1 00:00:01,910 --> 00:00:03,910 GILBERT STRONG: Well, you see spring has finally 2 00:00:03,910 --> 00:00:04,860 come to Boston. 3 00:00:04,860 --> 00:00:10,870 My sweater is gone and it's April the 16th, I think. 4 00:00:10,870 --> 00:00:13,980 It's getting late spring. 5 00:00:13,980 --> 00:00:21,580 So today, this video is the nice case, constant coefficient, 6 00:00:21,580 --> 00:00:26,180 linear equations, and the right hand side is an exponential. 7 00:00:26,180 --> 00:00:27,510 Those are the best. 8 00:00:27,510 --> 00:00:29,590 And we've seen that before. 9 00:00:29,590 --> 00:00:36,240 In fact, let me extend, we saw it for first order equations, 10 00:00:36,240 --> 00:00:38,700 here it is for second order equations, 11 00:00:38,700 --> 00:00:41,580 and it could be an nth order equation. 12 00:00:41,580 --> 00:00:45,840 We could have the nth derivative and all lower derivatives. 13 00:00:45,840 --> 00:00:48,330 The first derivative, the function itself, 14 00:00:48,330 --> 00:00:50,780 0-th derivative with coefficients, 15 00:00:50,780 --> 00:00:54,040 constant coefficients, equalling e 16 00:00:54,040 --> 00:00:56,773 to the st. That's what makes it easy. 17 00:00:56,773 --> 00:01:00,180 And what do we do when the right hand side is e to the st? 18 00:01:00,180 --> 00:01:03,490 We look for a solution, a multiple 19 00:01:03,490 --> 00:01:06,480 of e to the st. Capital Y times e 20 00:01:06,480 --> 00:01:09,440 to the st, that's going to work. 21 00:01:09,440 --> 00:01:14,650 We just plug into the equation to find that transfer function 22 00:01:14,650 --> 00:01:17,966 capital Y. Can I just do that? 23 00:01:17,966 --> 00:01:20,710 I'll do it for the nth degree. 24 00:01:20,710 --> 00:01:21,930 Why not? 25 00:01:21,930 --> 00:01:29,640 So will I plug this in for y, every derivative brings an s. 26 00:01:29,640 --> 00:01:33,050 Capital Y is still there, the exponential 27 00:01:33,050 --> 00:01:34,890 is going to be still there, and then 28 00:01:34,890 --> 00:01:39,620 there are all the s's that come down from the derivatives, 29 00:01:39,620 --> 00:01:43,810 and s's from the nth derivative. 30 00:01:43,810 --> 00:01:49,595 One s from the first derivative, no s from the constant term. 31 00:01:49,595 --> 00:01:52,600 Do you see that equation is exactly 32 00:01:52,600 --> 00:01:58,220 like what we had before with as squared plus Bs plus C. 33 00:01:58,220 --> 00:02:01,490 We have quadratic equation, the most important case. 34 00:02:01,490 --> 00:02:08,419 Now, I'm including that with any degree equation, nth degree 35 00:02:08,419 --> 00:02:10,050 equation. 36 00:02:10,050 --> 00:02:13,630 And what's the solution for Y? 37 00:02:13,630 --> 00:02:17,240 Because e to the st cancels e to the st. 38 00:02:17,240 --> 00:02:19,440 That whole thing equals 1. 39 00:02:19,440 --> 00:02:23,360 I divide by this and I get Y equal 1 40 00:02:23,360 --> 00:02:27,310 over that key polynomial. 41 00:02:27,310 --> 00:02:30,500 It's a nth degree polynomial. 42 00:02:30,500 --> 00:02:34,670 And the 1 over it is called the transfer function. 43 00:02:34,670 --> 00:02:41,000 And that transfer function transfers the input-- 44 00:02:41,000 --> 00:02:46,020 e to the st-- to the output-- Ye to the st. 45 00:02:46,020 --> 00:02:50,620 It gives the exponential response. 46 00:02:50,620 --> 00:02:52,030 Very nice formula. 47 00:02:52,030 --> 00:02:53,110 Couldn't be better. 48 00:02:53,110 --> 00:02:56,690 And you remember for second degree equations, 49 00:02:56,690 --> 00:03:04,510 our most important case is as squared Bs plus C. That's 50 00:03:04,510 --> 00:03:08,970 the solution almost every time. 51 00:03:08,970 --> 00:03:11,630 But one thing can go wrong. 52 00:03:11,630 --> 00:03:13,570 One thing can go wrong. 53 00:03:13,570 --> 00:03:18,440 Suppose for the particular s, the particular exponent, 54 00:03:18,440 --> 00:03:24,880 in the forcing function, suppose that s in the forcing function 55 00:03:24,880 --> 00:03:28,830 is also one of the s's in the no solutions. 56 00:03:28,830 --> 00:03:37,060 You remember the no solutions, there are two s's-- s1 and s2-- 57 00:03:37,060 --> 00:03:40,590 that make this 0, that make that 0. 58 00:03:40,590 --> 00:03:44,990 Those are the s1 and s2 that go into the no solutions. 59 00:03:44,990 --> 00:03:51,510 Now, if the forcing s is one of those no solution s's, we 60 00:03:51,510 --> 00:03:52,380 have a problem. 61 00:03:52,380 --> 00:03:59,520 Because this is 1 over 0 and we haven't got an answer yet. 62 00:03:59,520 --> 00:04:01,710 1 over 0 has no meaning. 63 00:04:01,710 --> 00:04:05,730 So I have to, this is called resonance. 64 00:04:05,730 --> 00:04:09,440 Resonance is when the forcing exponent 65 00:04:09,440 --> 00:04:13,940 is one of the no exponents that make this 0. 66 00:04:13,940 --> 00:04:18,390 And there are two of those for second order equations 67 00:04:18,390 --> 00:04:24,700 and there will be n different s's-- s1, s2, up to sn-- 68 00:04:24,700 --> 00:04:27,990 for nth degree equations. 69 00:04:27,990 --> 00:04:31,400 Those special s's, I could also call them 70 00:04:31,400 --> 00:04:34,000 poles of the transfer function. 71 00:04:34,000 --> 00:04:37,970 The transfer function has this in the denominator 72 00:04:37,970 --> 00:04:42,720 and when this is 0, that identifies a pole. 73 00:04:42,720 --> 00:04:47,540 So the s1, s2, to sn are the poles 74 00:04:47,540 --> 00:04:51,740 and we hope that this s is not one of those, but it could be. 75 00:04:51,740 --> 00:04:54,740 And if it is, we need a new formula. 76 00:04:54,740 --> 00:04:57,890 So that's the only remaining case. 77 00:04:57,890 --> 00:05:01,500 This is a completely nice picture. 78 00:05:01,500 --> 00:05:09,290 We just need this last case with some resonance 79 00:05:09,290 --> 00:05:12,610 when s equals say s1. 80 00:05:12,610 --> 00:05:17,490 I'll just pick s1 and when A, I know that As 1 81 00:05:17,490 --> 00:05:23,640 squared plus Bs 1 plus C is 0. 82 00:05:23,640 --> 00:05:27,600 So Y would be 1 over 0. 83 00:05:27,600 --> 00:05:29,830 And we can't live with that. 84 00:05:29,830 --> 00:05:33,300 I've written here for the second degree equation 85 00:05:33,300 --> 00:05:36,690 same possibility for the nth degree, 86 00:05:36,690 --> 00:05:43,130 An s1 to the nth plus A0 equals 0. 87 00:05:43,130 --> 00:05:46,260 That would be a problem of resonance. 88 00:05:46,260 --> 00:05:49,560 In the nth degree equation, this gives us resonance, you see, 89 00:05:49,560 --> 00:05:55,790 because remember, the no solutions were e to the s1t 90 00:05:55,790 --> 00:05:58,060 was a no solution. 91 00:05:58,060 --> 00:06:02,950 If I plug-in e to the s1t, the left side will give 0. 92 00:06:02,950 --> 00:06:07,050 So I can't get for equal to a forcing term on the right side. 93 00:06:07,050 --> 00:06:09,730 I need a new solution. 94 00:06:09,730 --> 00:06:16,805 I need a new y of t. 95 00:06:19,860 --> 00:06:21,130 Can I tell you what it is? 96 00:06:26,030 --> 00:06:30,660 It's a typical case of L'Hopital's rule from calculus 97 00:06:30,660 --> 00:06:35,800 when we approach this bad situation, 98 00:06:35,800 --> 00:06:40,460 and we are getting a 1 over 0. 99 00:06:40,460 --> 00:06:44,990 Well, you'll see a 0 over 0 and that's L'Hopital's aim for. 100 00:06:44,990 --> 00:06:47,060 So where do we start? 101 00:06:47,060 --> 00:06:52,060 A Y particular solution is this e 102 00:06:52,060 --> 00:07:12,950 to the st over this As squared plus Bs plus C. Right. 103 00:07:12,950 --> 00:07:16,520 That's our particular solution. 104 00:07:16,520 --> 00:07:18,140 If it works. 105 00:07:18,140 --> 00:07:23,170 Resonance is the case when this doesn't work because that's 0. 106 00:07:23,170 --> 00:07:26,060 Now, that's a particular solution. 107 00:07:26,060 --> 00:07:28,950 I'll subtract off a no solution. 108 00:07:28,950 --> 00:07:29,690 I could do that. 109 00:07:29,690 --> 00:07:31,350 I still have a solution. 110 00:07:31,350 --> 00:07:35,860 So I subtract off e to the s1t. 111 00:07:35,860 --> 00:07:41,600 So S1 is, e to the s1t is a no solution. 112 00:07:41,600 --> 00:07:45,490 This is what I would call a very particular solution. 113 00:07:45,490 --> 00:07:49,160 It's very particular because a t equals 0, it's 0. 114 00:07:51,690 --> 00:07:57,130 Do you see that, you see what's happening here. 115 00:07:57,130 --> 00:08:01,770 The question, resonance happens when s approaches s1. 116 00:08:01,770 --> 00:08:05,580 Resonance is s equal to, resonance 117 00:08:05,580 --> 00:08:07,290 itself is at the thing. 118 00:08:07,290 --> 00:08:14,230 Now, we let s sneak up on s1 and we ask 119 00:08:14,230 --> 00:08:16,720 what happens to that formula. 120 00:08:16,720 --> 00:08:21,100 You see, we're sneaking up on resonance. 121 00:08:21,100 --> 00:08:24,740 At resonance, when s equals s1, that will be 0 122 00:08:24,740 --> 00:08:27,220 and that will be 0. 123 00:08:27,220 --> 00:08:29,490 That's our problem. 124 00:08:29,490 --> 00:08:33,980 So approach it and you end up with the derivative 125 00:08:33,980 --> 00:08:36,590 of this divided by the derivative of this. 126 00:08:36,590 --> 00:08:38,750 Do you remember L'Hopital? 127 00:08:38,750 --> 00:08:43,460 It was a crazy rule in calculus, but here it's actually needed. 128 00:08:43,460 --> 00:08:49,610 So as s goes to s1 this goes to 0 over 0, 129 00:08:49,610 --> 00:08:52,000 so I have to take derivative over derivative. 130 00:08:52,000 --> 00:08:53,880 So let me write the answer. 131 00:08:53,880 --> 00:08:56,520 Y resonant. 132 00:08:56,520 --> 00:09:02,240 Can I call this the resonant solution when s equals s1. 133 00:09:02,240 --> 00:09:04,330 And what does it equal? 134 00:09:04,330 --> 00:09:08,940 Well, I take the s derivative of this 135 00:09:08,940 --> 00:09:11,170 and divide it by the s derivative of that. 136 00:09:11,170 --> 00:09:12,830 Derivative over derivative. 137 00:09:12,830 --> 00:09:21,490 The s derivative of that is te to the st. 138 00:09:21,490 --> 00:09:29,370 And the s derivative of this is 2As plus B. 139 00:09:29,370 --> 00:09:33,190 And now, I have derivative over derivative, 140 00:09:33,190 --> 00:09:35,620 I can let s go to s1. 141 00:09:35,620 --> 00:09:40,710 So s goes to s1, this goes to s1, and I get an answer. 142 00:09:40,710 --> 00:09:41,870 The right answer. 143 00:09:41,870 --> 00:09:47,310 This is the correct solution and you notice 144 00:09:47,310 --> 00:09:50,660 everybody spots this t factor. 145 00:09:50,660 --> 00:09:53,560 That t factor is a signal to everyone 146 00:09:53,560 --> 00:09:56,580 that we're in a special case when 147 00:09:56,580 --> 00:09:58,880 two things happen to be equal. 148 00:09:58,880 --> 00:10:02,760 Here the two things are the s and the s1. 149 00:10:02,760 --> 00:10:04,770 So that will work. 150 00:10:04,770 --> 00:10:06,870 So do you see the general picture? 151 00:10:06,870 --> 00:10:11,070 It's always this te to the st, t above. 152 00:10:11,070 --> 00:10:17,810 And down below we have the derivative of this polynomial 153 00:10:17,810 --> 00:10:19,970 at s equal s1. 154 00:10:19,970 --> 00:10:23,150 You know it's theoretically possible that we 155 00:10:23,150 --> 00:10:25,100 could have double resonance. 156 00:10:25,100 --> 00:10:28,360 We would have double resonance if that thing is 0. 157 00:10:31,580 --> 00:10:37,530 If s1 was a double root, if s1 was a double root, then, 158 00:10:37,530 --> 00:10:41,010 well, that's just absurd, but it could happen. 159 00:10:41,010 --> 00:10:45,550 Then not only is that denominator 0, but after one 160 00:10:45,550 --> 00:10:48,720 use of L'Hopital, so we have to drag L'Hopital back 161 00:10:48,720 --> 00:10:51,680 from the hospital and say do it again. 162 00:10:51,680 --> 00:10:53,350 So we would have a second derivative. 163 00:10:53,350 --> 00:10:57,390 I won't write down that solution because it's pretty rare. 164 00:10:57,390 --> 00:11:00,830 So what has this video done? 165 00:11:00,830 --> 00:11:07,270 Simply put on record the simplest case 166 00:11:07,270 --> 00:11:11,460 possible with a forcing function, e to the st. 167 00:11:11,460 --> 00:11:17,560 And above all, we've identified this transfer function. 168 00:11:17,560 --> 00:11:22,810 And let me just anticipate that if we need another way to solve 169 00:11:22,810 --> 00:11:27,100 these equations, instead of in the t domain, 170 00:11:27,100 --> 00:11:33,150 we could go to the Laplace transform in the s domain. 171 00:11:33,150 --> 00:11:35,120 We could solve it in the s domain. 172 00:11:35,120 --> 00:11:39,470 And this is exactly what we'll meet when 173 00:11:39,470 --> 00:11:41,150 we take the Laplace transform. 174 00:11:41,150 --> 00:11:43,660 That will be the Laplace transform, 175 00:11:43,660 --> 00:11:46,250 which we have to deal with. 176 00:11:46,250 --> 00:11:51,980 So that transfer function is a fundamental, 177 00:11:51,980 --> 00:11:57,210 this polynomial tells us, its roots tell us the frequencies, 178 00:11:57,210 --> 00:12:00,010 s1, s2, and the no solutions. 179 00:12:00,010 --> 00:12:03,810 And then 1 over that tells us the right multiplier 180 00:12:03,810 --> 00:12:06,140 in the force solution. 181 00:12:06,140 --> 00:12:11,250 So constant coefficients, exponential forcing, 182 00:12:11,250 --> 00:12:13,580 the best case possible. 183 00:12:13,580 --> 00:12:15,370 Thank you.