00:00:09,335 --> 00:00:14,290 The aim of today's lesson is to understand how linear systems react it to 4 00:00:14,290 --> 00:00:20,190 inputs with different frequencies. In the previous lesson we looked at AC 5 00:00:20,190 --> 00:00:22,830 analysis. So we were able to apply the techniques 6 00:00:22,830 --> 00:00:26,570 that we learned the beginning of this course to phasors systems where we have 7 00:00:26,570 --> 00:00:30,920 sinusoidal inputs. And using phasors those complex numbers 8 00:00:30,920 --> 00:00:33,496 to do basically the same things that we did before. 9 00:00:33,496 --> 00:00:36,410 And then there was the sinusoidal respond lab. 10 00:00:36,410 --> 00:00:40,410 Today we're going to take those tools and apply them to transfer functions. 11 00:00:40,410 --> 00:00:44,660 In transfer functions are, well first of all we're going to describe, describe 12 00:00:44,660 --> 00:00:48,310 what they are and how they're used and be able to plot them. 13 00:00:48,310 --> 00:00:52,210 Whenever we have a linear system with a sinusoid input we're going to have a 14 00:00:52,210 --> 00:00:55,460 sinusoid output and output will be at that same frequency. 15 00:00:56,470 --> 00:01:02,700 So here if I take my input equal to this Ain times the cosine of Omega t plus 16 00:01:02,700 --> 00:01:06,070 Theta n. We put it in phasor format we get this 17 00:01:06,070 --> 00:01:09,195 and for the output we get this for our phaser for the output. 18 00:01:09,195 --> 00:01:16,020 But the thing about linear systems is that as the frequency changes so will 19 00:01:16,020 --> 00:01:20,640 these quantities. So if I look at this example I'm going to 20 00:01:20,640 --> 00:01:24,860 instead of using Vs for voltages I'm going to let them be arbitrary inputs of 21 00:01:24,860 --> 00:01:30,400 X and outputs Y. And just for simplicity we're going to 22 00:01:30,400 --> 00:01:36,400 let the input phase be 0 since what we're really interested in is the difference 23 00:01:36,400 --> 00:01:38,471 between the input phase and the output phase. 24 00:01:38,471 --> 00:01:49,547 So here I have set up an input function, what's particular frequency that has this 25 00:01:49,547 --> 00:01:54,930 amplitude and some phase, which will be 0. 26 00:01:54,930 --> 00:01:59,045 And then here I have an output so you can see that it has a certain phase to it and 27 00:01:59,045 --> 00:02:01,590 then a certain amplitude associated with it as well. 28 00:02:01,590 --> 00:02:04,540 Now suppose that I change the frequency, let's see what changes. 29 00:02:04,540 --> 00:02:08,230 So now I've doubled the frequency and in this particular case we see that the 30 00:02:08,230 --> 00:02:16,730 output now has a smaller amplitude than the previous example. 31 00:02:16,730 --> 00:02:20,710 So even though my input has the same amplitude as the previous input the 32 00:02:20,710 --> 00:02:24,330 change in frequency lead to a different output amplitude. 33 00:02:24,330 --> 00:02:30,100 And we also we that where the previous example the output started down, this one 34 00:02:30,100 --> 00:02:31,950 starts up. So we see that it has a different phase 35 00:02:31,950 --> 00:02:34,540 as well. Transfer functions is the way that we 36 00:02:34,540 --> 00:02:37,510 describe this change in amplitude and phase. 37 00:02:38,520 --> 00:02:40,045 And so the way we're going to define it is. 38 00:02:40,045 --> 00:02:44,920 Lets say that the transfer function with respect to a func, a frequency f is going 39 00:02:44,920 --> 00:02:50,520 to be equal to the output phasor divided by the input phasor at a particular 40 00:02:50,520 --> 00:02:53,124 frequency. So that's what this little subscript f 41 00:02:53,124 --> 00:02:57,220 indicates. If you use complex numbers and calculate 42 00:02:57,220 --> 00:03:02,670 using these formats. You see this, it's essentially just going 43 00:03:02,670 --> 00:03:08,020 to be the the amplitude's going to be the relationship between the output amplitude 44 00:03:08,020 --> 00:03:11,430 and the input amplitude. And then the phase angle is going to be 45 00:03:11,430 --> 00:03:14,230 the phase of the output minus the phase of the input. 46 00:03:17,630 --> 00:03:20,500 This might be a little bit confusing so we're going to do a few examples to kind 47 00:03:20,500 --> 00:03:25,500 of illustrate the point. In our AC analysis section we looked at 48 00:03:25,500 --> 00:03:37,592 this circuit and we discovered that at the frequency f = 780 Hz that the output 49 00:03:37,592 --> 00:03:44,645 Vc was equal to 1 over the square root of 2. 50 00:03:44,645 --> 00:03:52,520 [UNKNOWN] minus Pi fourths times our input in phasor format. 51 00:03:52,520 --> 00:04:01,840 And so if I wanted to know the the transfer function at that frequency if I 52 00:04:01,840 --> 00:04:07,868 take this Vc and divide it by V. We get 1 over the square root of 2 phasor 53 00:04:07,868 --> 00:04:16,210 angle minus pi fouths. And so if we go and look at the value of 54 00:04:16,210 --> 00:04:19,208 transfer function 780, we see that it's indeed the case. 55 00:04:19,208 --> 00:04:23,710 well actually this is approximate but it's pretty close. 56 00:04:25,850 --> 00:04:30,020 Now typically we're not going to want to do these calculations for each individual 57 00:04:30,020 --> 00:04:33,920 frequency and then compile them all together to get our final format. 58 00:04:33,920 --> 00:04:37,590 Normally what we're going to want to do is come up with a, a functional 59 00:04:37,590 --> 00:04:44,280 expression in terms of the frequency or possibly the angular frequency. 60 00:04:44,280 --> 00:04:46,785 To identify what the transfer function will be. 61 00:04:46,785 --> 00:04:49,740 So that's what we're going to do in this example. 62 00:04:49,740 --> 00:04:53,870 Here we have a simple RC circuit and it is basically going to use the exact same 63 00:04:53,870 --> 00:05:00,570 tools that we used in the RLC example. So if I let this be a phasor V, here the 64 00:05:00,570 --> 00:05:08,830 impedance for the resistor is just R. And the impedance for this capacitor is 65 00:05:08,830 --> 00:05:17,248 going to be 1 over j omega C. So to do our calculation here, we see 66 00:05:17,248 --> 00:05:27,888 that our phasor Vc is going to be equal to 1 over j omega C impedance this. 67 00:05:27,888 --> 00:05:32,254 Divided by R plus 1 over j omega C, which is the impedance of both of these 68 00:05:32,254 --> 00:05:36,297 together. And we're going to multiply all of that 69 00:05:36,297 --> 00:05:41,872 by our input voltage, the voltage here. But transfer function, which here since 70 00:05:41,872 --> 00:05:48,022 I'm using omegas I'll let them do the transfer function with respect to omega. 71 00:05:48,022 --> 00:05:51,865 This is going to be the ratio of this to V, which means these V's cancel out and 72 00:05:51,865 --> 00:05:55,830 we'll multiply the top and bottom by J omega C just to simplify the expression a 73 00:05:55,830 --> 00:06:05,378 little bit. So it becomes 1 over 1 plus J omega RC. 74 00:06:05,378 --> 00:06:11,275 Then you see that we have one expression in terms of an angular frequency here 75 00:06:11,275 --> 00:06:17,746 that explains the way that this transfer function will behave. 76 00:06:17,746 --> 00:06:21,760 Now I define the transfer function with respect to frequency so it's really to do 77 00:06:21,760 --> 00:06:25,885 the conversion there. We'll change our omega to 2 pi f so here, 78 00:06:25,885 --> 00:06:31,635 this would be j2pifRC. Let's all clear all of that and put up 79 00:06:31,635 --> 00:06:38,075 the expression again here. Typeset's what's nice and easy to read. 80 00:06:38,075 --> 00:06:44,515 Now typically we, typically one thing that we can do to simplify this 81 00:06:44,515 --> 00:06:51,990 expression further is to substitute it. So let Fb equal to 1 over let fB be equal 82 00:06:51,990 --> 00:06:56,074 1 over pi RC and when you do that substituion you get H of f equal to 1 83 00:06:56,074 --> 00:07:02,950 over 1 plus J, f over fB. But using some analysis with doing 84 00:07:02,950 --> 00:07:06,910 complex numbers, we can actually break this up into two pieces because it's 85 00:07:06,910 --> 00:07:11,810 really difficult to for example, graph this function of complex numbers. 86 00:07:11,810 --> 00:07:16,080 Because for each different f, each frequency f there's a complex number 87 00:07:16,080 --> 00:07:19,520 associated with the transfer function. So typically what we will do is we'll 88 00:07:19,520 --> 00:07:24,030 break it up into a magnitude and a phase. Since that gives us two real numbers and 89 00:07:24,030 --> 00:07:26,975 it will lead to us having two different plots. 90 00:07:26,975 --> 00:07:31,573 And we can actually calculate that magnitude in face fairly simply. 91 00:07:31,573 --> 00:07:37,040 That turns out the magnitude of h of f is going to be equal to 92 00:07:37,040 --> 00:07:44,775 Here the top is 1 and we think of this this bottom as maybe being some A arg 93 00:07:44,775 --> 00:07:48,660 theta. If I put 1 over that then the result is 94 00:07:48,660 --> 00:07:53,190 going to be a function with an output. Or a complex number with an amplitude of 95 00:07:53,190 --> 00:07:59,570 1 over A. And a phase of 0 minus theta. 96 00:07:59,570 --> 00:08:04,310 Or 1 over A with an amplitude of minus theta. 97 00:08:04,310 --> 00:08:09,260 And so using that we can discover that the magnitude of H of f is going to be 98 00:08:09,260 --> 00:08:18,320 one over the square root of one plus f over fV quantity squared. 99 00:08:18,320 --> 00:08:27,882 And the phase angle will be equal to the minus arc tangent of f over fB. 100 00:08:27,882 --> 00:08:33,910 Now this, you might be lost on some of the details of exactly how to get this. 101 00:08:33,910 --> 00:08:38,200 And one of the reasons for that is people do complex numbers in high school and 102 00:08:38,200 --> 00:08:40,170 then they don't really touch it again for a very long time. 103 00:08:40,170 --> 00:08:46,280 So we will provide a, an appendix lesson. It's not part of the normal course flow 104 00:08:46,280 --> 00:08:49,410 that covers complex numbers and doing these types of calculations in case you 105 00:08:49,410 --> 00:08:52,170 were getting lost. So I'll put all that and I'll put up our 106 00:08:52,170 --> 00:08:56,670 two solutions here. And you can go through and actually plot 107 00:08:56,670 --> 00:08:59,745 these two functions with respect to frequencies. 108 00:08:59,745 --> 00:09:04,570 And so you notice that, that the magnitude of our input really didn't 109 00:09:04,570 --> 00:09:08,290 matter because it basically gets factored out in the end. 110 00:09:09,310 --> 00:09:14,620 So if I plot this out, we see that I get plots that kind of look like this. 111 00:09:14,620 --> 00:09:19,570 Now the first one for small values of fB, this first term dominates so it's going 112 00:09:19,570 --> 00:09:22,430 to be about 1. But as our frequency here increases, this 113 00:09:22,430 --> 00:09:26,130 one becomes less and less important. And this is going to start dropping down. 114 00:09:26,130 --> 00:09:31,285 This is going to start basically looking like 1 over, well fB over f kind of thing 115 00:09:31,285 --> 00:09:37,840 and in this example this is just our minus arc tangent behavior here. 116 00:09:37,840 --> 00:09:44,160 This bottom line corresponds to the brake frequency since if I put in fB here for 117 00:09:44,160 --> 00:09:48,660 the f, we discover that the magnitude is going to be 1 over the square root of 2. 118 00:09:48,660 --> 00:09:50,584 And the phase is going to be minus pi fourths for our transfer function. 119 00:09:50,584 --> 00:09:54,727 If instead we wanted to know what the transfer function was for a functioin 120 00:09:54,727 --> 00:10:03,675 that were operting at a 4 kilohertz frequency we could come to our plot here. 121 00:10:03,675 --> 00:10:10,430 And find 4 kilohertz per frequency range, find out what this point happens to be in 122 00:10:10,430 --> 00:10:12,742 so maybe about 0.4 and this is maybe around minus 3 pi eighths approximately. 123 00:10:12,742 --> 00:10:23,365 So we can quickly see what kind of magnitude and phase response that we will 124 00:10:23,365 --> 00:10:26,965 see for a particular input using these methods. 125 00:10:26,965 --> 00:10:34,200 In a final example we will again look at this RLC circuit and to do an analysis to 126 00:10:34,200 --> 00:10:37,030 see what the transfer function of this would be like. 127 00:10:37,030 --> 00:10:42,110 If instead of calculating it for a single frequency we do it as an expression of 128 00:10:42,110 --> 00:10:45,420 all frequencies. And in this case I will leave my transfer 129 00:10:45,420 --> 00:10:50,230 function in terms of omega's angular frequency just because it's going to be a 130 00:10:50,230 --> 00:10:54,010 little bit easier to write. Again we're going to be using the voltage 131 00:10:54,010 --> 00:11:02,880 divider so our Vc is going to be equal to 1 over j omega C. 132 00:11:02,880 --> 00:11:11,200 Divided by R plus j omega L plus 1 over j omega C multiply top and oh, almost 133 00:11:11,200 --> 00:11:13,954 forgot. Times Vs. 134 00:11:13,954 --> 00:11:20,305 If I take this Vc and divide it by Vs and multiply the top and bottom by j omega c. 135 00:11:20,305 --> 00:11:32,157 It's discovered that H of omega is going to be equal to 1 divided by things around 136 00:11:32,157 --> 00:11:46,552 a little bit here. 1 minus omega squared LC plus j omega RC. 137 00:11:46,552 --> 00:11:54,790 I'll go ahead and put this up here. And again we can use the same basic 138 00:11:54,790 --> 00:11:58,715 technique as we used before to find what the magnitude and phase of this will be. 139 00:11:58,715 --> 00:12:03,620 And so here the magnitude is going to be this expression, 1 over all of this in 140 00:12:03,620 --> 00:12:08,070 the square root and the phase is going to the minus arctangent of this ratio. 141 00:12:09,160 --> 00:12:15,170 You can actually put in different values for R, L and C and then plot out what 142 00:12:15,170 --> 00:12:18,750 this response will look like, what the magnitude and phase response will be 143 00:12:18,750 --> 00:12:21,260 like. Are using any kind of plotting software. 144 00:12:21,260 --> 00:12:23,965 And you'll discover that it kind of has some interesting behavior. 145 00:12:23,965 --> 00:12:30,340 Doctor Perry will in a later lecture go over what these transfer functions will 146 00:12:30,340 --> 00:12:35,550 look like and how to use them. In summary we introduced the concept of a 147 00:12:35,550 --> 00:12:37,510 transfer function. And we showed how to calculate the 148 00:12:37,510 --> 00:12:39,510 transfer function for a few different systems. 149 00:12:39,510 --> 00:12:43,490 And then we demonstrated the graph of the magnitude and the base response. 150 00:12:43,490 --> 00:12:49,290 And a little bit about how to read it. In the next lesson we will cover time and 151 00:12:49,290 --> 00:12:51,770 frequency domain. And then there will be a demo with a 152 00:12:51,770 --> 00:12:54,430 guitar string to show the frequency spectrum. 153 00:12:54,430 --> 00:12:57,980 When you pluck a guitar string it has a few different frequency components that 154 00:12:57,980 --> 00:13:01,180 contribute to give it the, that particular sound. 155 00:13:01,180 --> 00:13:05,480 And we'll start to see how we can then take these ideas of transfer functions 156 00:13:05,480 --> 00:13:08,750 and apply them to real world situations. Until next time.