1 00:00:03,250 --> 00:00:05,660 Welcome back to Linear Circuits. This is Doctor Ferri. 2 00:00:07,240 --> 00:00:09,920 This lesson is on the Frequency Response. 3 00:00:09,920 --> 00:00:13,500 We're actually breaking the idea of frequency response into 2 lessons, one 4 00:00:13,500 --> 00:00:17,400 that contains linear plots and the other that will be on boaty plots. 5 00:00:19,420 --> 00:00:22,329 So, this topic is the Frequency Response. Topic. 6 00:00:22,329 --> 00:00:27,890 In the previous lesson, we introduce the transfer function. 7 00:00:27,890 --> 00:00:31,500 And we also introduce the frequency re, the frequency spectrum. 8 00:00:32,830 --> 00:00:35,950 This particular lesson is aimed at combining 9 00:00:35,950 --> 00:00:38,179 those two concepts and using them together. 10 00:00:39,340 --> 00:00:41,510 To introduce the frequency response, it's a way 11 00:00:41,510 --> 00:00:44,490 of showing how a system processes signals of 12 00:00:44,490 --> 00:00:46,289 different frequencies, and understanding the 13 00:00:46,289 --> 00:00:50,040 frequency spectrum helps in understanding transfer 14 00:00:50,040 --> 00:00:52,880 function helps in order to be able to do this lesson. 15 00:00:55,650 --> 00:00:59,450 So, let's go back to the transfer functions. 16 00:01:03,978 --> 00:01:08,770 A, an RLC circuit in this particular configuration has this transfer function. 17 00:01:08,770 --> 00:01:11,100 Remember we got this transfer function, for 18 00:01:11,100 --> 00:01:14,610 example, by doing a voltage divider law. 19 00:01:14,610 --> 00:01:20,890 Finding the voltage v sub c as a function of v sub s. 20 00:01:20,890 --> 00:01:27,100 So it turns out that v sub c is equal h times v sub s. 21 00:01:27,100 --> 00:01:28,790 Where this is the phaser for v sub s. 22 00:01:28,790 --> 00:01:32,080 This is the phaser for v sub c, and this is the relationship. 23 00:01:32,080 --> 00:01:33,910 That's the transfer function. 24 00:01:33,910 --> 00:01:37,330 So the transfer function, we've calculated already to be this. 25 00:01:37,330 --> 00:01:39,209 If I want to find the magnitude of it. 26 00:01:43,300 --> 00:01:43,300 [SOUND] 27 00:01:43,300 --> 00:01:46,042 It's going to be the magnitude of the numerator, 28 00:01:46,042 --> 00:01:48,715 which is 1, divided by the magnitude of the denominator. 29 00:01:48,715 --> 00:01:52,110 Well, the magnitude of the denominator, this is a complex number. 30 00:01:52,110 --> 00:01:54,980 It's the square root of the sum of the squares of the 31 00:01:54,980 --> 00:01:59,670 real parts squared, which is 1 squared, plus the imaginary part squared. 32 00:02:02,680 --> 00:02:08,172 So now I have a function which is the magnitude of h as a function of omega. 33 00:02:08,172 --> 00:02:15,320 The angle of h, can also be written as a function of, of omega. 34 00:02:15,320 --> 00:02:20,240 In this case, the angle of the numerator is one, 35 00:02:20,240 --> 00:02:23,840 the angle of the denominator is the angle of this. 36 00:02:23,840 --> 00:02:26,976 Since it's in the denominator, we're going to have a negative sign there. 37 00:02:26,976 --> 00:02:28,844 And it's 38 00:02:28,844 --> 00:02:34,610 negative arctan, of the imaginary part over the real part. 39 00:02:34,610 --> 00:02:35,410 So it's mega RC over one. 40 00:02:35,410 --> 00:02:35,970 And the 41 00:02:38,800 --> 00:02:41,230 negative is because this part is in the denominator. 42 00:02:46,000 --> 00:02:47,570 So just make it a little bit clearer there. 43 00:02:47,570 --> 00:02:48,070 Since 44 00:02:50,010 --> 00:02:54,530 I now have these functions of a mega. I can plot this. 45 00:02:54,530 --> 00:02:58,620 This is a real function of omega. I can plot it versus omega. 46 00:02:58,620 --> 00:03:02,010 And for this particular case, with particular values of r and 47 00:03:02,010 --> 00:03:04,770 c, I would get a plot that looks sort of like this. 48 00:03:04,770 --> 00:03:06,885 A magnitude here versus omega. 49 00:03:06,885 --> 00:03:09,650 The angle and degrees versus omega. 50 00:03:09,650 --> 00:03:11,900 And this is a frequency response, so 51 00:03:11,900 --> 00:03:15,478 we're using terms a little bit interchangeably here. 52 00:03:15,478 --> 00:03:20,160 when people use the word transfer function, sometimes they'll, they'll talk 53 00:03:20,160 --> 00:03:24,930 about transfer function as a little bit more general term then this. 54 00:03:24,930 --> 00:03:27,410 And in our case in as far as this course is 55 00:03:27,410 --> 00:03:31,855 concerned, we're using the term transfer function to mean this function here. 56 00:03:31,855 --> 00:03:35,190 This H of Omega, and then the, when we go 57 00:03:35,190 --> 00:03:38,410 to the plots I'm going to call the plots the frequency response. 58 00:03:41,910 --> 00:03:47,240 So an example of how we use the frequency response, is shown here when we look at 59 00:03:47,240 --> 00:03:52,680 an RC circuit. The input voltage to this RC 60 00:03:52,680 --> 00:03:58,880 circuit, is this signal right here. And this signal is clearly a sine wave. 61 00:03:58,880 --> 00:04:01,340 Super imposed with a high frequency sign wave, so low 62 00:04:01,340 --> 00:04:05,210 frequency sine wave added together with the high frequency sine wave. 63 00:04:05,210 --> 00:04:06,880 If that's my input voltage, 64 00:04:06,880 --> 00:04:11,340 then my output voltage to this circuit looks like this, if I plot it. 65 00:04:11,340 --> 00:04:13,580 Now clearly you can see something's happened. 66 00:04:13,580 --> 00:04:16,540 My low frequency signal has not changed very much. 67 00:04:16,540 --> 00:04:20,000 But my high frequency signal is now much smaller in amplitude. 68 00:04:21,820 --> 00:04:25,700 And it helps for us to understand this, better in the frequency domain. 69 00:04:29,250 --> 00:04:33,410 So in the frequency domain, I can plot the spectrum of this input signal. 70 00:04:33,410 --> 00:04:36,290 So I've got a low frequency signal, that's at 71 00:04:36,290 --> 00:04:39,660 50 radians per second, and an amplitude of 1. 72 00:04:39,660 --> 00:04:41,155 If you look at this, you can see. 73 00:04:41,155 --> 00:04:46,614 That the average value here is a 74 00:04:46,614 --> 00:04:50,630 low-frequency signal with an amplitude of 1. 75 00:04:50,630 --> 00:04:53,020 And then the high-frequency signal so on top of 76 00:04:53,020 --> 00:04:54,540 that, is another sign wave again, with an amplitude 77 00:04:54,540 --> 00:05:02,392 of 1. So I plot those to the frequency spectrum. 78 00:05:02,392 --> 00:05:07,210 This is the frequency response of this system. 79 00:05:07,210 --> 00:05:08,920 Of this circuit. 80 00:05:08,920 --> 00:05:10,850 For a particular value in R and C. 81 00:05:10,850 --> 00:05:13,470 To figure out what this output voltage is 82 00:05:13,470 --> 00:05:17,106 going to look like, I take the input voltage. 83 00:05:17,106 --> 00:05:20,320 Spectrum at 50 radians 84 00:05:20,320 --> 00:05:20,800 per second. 85 00:05:20,800 --> 00:05:25,240 Now go to this plot and look at 50 radians per second, which is right here, 86 00:05:25,240 --> 00:05:30,140 and I go up, follow up this, and I find out what this plot, this value is. 87 00:05:30,140 --> 00:05:34,866 That's about point, say 0.95. 88 00:05:34,866 --> 00:05:40,140 I take 0.95 and multiply it 89 00:05:40,140 --> 00:05:45,340 by this amplitude of 1. And then I 90 00:05:45,340 --> 00:05:48,975 get the spectrum of the output. If that's 1, this 91 00:05:48,975 --> 00:05:52,732 is 0.95. Now at 800 hertz, or 92 00:05:52,732 --> 00:05:56,850 800 radians per second I have an amplitude of 1. 93 00:05:56,850 --> 00:06:03,798 And, in this case, I get a value of about, I'd say, 0.1 0.13. 94 00:06:03,798 --> 00:06:10,638 So I take 0.13, multiply it by this input amplitude, 95 00:06:10,638 --> 00:06:16,262 to give give me the output amplitude, which is 96 00:06:16,262 --> 00:06:22,240 about 0.13. And what we found in the output, 97 00:06:22,240 --> 00:06:28,000 I've, the input amplitude is almost the same for the low frequency signal. 98 00:06:29,170 --> 00:06:32,460 But the high frequency signal has a much smaller amplitude. 99 00:06:32,460 --> 00:06:35,840 So what's happened is, I've used the frequency response 100 00:06:35,840 --> 00:06:38,690 to tell me how this circuit. 101 00:06:38,690 --> 00:06:40,880 Processes sinusoidal at different frequencies. 102 00:06:40,880 --> 00:06:45,090 At low frequencies, it passes them through without much change. 103 00:06:45,090 --> 00:06:48,965 At high frequencies, it, it attenuates them, or makes the amplitude much smaller. 104 00:06:48,965 --> 00:06:51,830 And that's a big benefit for the frequency response. 105 00:06:54,770 --> 00:06:56,260 Now let's look at an example of how to 106 00:06:56,260 --> 00:06:59,325 use a frequency response in solving an actual problem. 107 00:06:59,325 --> 00:07:03,580 Suppose the circuit has the frequency response plot shown here. 108 00:07:03,580 --> 00:07:06,230 Where I'm showing both the magnitude and the angle plot. 109 00:07:06,230 --> 00:07:12,555 And the question is, what is the steady state response v0 to input of this. 110 00:07:12,555 --> 00:07:16,560 Okay to, in order to be able to analyze this, 111 00:07:16,560 --> 00:07:19,480 it really helps us to go back to the transfer function. 112 00:07:19,480 --> 00:07:19,910 I'm doing 113 00:07:19,910 --> 00:07:22,950 a little bit more detail here than I did in the previous slide. 114 00:07:22,950 --> 00:07:26,250 Because, I'm also including the effect of the angle. 115 00:07:27,690 --> 00:07:32,950 So, going back to the, what we learned in transfer functions, we said if an in, if a 116 00:07:32,950 --> 00:07:39,070 there's a input Ai cosine of Omega 1 T 117 00:07:39,070 --> 00:07:44,950 into my circuit, the output of my circuit will have the form in steady state 118 00:07:44,950 --> 00:07:49,190 of A out cosine at that same frequency. 119 00:07:51,600 --> 00:07:54,600 But shifted in phase and a change in amplitude. 120 00:07:54,600 --> 00:07:58,540 So cosine same frequency, different amplitude, different phase. 121 00:07:58,540 --> 00:08:02,485 And we calculate the amplitude and the phase through this transfer function. 122 00:08:02,485 --> 00:08:09,330 Out is equal the magnitude of h. The transfer function, 123 00:08:09,330 --> 00:08:15,695 evaluated at that frequency times A n /g. And the angle, 124 00:08:15,695 --> 00:08:21,720 beta is simply the angle of h at that frequency. 125 00:08:23,630 --> 00:08:26,940 So if I look at this particular example. Now we've got numbers here. 126 00:08:26,940 --> 00:08:34,300 We, we've got an input of two which is constant, and we've got this input, 200. 127 00:08:34,300 --> 00:08:40,040 Now at two, that corresponds to a DC input. 128 00:08:40,040 --> 00:08:40,840 Which corresponds 129 00:08:40,840 --> 00:08:46,390 to a frequency, let's look at here, a frequency of 0. 130 00:08:46,390 --> 00:08:51,990 So at omega equals 0, at omega 131 00:08:51,990 --> 00:08:57,330 equals 0, and I'll mark this is DC input. 132 00:08:57,330 --> 00:09:01,185 At omega equals 0, I'll look at this plot at omega equals 133 00:09:01,185 --> 00:09:05,892 0 I have a magnitude of 1 and an angle of 0. 134 00:09:05,892 --> 00:09:10,174 So h at 0 is 1 angle of h at 0 is 0 degrees. 135 00:09:10,174 --> 00:09:16,550 So the corresponding output to an input of 2, so an input of 2 going 136 00:09:16,550 --> 00:09:22,110 into my system. 137 00:09:22,110 --> 00:09:26,145 Gives me output of 2, because it scales by 1. 138 00:09:28,770 --> 00:09:35,880 At omega equals 200, and this is in radians per second, 139 00:09:35,880 --> 00:09:42,436 I look over at 200, and I follow this up. 140 00:09:46,130 --> 00:09:51,163 And calculate that value. That's at 141 00:09:51,163 --> 00:09:57,193 approximately 142 00:09:57,193 --> 00:10:01,760 0.45. And that is an angle 143 00:10:01,760 --> 00:10:07,560 of about minus 65. So the corresponding 144 00:10:07,560 --> 00:10:11,815 value, if I put, the corresponding output 145 00:10:11,815 --> 00:10:18,070 the, A out, is equal to 0.45 146 00:10:18,070 --> 00:10:21,790 times the input amplitude, which is one, and the 147 00:10:21,790 --> 00:10:25,688 angle Is this angle minus 65°. So let me put it 148 00:10:25,688 --> 00:10:31,280 altogether. V out is 149 00:10:31,280 --> 00:10:37,145 equal to the output due to this part by itself to the 150 00:10:37,145 --> 00:10:38,194 DC part. 151 00:10:38,194 --> 00:10:44,350 And that output is 2 plus the output corresponding to this part by itself. 152 00:10:44,350 --> 00:10:46,500 And that output would be A out 0.45 times the cosine of that frequency is 200 t. 153 00:10:46,500 --> 00:10:52,166 The angle is minus 65 degrees. So, what we're able to do is use the 154 00:10:52,166 --> 00:10:58,538 frequency response to be able to determine 155 00:10:58,538 --> 00:11:03,494 what the output is to a particular 156 00:11:03,494 --> 00:11:08,140 input involving a DC component. 157 00:11:08,140 --> 00:11:11,890 Or sine wave and it could be we could have multiple sine waves and we do the same 158 00:11:11,890 --> 00:11:20,570 procedure for every sine wave and then we just add them altogether at the end. 159 00:11:20,570 --> 00:11:23,240 So, in summary a frequency response is 160 00:11:23,240 --> 00:11:25,445 a plot of the transfer function versus frequency. 161 00:11:25,445 --> 00:11:29,420 A frequency response can be used to determine the steady-state sinusoidal 162 00:11:29,420 --> 00:11:31,953 response of a circuit at different frequencies. 163 00:11:31,953 --> 00:11:39,340 So for our next lesson, we will do the same sort of thing in analyzing 164 00:11:39,340 --> 00:11:43,390 the frequency response, but we will be using Bode plots rather than linear plots.