WEBVTT 1 00:00:00.000 --> 00:00:06.960 Well hello everyone and welcome to our sixth and final lecture on Laplace transforms. 2 00:00:06.960 --> 00:00:13.600 Today we're going to continue working with the second shifting theorem and we'll see how to take 3 00:00:13.600 --> 00:00:21.120 inverse Laplace transforms with the second shifting theorem. We then move on to look at an application 4 00:00:21.120 --> 00:00:27.680 of the second shifting theorem where we've got a differential equation that involves a step function 5 00:00:28.320 --> 00:00:32.880 as its portion function that's the function on the right hand side of the differential equation. 6 00:00:34.080 --> 00:00:40.640 Then we introduce another widely used function in engineering called an impulse function and it's 7 00:00:40.640 --> 00:00:46.960 also known as a Dirac delta function and finally we'll solve a differential equation 8 00:00:46.960 --> 00:00:50.560 that involves an impulse function as its portion term. 9 00:00:51.280 --> 00:01:04.720 Okay so in our last class we introduced the second shifting theorem and we saw how the second 10 00:01:04.720 --> 00:01:12.880 shifting theorem can be used to take Laplace transforms of time shifted or delayed functions 11 00:01:13.040 --> 00:01:21.920 involve step functions or products of step functions. Today we're going to look at how 12 00:01:22.960 --> 00:01:33.280 we calculate the inverse Laplace transform of functions that have been shifted and this diagram 13 00:01:33.280 --> 00:01:39.680 from our last lecture shows a typical time delayed we call them shifting truncated functions without 14 00:01:39.680 --> 00:01:46.320 function it's given by f of t minus capital T, capital T corresponds to the shift the minus 15 00:01:46.320 --> 00:01:52.160 represents a shift to the right and the step function truncates our function you can see that 16 00:01:52.160 --> 00:01:58.960 in the diagram here here's our function f of t we're multiplied by the step function with critical 17 00:01:58.960 --> 00:02:06.400 value zero just to set it to zero before time t equals zero so that nothing's happening up to time 18 00:02:06.400 --> 00:02:12.720 t equals zero and then our step function with critical value t equals zero that's u of t takes 19 00:02:12.720 --> 00:02:21.520 the value one from zero onwards so we've just got our function f of t here. Now if we want to model 20 00:02:21.520 --> 00:02:28.960 say a delay in a system we can shift the function in the time domain let's say by capital T units 21 00:02:28.960 --> 00:02:38.240 and to the right so we we introduce the function f of t minus capital T and that shifts this function 22 00:02:38.240 --> 00:02:46.560 by t units to the right now to ensure that the function is zero up until this critical value 23 00:02:46.560 --> 00:02:53.520 we simply multiply our f of t minus capital T or shift function by our step function with our 24 00:02:53.520 --> 00:03:00.960 critical value capital T which equals the shift in our function and we obtain this 25 00:03:02.480 --> 00:03:10.560 time delayed function or shifted and truncated function if you like and we saw that we can take 26 00:03:10.560 --> 00:03:16.720 the Laplace transforms of such function so the Laplace transform of a shifted and truncated 27 00:03:16.720 --> 00:03:27.920 function is simply we have the Laplace transform of the unshifted function that's f of s and we've 28 00:03:27.920 --> 00:03:35.520 incorporate the shift directly in the Laplace transform in this term here e to the minus s 29 00:03:35.520 --> 00:03:46.000 times capital T where capital T represents the shift or the delay if you prefer. So just an 30 00:03:46.000 --> 00:03:53.920 example to remind us how we went about calculating Laplace transforms of these delayed functions so 31 00:03:53.920 --> 00:04:00.960 here's our example to determine the Laplace transform of g of t equal to t minus pi to the power four 32 00:04:01.840 --> 00:04:08.240 and that's multiplied by the unit step function whose critical value is pi if you remember we use 33 00:04:08.240 --> 00:04:15.280 this diagram here that allows us to go from the time domain over here with our shifted and truncated 34 00:04:15.280 --> 00:04:25.280 function our delayed function given here over to the estimate to get its Laplace transform we would 35 00:04:25.280 --> 00:04:31.200 we'd like to be able to go directly but we can't quite do that so we've got to use this diagram 36 00:04:31.200 --> 00:04:37.920 here to get from time domain to estimate so the first thing we do following this diagram here we 37 00:04:37.920 --> 00:04:44.160 ignore the step function and we ignore the shift to give us leave us the function f of t we take 38 00:04:44.160 --> 00:04:51.360 its Laplace transform to get f of s and then we multiply the Laplace transform f of s big f of s 39 00:04:51.360 --> 00:05:00.000 by e to the minus s times capital T where capital T represents the time shift or the delay depending 40 00:05:00.000 --> 00:05:06.400 how you want to look at it and it also equals the critical value of the step function so let's have a 41 00:05:06.400 --> 00:05:12.240 look at this example here so here is the example we've got to ignore the step function ignore the 42 00:05:12.240 --> 00:05:19.280 shift and that just leaves us that f of t is equal to t to the power four i've ignored the shift 43 00:05:19.280 --> 00:05:26.080 the step function and i've ignored the shift giving me f of t is equal to t to the power four the Laplace 44 00:05:26.080 --> 00:05:32.320 transform of t to the power four i get from number four of the tables remember it's um 45 00:05:34.000 --> 00:05:42.160 it says that t to the n has Laplace transform n factorial or s to the n plus one and with 46 00:05:42.240 --> 00:05:49.040 n equal to four we find that that's four factorial over s to the five four factorial is four times 47 00:05:49.040 --> 00:05:57.200 three times two times one which is 24 so the Laplace transform big f of s is equal to 24 over s 48 00:05:57.200 --> 00:06:05.120 to the five and then in our final step to come down this side of our diagram we simply multiply 49 00:06:05.200 --> 00:06:13.360 f of s obtained in step two by e to the minus s t that's capital t where that corresponds to 50 00:06:13.360 --> 00:06:18.480 the critical value of the step function and it also equals the shift which and in this case that's 51 00:06:18.480 --> 00:06:27.360 equal to pi so we multiply 24 over s to the five from step two by e to the minus pi s and that gives 52 00:06:28.000 --> 00:06:37.920 our Laplace transform of this time shifted function here this delayed function down here 53 00:06:39.040 --> 00:06:45.440 okay so uh just quickly just to remind you of what kind of what that all means here with here's 54 00:06:45.440 --> 00:06:52.640 the function t to the power four shown in blue we shift it by pi units and it's a shift to the 55 00:06:52.640 --> 00:07:01.040 right the t minus pi to power four is simply the graph of t the power four shifted pi units to the 56 00:07:01.040 --> 00:07:07.680 right and then we truncate this function by multiplying it by the unit step function whose 57 00:07:07.680 --> 00:07:15.200 critical value is pi and that has the effect of setting everything in this red function up to pi 58 00:07:16.160 --> 00:07:22.320 sets everything equal to zero and that's what we've got here you can see everything along here is set 59 00:07:22.320 --> 00:07:29.760 zero and the function only kicks in when t is equal to pi and that's simply because the step 60 00:07:29.760 --> 00:07:38.400 function the unit step function u t minus pi is equal to zero up to pi so you're multiplying t minus 61 00:07:38.400 --> 00:07:46.320 pi to the power four by zero all the way up to pi so you get zero and then f from pi onwards 62 00:07:46.320 --> 00:07:51.520 the unit step function takes the value one so all we have is the graph of t minus pi to the 63 00:07:51.520 --> 00:07:58.720 power four and that's just the red graph from pi onwards and that's what we see down here so that's 64 00:07:58.720 --> 00:08:06.480 what's going on that is our delayed function our truncated function whatever you want to call it 65 00:08:07.520 --> 00:08:14.480 so we can apply the second shifting theorem in reverse to allow us to calculate 66 00:08:15.440 --> 00:08:22.480 inverse Laplace transforms and you know as we saw we took the Laplace transform of this 67 00:08:22.480 --> 00:08:29.680 shift truncated function and we obtained this expression here we've got the Laplace transform 68 00:08:29.680 --> 00:08:35.600 of the unshifted function the unshifted function is f of t it's Laplace transform is f of s and 69 00:08:35.600 --> 00:08:42.480 we include the shift directly in the transform in this exponential term here e to the minus s t 70 00:08:42.560 --> 00:08:47.520 where t is capital t represent the shift now we want to go the other way now suppose we've 71 00:08:47.520 --> 00:08:53.280 got this expression here we want to retrieve the time domain function that would have produced this 72 00:08:53.840 --> 00:09:02.560 so we take inverse Laplace transforms so the inverse Laplace transform of this function 73 00:09:02.560 --> 00:09:08.320 is just given by this shifted untruncated function so the second shifting theorem states that 74 00:09:08.320 --> 00:09:16.320 multiplying Laplace transform f of s in the s domain by an exponential function e to the minus s 75 00:09:16.320 --> 00:09:24.400 t corresponds to a shift in the inverse transfer shift of the inverse transform in the time domain 76 00:09:24.400 --> 00:09:32.800 by t units so that's exactly what's happening here there's the shift f of t minus t it's a shift 77 00:09:32.800 --> 00:09:37.840 to the right and of course we multiply by the step function to kill everything off up to that 78 00:09:38.320 --> 00:09:44.880 critical value so that the function only starts at capital t hence you can think of as a delayed 79 00:09:44.880 --> 00:09:54.800 function it only starts at this critical value capital t so in a similar manner to the using 80 00:09:54.800 --> 00:10:01.760 the diagram to calculate Laplace transforms using the second shifting theorem in our last class 81 00:10:01.760 --> 00:10:07.440 when we went round this diagram in the clockwise direction and as we saw in the example I just did 82 00:10:08.720 --> 00:10:14.000 I'm going from the time domain to the s domain calculating the Laplace transform we're now going 83 00:10:14.000 --> 00:10:18.560 to go the other way we're going to calculate inverse Laplace transforms we want to go from 84 00:10:18.560 --> 00:10:26.400 the s domain to the t domain and we're going to do that by going anti-clockwise round this diagram 85 00:10:28.880 --> 00:10:34.320 we would of course ideally like to go just directly across down here down the bottom of 86 00:10:34.320 --> 00:10:39.600 my diagram here but it's a bit complicated to do that sometimes too complicated so we use our 87 00:10:39.600 --> 00:10:45.680 diagram here to travel round in an anti-clockwise direction from the s domain to the t domain 88 00:10:46.640 --> 00:10:53.200 so we'll just look at some examples of how this is applied so the question here is determine the 89 00:10:53.200 --> 00:11:01.200 inverse Laplace transform of g of s is equal to 1 over s plus 6 times e to the minus 8s now 90 00:11:01.440 --> 00:11:08.160 how would I know to use the second shifting theorem here well the giveaway is that we're in 91 00:11:08.800 --> 00:11:16.640 the s domain and we're multiplying a function of s this one here that's called that big f of s 92 00:11:16.640 --> 00:11:23.520 we're multiplying it by an exponential function that suggests to me that I should use the second 93 00:11:23.520 --> 00:11:30.160 shifting theorem the fact that I'm multiplying some function of s by an exponential that will 94 00:11:30.240 --> 00:11:37.200 suggest to me to use the second shifting theorem so I'm going to start in the bottom right of my 95 00:11:37.200 --> 00:11:44.080 diagram and travel round in an anti-clockwise direction so the first thing I have to do 96 00:11:44.080 --> 00:11:50.720 according to my diagram here I've got my function f of s times e to the minus st so if I just look 97 00:11:50.720 --> 00:11:58.080 at the function here I can easily identify that f of s will be 1 over s plus 6 and e to the minus 98 00:11:58.080 --> 00:12:06.000 st will simply be this exponential term e to the minus 8s so that capital T the shift the delay 99 00:12:06.000 --> 00:12:12.640 whatever you want to call it is equal to 8 in this case just simply pattern matching e to the minus 100 00:12:12.640 --> 00:12:20.800 8s with e to the minus st so capital T must equal 8 and we'll use that a little later on in this 101 00:12:20.800 --> 00:12:28.000 question so for the moment I've identified that f of s is equal to 1 over s plus 6 so we're here 102 00:12:28.000 --> 00:12:34.160 in the diagram we want to go across to the time domain and we do that by taking inverse Laplace 103 00:12:34.160 --> 00:12:43.040 transforms of f of s so here we are we want to get f of t it's the inverse Laplace transform of f of s 104 00:12:43.040 --> 00:12:50.400 which is 1 over s plus 6 we know that one that's we've seen it many times now it's number 5 1 over 105 00:12:50.400 --> 00:12:57.360 s plus alpha with alpha equal to 6 from our tables we find that the inverse Laplace transform 106 00:12:57.360 --> 00:13:07.040 or f of t in other words is e to the minus 6t that's just using number 5 so we're here now we're 107 00:13:07.040 --> 00:13:16.400 we've got f of t and we want to find our shifted function shift and truncated function the delayed 108 00:13:16.400 --> 00:13:24.480 function so what do we do well wherever we see a t in f of t we replace it by t minus capital T 109 00:13:24.480 --> 00:13:31.920 where capital T represented the shift from our question up here sorry in the estimate from 110 00:13:31.920 --> 00:13:37.200 earlier on we identified on the last slide we identified that was equal to 8 so we want to 111 00:13:37.200 --> 00:13:43.360 replace every occurrence of t in f of t by t minus 8 and then we want to multiply it 112 00:13:43.360 --> 00:13:49.520 by the step function is critical value is 8 and that's exactly what I say here so f of t 113 00:13:49.520 --> 00:13:57.120 was obtained at step 2 it was e to the minus 6t so wherever I see a t I replace it by t minus 114 00:13:57.200 --> 00:14:06.080 capital T capital T our shift was equal to 8 so I replace t by t minus 8 and that's what I do down 115 00:14:06.080 --> 00:14:13.840 here so I get e to the minus 6 times t minus 8 and I multiply by the step function whose critical 116 00:14:13.840 --> 00:14:22.240 value is 8 so that gives me my inverse Laplace transform of my original function shown up here 117 00:14:22.320 --> 00:14:29.360 in the green box is g of s and the time domain it's inverse Laplace transform is g of t and that's 118 00:14:29.360 --> 00:14:38.560 it given down here so you can see it is indeed a shifted function we've got t minus 8 here that's 119 00:14:38.560 --> 00:14:42.960 the shift in the function and it's truncated because we're multiplying by the step function 120 00:14:42.960 --> 00:14:51.040 whose critical value is 8 so if you remember in our last lecture we went the other way if we were 121 00:14:51.040 --> 00:14:58.320 given a function like this we would use the second shifting theorem in reverse to obtain g of s 122 00:14:59.360 --> 00:15:06.720 okay so that's one example so let's have a look at another example of applying the second shifted 123 00:15:06.720 --> 00:15:15.520 theorem to obtain the inverse Laplace transform of a function and here's my function here 1 over s 124 00:15:16.240 --> 00:15:23.840 into s plus 2 times e to the minus 4s so that is certainly of the form f of s times e to the 125 00:15:23.840 --> 00:15:30.080 minus st and that's multiplied by the exponential is the clue for me to use the second shift in 126 00:15:30.080 --> 00:15:37.440 theorem so again I'll go round my diagram anti-clockwise so I've got to identify f of s well 127 00:15:37.440 --> 00:15:44.640 that's pretty easy I do that just by ignoring the exponential I get 1 over s into s plus 2 now 128 00:15:45.600 --> 00:15:52.000 I need to find a candidate or candidates from my table Laplace transform so I can 129 00:15:52.720 --> 00:15:59.760 invert this function here now if I look if you remember I focus on the denominator at first 130 00:15:59.760 --> 00:16:06.720 to identify one or maybe more than one function from the table that I can use so 131 00:16:06.720 --> 00:16:12.400 do I have anything in the table that has maybe s times s plus a number well if I look down 132 00:16:12.400 --> 00:16:19.760 in my table I would eventually get to number seven and here I've got s times s plus alpha and that's 133 00:16:19.760 --> 00:16:25.840 certainly of the form that I've got here if I set alpha equal to two if I said alpha equal to two here 134 00:16:25.840 --> 00:16:32.160 I get s into s plus two and that's exactly what I've got here so number seven is definitely 135 00:16:32.960 --> 00:16:41.360 the one to use here one small problem though at the moment is that the numerator also has an alpha 136 00:16:41.600 --> 00:16:47.920 in it so that puts in the numerator must match this term in the denominator at the moment it doesn't 137 00:16:47.920 --> 00:16:53.600 I've got a one in the numerator and there's a two in the denominator here so what I need to do is a 138 00:16:53.600 --> 00:17:01.360 little bit of manipulation so that I can write down exactly what the table wants me to have before 139 00:17:01.360 --> 00:17:09.200 I can use number seven and we've seen how to do this in previous lectures so my function is here 140 00:17:09.200 --> 00:17:15.840 my f of s is here so I'm going to write down exactly what the table would have me use if I'm 141 00:17:15.840 --> 00:17:22.240 going to use number seven I must have an alpha in the numerator and in this case alpha is equal to 142 00:17:22.240 --> 00:17:31.120 two so I write down two over s into s plus two that's exactly of the form alpha over s into s 143 00:17:31.120 --> 00:17:38.080 plus alpha but I've changed things a little because I didn't have a two in my original numerator it was 144 00:17:38.080 --> 00:17:45.760 actually a one I had but I can easily fix this I multiply by a fraction where the fraction has 145 00:17:45.760 --> 00:17:51.760 the thing that I want in its top line that's a one because I want a one in the top line that's what 146 00:17:51.760 --> 00:17:58.880 I had and I divide by what the table forced me to introduce the table forced me to introduce a two 147 00:17:58.880 --> 00:18:07.120 because I needed this value the alpha on top to match the alpha below so the fraction just once again 148 00:18:07.120 --> 00:18:14.400 on the top line includes the thing that I want in the numerator that's a one down below I have what 149 00:18:14.400 --> 00:18:22.400 the table forced me to introduce and that was a two so once I've done that it's a fairly straightforward 150 00:18:22.400 --> 00:18:28.720 process to invert because the thing in the square bracket will just be one minus e to the minus two 151 00:18:29.280 --> 00:18:37.600 and don't forget you multiply by a half so that is the inverse Laplace of f of s which is f of t 152 00:18:37.600 --> 00:18:44.160 so I'm now up here in my diagram and the final step requires me to replace every occurrence of t 153 00:18:44.880 --> 00:18:54.160 in f of t by t minus capital T where capital T is the shift or the critical value in the step 154 00:18:54.160 --> 00:19:03.600 function the criteria where capital T is the shift or the delay in other words so what is it 155 00:19:03.600 --> 00:19:10.640 equal to in this case well just reading off it's equal to four so what I do now is I replace 156 00:19:11.520 --> 00:19:20.000 t in f of t by t minus four and I multiply by the step function whose critical value is four 157 00:19:20.000 --> 00:19:30.240 and that gives you my answer I have a half one minus e to the minus two brackets t minus four 158 00:19:30.240 --> 00:19:37.360 because I'm replacing t at step two and f of s by t minus four so the whole thing gets multiplied 159 00:19:37.360 --> 00:19:45.120 by two so I've got minus two brackets t minus four close brackets and all of that gets multiplied 160 00:19:45.120 --> 00:19:49.760 by the unit step function as you can see in the diagram here the unit step function whose 161 00:19:49.760 --> 00:19:56.960 critical value is four because big t was equal to four and that was given it appears in the 162 00:19:56.960 --> 00:20:03.040 exponential term in our Laplace transform that we started out with so that's another example 163 00:20:04.320 --> 00:20:11.440 moving on let's have a look at another one so what do we got here we've got our function g of s 164 00:20:11.440 --> 00:20:20.320 c equal to one over s into s squared plus four times e to the minus three s okay so is this in 165 00:20:20.320 --> 00:20:26.720 the form for the second shift in theorem well yes it is we've got some function f of s there it is 166 00:20:26.720 --> 00:20:32.560 one over s into s squared plus four that's our f of s I will multiply that by an exponential 167 00:20:33.360 --> 00:20:37.920 that's what tells me that I want to use the second shifted theorem the fact that I'm multiplying 168 00:20:37.920 --> 00:20:45.520 some function of s by an exponential e to the minus s t that's capital t and you can easily see 169 00:20:45.520 --> 00:20:52.000 here by comparison that capital t is equal to three we'll use that in the last step so first thing 170 00:20:52.000 --> 00:20:59.760 we have to do is we have to according to our diagram we're down here that's our function here down 171 00:20:59.760 --> 00:21:06.320 the bottom right we first of all ignore the step function to leave me big f of s so big f of s is 172 00:21:06.320 --> 00:21:14.160 simply one over s into s squared plus four now I would check my table I would focus on the denominator 173 00:21:14.160 --> 00:21:22.080 and look for possible candidates from my table eventually I would reach number 10 and that would 174 00:21:22.080 --> 00:21:28.880 certainly be look quite promising because I've got s into s squared plus a number and this number 10 175 00:21:28.880 --> 00:21:34.800 has s into s squared plus a number they write the number as omega squared but that's not a problem 176 00:21:34.800 --> 00:21:42.640 if omega squared is equal to four then omega must be equal to two in this case but I must have 177 00:21:42.640 --> 00:21:50.560 an omega squared in the numerator at the moment these two don't match but according to the table 178 00:21:50.560 --> 00:21:56.960 they must have I've got a four down here I must have a four up top so we'll write this we'll do what 179 00:21:56.960 --> 00:22:04.560 we always do we write down exactly what the table would have us use so it'll be four over s into s 180 00:22:04.560 --> 00:22:12.160 squared plus four and that's exactly of the form omega squared over s into s squared plus omega squared 181 00:22:12.160 --> 00:22:17.920 however we've changed the function this function the square brackets is not exactly the same as 182 00:22:17.920 --> 00:22:24.320 we started with because this one's a four in the numerator whereas we started with a one in the numerator 183 00:22:24.320 --> 00:22:33.120 so we must have a multiplying factor to compensate for that and it's the same thing as we've done 184 00:22:33.120 --> 00:22:39.680 several times in the past we have a fraction where the top line is the thing that we want 185 00:22:39.680 --> 00:22:47.040 which is one and in the bottom line we introduced what the table forced us to introduce which was 186 00:22:47.040 --> 00:22:53.680 the four so a fraction will be one what we want divided by four which we're forced to introduce 187 00:22:53.680 --> 00:23:00.320 so it's a quarter we can then take the Laplace transform or the inverse Laplace transform sorry 188 00:23:00.400 --> 00:23:07.760 of the term in the square brackets that is just with omega squared equal to four that means omega 189 00:23:07.760 --> 00:23:14.240 is equal to two so just be careful it's omega that you want here not omega squared so omega will be 190 00:23:14.240 --> 00:23:20.720 equal to two it's number ten with omega equal to two and for the term of the square brackets it becomes 191 00:23:20.720 --> 00:23:27.520 one minus cos two t of course don't forget we're multiplying by a quarter so we find that our 192 00:23:27.520 --> 00:23:33.520 function take the inverse Laplace transform of big f of s our function f of t would be equal to a 193 00:23:33.520 --> 00:23:43.840 quarter times one minus cos two t the final step will involve us replacing every occurrence of t 194 00:23:43.840 --> 00:23:50.880 in f of t that we've just found at step two replace every occurrence of t by t minus capital t 195 00:23:50.880 --> 00:23:57.360 capital t is the shift value that appears in the exponential in our original function 196 00:23:58.640 --> 00:24:04.880 and in this case as we said at the beginning we identified that to be equal to three so that 197 00:24:04.880 --> 00:24:14.160 means that we replace every occurrence of t in in f of t by t minus three and then we'll also have to 198 00:24:14.160 --> 00:24:22.320 multiply by the step function whose critical value is capital t in other words critical value is three 199 00:24:22.320 --> 00:24:29.840 so let's just do that so i'm going to replace every t in f of t by t minus three and i'm going to 200 00:24:29.840 --> 00:24:36.560 multiply by the step function ut minus three so and that'll give me the inverse Laplace transform 201 00:24:37.200 --> 00:24:44.240 of my original function g of s so g of t that's the inverse Laplace transform will therefore 202 00:24:44.240 --> 00:24:53.200 equal we've got a quarter we'll have one minus cos two times now it'll be t minus three so that's 203 00:24:53.200 --> 00:25:00.880 what i've got here and all of that gets multiplied by our step function ut minus three and that's 204 00:25:00.880 --> 00:25:10.720 exactly what the bottom left statement in our box tells us to do that's this function here 205 00:25:10.720 --> 00:25:18.480 will match this one so that's the inverse Laplace transform of that function g of s and to do that 206 00:25:18.480 --> 00:25:29.120 we used our second shifting theorem so here's another example um what to find the inverse Laplace 207 00:25:29.120 --> 00:25:36.880 transform of g of s equals three over s squared plus 49 e to the minus nine s and you know by now 208 00:25:36.880 --> 00:25:44.400 we realize this takes the form of a function of s f of s times an exponential involving s so that 209 00:25:44.400 --> 00:25:52.720 tells me to use the second shifted theorem so um let's have a look at this so let's take f of s 210 00:25:52.720 --> 00:25:59.600 first of all that's what our diagram tells us ignore the exponential and identify f of s f of s 211 00:25:59.600 --> 00:26:07.760 is three over s squared plus 49 there we are down here um i would look in my table and i would see 212 00:26:07.760 --> 00:26:16.640 do i have any terms that look like this and i would reach number eight and number nine also 213 00:26:16.720 --> 00:26:22.640 would have s squared plus some number number eight you can see it's s squared plus a number 214 00:26:22.640 --> 00:26:28.000 where they've written the number as omega squared but that's very easy 49 it's just seven squares so 215 00:26:28.000 --> 00:26:35.040 we can write that in this form s squared plus seven squared now we've seen number number nine we've 216 00:26:35.040 --> 00:26:40.400 seen numbers eight and nine before and if you remember number nine is just the cos omega t and 217 00:26:40.400 --> 00:26:46.560 number nine has the Laplace variable s in the numerator so that will not work here because 218 00:26:46.560 --> 00:26:52.240 we do not have an s in the numerator in this case we've got a constant that's three in the numerator 219 00:26:52.960 --> 00:26:58.800 and that means we use number eight because that also has a constant in the numerator so if you're 220 00:26:58.800 --> 00:27:03.520 doing this with a full set of tables in front of you you would narrow by looking at the denominator 221 00:27:03.520 --> 00:27:08.160 you would narrow the field down to numbers eight and nine you would then look at the numerator and 222 00:27:08.160 --> 00:27:13.360 you would spot that number nine involves the Laplace variable s in the numerator so you'd 223 00:27:13.360 --> 00:27:19.840 discount that one and you would use number eight but before we can use number eight directly we 224 00:27:19.840 --> 00:27:28.400 notice one thing that in the table we've got s squared plus omega squared down below and in the 225 00:27:28.400 --> 00:27:36.320 numerator we've got omega so we've written this as s squared plus seven squared so omega in our 226 00:27:36.320 --> 00:27:42.560 case is seven but we don't have a seven in the numerator but we can easily fix that the way we've 227 00:27:42.560 --> 00:27:50.800 been doing in the past we can write down exactly what the table would have us use it's omega over s 228 00:27:50.800 --> 00:27:58.640 squared plus omega squared and here omega c could be seven we correct this by multiplying by a fraction 229 00:27:58.640 --> 00:28:06.800 whose numerator is exactly what we want and we want a three and we divide by what the table 230 00:28:06.800 --> 00:28:13.280 forces to introduce in the top line here which is a seven so a fraction that we multiply by 231 00:28:13.280 --> 00:28:23.520 would be three over seven okay so we can now take the inverse Laplace transform of f of s to give us 232 00:28:23.600 --> 00:28:32.880 f of t and that's fairly straightforward because it's number eight as we've said with omega over s 233 00:28:32.880 --> 00:28:38.480 squared plus omega squared with omega equal to seven and by use number eight I find that the inverse 234 00:28:38.480 --> 00:28:46.240 Laplace transform is sine seven t but don't forget we've got to multiply by our factor three over seven 235 00:28:46.240 --> 00:28:53.040 by this fraction here three over seven so I've found f of t so that all that that leads me to do now 236 00:28:53.040 --> 00:29:03.680 is the last step replace t in f of t by t minus capital T where capital T corresponds to the shift 237 00:29:04.720 --> 00:29:12.160 which we can identify easily from our original function g of s remember e to the minus s t 238 00:29:12.160 --> 00:29:21.920 so capital T is equal to nine so we replace t in f of t by t minus nine and we multiply 239 00:29:22.000 --> 00:29:28.480 everything by our step function with critical value nine so that's what I do down here finally 240 00:29:29.520 --> 00:29:37.200 my f of t is up here so I've got three over seven sine seven we place t by t minus nine so there you 241 00:29:37.200 --> 00:29:44.480 are so sine seven brackets t minus nine close brackets and multiply everything by the unit 242 00:29:44.480 --> 00:29:52.960 step function whose critical value is equal to nine remember the critical value matches the shift 243 00:29:53.680 --> 00:30:02.880 so there we are so that is the solution okay move on now and let's take a mobius question on 244 00:30:04.720 --> 00:30:09.040 the second shifting theorem find the inverse Laplace transforms using the second shifting 245 00:30:09.120 --> 00:30:15.600 theorem so here's our function given over here it's g of s is given here and we're told we're 246 00:30:15.600 --> 00:30:21.040 required to complete this question in three steps well that's exactly what we've been doing 247 00:30:22.960 --> 00:30:33.360 I should say here though that that that mobius uses alpha as the shift variable rather than capital 248 00:30:33.360 --> 00:30:39.760 t which we use and that's that's slightly unfortunate because alpha also has another meaning 249 00:30:39.760 --> 00:30:47.920 in the tables so I'll tend to keep using capital t and I'll just tell you wherever 250 00:30:47.920 --> 00:30:56.320 it's representing alpha so here's my function here here's my function my first question says identify 251 00:30:56.400 --> 00:31:04.000 the value of alpha which is equal to capital t so that's very easy just by comparison I can 252 00:31:04.640 --> 00:31:10.800 also identify f of s f of s is the thing that's multiplying the exponential it's s minus two 253 00:31:10.800 --> 00:31:18.000 over s minus two all squared plus nine so that's my f of s and my exponential e to the minus three s 254 00:31:18.000 --> 00:31:25.280 comparing it with my table e to the minus s capital t capital t must be equal to three or as 255 00:31:25.280 --> 00:31:32.400 mobius calls they identify the alpha but it's our capital t so it's equal to three f of s already 256 00:31:32.400 --> 00:31:39.040 identified it's the term that multiplies the exponential so it's s minus two or s minus two 257 00:31:39.040 --> 00:31:44.480 all squared plus nine so that's the first part of the question the next part of the question 258 00:31:45.760 --> 00:31:52.880 we're asked to determine f of t so that just going by our diagram here that's exactly what we want 259 00:31:52.880 --> 00:31:59.520 to do maybe I can carry this over to the next slide if there's room there yeah so according to our 260 00:31:59.520 --> 00:32:05.920 diagram that's exactly what we want to do we want to invert f of s that we found in step one to get 261 00:32:05.920 --> 00:32:15.280 f of t so if I do that there's my function f of t there it is in here in the brackets here we want 262 00:32:15.600 --> 00:32:25.280 to inverse the plus transform of that if I look at my Laplace tables I'll eventually find number 15 263 00:32:25.840 --> 00:32:32.800 this of this form s plus alpha over s plus alpha all squared plus omega squared so that's certainly 264 00:32:32.800 --> 00:32:42.000 matches this function here my f of s function if I set omega equal to three and alpha equal to minus 265 00:32:42.000 --> 00:32:50.000 two I've got a perfect match so I can invert them to give me e to the minus minus two t or e to the 266 00:32:50.000 --> 00:33:00.080 two t cos three t because omega is equal to three so that gives me my function in my diagram f of t 267 00:33:00.080 --> 00:33:10.640 and you can see I've entered that in here so e to the two t cos three t okay so I'm moving on to step 268 00:33:10.640 --> 00:33:18.000 three that finally asks us to obtain the inverse Laplace transform of my original g of s 269 00:33:19.360 --> 00:33:28.480 so my original g of s is given here we know what our function f of t is we identified that at our 270 00:33:28.480 --> 00:33:36.720 last step there it is so all we want to do now is replace every occurrence of t in f of t by t 271 00:33:36.800 --> 00:33:43.040 minus capital t capital t we identified earlier it's been equal to three it's whatever the value 272 00:33:43.680 --> 00:33:50.960 appeared in the exponential function is and it's equal to three here so we replace every 273 00:33:50.960 --> 00:33:57.760 occurrence of t in f of t by t minus three and we multiply by a step function whose critical value 274 00:33:57.760 --> 00:34:06.640 is three so now if I do that there f of t there so every t are replaced by t minus three so again 275 00:34:06.960 --> 00:34:17.920 e to the two brackets t minus three times cos of three brackets t minus three multiplied by the 276 00:34:17.920 --> 00:34:25.360 step function whose critical value is three so that's ut minus three so that is my solution that is 277 00:34:25.360 --> 00:34:33.360 the inverse Laplace transform of this function here and again you know you can see it's a shifted 278 00:34:33.360 --> 00:34:40.800 function for every t you've got t minus three and it's truncated because it's multiplied by the 279 00:34:40.800 --> 00:34:47.840 unit step function and again if we wanted to go from the time domain up to here to the s domain 280 00:34:47.840 --> 00:34:53.840 by taking Laplace transforms we would use the methods we saw last week it's just basically 281 00:34:53.840 --> 00:35:03.280 the process we're using today in reverse okay so here's another example here now so we've 282 00:35:03.280 --> 00:35:12.320 been asked to determine the inverse Laplace transform of this function here okay so we need 283 00:35:12.320 --> 00:35:16.560 to do a little bit of manipulation first because if we look in our tables we're not going to find 284 00:35:16.560 --> 00:35:24.000 anything that looks anything like this so what I've noticed is I can split this into two fractions 285 00:35:24.720 --> 00:35:31.120 inverse Laplace over s squared minus the inverse Laplace of e to the minus two s over s squared 286 00:35:31.840 --> 00:35:38.080 now this one is easy this will just be number three in our tables the inverse Laplace of one 287 00:35:38.080 --> 00:35:45.040 over s squared is simply t this one on the other hand it's a bit more complicated because I'm 288 00:35:45.040 --> 00:35:50.720 multiplying one over s squared by e to the minus two s and the fact that I'm multiplied by an 289 00:35:50.720 --> 00:35:57.760 exponential in the s domain would suggest to me that I need the second shifting theorem so I'm just 290 00:35:57.760 --> 00:36:02.960 marking this with a star here because I'm going to refer to it later on so I'm now going to just 291 00:36:02.960 --> 00:36:11.200 consider this term here the inverse Laplace transform of e to the minus two s over s squared 292 00:36:11.920 --> 00:36:18.960 so we know as I've just said that we need to use the second shifting theorem so we'll go 293 00:36:18.960 --> 00:36:25.120 round our diagram in an anticlockwise direction my first there's my function one over s squared 294 00:36:25.120 --> 00:36:33.280 e to the minus two s so the first thing I do is I ignore the exponential and identify f of s 295 00:36:33.280 --> 00:36:41.200 so f of s is equal to one over s squared there we are step two take the inverse Laplace transform 296 00:36:41.200 --> 00:36:48.720 of f of s to give me f of t well that's very easy it's number three in the tables invert one over s 297 00:36:48.720 --> 00:36:59.840 squared you'll get that f of t is equal to t and then I need to identify capital t what's the shift 298 00:36:59.840 --> 00:37:05.760 well the shift is the value that appears in the exponential if you just match them up here e to 299 00:37:05.760 --> 00:37:12.560 the minus s times capital t would be equal to e to the minus two s meaning that capital t is equal 300 00:37:12.560 --> 00:37:19.760 to two so that gives me my shift and also gives me the critical value of my step function so what I 301 00:37:19.760 --> 00:37:26.160 want to do now that I've identified t is equal to two I've got to replace every occurrence of t 302 00:37:27.280 --> 00:37:35.760 in f of t that I found at step two I've got to replace that by t minus two and I've got to multiply 303 00:37:35.760 --> 00:37:44.560 by step function whose critical value is two such just u t minus two so that expression down here 304 00:37:45.120 --> 00:37:51.680 is the inverse Laplace transform of one over s squared e to the minus two s so we do indeed have 305 00:37:51.680 --> 00:37:59.440 a shifted function so t has been shifted t minus two two units to the right and we truncated it 306 00:37:59.440 --> 00:38:05.280 we've killed it off up to t equals two by multiplied by a step function whose critical value is two 307 00:38:05.280 --> 00:38:12.640 so I mark that with a double star I'm going to combine that with this one on previous slides where 308 00:38:12.640 --> 00:38:20.000 I've got t minus this quantity which is just the double star quantity so it's t minus all of that 309 00:38:20.800 --> 00:38:28.160 and that gives me the inverse Laplace transform of the function g of s that I started out with 310 00:38:28.160 --> 00:38:40.080 so that's it down here at the foot of the slide okay so we've seen then how to obtain inverse Laplace 311 00:38:40.960 --> 00:38:51.440 transforms that we needed the second shift in theorem so we're now going to apply this method 312 00:38:51.440 --> 00:38:57.840 to solving differential equations and these differential equations will typically involve 313 00:38:57.840 --> 00:39:04.640 step functions on the right hand side or even products of step functions and other functions 314 00:39:04.640 --> 00:39:12.240 going back to if we remember I think maybe it was in our fourth lecture when we looked at products 315 00:39:12.240 --> 00:39:19.760 of the step function and other functions we we had maybe f of t times the step function that would 316 00:39:19.760 --> 00:39:25.440 have the effect of killing f of t off at a certain point whatever the critical value the step function 317 00:39:25.440 --> 00:39:32.480 was and then we had these window functions where a function could only exist inside a window between 318 00:39:33.680 --> 00:39:40.480 two critical values in step functions so these are the type of things that you do come across and 319 00:39:40.480 --> 00:39:46.480 these are the type of functions that you need to use the second shift in theorem for solving the 320 00:39:46.480 --> 00:39:50.720 associate differential equations these these differential equations appear in electrical 321 00:39:50.720 --> 00:39:56.160 mechanical engineering in lots and lots of areas so let's just see how we go about doing this so 322 00:39:56.160 --> 00:40:01.120 here's a first order differential equation where the fortune term the fortune function that's a 323 00:40:01.120 --> 00:40:08.320 function on the right hand side is given by this step function here whose critical value is at t 324 00:40:08.320 --> 00:40:14.400 equals three this could represent maybe a voltage for example that's been delayed by three seconds 325 00:40:15.360 --> 00:40:24.160 nothing happens until let's say a switch is flicked after three seconds and this comes into play here 326 00:40:24.160 --> 00:40:31.120 so the voltage of maybe one volt will be applied and it will stay on forever and ever so these type 327 00:40:31.120 --> 00:40:36.800 of functions as you can see there what are known as discontinuous functions because you reach three 328 00:40:36.800 --> 00:40:44.880 and it suddenly instantaneously jumps from zero up to one these these are often called 329 00:40:44.880 --> 00:40:52.000 discontinuous functions in the in the sense that they are not continuous very very roughly speaking 330 00:40:52.000 --> 00:40:58.160 are continuous functions a function that you could draw without lifting your pen or the paper 331 00:40:58.160 --> 00:41:04.240 so in reality this dotted line is in here so you know you couldn't go from down here to here 332 00:41:04.320 --> 00:41:10.560 without lifting your pen or the paper the dotted line is just to show that there is a sudden jump 333 00:41:10.560 --> 00:41:16.000 at t equals three so here's our differential equation here's our initial condition which is a zero 334 00:41:16.000 --> 00:41:23.040 initial condition which will make things a little bit easier for us and i've rewritten the differential 335 00:41:23.040 --> 00:41:29.040 equation over here along with the initial condition so we saw how to solve differential equations maybe 336 00:41:29.040 --> 00:41:34.480 in our third lecture of a member rightly so we need to take Laplace transform so both sides of 337 00:41:34.480 --> 00:41:42.160 the differential equation so you can think of this as being minus one x dot so the minus one if you 338 00:41:42.160 --> 00:41:48.240 like to constant the constant linearity property it can get taken outside the Laplace operator so 339 00:41:48.240 --> 00:41:55.200 it's minus the Laplace transform of x dot plus the Laplace transform of x plus the Laplace transform 340 00:41:55.280 --> 00:42:03.120 of the step function with critical value three so if i do that we saw how to do this that the Laplace 341 00:42:03.120 --> 00:42:11.360 transform of x dot was simply number 24 in the table it was equal to s x bar minus x of zero 342 00:42:11.360 --> 00:42:16.880 where x zero is the initial condition and of course in this case don't forget there's a minus 343 00:42:16.880 --> 00:42:23.760 outside here okay and then we've got the Laplace transform of x which number 23 tells us that 344 00:42:23.840 --> 00:42:31.520 that's x bar there we are and the Laplace transform of our step function of critical value three 345 00:42:31.520 --> 00:42:40.640 will simply be 10 it's number 26 1 over s e to the minus s times three because three is playing 346 00:42:40.640 --> 00:42:49.440 the role of the capital t and there we are it's 1 over s e to the minus 3 s then we apply the 347 00:42:49.440 --> 00:42:55.920 initial condition x of zero equals zero well that's pretty easy to do so and simplify that you get 348 00:42:55.920 --> 00:43:01.920 minus s x bar plus x bar is equal to 1 over s e to the minus 3 s so that's pretty simple just 349 00:43:01.920 --> 00:43:10.640 applying that zero initial condition doesn't really change anything so my next step so this is just 350 00:43:10.640 --> 00:43:16.640 copied from the previous slide i now want to solve for x bar this is solving the algebra problem so 351 00:43:16.640 --> 00:43:24.800 let's just remind ourselves the calculus t domain time domain problem which was a differential equation 352 00:43:26.000 --> 00:43:32.560 we took Laplace transforms converter time domain differential equation to an s domain 353 00:43:33.200 --> 00:43:40.640 algebra problem so the calculus problem now becomes a much easier algebra problem we solve 354 00:43:40.640 --> 00:43:45.600 the algebra problem that's what we're going to do here we're going to get an expression for x bar 355 00:43:45.600 --> 00:43:52.000 that'll be the solution of the algebra problem and then at the next step we'll have to invert to go 356 00:43:52.000 --> 00:43:58.160 from the s domain to the t domain because our original problem our original differential equation 357 00:43:58.800 --> 00:44:04.080 was in the time domain so our solution should be expressed in the time domain as well 358 00:44:06.480 --> 00:44:12.160 so as i said from the previous slide after applying the initial condition this is what we 359 00:44:12.240 --> 00:44:21.600 obtain now i just find it easier i want to isolate x bar i want to get an expression for x bar so 360 00:44:21.600 --> 00:44:26.960 what i'm just to make things a little easier i'm going to multiply this equation through by minus 361 00:44:26.960 --> 00:44:35.920 one so that's exactly what i've done here that's all i've done here and then i can take x bar out 362 00:44:35.920 --> 00:44:42.320 it's a common factor because it appears in both terms on the left hand side so i get s minus one 363 00:44:43.040 --> 00:44:47.600 x bar because really there's a one here even if we don't write it there's a one here and if you 364 00:44:47.600 --> 00:44:54.880 expand this term you should get the line above s times x bar yes minus one times x bar so that's 365 00:44:54.880 --> 00:45:05.280 absolutely fine and then i divide through by s minus one and that gives me my time my estimate 366 00:45:05.280 --> 00:45:12.560 solution so x bar remember it's a function of s and i divide through by s minus one i get that x bar 367 00:45:12.560 --> 00:45:22.800 is equal to minus one over s into s minus one times e to the minus three s now my final step of 368 00:45:22.800 --> 00:45:30.560 course is to take inverse Laplace transforms this is x bar of s inverting that would give me x of t 369 00:45:30.640 --> 00:45:36.640 my solution but i'm also going to of course invert this side and there's a little bit of work here 370 00:45:36.640 --> 00:45:42.000 because i notes that i'm in the estimate and i'm multiplying by an exponential that suggests to me 371 00:45:42.000 --> 00:45:47.840 that i need the second shifted theorem so i'll need the second shifted theorem to invert this one here 372 00:45:49.760 --> 00:45:59.680 furthermore i will need um partial fractions this i think this is actually given the table 373 00:45:59.680 --> 00:46:06.960 but mobius always requires the use of partial fractions so just for practice we'll do that here 374 00:46:06.960 --> 00:46:13.600 so we've got two linear two distinct linear terms in the denominator so we need two fractions one 375 00:46:13.600 --> 00:46:20.560 for each term we'll need an a over s plus b over s minus one that's one for each of these two terms 376 00:46:20.560 --> 00:46:28.320 we multiply both sides by the denominator of the left hand side so if i just take my tablet 377 00:46:28.320 --> 00:46:36.640 just very quickly the denominator of the left hand side is s into s minus one so if i multiply 378 00:46:36.640 --> 00:46:42.880 the left hand side well pretty obviously by zone denominator they'll cancel and that just leaves me 379 00:46:42.880 --> 00:46:53.360 minus one on the left hand side so then i would take this term if i can and i would multiply the 380 00:46:53.360 --> 00:47:00.000 first term on the left hand side by that on the right hand side by this denominator the s is 381 00:47:00.000 --> 00:47:06.880 cancelled give me a times s minus one and then finally i multiply the second term the right 382 00:47:06.880 --> 00:47:15.360 hand side the s minus one's cancel and it gives me b times s so i've now got one equation with two 383 00:47:15.360 --> 00:47:23.040 unknowns but i can simplify this a little just by setting choosing values of s that will make 384 00:47:23.600 --> 00:47:32.320 one of the unknown variables disappear so for example if i set s equal to zero the b disappears 385 00:47:32.320 --> 00:47:39.440 and just these many equation for a so if s is zero i'll get minus one is equal to minus a or in 386 00:47:39.440 --> 00:47:45.840 other words a is equal to one so then i'll choose s equal to one so that a will disappear and it'll 387 00:47:45.840 --> 00:47:52.400 give me an equation for b so with b equal to one i'll have a times zero which is zero this b times 388 00:47:52.400 --> 00:47:59.280 one which is just b so b will be equal to minus one so i've got a i've got b i can plug these back 389 00:47:59.280 --> 00:48:06.960 in here and i find that the partial fraction representation of my expression for x bar is 390 00:48:06.960 --> 00:48:15.120 given down here now these are two very easy terms to invert these are very easy that's just number one 391 00:48:15.120 --> 00:48:22.480 and number five in our tables so that's what i'm saying here so if x bar was equal to this term 392 00:48:22.480 --> 00:48:31.600 here we used partial fractions on it to then obtain this expression here so that if you like i'm now 393 00:48:31.600 --> 00:48:39.040 going to use the second shifting theorem f of s times e to the minus s times capital t so f of s 394 00:48:39.040 --> 00:48:47.840 is this term here in these round brackets so that's f of s and i can match my exponentials together to 395 00:48:48.880 --> 00:48:56.960 t the shift value is equal to three so first of all though step one i'm using the second 396 00:48:56.960 --> 00:49:04.720 shifting theorem says ignore the exponential to leave me f of s so f of s is simply one over s minus 397 00:49:04.800 --> 00:49:11.920 one over s minus one according to my diagram i then have to invert big f of s to go to the time 398 00:49:11.920 --> 00:49:21.760 domain and find f of t so if i do that what do i obtain well i'll obtain this is just number one so 399 00:49:21.760 --> 00:49:28.240 the inverse Laplace of one over s is just one and the inverse Laplace of one over s minus one it's 400 00:49:28.240 --> 00:49:35.120 number five with alpha equal to minus one i've got s plus alpha here i've got s minus one there just 401 00:49:35.120 --> 00:49:41.760 match them meaning that alpha's minus one so with alpha minus one i get e to the minus minus one t 402 00:49:41.760 --> 00:49:49.680 or e to the t so there we are so putting that all together f of t is equal to one minus e to the t 403 00:49:49.760 --> 00:49:59.600 there we are there okay so then i've got f of t now i have my final step involves me identifying 404 00:49:59.600 --> 00:50:05.360 the shift value the capital t value which i've already done just by comparing the exponential 405 00:50:05.360 --> 00:50:15.920 terms in my function and in the table as we said that capital t was equal to three so every occurrence 406 00:50:15.920 --> 00:50:22.560 of t and f of t are replaced by t minus three and i multiply it by a step function whose critical 407 00:50:22.560 --> 00:50:33.680 value is three and that will give me my answer so finally i'd say that the inverse Laplace transform 408 00:50:33.680 --> 00:50:42.240 of this term here which i used partial fractions on will be using f of t here replacing t by t minus 409 00:50:42.320 --> 00:50:51.440 three it'd be one minus e to the t minus three times the step function u t minus three step 410 00:50:51.440 --> 00:51:00.160 function of critical value three so that is my solution and of course if i invert x bar of s 411 00:51:00.160 --> 00:51:06.400 x bar is our function of s i get x of t and there's the solution to the differential equation 412 00:51:06.400 --> 00:51:15.360 at the foot of the slide there it is there so that is my if you like my shifted and truncated 413 00:51:15.360 --> 00:51:23.040 function my delayed function here x of t okay and that was the solution of this differential 414 00:51:23.040 --> 00:51:27.200 equation there we go that was the differential equation i was given with the initial condition 415 00:51:27.840 --> 00:51:34.880 we um took a plus transform both sides differential equation we've got our algebraic solution here 416 00:51:34.880 --> 00:51:42.000 and we saw that because of some exponential multiplying some function of s uh in the s domain 417 00:51:42.000 --> 00:51:48.400 we've we know that we've got to use the second shift theorem and we did that we needed to use 418 00:51:48.400 --> 00:51:56.160 partial fractions along the way we did that and we found our time domain solution okay so um 419 00:51:56.560 --> 00:52:05.600 let's have a look at another example now um this time we're going to do a second order differential 420 00:52:05.600 --> 00:52:12.960 equation again with a step function on the right hand side so this here of course should be a four 421 00:52:14.960 --> 00:52:21.680 so yeah so here we are so x double dot plus nine x is equal to u t minus four a step function of 422 00:52:21.680 --> 00:52:27.520 critical value four there you go that's it's just um it's just been delayed until t equals four 423 00:52:27.520 --> 00:52:33.360 it switches on at t equals four takes the value one and stays at one forever up until t equals four 424 00:52:33.360 --> 00:52:39.120 it's got the value of zero that's just a standard step function so that's our differential equation 425 00:52:39.120 --> 00:52:45.280 we've got zero initial conditions second order equation so we need two conditions one tells us 426 00:52:45.280 --> 00:52:51.600 that x zero is zero and the other one tells us that x dot of zero is equal to zero so we have to 427 00:52:51.600 --> 00:52:58.640 take Laplace transforms of our differential equation well we can do that and we've seen this before so 428 00:52:58.640 --> 00:53:05.280 this um I think it's maybe number 25 in the table to get the x double dot Laplace transform there you 429 00:53:05.280 --> 00:53:12.720 are plus nine times the Laplace transform of x gives me nine x bar and we want the Laplace transform 430 00:53:12.720 --> 00:53:19.280 of a step function with critical value four it's number 26 with capital t equal to four so it 431 00:53:19.280 --> 00:53:26.720 becomes one over s e to the minus four s so that's it so that's the Laplace transform of our 432 00:53:26.720 --> 00:53:32.800 differential equation the next step is to apply the initial conditions well there's zero initial 433 00:53:32.800 --> 00:53:39.680 conditions so it goes in here and in the next term so that's what I'll do on the next slide so 434 00:53:39.680 --> 00:53:44.800 that's I'm just carrying this over from the previous slide my initial conditions are both zero so in 435 00:53:44.800 --> 00:53:55.440 the go and if I simplified I obtain this expression here I then of course want to find the solution 436 00:53:55.440 --> 00:54:02.960 to the algebra problem and that's fairly easy to do x bar appears in both terms on the left hand 437 00:54:02.960 --> 00:54:10.240 side so I take it out a common factor and I can s squared plus nine times x bar is equal to one over s 438 00:54:10.240 --> 00:54:18.800 e to the minus four s divide both sides by s squared plus nine and I've got my expression for x bar 439 00:54:18.800 --> 00:54:25.200 when I look at it I notice that I'm multiplied by an exponential in the s domain so that would 440 00:54:25.200 --> 00:54:31.440 suggest that I use the second shifting theorem I can easily identify my function f of s it's 441 00:54:31.440 --> 00:54:38.400 whatever multiplying the exponential there it is and I can identify my shift value which is 442 00:54:38.400 --> 00:54:46.560 capital T that appears in the exponential so clearly T is equal to four okay so that's what 443 00:54:46.560 --> 00:54:54.480 we'll do so I've been given x bar is equal to this or rather I calculated that x bar was equal to this 444 00:54:54.480 --> 00:55:02.160 I've already just stated f of s is this function here the shift value T is appearing in the exponential 445 00:55:02.160 --> 00:55:10.800 it's equal to four so I now need if that is my value here I can match it in the table 446 00:55:11.760 --> 00:55:18.960 f of s is equal to one over s into s squared plus nine I would look in my table and see 447 00:55:19.600 --> 00:55:25.760 looking at focusing on the denominator s into s squared plus a number I would eventually hit 448 00:55:25.760 --> 00:55:32.160 number 10 and we've seen number 10 before it's s into s squared plus omega squared well that's pretty 449 00:55:32.160 --> 00:55:40.240 easy if omega squared is nine omega s equal to three so it's not so I'm using number 10 with omega 450 00:55:40.240 --> 00:55:46.240 equal to three but I've got to be careful because if I look at number 10 just a little bit closer 451 00:55:46.240 --> 00:55:53.280 I've got s into s squared plus omega squared but the top line also has an omega squared so the top 452 00:55:53.280 --> 00:55:59.920 value must match this bottom value the omega squared so what do I go well I've got a nine here 453 00:55:59.920 --> 00:56:06.400 so I need a nine up top and we know how to do that so we basically multiply by a fraction 454 00:56:06.960 --> 00:56:13.360 whose top line has the value that we want which is a one and whose bottom line the 455 00:56:13.360 --> 00:56:19.920 denominator includes the value that the table forces to introduce here and that was nine so I'm 456 00:56:19.920 --> 00:56:26.560 going to multiply by one over nine it's then a fairly straightforward process just been just a little 457 00:56:26.560 --> 00:56:34.080 bit careful here omega squared we've got here is equal to nine but the initial plus transform it's 458 00:56:34.080 --> 00:56:40.720 omega that appears here not omega squared so if omega squared is nine then omega is equal to three 459 00:56:40.720 --> 00:56:47.040 so my function the initial plus transform of that function there would be one minus 460 00:56:47.040 --> 00:56:53.920 cos three t but don't forget you've got to multiply it by one over nine so I've got one over nine times 461 00:56:53.920 --> 00:57:02.800 one minus cos three t my final step looking up at my diagram that's my f of t now I just found 462 00:57:02.880 --> 00:57:12.800 my final step involves me replacing every occurrence of t by t minus the shift value capital t which I 463 00:57:12.800 --> 00:57:20.160 can easily get from the exponential term here so we found that capital t was equal to four so I've 464 00:57:20.160 --> 00:57:28.080 got to replace every occurrence of t in f of t by t minus four and I've got to multiply by a step 465 00:57:28.080 --> 00:57:34.320 function whose critical value is four and that's exactly what I've done down below here so I'll have 466 00:57:34.320 --> 00:57:43.680 one over nine times in brackets one minus cos three t gets replaced by t minus four so I've got to be 467 00:57:43.680 --> 00:57:50.320 t minus four in brackets multiplying the three and all of that all of that term I've got to multiply 468 00:57:50.880 --> 00:57:58.000 by a step function whose critical value is four and that's exactly what I've got in the answer here 469 00:57:58.080 --> 00:58:05.600 and of course I should invert the left hand side as well because x bar is an s domain function 470 00:58:07.280 --> 00:58:16.560 I get x of t which will be my time domain solution to the differential equation okay so 471 00:58:17.680 --> 00:58:26.080 that's an example of using the second shifting theorem to solve first and second order differential 472 00:58:26.080 --> 00:58:31.520 equations where the right hand side term that's a forciant term in the differential equation 473 00:58:32.480 --> 00:58:37.360 is a step function and that of course as we've said can be used to model 474 00:58:38.400 --> 00:58:45.200 voltages being applied or in mechanical context could be a force that's been applied 475 00:58:45.200 --> 00:58:50.240 we're now going to move on and look at another type of function that when it appears in differential 476 00:58:50.240 --> 00:58:57.840 equations will often involve the use of the second shifted theorem that function is the 477 00:58:57.840 --> 00:59:03.840 Dirac delta function in engineering terms it's often referred to as the impulse function so 478 00:59:03.840 --> 00:59:14.480 suppose let's just consider this graph of a function we'll call d of t below here so basically 479 00:59:15.200 --> 00:59:23.600 it's a square pulse or a heart function whichever one you want to call it and this is the Greek 480 00:59:23.600 --> 00:59:33.520 letter epsilon the Greek letter epsilon I'll just try to do it here is usually used to represent 481 00:59:33.520 --> 00:59:42.880 a very small amount so if you've got t minus you've got our pulse a rectangular pulse centered on t 482 00:59:42.880 --> 00:59:51.200 here and you've got to the left you could t minus epsilon and to the right you could t plus epsilon 483 00:59:51.200 --> 00:59:58.960 so basically it's t subtract a very very small amount and t add on a very very small amount so 484 00:59:58.960 --> 01:00:09.520 this is actually a very narrow function really so I say here the graph may represent a force 485 01:00:09.520 --> 01:00:15.760 that's applied and then removed or a voltage that's switched off and then on just very quickly only 486 01:00:15.760 --> 01:00:23.280 applied for a very short time so I'm not really going to go into great detail about the Dirac 487 01:00:23.280 --> 01:00:30.560 delta function I'll just give you a very brief outline of it really before we apply it in a 488 01:00:30.560 --> 01:00:39.200 differential equation so let's just suppose that the total area all under this graph here this 489 01:00:39.200 --> 01:00:47.040 this rectangular pulse here is equal to one square unit well suppose we make this into the 490 01:00:47.040 --> 01:00:55.120 even smaller even narrower but we require that the area must remain the same at one square unit 491 01:00:55.120 --> 01:01:03.280 then obviously if we're narrowing the base then the height must increase and some books will refer 492 01:01:03.280 --> 01:01:10.320 to this as a blip this is just a blip here and as we get this narrower and narrower and narrower 493 01:01:10.320 --> 01:01:20.080 we can think of it becoming just infantescently small the width of this pulse here so it'll become 494 01:01:20.800 --> 01:01:28.800 it'll be incredibly tall if we wanted the area to remain as one square unit and this infinite 495 01:01:29.440 --> 01:01:37.600 at some value t equals t is called the Dirac delta function or impulse function and denoted 496 01:01:37.600 --> 01:01:47.760 like so delta is a Greek letter d delta of t minus capital t and capital t is where this 497 01:01:48.160 --> 01:01:58.240 delta function occurs okay so that's it that's it there so it's why this this function 498 01:01:59.840 --> 01:02:05.280 Dirac delta function um in engineering terms it's often referred to as the impulse function 499 01:02:05.280 --> 01:02:13.440 it's widely used um in mechanical and electrical systems and it's used to represent a situation 500 01:02:13.440 --> 01:02:22.720 when a huge force is suddenly applied over a very short time period so in the electrical context you 501 01:02:22.720 --> 01:02:29.920 might have a short circuit in an electrical circuit you might have a lightning strike or mechanical 502 01:02:29.920 --> 01:02:36.400 terms you might have a hammer striking an object anywhere where you get a huge large force being 503 01:02:36.400 --> 01:02:42.560 applied over a very short time interval the impulse function is a good way to model it 504 01:02:42.560 --> 01:02:50.480 mathematically also um in let's say a civil engineering context perhaps if you've got a 505 01:02:50.480 --> 01:02:58.720 body of water and suddenly a large quantity of pollutant is emptied into that body of water 506 01:02:59.520 --> 01:03:05.680 that's the equivalent of a large force being applied over a very short time interval and 507 01:03:05.760 --> 01:03:14.160 civil engineers would use the impulse function to model that situation so we can solve differential 508 01:03:14.160 --> 01:03:23.440 equations that involve the delta function um using the second shifting theorem so here for example 509 01:03:23.440 --> 01:03:30.960 is a first-order differential equation x dot minus 6x is equal to three times delta t minus eight so 510 01:03:30.960 --> 01:03:38.400 this is our delta function our infinitely tall function whose area underneath this function is 511 01:03:38.400 --> 01:03:44.640 equal to one but it's centered at t is equal to eight so that diagram there here the critical value 512 01:03:44.640 --> 01:03:51.840 if you want to call it that is equal to eight so we've got a zero initial condition so this is 513 01:03:52.480 --> 01:03:57.920 not far off it doesn't look totally unlike the differential equations we're solving for 514 01:03:58.000 --> 01:04:06.160 unit step functions and the process is fairly similar really so so what do we go we've got 515 01:04:06.160 --> 01:04:11.600 to take Laplace transform so both sides of the differential equation and I do that Laplace transform 516 01:04:11.600 --> 01:04:17.280 of x dot minus six times the Laplace transform of x and that's going to equal three times the Laplace 517 01:04:17.280 --> 01:04:23.440 transform of the delta function here so that's what I've got here so x dot well we know that that's 518 01:04:23.440 --> 01:04:33.120 number 24 the tables sx bar minus x is zero minus six times x bar the Laplace of x and now we've got 519 01:04:33.120 --> 01:04:39.600 to do three times the Laplace transform of this delta function now that's very easy because as you 520 01:04:39.600 --> 01:04:50.320 can see the Laplace transform of her delta function centered on big t it's just e to the minus st so 521 01:04:50.800 --> 01:04:57.840 it's centered on eight the Laplace transform is just three e to the minus eight s okay so I'll just 522 01:04:57.840 --> 01:05:05.920 say something before we move on if your delta function is centered at the origin so the capital 523 01:05:05.920 --> 01:05:14.080 t is equal to zero then it's Laplace transform is e to the zero which is just one and you can see that 524 01:05:14.080 --> 01:05:24.000 in the line above now that would you know it would Laplace transforms and the second shift in 525 01:05:24.000 --> 01:05:31.040 theorem make it so much easier to solve equations like this than you would then you would if you 526 01:05:31.040 --> 01:05:36.560 are solving it in the time domain basically for the methods that we know for the methods we see in 527 01:05:36.560 --> 01:05:43.440 this course in fact in in second year university mathematics in general in engineering mathematics 528 01:05:43.440 --> 01:05:49.760 the time domain methods are solved in differential equations would not be able to cope with this type 529 01:05:49.760 --> 01:05:56.640 of equation simply because of the function on the right hand side but with Laplace transforms 530 01:05:56.640 --> 01:06:04.800 we can solve it so we've taken Laplace transforms of our differential equation we'll now apply the 531 01:06:04.800 --> 01:06:10.720 initial condition which is zero of course we'll simplify this a little there we are and now we've 532 01:06:10.720 --> 01:06:18.640 got to solve the algebra problem we've got to get an expression for x bar so I look at the left 533 01:06:18.640 --> 01:06:25.280 hand side x bar appears in both terms so I can take it out as a common factor so I get s minus 534 01:06:25.280 --> 01:06:33.920 six times x bar is equal to three e minus eight s I can solve for x bar and it simply becomes x bar 535 01:06:33.920 --> 01:06:40.880 is equal to three over s minus six e to the minus eight s and again I look at this function I know 536 01:06:40.880 --> 01:06:48.400 I've got to get the initial Laplace transform of this both sides of this equation well x bar is a 537 01:06:48.400 --> 01:06:55.360 function of s it's that's easy to invert it just becomes x of t but this here just as in the examples 538 01:06:56.320 --> 01:07:02.080 we finished doing with the unit step functions on the right hand side of the differential equation 539 01:07:02.080 --> 01:07:10.320 I notice that I've got an exponential function here and that tells me that I will need the 540 01:07:10.320 --> 01:07:16.080 second shifted theorem to calculate the inverse Laplace transform so I've got my diagram here 541 01:07:16.080 --> 01:07:22.240 for the second shifting theorem I'm going round anti-clockwise I've got f of s which I can easily 542 01:07:22.240 --> 01:07:29.040 identify as three over s minus six times my exponential function and the way the shift 543 01:07:29.040 --> 01:07:35.840 value capital t is equal to eight can easily just by comparison I can easily identify that 544 01:07:35.840 --> 01:07:43.680 so f of s is equal to three s minus six we're quite familiar with this sort of expression we know 545 01:07:43.680 --> 01:07:50.240 it's number five in the tables with alpha equal to minus six so the inverse Laplace of that is e 546 01:07:50.320 --> 01:07:58.720 to the minus minus six t or e to the six t and don't forget to multiply by three so f of t that's 547 01:07:58.720 --> 01:08:07.440 the inverse Laplace of big f of s f of t is three e to the six t my final step is now to replace 548 01:08:07.440 --> 01:08:17.760 every occurrence of t and f of t by t minus eight because eight is the shift value we saw that earlier 549 01:08:17.840 --> 01:08:25.680 when we just compared these expressions so I replace every occurrence of t and f of t by t 550 01:08:25.680 --> 01:08:32.400 minus eight and I multiply by the step function that's what the table tells us to do whose critical 551 01:08:32.400 --> 01:08:38.880 value is capital t in other words the step function with a critical value eight and when I do that I 552 01:08:38.880 --> 01:08:47.680 obtain the expression down below I invert x bar of s to get x of t and I've got three e to the 553 01:08:47.840 --> 01:08:54.960 six replace t by t minus eight there we are so I've got three e to the six brackets t minus eight 554 01:08:54.960 --> 01:09:02.080 close brackets multiplied by a unit step function whose critical value is eight that's ut minus eight 555 01:09:02.080 --> 01:09:11.520 so that is my solution to the differential equation um there we are so that's that's it 556 01:09:11.600 --> 01:09:16.080 given there that's it that's it given there we can ignore this I'll have to remove this 557 01:09:17.600 --> 01:09:22.880 on this slide I've listed some properties of the the delta function we know that 558 01:09:25.280 --> 01:09:31.520 you know we're outside if you're not away from where the function is sitting on let's see if we go 559 01:09:32.160 --> 01:09:38.000 back to here outside away from t either to the left or right the delta function is zero so it 560 01:09:38.000 --> 01:09:47.520 takes now value here at capital t the area under the curve is equal to one and there's some other 561 01:09:47.520 --> 01:09:54.240 properties here this fourth property says that the delta function is actually the derivative 562 01:09:54.240 --> 01:10:01.520 of the unit step function okay so in other words the if you look at it the other way the unit step 563 01:10:01.520 --> 01:10:09.120 function is the integral of the delta function okay so finally then just to summarize what have 564 01:10:09.120 --> 01:10:14.960 we done today so in this lecture we've looked at the second shifting theorem to calculate 565 01:10:15.680 --> 01:10:25.120 inverse Laplace transforms and that was typically where we saw that our function in the estimate 566 01:10:25.120 --> 01:10:33.200 involved multiplication by an exponential you had a term that was of form f of s times e to the minus s 567 01:10:33.760 --> 01:10:37.520 t where t is capital t representing a shift 568 01:10:39.920 --> 01:10:45.840 so we used the second shifting theorem then to solve a couple of differential equations the first 569 01:10:45.840 --> 01:10:51.040 order one and the second order one with a forcing function at the right hand side function involved 570 01:10:51.280 --> 01:10:59.360 step functions we moved on then to introduce the Dirac delta function or also known as the 571 01:10:59.360 --> 01:11:08.080 impulse function which as I said is used to model large forces being applied over very short time 572 01:11:08.080 --> 01:11:15.840 intervals and we solved a differential equation involving the impulse function and that of course 573 01:11:15.840 --> 01:11:26.080 needed us to use the second shifting theorem so on this slide I've listed some questions that you 574 01:11:26.080 --> 01:11:31.520 should be able to attempt basically you should now be able to attempt all the Mobius questions as 575 01:11:31.520 --> 01:11:36.960 questions 1 to 16 I haven't included any more questions from the notes because I think there are 576 01:11:36.960 --> 01:11:44.560 really plenty Mobius questions to keep you busy really plenty practice with with all the Mobius 577 01:11:44.560 --> 01:11:51.920 questions that are available so that's the end of Laplace transforms in the next section we're 578 01:11:51.920 --> 01:12:02.080 going to look at solving linear systems of ordinary differential equations that's what we've got more 579 01:12:02.080 --> 01:12:10.960 than one differential equation and these arise in mechanical and electrical systems and many other 580 01:12:10.960 --> 01:12:17.520 systems as well so for the time being then we'll just stop the recording with that