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0:00
okay let's do some math for fun today I have the Laplace transforms for you guys we have a total of 24 of them and of course here's the file you guys can go ahead download that please see the description for the file and the notes and also the timestamps for your convenience and here is the definition of the Laplace transform as you can see is just an improper integral and you have e to
0:21
the negative T times the function right here and hopefully each converge and to guarantee that this work here converges you just have to make sure that F of he has exponential water what that means is that F of T cannot be too big it is you can't just think about it if you have e to the positive T squared in that case that will make this improper integral diverge right so I just have to
0:46
make sure that happens and if you ever wonder why that last transport can help to solve differential equations go ahead and check our mu prime masters video I have that lot I will have that link in the description for you as well man that was so good because I actually have never seen that before when I was taking Laplace transforms backing that the days I mean taking the differential equation
1:08
cast back in the days all right so now here we go this right here it's the first one of course we'll be doing a lot of proofs so here we go now hands transform of e to the 8th Yi well that's the integral from 0 to infinity and we'll just use this formula so we have e to the negative as T times e to the 8th he and then he so you just apply that a phoenician and in the end we're just
1:34
going to be working a lot of integrals pretty much all right so now this right here and here they are just Yi so it's good we can just integrate this from 0 to infinity and we can just come by this as a - yes yeah so I'll just read yes a minus s and then what fact how that he right here and then I will have the TV at you so now to tend out this improper integral you should technically take
2:03
the limit and then you can say B goes to infinity and then the interval corresponds hero to infinity all that stuff but in this video I'm just going to skip that just write down the infinity symbol you just remember we will be taking limits okay here we go to do this first of course just note your integration this is just e to a number times T to the first power so the integral of this is just
2:28
e to the a minus s T but remember the duet the reverse chandu right the derivative of this is just that when we are doing the integral be sure divided by that so we will have to do 1 over a minus s so that's the integral part and then we will have to plug in plug in from 0 to infinity under stuff now here is the deal when we plug in infinity to the T you will see that we'll end up with something like this 1 over a minus s and then e to the a minus s and here you have past infinity
3:07
right and then minus and again I know this is a penetration but as I said it's just a short notation it's still ok just remember what he can limit and then you put the wrong here so we have 1 over a minus s and then we have e tired to the a minus s 2 times 0 like this right so that's the first part this is the second part and now here is the deal in water for this right here to converge we have to make sure each to this power and she end up to be 0 because we have infinity times
3:43
this already the idea is that we need to make sure this part has to be negative and 1 for this to be negative of course that means I will have to have a minus s to be less than 0 and of course we can solve this inequality move the s to the other side meaning that s has to be bigger than a we need to have this condition here in order to make this negative so that you can see we will have e to the
4:10
negative infinity and then that will be 0 times all this that will be 0 so with this condition we will see that the first part is just nice T equal to 0 and then the second part were just - and you see this is just over a minus s and that e to the zero power right here is just one so we just have this right here and finally what we can do is you can also just distribute the negative here
4:39
and here meaning just switch the water so get 1 over s minus a like this so this right here is it and again until you finish all that you should technically include this condition so you will have to say s should be greater than a right so that's it that's it that's it very good huh that's the first one not so bad alright so now continue and we'll be using this for the next few so look
5:08
at this look at this ok so again how are you guys doing let me know leave a comment down below let me know I'm doing ok and doing that for math lady how about you guys and hopefully everybody is doing well of course stay safe stay healthy all right number two right here and of course I will be doing this in water in the sense that one is based on the other blah blah blah right so here is the second one who are between the power function instead so we have this our goal
5:42
is to get a Laplace transform of T to the N power and here I'm going to assume that and it's just a non-negative whole number right so I want to know what this is I would like to have 0 1 2 so on and so on in fact later on we can actually extend this to like some like weird numbers such as n is equal to 1/2 but that would be for question number 23 so nah okay this is how you can do it first you can
6:17
use the definition so you can put a T to the nth power right here instead right and you can just work that out and you can use integration by parts and times actually technically you have to do that a lot of times and you how to do a lot of math or you can do the following check this out this is how we are going to do it I will use this right here to help us so have a look right here we know
6:44
the Laplace of e to the a T is equal to that and if you have e to the HT in fact we know the series expansion for that isn't it so I will read this down right here for you guys so this is how we are going to do it so notice that we have this already a Laplace transform e to the 8th he which we know this is equal to 1 over s minus a like this and in this particular case I'm going to choose a to be
7:16
greater than 0 I can choose Y for either 1 I can say love has off into dy over positive a times T right so of course this is still a yet however you still have to include this condition so you just have to make sure that both s then a are greater than 0 so that's pretty much why I hat and again I mentioned about the series we know how to handle the series right here it's already style Rico let
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me just put on Rico and I will just put this down here for you guys e to the X power this is just a power series expansion for e to the X this is the series as n goes from 0 to infinity and we have X to the nth power over n factorial and if you would like to recall and review your series you can check out my powers use marathon for you guys right so that's what we have and this right
8:14
here is true for all X so let me also just write that down right here so what we are going to do is you see here we have e to the 80s so let's go ahead and just focus on this part first stuff when I have e to the 8000 I have to do is just put the 80 into this egg's right here and this is the jet because it's true for all the inputs so I can just open a parenthesis with a t inside raised to
8:42
the nth power over factorial like that so that's pretty nice and of course you can just work out this right here a little bit and you see this year's the series as n goes from 0 to infinity and this is how I'm gonna do it I will put down t3 and power first like this over n factorial and then we have a to the nth power like that so this is what we have all right okay then another thing I want to do this that's to get this part and I will just do a little divider here well we have
9:18
one over topic so recall that 1 over 1 minus X yes this is our best spread this right here is equal to the sum as n goes from 0 to infinity just X to the nth power that it's however this is only true if the absolute value of x is less than 1 right so be really careful on this right here however it's actually not so bad oh now let's look at a series expansion for this so we have 1 over s
9:51
minus a like this yeah well in order for us to use this we actually have to have a 1 right here but this right here is an S that's not good that's ok because we can factor out the s so it looks that we have 1 over s times 1 over R factorial yes so we have one right here and then factor out the S so we have minus a over s like this and of course you can put on parenthesis like that now here is a deal because we do have this condition have a look as it's greater than a and they are
10:26
both positive and right here we have a over s so of course in this case here notice that the absolute value of a all 4s the top is smaller than the bottom right this right here has to be less than 1 for sure so we'll test said we know we can use our best friend right here in the front oh we still have 1 over s and then right here we can put this into here so we have the series as n goes from 0 to infinity and we just need to write this down here to open the parentheses with a over
11:03
s and then raised to the and power like that alright so that's pretty much it and now I'm just going to erase this and then we will see how we can make the connection between one and the other okay oh of course maybe that's simplify this a little bit early so this right here I will have the summation n goes from 0 to infinity on the top we have a two ends power but let me put that aside like this a to s power on the bottom we have s to the nth power but we have another one on
11:38
the bottom as well so on o we have 1 over s to the n plus 1 like that ok now have a look right here this is true that we know so here is the deal let's go ahead instead of looking at the a that means eat the e to the a T and also this have a look e to the HT is nicely equal to this isn't it yes yes so I can just actually put that down right there so we have the Laplace transform and actually let me put this down blue we have the sum from n goes from 0 to infinity of all that
12:24
so we have P to the N over factorial a to the nth power like this and then on the right hand side we have 1 over s minus a and again in terms of a series it's equal to this isn't it yes yes so we can just write that down the series as n goes from 0 to infinity and we have 1 over s to the n plus 1 times a to the N power now here is the thing you had to just believe me in this case where the plus is actually linear meaning that the Laplace of a sum is the sum of the other plus and
12:59
even though inside it's an infinite sum but let me just tell you it will still work in this case all right because this right here converges so what we can do right here is that we can switch the water of the laplace and also the summation and if you want to see a teacher can just change that to improper integral and what the appa okay let's just do it right we have the sum as n goes from 0 to e that he let's put that down first and then we have the Laplace transport of our we won
13:27
inside now here is the deal have a look right here inside here what I want to do is you see we have heat and power I'm just going to have that inside of my Laplace transform so we have a transform of T to the and power like this right and then on the outside we have the one over N factorial let me just put that down right here and then at the end we still have the a to the nth power like this so have a look this is a summation of all that again technically every single being said
14:04
the Laplace but I took a 1 over n factorial and also to put that a to the nth power on outside and the reason I can do that because in a la classe world only T matters T is the variable and the N is just like a constant so just take that out you can just read out so that's that here all right and then on the right hand side we have this is equal to the sum as n goes from 0 to infinity of 1 over s 2 n plus 1 MJ we have a to the N power and here is the punchline maybe you
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can see already which is very nice have a look left hand side on the right hand side the above series and you can just look at this part this is a 2n power likewise this is also a power so it's like infinite polynomial right warp infinite power series namely though you see that this and that they have to be the coefficients and water for the polynomials to equal what the infinite series to be equal to each other the coefficients has to equal to each other so we can conclude that
15:09
this and that must be the same so we started you see here we actually just have 1 over n factorial the Laplace transform of T to the nth power this has to be the same as 1 over s to the n plus 1 so as you can see in the end all we have to do is just multiply the N factorial ample sites so you can just see that if you would like this is so nice they can so nicely and then in the end you get a Laplace transform of T to ins power that's equal to factorial over s to the n plus 1 power
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so here we have it this is it and technically right here as I said and is going from 0 to 0 1 2 3 and so on to infinity and that's exactly how you stress all that good stuff so again this is factorial over s very nice right so that's it alright so you can see that this is actually really useful and then we'll be using that for the coming up ones as well very cool the next one is sine and cosine yes actually question student question poor will be doing two things one so this
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is like turning two parts it's with one stone but we actually happen to kill parts will cure Laplace transforms all right so here we go this is question number three I just want to make sure that I'm still recording so right here question three we want to have the Laplace transform of sine D and B is just our number of course so we need to know what this is and of course when we have sine that's also talked about the cosine as well so I'll put down number four right here
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number four Laplace transform of cosine of BT and if I can do this a few ways first way is you can use integration by parts and again will be similar things to the things that you have to do with T to the nth power if you use integration by parts and you have to plug in the limit and all that stuff in this video what a [ __ ] I'm just making my video here I will show you guys how you can use
17:34
complex numbers to take your both of them with just one computations and not a thing you can do is you can figure out sine of T and cosine T and later use one of the properties I'll show you and you can generalize it sound in Campeche by anyway here is how we are going to do it again I'm going to write this down here for you guys we are going to use the fact hat with it earlier the Laplace
18:03
transform of e to the a T this is being equal to 1 over what yes s minus a yeah all right so this is what we are going to do we are going to leave dangers these sometimes meaning that we will have to go to a complex world sometimes so have a look with the a right here right what we are going to do is I'm going to let a equal to well we have to have the complex number so we have to introduce the eye and then Oh buddy I'm using the P so I will just have I be like this so by using
18:44
this you'll see that the next line will have the following so I will just put this down mmm black yeah so this is what we have and now we will just use their so here we have the Laplace transform excuse my little brace it's really hard for me to do that by anyway easy and then the a is now I P and then we still have the T and then this is equal to 1 over s minus I B like this right so we can just of course do that and now another thing that you have to remember is the order spoon
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lassoed let me also write it down right here for you guys so this is the Euler's formula which says when you have e to the I theta this right here it's nice to equal to cosine theta plus I sine theta it's amazing formula right and how we are going to use it is will the left hand side presents the cosine and sine already yeah and then the right hand side is a rational expression you just have to fix that little bit because remember I don't my to be on the bottom I like to be on top
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in that we can just worked out and that's pretty much the idea so here we go this is how we are going to do it the left side have a look we have the theta being the PG right here right so the left hand size can easily equal to the Laplace transform of sine well actually cosine first of PT so we put that down right here and then we add I times sine of BT so this is what we have on the left hand side and now on the right hand side to fix that situation when you don't like to
20:38
be on the bottom cool I had multiply that top and bottom by its conjugate so we KS plus IB and then s plus I like this and like that all right on a pattern of do this in your head when you multiply this I'll just get s times s which is a square and there - and I times I is I squared so it becomes negative 1 so it becomes a plus and then we have e squared so this is what we have on the bottom and
21:08
on the top you see that we just have s plus I B just like that isn't it now here is the cool part again Laplace is linear meaning that the Laplace of a sum is the sum of the Laplace and when you have a constant multiple you can also put that in the front of the Laplace so have a look I'm going to break this down as the Laplace transform of cosine of BT now close that and that's added
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words I will put the eye to the front and then we have the Laplace of sine of BT like this right so I just put you know I just work at them apart like this and that and again just really excuse my little brace right all right and I all right so we can do the same thing we have the real part and also let me imaginary part the real parts just this which is the following s over s square plus B Square and then plus I times B over s square plus B Square now we'll play the game the game
22:17
is called me to match very fun because this and that yeah both the real part so of course this has to be equal to that isn't it and then this is same a generic part though is that so this has to be the same as that and with that we are done because you see that the Laplace of Stein PT is nicely equal to this we have the PIA mater over a square plus B Square again we are done what's that and then for this will have the SS on the table as over a square plus B squared like this
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there's no v in my video right so just s this is now five anyway that's it very cool huh so you don't have to do the integration by part X so many times so this is so wonderful it's awful look this is it alright now we did full already so this is going pretty well and I have some surprising intercourse for you guys as well so be sure you watch that as well I say that will wear will now what's the next one Oh remember the distinction cost which is like like the knockoff version of
23:35
North here we go number 5 number 6 well here let me just write that down number 5 we have the following the Laplace transform of thing T right since T which is the hyperbolic sine function well again in this case if you allow you can go and use the integration by parts which the definition of the Laplace transform but which end to the following here we go you have to remember the definition of the sinc function well this is equal to the following this is nice the equal to
24:17
e to the path the input right here which is T and a minus e to the negative T all over 2 like that and we know how to hand off the Laplace transform into this number T already yes we do so we can't you just break them apart and let me show you how first we have the one half let's go ahead and put that all the way in the front and then again we can just do the following now class of the first
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which is the Laplace of e to the T and there - the Laplace of the second and then we have the 1/2 all the way in your front already so this is what we have yeah yeah put oh I put on the beach heat I'm sorry I put on the beach things off bTW I'm sorry so PT is actually our variable of our input right so I will have to change this this this this to the PT that's all so easy fix here we have e to
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the PT and then here we have e to the negative PT and then this is e to the PG and this is e to the negative B okay so just easy fix now we can continue for this right here of course again the 1/2 it's just all the way in the front like this and then for this one of course we can just recall the formula which is just 1 over s goes first - beat here right here so you actually just put on
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be like this and then minus 1 over and we have s - this is negative PG so we will actually have going to give that class later on so this is 1/2 and now of course we can just combine the little fractions order stuff so right here we will just multiply the pot top and bottom by s plus B and then as plus B and I will just multiply this by s minus B and then multiply the bottom by s minus B so all
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that good stuff now we have 1/2 in the front and then on the bottom you just have s minus B times s plus B which is s square minus B like this and I be square sir and on the top though we have s minus we have s plus B so let me just put on s plus B minus s and then minus minus becomes plus so we have a plus B like this and of course just work that out you can see that the SNS can so and
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on top B and BS just go ahead and you have to be and then in the end of course you can cancel the 2 and the 2 right here so this is very similar to the original sine function yeah you actually end up with B on the top as well over the bottom is s square minus B Square like this so that's it however do we need to put on any conditions well Apodaca from here to here remember we had to
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put on s is greater than what s has to be greater than B so I would actually have to say S has to be greater than B the me at you just write that down right here has has to be greater than just a hole we did earlier and now this right here you will just have to make sure that S has to be greater than this improve which is negative P and depends on what B is so you can just kind of include that
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and all that stuff but I'm just going to do that to you guys because I don't like B is negative or pasty and all that but yeah you can just say s is greater than B that's actually pretty good okay all righty anyway this right here yes P over s square minus B Square like this and number six very similar that last transport of cosh but the input here is P again actually this is cosh SM
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input now remember the definition of cosh is the following gosh is nicely equal to e to the improve of speed he and then plus e to the negative input which is negative PT o / - like this then again we can just do what we do over there so this is the same as one Laplace transform of e to the PT and then we have to add the Laplace transform of e to the negative B T like this thing like that okay let's just go ahead and do this this is one-half times this is 1 over 1 so
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1 over s minus the P right here and in this case we are adding 1 over s negative and then this is negative P just like that and again let me just show you guys this real quick for the first one we will just get multiplied by s plus B as plus B and then for this one we multiplied by s minus B and then s minus B and that's the all the way in the front we still have the 1/2 on the bottom
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again it's going to be a square minus B Square on the top is s plus B and then this is plus and this is s and then this is minus B so it's like that in this case the s I mean the PMB can so SNS is twist so the 2 and then to cancel so in the end we end up with an S on the top so this is the S portion and again very similar to the original cost original cosine of PG so this right here is
30:27
s on top but you have s squared minus B squared on the bottom like this right and again you can tell me what the condition is just going to write that down that s should be equated and yeah because it does already matting yeah because you square that anyway so I'll just have to make sure a square has to be greater than B squared then yeah just alright so that's pretty much it and continue doing pretty well for this one better than a differential equation so we have the next
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one which is unit step function by this okay okay so let me erase all this so again as I said who will be using definitions whenever we have to use whenever I do two proofs sometimes if we can see the connection between the things would have done already you don't have to use the definition all right number seven I'll put this down here for you guys seven that last transform and here we
31:45
have the unit step function so I'll just put this down like that like this is the best I can write t minus some number a like this yeah all right and right here I have to remind you guys for the unit step function unit step function is this right here it's actually a piecewise function so I will actually 20 for you guys first the idea of the unit step function is that if you have the following look at what the a is first so let's say I'm serious my a and anything before a when
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you are doing a when you when you are doing t minus a anything before a is zero so you have this and then you put an open circle and then after a at a you have to jump to one right and then third open circle as well like this right so the idea of the unit step function is that if the input is passed did they will give one if the inputs naked heat you get 0 and if the input is 0 some book defines you to be 0 some books defined it to be not undefined so I will use
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the open circle in this case so I already start here for you guys you are t minus a is equal to either 0 again if the input so I will write down t minus a is less than 0 right it's equal to 1 if the input name eg minus a is greater than 0 and the reason I want to do this it's because I would like to tell you if you'd reverse the t minus a with a minus T which would reverse the picture but
33:40
that's happening later on so no so keep that in mind if the insides positive then you get one if the incise negative you get 0 if the inside she wrote some books define it to be nothing sample defined it to be 0 or 1 depends doesn't really matter because we are doing integrals so one point it's not going to change the whole change anything anyway how can we do this yes definition of Laplace transform so let's write it down this is the integral going from 0 to infinity e to the
34:16
negative as T times u of t minus a like so and we have the did he do this so well we'll have a look this is what's going to happen when you multiply a function with a unit step function so let me graph e to the S negatives people you guys which is just an exponential decay function like this assuming s is positive so this is e to the negative T right and originally this integral we have to go from
34:55
zero all the way to infinity yeah but now after we multiply by the unit step function so I will write this down if you have e to the negative T unit step function of t minus a what's going to happen is that we'll pick a where a is that says right there anything before a this is 0 0 times this that will make everything 0 right so actually you scared zero right here and you have an open circle at a there's a jump to one at one times this is just just that so you will
35:33
just jump to here and then you have this okay so instead of integrating the whole thing have a look you actually do not have any area from 0 to a at all all we have to integrate is just from a to infinity just this part and again at the end points right here it's just the point so if you're talking about integral doesn't matter all right so here we go we just have to integrate from A to infinity and again this right here changed the zero to a and from a to infinity this portion is
36:12
just the same as the original so I can just write down e to the negative as you like so and then we have like that and then again right you can just go ahead and integrate out of this so integrating this it's just e to the negative T and then we divide it by this right which is just going to be 1 over take TS then we plug in plug in going from a to infinity well well when you put infinity here we have 1 over negative e to the negative s and we have infinity right
36:47
here and that's the first part and out of - the second part is 1 over negative e to the negative and then you have a like right here all right if this is going to be 0 depends what I see is this is negative already this is positive infinity s has to be positive so that we can actually get a negative infinity out of this right so with that being said s has to be greater than 0 so with that condition though we can actually say the first part has to be 0 and again we need we
37:32
need to have that condition and it's in there right this time right yes that we need to have that in order to make this work and what are the rest negative negative becomes positive and then we just have that that's pretty much it so in the end we have the plus 1 over s e to the a so finally we were just ready Stan s e to the negative as a and some people write to ride yes e to the negative is it depends on how you want to ride it but this is okay and then / s yeah again
38:08
we have the condition so that's right there down s has to be greater than zero in order for that to work just like that right so this right here it's a unit step function and now let me introduce you guys something that's very similar to the unit step function they need a window function so it's the PI sub apt function I should be able to do that right here so I will do that right here
38:35
number eight so here we have the Laplace transform and inside the capital PI a comma P this is not the original type function it's just the I think Scott window function over oh by the way this is also called the Heaviside function but it's not because this sides happier that's why it's zero happy sighs just a hangover the person right I thought it was only funny anyway so we have pi AP and input is just usually with a sheet like this right so that's what we have and again I
39:18
will have to tell you guess what PI ap years well PI ap it's actually not so bad Oh PI a P P this right here again as I said it's very similar to that perhaps I'll show you guys a picture first well very similar you have to tell me two numbers a and B suppose P speaker than eight right so let's say we have a here and B here the idea of this right here is that anything outside of a and B is going to be zero and anything between of a and B is equal to one that's pretty much it
39:55
so you can just grab it right here open open like this anything else a zero anything between is one and again we do not worry about the end points right I'll just have that and then here is the property of the pie a B of T so you can write it down like this zero should write down one purse 1 if the input is in between of a and B and again to not worry about the endpoints 0 otherwise okay so hopefully because it's the problem zero if otherwise no that doesn't make sense
40:49
if otherwise alright so just I'm sorry about words otherwise meaning that yeah all right so that's pretty much it however there's a really nice connection between the unit step function and PI a B I will also tell you the PI ap is actually just nicely equal to the unit step function of t minus a which is this part and then you - are you - you t - be like this
41:29
yeah all right so just like that all right we're stopping said in order to do the Laplace of this we can actually just use this right here which is very nice so we actually just have to do the Laplace transform again Paille be is just going to be the unit step function of t minus a and then minus the unit step function of t minus b like this which is very nice because now we can just do a Laplace transform the first one - Laplace transform the second one not part of the
42:05
flash transform the first one is what it's kind of hard no no no it is open here so we actually just get e to the negative s a over s right that's pretty much it that's all we have to do and s has to be greater than zero just like that and then in between so - so just go and put our - and then we have this so all we have to do is change the a to P so we have e to the negative P over s in the end
42:35
of course that's right things done together e to the negative a minus e to the negative P or over s s has to be greater than zero and with that we are done Jessica okay so this is the window function I believe that's pending of that so it's a PI a business all right next one's going to be pretty crazy it's still kind of like a piecewise but it's not as no more looking as the one that we just did
43:09
so have a look all right have a look and let me know if you guys are taking differential equations if you are man I've missed that class I'm not teaching differential equation this semester nope I haven't taught you in three years but you know I just feel like doing some applause transport so why not all right so this is the number nine and this is the Laplace transform of the Dirac Delta
43:47
function so you just put down the Delta this is the Delta and then t minus a like this right so again I will have to explain to you guys what the Delta function is well to a picture tell me where a is anything besides a yes zero so you have this open circle so happen at a the function goes up to infinity all the way to infinity all the way to infinity you can even see it you can even see it so just yeah just cannot see it so it doesn't really make sense for me to graph that
44:33
you just have to grab it from the definition and I'm going to read it down alright so here is the deal I will just have to write it down for you guys the Delta t minus a again I would write on a name later but this is 0 if T is equal to if T is not equal to a and this is equal to infinity if t is equal to a so again the pictures everything is 0 except for a goes out to infinity right so there's no way for me to give you guys a legitimate graph of that and again this is Cody
45:12
what I should write down the name this is the red which I think that witty Delta Tom [ __ ] right
45:26
maybe without he I don't know what city I didn't ready down by anyway I'm sorry let me actually just not write down the name alright so this right here is what we have now this is how the Laplace transform is going to look this right here we are continues at the Venetian and see if Abbas so integral from 0 to infinity again we have e to the negative T and then we are going to multiply this with the outer of t minus a like that like this hmm how in the
46:12
world can we make the hospital where I have to tell you get a few more scenes about the Delta function right here right so one property that you have to know it's a following right if you integrate from negative infinity to positive infinity because the domain is all real numbers so you go from negative infinity to positive infinity it doesn't really matter action and if you integrate the out of t minus a right here let me just ask you guys this question
46:43
first what do you think what the answer is and again you can just really imagine that you have the graph by it's not ideal to have a graph is right here put the outer is everything zero but when you have a yes going to be infinity and remember when you up integral you're trying to find the area the area right here is zero the area right here is zero what's the area like let's say this right here is the infinity this is that infinity up here right we have
47:20
one point what's the area from here to here is there any width no the width is zero but what's the height the height infinity 1 0 times infinity technical is indeterminate right however by definition to solve the the best part what is right here is that this right here is defined to be 1 this is the property of reading click with the Delta function alright so you have to just kind of take my word for that if you were like hey so that's the idea that's the idea number
47:51
one right and in fact you also have idea number two right here if you have the following let me erase my picture and I'm arresting you here idea number two if you have the integral from negative infinity to positive infinity if you have a function let's say f of T times the outer of t minus a like this well well again let me just let's just think about as a picture yes a-anything by a is zero like this so this right here it's going to give you infinity when X is when T is
48:29
equal to a but this time we have infinity times F of T f of T at a is just F of a right so of course when we have F of T at a is equal to F of a that's clear right that's the value of the function at a of course but you have F of a times infinity so earlier infinity was right here isn't it and suppose you have F of a times the Infinity so of course it's like another motor hole you dam multiple just multiply by one that's all so whenever you have this all you have to do is
49:10
put a into the function so actually just get F or hey that's it again it's just like the Infinity times y but that number is that's it and then you integrate that so of course you know just take that number to be the answer so this is like the craziest part yeah well with this being said in fact this right here is super easy because what we had to do is notice this right here is the Delta function and again this right here even though it's going from 0 to infinity it's the same as
49:46
going from negative infinity to positive infinity because everything is 0 only when a yes only 20 today right so earlier heart negative infinity to infinity that's why I said doesn't matter doesn't matter all you have to do is the following pull a into the variable T right here and that's it so finally we'll just have ye to the neck is hey seriously that's it as crazy as it looks like and just really hard to take the definition as how D is because 0 times infinity from your calc
50:30
want how to classes you should know that it's indeterminate meaning you have to do more work in order to figure out the answer but we take that to be 1 in this case all right pretty cool
50:46
okay that's number nine okay now this is somewhat crazier so we'll be doing number 10 and this time I will be doing two of them for number ten so here is number ten now class transform of F of T minus a sorry wrong color Laplace transform of F of T minus a and you see usually is just F of T but when you have F of t minus a that's how you shift the function in turn left foot right depends on
51:25
what a is and then not only this though we also have a unit step function after that so this is the first one from number ten meanwhile you might be thinking that if we just have f of T right so I will also have to show you what's the Laplace transform of F of T only and multiplied by U of T minus eight like this so it's some like the unit step function business again alright let's figure
51:59
this out first to do so let's go ahead and use our definition so integral from 0 to infinity e to the negative T oh that which is f of t minus a times the unit step function of t minus a and then we have the TD of course again whenever you multiply this you just have to observe what's happening right here right so I'll draw the picture one more time for you guys just in case it you skipped what we talked about over you so this right here we are trying to find the area from 0
52:41
to infinity under this function let me graph this right here first and I would do that with flu so yeah e to the negative T times that let's say the function looks like this I don't know though e to the negative T F of t minus a right so let's say the function looks like that and again you have to tell me what a is that say it's right there so this is how the function looks like it's just this
53:11
now I will graph it to the negative as t f of t minus a times u of t minus a so here is going to be the change after we multiply by that and the only change is that well anything before a it's going to be zero so we have this right here like that and then you jump to 1 and 1 times this is of courses that so anything after a it's just the original blue curve like that so it's similar to what we did earlier and if you want to find area for o to infinity of this you can just find the
53:53
area from a to infinity of the padule function so that's pretty much the idea so here is going to be the integral from a to infinity so I will actually put this time blue why not a to infinity and then the insides just going to be the following e to the negative T f of t minus a that's it tgd that guy yeah now how can we handle this though we haven't done some the gentlemen integration
54:31
for while huh especially for this video let's do some fundamentals namely u substitution so let's go ahead and do you substitute but we use the you already write here so let me use a V right so I'll say V is equal to the input which is t minus a and until V is equal to DT so that's pretty good now take this integral to the world well keep in mind this right here is T goes from a to infinity now take this integral to the feel world he will go to where studying where you put a
55:15
right here a - a with your zero and then you pull energy right here infinitely - the final number is during infinity so P goes from 0 to infinity and here we have E and then we have negative s T is photo well we know P is equal to t minus a so of course that means P is equal to D plus a like that so I will just put that down this is P plus a parenthesis that guy and then we have F of P and
55:56
then P P and T G are the same so I can just put that down right here so that's pretty nice and now we will finish everything in the world P is just a dummy variable and we just have to figure it out well right here of course we can just work down our little bit this is e to the power of negative s t minus s a yeah now have a look into this power it's a same as e to this power times e to the power e to the negative a so that's just a constant in the v-world so we can just put
56:33
that to the outside so this is going to be the following let's write down e to the negative s a first right and then for the inside here have a look this is going to be the integral going from 0 to infinity and we have e to the negative fee and then here we have this namely F of P DV like this
57:03
what's this well here it's something really nice in the front we still have that e to the negative si but this ratio is well this is just nicely equal to our definition of the Laplace transform of F of T isn't it even though I'm using P V and V right here but these just a dummy variable as long as you have this this than that being the same well is still co-sponsor to infinity this is still going to be the Laplace transform of our function which is f of T like that right so
57:41
this right here is it so this right here is just translation formula so all you have to do is when you have e to the negative a when you have the function multiplied by the unit step function if they have the same input all you have to do is in the S world be sure you multiply by e to the negative a and then you do the Laplace transform of the original function and that's pretty much it
58:11
and now I just recall that I happened right down this for you guys usually we can write this down up is e to the negative a and this right here we can put it as capital F of s like this right so this and that just extend the notation the output of the Laplace transform of F of T tot he is just F of s I so you can just keep dying mine I'm going to write this down right here though I think and
58:39
I hope this more clear this way so I will put this down here e to the negative F a and Laplace transform of better for you guys have he all right so that's the first one I will show you guys two examples don't worry but when not on your number ten because what if the input right here they don't match no good don't worry I got you so have a look here well if the input don't match you can
59:19
do the following make the match first so I'm just going to start by saying that so this right here is the second question yeah kind of this like camping file let G of t minus a be this right and why do you want to do that because now I can definitely put this down right here isn't it so here we go I will just write this down here for you guys this is going to be the Laplace transform F of T is going to me just write it down again F of T and we have the unit step function of t minus
1:00:00
a this now I can just put it down as G of T minus a so we can say this is the Laplace transform of can we still see yes G of T minus a multiplied by U of T minus a like this right and do we know how to handle this yes we do just that by instead of using f we have to use G right because I'm saying G t minus a you t minus a so this right here is nice to equal to you just do the formula namely
1:00:36
e to the negative s a Laplace transform of G of T like this yeah hmm so that's pretty much it but the promise that man I'm trying to show you guys the function was f and now I have a formula was G so what the heck is G right don't worry I got you again if you look at this G of t minus a which we set it to be f of T well all we have to do is kind of ship it back because I just had to figure out what's G of T in terms of F that's all and to do so all we had to do is change the t to
1:01:20
t plus a you see t plus a minus h of t likewise we've just had to do the same right here t plus a so you will see G of plugging you actually just get T and plugging you just get F of T plus a so in other word this is the same as that well finally you can just write this down right here as e to the negative a times the Laplace transform of this guy which is our F of T plus a just like this right well here is the power behind the story if this right here is just
1:02:05
F of T which is not with the same impress that all we have to do is the following you still multiply by e to the negative 8 whenever you have the unit step function just go ahead and do that let's just go ahead to that but if they don't have the same clue all you have to do is do the Laplace transform of F and just have to compromise it this is minus a you go ahead and do f of T plus a so this right here is yet you can we see yes all right so that's that
1:02:41
yeah and that's exactly why I just show you right here right started two in one and now I will show you guys two examples for the next question so let's go ahead and do the following
1:03:05
yeah and to be very honest with you guys I do miss my students haven't seen them for a while and hopefully everybody is doing well and hopefully my students you guys are watching and you guys are doing well too all right so tell number 11 you haven't you surf a happy face yeah all right so we have two examples right here for you guys first one let's see the Laplace transform of t minus 2
1:03:38
square times sorry sorry I forgot the input show period all right number 11 Laplace transform of t minus 2 square t minus 2 square times the unit step function of t minus 2 so in this case the a is curly Rd equal to 2 right versus the second one the Laplace transform of just T squared times the unit step function of t minus 2 right which one is easier this one slightly easier not so paddle
1:04:21
because again just the computational inside easy it's not too bad alright all we have to do is the following they do have the same input and the aid right here is 2 so we have to do this first you go ahead and multiply by e to the negative as and the a is 2 so twist like this yeah and it's really easy if you ask that rational questions and you have like t minus 5 because you have to proof 5
1:04:49
right here and becomes e to the negative s and then 5 so I usually you know try to avoid not just 5 so this is a stuff too yeah that's pretty clear anyway you do this part first and then the next part is you go ahead and do that you shall not pass of f of T which you focus on this part just in order input just do can impress t only so you have to do T squared like that right now how do you do that well we did that earlier for question number two already right so you have to
1:05:29
remember this kind of things I will write this down right here for you guys we call that for question number two the class transformed T to the nth power this is equal to n factorial over s to the n plus one so all we have to do is the following here we have e to the negative 2 s this is not 25 this is 2 s and negative 2 s and then for this and these two two factorial is 2 so we just multiply by 2 over 2 plus 1 is 3 so you have s to the third power like that but that's pretty
1:06:06
much it so in the end of course you can register in either way maybe you can read yes to e to the negative 2 s over s to the third power like that so this right here this is s right and again this is meant to be 2 factorial by 2 Phi 2 is 2 times 1 which is the same as 2 so no worry about that now have a look here how do you do it the inputs are not the same anymore but we can use this right here now so what we have to do is first notice that our a is again 2 so go
1:06:45
ahead and do this which is e to the negative s times 2 then what we are going to do is multiply by the Laplace transform the function here is T Square but this case you have to do t plus a of the function instead of the function so we will have to put t plus 2 so I'll just write on T plus 2 inside of the function and then we have to work this out how can we work out a lot faster and small G + 2 squared oh don't worry just do the algebra so here we go this thing is just e
1:07:20
to the negative 2 s multiply this out the usual way so we actually get a lot class transform this is going to be T square plus 2 of this that which is going to be for tea and the nasty two squared which is +4 like this yeah so just like this and then oh no we can just do this please than that by using that this case is when T is when n is equal to zero right 20 to the zeroth power all right so here is e to the negative 2 s see I make the mistake again this is e to the negative 2 s
1:07:59
all righty square which is the semester I'll put on the parenthesis for the result of the Laplace transform here we have two factorial which is just 2 over s to the third power double check good no it's right here so yeah obviously next what is just four and then the Laplace transform of T is going to be 1 factorial over T squared right and it's 1 so on the top is 4 times 1 which is 4 over s squared like this and lastly this is going to be plus 4 and again this is T to this
1:08:41
roots power and 0 factorial is equal to 1 and then put this over here you have X to the first power namely for over s like this and usually nobody oil made you combine fractions for a lot of last transport so I think this is totally ok right so just leave it like that if you combine the fractions I don't think impress anyone anyway so just leave it like that anyway number 11 that's it
1:09:18
okay all right so now we will be doing some more proofs that kind of things right so continued number 12 here is the Laplace transform of F 80 F of 8 he please do not call this to be fattens nanites you say f of 8 yeah alright so here is the deal I am going to again I will write this down here for now just make a note that when we have the Laplace transform of our function f of T this right here can Gradius capital f of s right that's what we mentioned the earlier and
1:10:07
for number 12 if we have the Laplace transform but the inside is not just a stop T but rather is the Laplace transform of F of a T like this first thing first yes the increased a T is just a x of T but right here it's not going to be the same as f of a s it's not like this this is not true in general right and this is equal to one but that's redundant so we actually have to do this carefully and now here with how we are going to do it carefully use definitions of
1:10:50
course so integrate from 0 to infinity e to the negative T of F of T so I'm sorry again I forgot to make things I read so like this and then we have the right hmm so again let me just cross this out no how can we make this happen though I can u substitution so here I think the best choice is to have the 8 he to be our u and we haven't used you in this case over just you so that u equal to 80 and of course it has to be a legitimate number on that stuff and oh I forgot
1:11:40
let me just prove the case that when a is greater than zero let me just say that a case where a is greater than zero and yeah you see cleaner this way yeah yeah anyway U is equal to 80 sometimes it is not greater than zero you can use some property saying thanks to anyway by the way D you if you go to a little root of this which is a DT DT is equal to D you offer a right and again we have the condition a is greater than zero and this is why we had to have that condition to make
1:12:20
our proof easier alright for our inter going that you world it goes from 0 to infinity so that means T goes from 0 to infinity yeah now let's go ahead and take this integral to the new world well you is the same as 80 a is passed deep and when T is 0 we get u being 0 and when he is infinitely you put here a times infinity a is positive you get infinity right so that's pretty much it and then
1:12:57
you just continue e and then you have negative as T and T we have to stop that so let's go ahead to that justify process by a we care T equals a you - you obey you operate anyway so s is 2 s but he is you over a and then here we have F and the input is that you now and then did he is equal to this so I will write this down here for you guys this right here is my U and then a DT is this so the T you right so now let's just go ahead and do all the rest so here's the usual business
1:13:46
we have the 1 over a we can put that all the way in the front and we'll finish everything that you world so we have 1 over a integral 0 to infinity they are in terms of you right it's pretty nice now here is the deal I'm going to rewrite this as follows is a negative and have a look let me write this down as s all for a and then times you right here times you on the side yeah right and then this right here already has f of U and then an outside here of course we still harder to
1:14:27
you so this is what we have now have a look one of race all the way in the front so that's nice what's this dog well you you a new match and this is integral going from 0 to infinity so this is almost like our Laplace transform and you have to remember this right here the definition was interpoc going from 0 to infinity and we have e to the negative whatever that we have here s T and then F of T DT right you see F of s as in that match and then a GTG match and we have two
1:15:11
negative and also the infinity is 0 on the start here this right here is our new input in the s world so all we have to do this put that down into capital F so we have capital F of s over a just like that therefore the Laplace transform of D to add of a T is equal to 1 over a capital F of s over a like this right and again I only prove the case for a is greater than 0 you can go and
1:15:46
try it is less than 0 just you know somebody's are part I could limits of the integration you have to fix that you'll still work and sometimes if you don't go through all this you can just use the property of some function to make the naked eco where if necessary by the way this right here is it point number two all right so this is how you change that oh by the way let me just do a quick
1:16:13
example for you guys how you can make this work this is actually a little bonus I didn't prepare data I'll show you guys how cool this is so real quick example suppose you know this the integral of sine T is equal to which is with the one on top 1 over s squared plus 1 squared so just like this s is the sign is which the beyond what happened this is piece 1 now if you know this this is what
1:16:48
we know and if you want the Laplace transform of s of T DT well this right here is precisely our F of s I have a look here is the S here is the S yeah here we can use that formula this right here is the number multiple early I used a G but iam using VT because I can match with the formula that we got earlier anyway what we can do is just go ahead and do 1 over Y but this number is yeah and then you put down s over P in here so just go ahead to F of s over B so we get 1 over B F
1:17:40
of s over B is just 2 s over B right here so we multiply by 1 over parenthesis as over B and you square that guy and you plus 1 after that and then you will see right here we can just get a common denominator which is going to be P squared so go ahead and do that and multiply on the top so in get actually 1 over B B Square over this and that is just a square this and that is B Square and of course this and Arkansas doesn't this look familiar yes it does it's just the over
1:18:17
s square plus B square so that's how you can use this formula to generalize sometimes don't worry about the constant multiple just put on one and after that you can generalize it by doing that okay that's it all right so numbers there think and again the capital F of S is just the result of the Laplace transform of the need to adopt number 13 we are going to multiply by something else Laplace transform of e to the HT e to the 8th so it's eat this is each f of T so here is
1:19:27
the tio you notice that when you have a Laplace transform of a product two functions in fact there's not a good formula in general only if you know a function it's like a special one such as this one if you have e to the eighty times F of T this right here it's actually rather special and I'll show you guys how can actually you know come with a nice to know for this right so here we go use definition so here we have integral going from 0 to infinity e to the negative T yes and
1:20:01
then you just put this down right here e to the 80 times F of T and then we have the TG after that of course now just do the integral pretty much these are not we can put them together so we get integral from zero cannot eat woody and let me just read yes a minus s a minus s and we have actually no no no no no no well technically this is negative s plus a the N times T yeah if you factor out that he and you have like this right good now this is what we're going to do right here
1:20:49
I'm going to factor out a negative I'll do this this water right so I have the integral going from 0 to infinity e I was factor out negative and then input will be the following to the stump blue stencil it'll be better in Kobe s and then minus a you can just work out algebra and then you have that here to that and this is f of T G do we need to use any use substitution the answer to that is no we don't have to because again TT and he match and this is negative and then you
1:21:28
have 0 to infinity already so just to remind you guys one more time if you have the integral going from 0 to infinity of e to the negative s T f of B thought he did he this right here is nicely equal to F of s yep this match and now you have this so what's that that's all you include in the s world that's it so all you have to do right here is just write down capital F oh - hey like this
1:22:05
very very very nice so this right here is how can translate in the s world earlier when we translate it has to be in the T world and use the unit step function and if you want to make any shipment new or just have to multiply the F of T with E to the 18 right so that's the idea and now of course I'll show you guys an example and I'll show you guys two examples so let's do number 14 so the first
1:22:37
one I have this for you guys and then Val will just give you guys another one because I think I would give you guys two examples right first one the Laplace transform T to the third power e to the 2t I forgot to write number 30 T to the third power e to the 2t like this right in fact you can do this with two ways depends on which one you take to be the original function f of T well I'm going to do it with that way so that means I will take this to be the e to the 8th he
1:23:19
of course I'll take this to PDF of T yeah so this is how we are going to do it first of all if you would like you can put this into that form it's not necessary about to be for you guys anyway this is the Laplace transform it will be 2 T and then we have T to the third power right just like this then what we are going to do is we have to figure out what's the Laplace transform of T to the third
1:23:45
power first so I will just read this down here for you guys now flash transform of T to the third power this right here is you go to what you two three factorial so that's put on some color so yes sweet so it will last three factorial over s raised to the three plus one power and what's 2 factorial 3 times 2 times 1 that's going to be 6 so this is nicely 6 over s to the fourth power
1:24:13
isn't it and in this case this is our s right here ma attention this right here is precisely our F of s right so what's the answer for this well this right here it's just going to be f of s minus 2 all we have to do is instead of the s right here you just have two stops at us - - because in here too and all you have to do this six on the top over s minus two in here and then the great start to the fourth power so here we have the s minus 2 raised to the fourth power and you
1:24:58
are done again this right here is precisely our F or as minus two right here it's regular the two where F of S is equal to this and then we know F of S is equal to that so that's all we have to do very very nice right so again another number 14 as a little bonus because I don't think I am doing enough for the marathon so number 14 again so put it a star next to maybe let's do another one how about let's do a movie star what that last transformed and let's do this I would like
1:25:36
to do e to the 3t and then let's do a cosine one so that's the cosine of T like this well in this case again we do the same thing first we have to figure out what's the Laplace transform of the cosine T actually let's do cosine of 2t why not study more numbers more better all right so that's going to figure out what's that plus transform this part which is the last transform of cosine of 2t and remember cosine is the one that has the s down the top so this is equal to
1:26:13
s on the top over and you have s square minus P which is 2 right so you have to square like this so that's pretty much it two squares for pi i'm not going to write it down because this is not the final answer anyway here is yes so you know this right here is capital F of s yeah and right here you notice we are multiplying this with e to the 3t so what should the answer be well we just have
1:26:45
to do s minus 3 in the capital F that's all so we chose how to make sure we end up with capital F of - three so now here we have s here you have s so go ahead and just tap s minus three here over put the s minus three here oops wrong color and then you square that and then here you have the - sorry cosines with plus sorry sorry cosine sweets plus right consensus with plus so you have this and then in the end you just add two squared which is four like this and then you are done so
1:27:30
just like that so this is not so bad right one if you have to go backwards well maybe in the future not today today is just a lot of class maybe in the future next time was so I don't think I could come with that Tony for that class transforms for the inverse I'm not sure by the way this right here I see it right maybe you guys can do it or make that into a contest if you guys would like to like in first up last transform was on the other if you're interested let me know and let
1:28:02
me see what you guys related to all my stuff and I'll try to work things out with you guys
1:28:13
oh yeah next one is number 15 Alice are looking what number what should I be doing number 15 here we have Laplace transform and again earlier I said a function times f not a function usually need to have a general formula but today if you have a function as the f of T so that's a general case but if you have this multiplied with just T to the first power this actually gives you a pretty nice result right this actually gives you a pretty nice result so let me see how we
1:28:58
can actually make this happen of course this right here if calling out here we shall use the definition it's the integral going from 0 to infinity and e to the negative T and you multiply with T F of T DT or the stuff inside right okay but in this case I don't think we can integrate this nicely though I don't know so we have to do this from a different point of view right let me just write this down here for you guys so let's see this is what we know again if you
1:29:44
just have f of teams that that's capital F of s so let me put this down like this we know that let me write down which one should I write on first I want to write down F first what should I write on the claspers don't just in one stack just be picky where it's not on other it's just pink
1:30:12
oh yeah let me just write on cap that first that's a really matter that much we know that capital F of s this right here is just a definition of the Laplace transform meaning the integral going from 0 to infinity and we will write this down better ah okay I'm just being picky I'm sorry okay all right this is just the definition of the Laplace transform Intel quad coming from 0 to infinity
1:30:42
and we have e to the negative T and we have little F of T like this right so that's pretty good now let's make a computer this than that this right here has an extra T that's all that's the only thing that's happening right I really want to just multiply the two inside what's the integral but we had to do that really carefully because we are in the key world in the integral right hmm I have
1:31:16
this T right here and this is negative s imagine if I can differentiate this with respect to s if I treat this as the variable differentiating e to the negative T where s is the variable so we just had to do the chandu and we get negative T the front that's very nice that's how I can squeeze out that he instead of the integral right so that's actually how we are going to do it we are just going to look at this equation and you see the left-hand side is a function in
1:31:50
terms of X so we are just going to differentiate this with respect to s like this right so on the left hand side of course this is just going to be the derivative with respect to s of our f of s like this so that's just going to be some calc 1 derivative or whatever but only right-hand side this is how we are going to do the derivative of integral you have to use what we call the Leibniz rule and sometimes that's also called the famous taking of integration but anyway if
1:32:23
you do the derivative of this you will just have to do the partial derivative of the inverse that's okay that's true so let me write this down here for you guys where you will first write down the integral from 0 to infinity instead of putting this on outside we are going to put this inside the integral and we will do the partial with respect to s and the reason will change to this to the partial notation it's because the inside here technically we have two variables by anyway
1:32:53
to the partial of this expression with respect to s so I will write this time red we get e to the negative as T and then we have F of T in order to focus on that and then on the outside here was to half that eg another focus on this part so do this first that means we can write an integral from 0 to infinity and then we still have to did heat on the very outside now here we are taking the derivative of this expression in terms of the s in the S world so as is the constant T is
1:33:29
just the variable S is the variable XI is just a constant there we go I'm sorry all right so I can just write down the constant multiple first which is the F of T no BTO and then Teddy the Rope T of e to the negative T well me the root of this always repeat part so we have e to the negative T then use the chandu as how efficient the area multiplied by DT root of this the review of negative as T in the s world is negative T so you just multiply by negative right here and
1:34:04
that's not wonderful isn't it because this matchu is that very nice except for the negative sign well what can we do with a negative sign this is just a negative 1 we can kick that to the other side right so happy dog this is showing us the integral from 0 to infinity of positive T and our actually just weather everything right this e to the negative T to the T right here he F of T
1:34:36
DT this right here again we can keep the negative to the other side so just have to multiply the negative with the derivative in a swirl of the capital of F of s like this so that's pretty much it all you have to do is look at what f of S is look at that first and then go ahead differentiate that with respect to s right sorry let me just write down better for you guys look here what F of s is first and then depreciate is with respect to s but not thanked yet because you still have
1:35:15
to multiply by a negative so that's it just like that very nice now I would demonstrate an example for you guys that's number 16 I'll show just put it down right here should work okay number 16 is equal to the follow a Laplace transform of T times sine oh like this well if you're outgoing now we can just run through the Laplace the top finish it again but we will just use the result now here is
1:35:53
the deal you see we multiply this with t to the first power so I just have to figure out what's the Laplace transform of the sine of BT so I will just rewrite we know F of s which is the Laplace transform of sine of BT this right here is equal to B over s square plus B Square I am NOT making a mistake like earlier so this is good so this is our F of s well this right here is equal to the following we just have to go ahead and do the negative and differentiate this with respect to
1:36:32
s of the function which is f of s which is just this I'll just put this down half of s yeah and of course this is the same same negative differentiating and to differentiate F of S which is that remember P is a constant is this the variable we can write this as the following put this up here with a negative exponent the clock will start right so we can write this as B times s squared plus B squared to the negative 1 like this pretty nice huh now of course bring the power
1:37:13
to the front minus one and then that's ready down negative times negative is going to give you past if so that's good and then we'll just have the parenthesis a square plus B Square to the negative two power and of course don't forget your chandu multiplied by the derivative inside the routier of two that's the relative a square plus B Square is 2's like this and now we can of course just put everything down together this done with ha so we have 2's with the be on top
1:37:48
again this is the S right over this right here is the parenthesis a square plus B Square and then raised to the square power on the bottom like this so this right here is the Laplace transform of T times sine of BT right just like that very cool huh and now as an extension what if you motor perfectly square yes I said just go ahead just stand the idea if you want if you have G Square here you to the second derivative and you multiply negative 1 twice if you have t third
1:38:28
power just go ahead and do what differentiate the three times and then you multiply by negative one to the third power and all that good stuff alright so that's pretty much it for number 16 and I were actually like to go back to number of working for you guys a little bit
1:38:51
arrested that not good actually let me just put down what I have right here on the notes I wrote this Tanis honorable mention so for the idea so this is just the honorable mention well the idea is that if you have the Laplace transform and originally if you have the function right here already yeah but if you have to multiply this with T to some power and assuming any just from 0 1 2 3 4 5 and so on maybe they talkin to Sun like fractional to row that here but that they
1:39:40
turn anyway though this right here all you have to do is the following first do care what Laplace transform of F of T s as a result just ready as F of s yeah this is T 3 and power all you had to do this duty and the rabbit here in the s world like so I to the nth derivative in the s world so let me just read it some nicer to you guys the power over like that but you see each time to knit a row but he not only you to need a robot heap I also had to multiply by negative 1 yeah
1:40:21
so you actually have to do negative 1 raised to power just make sure that you have the right sign in all that stuff I saw this is my honorable mentioned for you guys this is just doesn't look nice sorry one better maybe alright so this right here is just an honorable example no no no no I'm not doing that I wrote to you I'm going to do the following the Laplace transform of T to the third power e to the 2t does this look familiar he should that's question number 14
1:41:05
anyway earlier the way that we did it is we took this as F of L F of T right that was actually easy way now I'll show you gets the hardware to do this now I would have to take this as my F of T the little F of T and this is T to the third power because here we have T to the third power so that means I will have to multiply the negative one three times and then I will have to find his third row but here which is the jerk in this case the third row but here oh the following
1:41:43
well let me just remind you guys let me write it down right here if you look at the Laplace transform of e to the 2t if you just look at this party the 8th values right here is 2 so this right here is 1 over s minus 2 s has to be greater than 2 pi don't write that down too much and this right here is precisely our F oh right so to desert the rubber tip of F of s which is just a 1 4 s minus 2 like this but again if you want to to deter operty of this it's easier if you'd
1:42:21
write es s minus 2 to the negative 1 power and then just enjoy your derivative all right here we go the trick is to do this carefully with the negative signs here you can negative 1 to the first derivative so I will have just D to D s to write to the first erupted heat right here I'll bring that one negative 1 to the 4 minus 1 the Chandu is just 1 so doesn't really matter so you get negative s minus 2 like this when I minus 2 yeah like that so that's the first
1:42:58
derivative and then let's just have the negative 1 all the way near at next you do this again put this to the front minus 1 this kind of fund you just wander over here because you just did want a rubber tip already here you again negative negative becomes positive 2 and then you have s minus 2 and that's negative 3 and then what do you do next yes you do you again that's the last one put this to the front minus 1 so that's going to be negative 6 times negative 1 oh no it's
1:43:35
going to possibly 6 and then you have parenthesis s minus 2 to the fourth power and of course we can write this down as 6 over s minus 2 to the 4th power like that right just like this very good huh and of course don't do it this way don't throw it this way this longer this way you should take to the 3rd power as you are f of T and then you do the shift right that's much better but you're not
1:44:07
sure you guys both right so no complex alright so that was number 16 it's my honorable mention
1:44:25
earlier we multiplied by T this time we are going to fresh welcome to / T God and you can expect it we will be getting into gross of course so let me just write this down right here for you guys alright so number 17 but I mean you actually register reversed again Laplace transform he actually let me register because it's really cool you put it on like this first Laplace transform of
1:45:01
F of T this is just f of s good and as we all know if you have the Laplace transform if you multiply by G of f of T this right here all you have to do is you depreciate in the S world of that so you have the capital F of s right here but don't forget you multiply by negative one and now have you thought about what if we have to divide it by T so let's go ahead and look at question number 17
1:45:37
here is the question what's the Laplace transform of F of T let me put it down on the top and then with the PI D PI instead so how will we get if we do like that well this is like T to the negative one power it's just that you do the integral huh yes I agree but we have to do it carefully because you cannot just integrate f of s because if you do this you forgot the constant you forgot that yes well no joke aside if you do this you may end up with some constants yeah because this
1:46:22
is a definite integral and maybe you say maybe just go from zero to infinity that's actually not true so you have to do this one carefully and this is how so have a look let me write down the definition first this right here is the integral going from zero to infinity and we still have e to the negative T and O this I will just put it down right here for you guys in red I have up T over T
1:46:52
like so and then we have the DT so that's exactly what we are right hmm let's take bow it this is the F of T and then we have the F of T here how do we prove this earlier we differentiate this yeah we differentiate that with respect to s where we use the fineman's technique or the librarians rule derivative of e to the negative as T that's how we can produce an extra G that's how we get out right
1:47:39
this time we want to divide so maybe this guy okay I'm not going to differentiate I should be integrate because if I integrate e to the negative T I will have to get what I will end up with D by D by this right so that's the hint and I will just have to make some observation and perhaps that do the following let me put this down right here so first put on some observation right if we
1:48:13
integrate e to the negative s just integrate this part well what's the answer for this this is going to be e to the negative as T that works that's great but the reason I put down dgod forever it's because of the following I'm trying to produce the over T on the bottom so I'm not going to integrate this in the T world instead I will be integrating this in the s world so T is that why because in
1:48:47
the s world T is the constant so when I do this integral in the S word we will have to divide it by negative T so I actually end up with 1 over negative T like this so we had to integrate this in the s world really carefully we really kept on that so that's nice though but the promise that I was to end up with some bizarre things because I will have the constant plus C for example right so that's no good so we should somehow use a definite integral to help us out so maybe it's a
1:49:27
good idea to still have infinity because he popped infinite he put it here and you just need to write down s has to be greater than 0 and that would become 0 that's nice but I want to end up with hmm well let's see so I just integrate this from 0 no actually I'm integrating this in the as well I get that and then he no cannot inter cannot put this down this is 0 if I put on 0 again were integrating in this world I will have to put a 0 in here sorry I should have said the infinity is
1:50:24
for the S right here because what we in this in the S world if I put a 0 in the S that part goes away I don't have to eat with negati as T anymore so be really careful I actually have to integrate this from s to infinity check this out right so again let's integrate e to the negative T from s to infinity of in the S world so of course you draw into equation and we got that and then what
1:50:55
we do is we have to plug in plug in I will just go ahead and plug in this in black I'll say plug in plug in as two but the idea is that again we are doing the s world okay now here is the piece of art you can plunking s is equal to s s is equal to infinity the promise that nobody does that some people are okay with that but some people are not what you had to do is right here instead of using the s you will have to change that to a different variable because the dummy variable
1:51:34
and that cannot be the same and the promise that only at the end you want to end up with the S so you just have to change the S earlier so earlier I will just have to change the OD s inside there to be you just say that you and the same right here this is still going for ice to infinity that's good but now this s it's actually you and again this is just how you can be savvy technical and this is again just the observation that were making I don't want to tell you guys just do
1:52:05
that and you'll get the answer but this is the thing that you have to do a little bit on your sometimes to make sure that you can actually get this to work my duet not continue you ghost use infinity here you put it here and of course T is greater than zero because originally this right here T is going from zero to infinity so that will take care of the business you push this right here the first part will be 0 and then you will have to - you put s right here so and
1:52:34
outwards 1 over negative T and then you have e to the negative 2 put s right here ok s T like that and all you end up with each will leave negative as T over T this is so wonderful isn't it because now this portion you can just have a intercooling set that's the idea all right so why exactly do we have to do with this though let me show you so this is the observation it took us a while but now this is what we know again let me just write down the original let's see where is that
1:53:21
thing we know this yeah so we know the integral of the Laplace transform of F of T is equal to f of s this is what we know now for this right here I can change that to the definition namely the integral from 0 to e e to the negative s right T and you have F of T and like this and this is equal to the integral and you can have that to be - you know that but really we want to integrate and what to change the time we variable to be you so again this variable and that they will
1:54:11
match so instead of putting down s right here I will actually use U and then right here I will actually have F owe you and this right here it's actually tu because what integral is in a new world and then no sorry not integrating yet because the Laplace it's already a let me do again all right this is probably not yeah so let's put on the definition of Laplace transform of papers which is the integral going from 0 to infinity e to the negative s T and then you have
1:54:52
F of T DT and that's equal to F of s yeah so this is just a definition and of course the right hand side to s now check this out this s and s are the same but in fact we are about to integrate and we still want to end up with s to be nice yeah so in started having the S right here I will change that to be you to be you and again this is just a dummy variable you could have used to fire you one other stuff right now what we are playing to is we are going to - what we did right here
1:55:33
we will integrate both sides from s to infinity so right here I'll integrate this from s to infinity and then from here are also integrate this on s to infinity and again you see the output of the integrating plaque would be in terms of U so I will have that EU right here likewise the output right here will be in terms of U so this is again in the new world like this right so now here is the deal we end up having to grow but not so bad at all because if you look at that
1:56:12
he and also that you I will write this down here for you guys let's say this right here is the T axis and this right here is that you access yeah can we see okay okay first of all T goes from 0 to infinity so it's just this portion yeah I then u goes from s to infinity and I say s is somewhere right here so the region that you're talking about is just this part so it's just like a rectangle yeah the best part is that we can just switch the water of integration and that's very
1:56:47
nice let me do that for you guys let me do the integral from 0 to infinity and then the blue integral from s to infinity and we'll have the tu first and then we have the TT after that and for the inside well I really want to have this to happen so what I will do is let me just read it down again let me just read on everything again so inside here is e to the negative u and we have
1:57:24
inside here we have e to the negative and then we have F of T like this now the right hand side is still equal to the integral from s to infinity f of U T you okay only that hand side have a look we are looking at the world first that means f of T is just a while it's just a constant so you can put that right here can I put it all the way in a bundle because not to deal with time that's the
1:57:53
world so have a look all right so let me put in straight here so we have two integral from zero to infinity and then the F of T right here and then we have the actually let me just put the F of T in black so we have the F of T right here and then we have the integral from s to infinity and then we have e to the sorry e to the negative UT e to the negative u T just copied it wrong e to the negative T and then T U and T right and now have a look we know this right here is equal to
1:58:43
this is the integral from s to infinity of f of U the U what's this in blue this is exactly what we came up with earlier so I can just put on a result right here I'm not going to work that out yet you can just you know look at that so we actually have the integral going from 0 to infinity and this is f of T and then we have this put the whole thing so we have downwards e to the negative T right so now we actually end up with this so that's why it was really sorry about you
1:59:21
know that's one of the things what we used to anyway this is equal to that we had a TT what's this this is exactly what we're trying to find yeah so I'll just put this tongue I know it's like a triangular equal sign by the way finally I'll write this down right here for you guys this is telling you that the Laplace transform of F of T if with the party pide here right if we divide it by G because this pressure represents that this is equal to the integral going from s to
2:00:05
infinity of F of U T U and this right here is it very nice huh so you can see our assumption was correct earlier when you multiply it by T you differentiate x negative 1 and just how to do it carefully when we divided by T we actually integrate but you had to do it from s to infinity and you don't actually multiply by a negative whatsoever so have a look have a look have a look pay attention to the observation especially that part and it's just how we can make things happen
2:00:43
that's wanted to tell you guess all right so I'm going to leave this on the board just remind you guys because for the next one it's going to be an example I can figure out crazy now class based on that all right so of course that's copper to black pen where pen events I used to do sometimes all right number 18 right here for you guys I'm pretty here let's look at the Laplace transform of sine G just sine T right and then I want to divide it by T oh my god
2:01:29
first thing first perhaps I will write it down right here we know that we know that what's the class of sine T this is just sang of 1p yeah so it's 1 over s square plus 1 that's all we know and again we are about to integrate so I'm not going to use s instead of the s right here we'll just change to you right so here we go because we're dividing by T right here all we have to do is we
2:02:16
will be looking at the Laplace transform of sine T which is this let me write down 1 over u square plus 1 for you guys so this right here is our f of u yeah and then you go ahead integrate from s to infinity so you look at this and you integrate that from s to infinity and you are in the new world and how can we continue just do the calculus of course all right so what's the integral 1 over u square plus 1 in the world in first tangent so for this we just have the inverse tangent
2:02:54
of U and then we're parting as an infinitely other stuff cloudy infinitely inverse tangent of infinity is precisely power to let me let me just write down everything for you guys inverse tangent of infinity and then - just like this and then - inverse tangent oh yeah this right here is what PI over 2 as I said the earlier so we get PI over 2 and a - inverse tangent of s look at that how cool is this yes this is the Laplace transform of the sign T over T very very nice
2:03:45
thanks to this formula some honorable mention right this is number 18 and I'm almost done but some honorable mention when you have that you can actually end up with some very nice integrals let me show you remember what's this this right here it's a cemetery integrating and just to cut the whole thing inside as your function this represents the integral going from 0 to infinity and we write down e to the negative T and we have sine T over T like this right now with
2:04:36
this being done you just have to tell me ask value you can put it here it can come with a lot of cool integrals so let's see what's the value of lemon juice tear gas on abroad mention number one what's the value of the integral from 0 to infinity of sine T over T debt well this computes that this is the case when s is equal to 0 so I have to do is tell you guys this is equal to 0 and then put
2:05:08
0 into here and work that out so this right here it's going to give you PI over 2 minus inverse tangent of 0 and that's 0 so on y'all we end up with PI over 2 very nice isn't it and of course you can come with another one such as what's the integral from 0 to infinity if you end up with something else that's crazier let's say let's do this e to that he to the negative T e to the T actually let's put that that's just Twitter and then we have sine T over T tt what's this dad
2:05:53
well in this case s is equal to one and can s be negative one no P capital right if s is negative one this won't converge so this wouldn't make sense so you still have to do this with culture if you do that be sure you have to make sure that s has to be positive and so they mean just credits down here for you guys s has to be positive by the way that's not ruin the fun s is one in our case
2:06:29
so we're just plugging one into there so we get PI over 2 minus inverse tangent of 1 what's the inverse tangent of 1 well here we have PI over 2 this right here is minus inverse tangent of 1 is just PI over 4 and get our common denominator 2 & 2 so you know it's 2 pi minus pi is PI and 2 times 2 is 4 and now match so it's just PI over 4 so yeah we have another one very cool huh so thanks to this formula installed cool isn't it I have another one for you guys number 3 on there
2:07:09
Paul mentioned so let's look at the integral from 0 to infinity actually let's do it like this let's do the negative infinity to positive infinity and here I would like to ask you sine of e to the X Oh God hey this does not look that any of that how can we do this well do our usual calculus stuff let's do some substitution yeah so have a look here maybe that's do some use up that u equal to the input function which is e to the X and you see T you will see the same as e to the X
2:07:46
DX so that means DX is the same as T u over e to the X don't worry you see after the substitution you see that G u over u so if you put it back have a look in the X well X goes from negative infinity past infinity take this integral to the New World well when X is negative infinity put it here e to the negative energy is nice to equal to 0 to infinity here e to the infinity years so you goes from 0 to infinity like that ok inside here we have sine this is our u so assign u
2:08:26
DX is T u over u so I can put down tu over you does this look familiar now yes it does it's just nicely equal to PI over 2 isn't it so as you can see this is the reason why we say the 10-meter robot as a matter because the outputs just contain number so is it TT and he you you and you and that's not so let me match this right yes the same as that which is PI over 2 PI over 2 very very very very very nice right so some really cool stuff if you know some up
2:09:07
last transport some very nice intercourse I can also do for free very good so let me finish the things and then perhaps Aki because my other unwelcome mention later on but processing tell you see if you have co-signed oh right if you have cosine T over T in fact that the past does not just because the integral is actually type project by if you have one minus cosine T over T that actually will give you a convergent integral and you actually get nice one as
2:09:46
well I'll do that all the way at the end just a little pompous but this is really cool right
2:10:05
all right we are going to be taking Laplace transform of a derivative so have a look here number 19 on the board for you guys number 19 here we have the plas transform of F prime so it's no longer just the original yes DITA robot here all right all right all right use definitions of course so here we go this right here by definition is the integral from 0 to infinity e to the negative T times F prime of T and then yep very nice persona it's not so nice because
2:10:55
how can we deal with this empowering to code us that I wrote it well I want to have the original because you have the original we know that's just the original capital F of s this is F prime so how can I give to F of T from F prime will have to integrate this guy yeah you see where to integrate just one part inside the integral so what should we do with that go ahead and differentiate one
2:11:28
part to be integrated the output part the other part would be differentiated so what should we do yes integration by parts and yes di method so have a look oh my god this is so much fun d and an i and now put on the plus minus just two rows is enough don't go too crazy of course we want to integrate F prime of T because if we do that sorry that's a horrible F because if we do this we'll just get the original F of T very nice and of course in the meantime that differentiate e
2:12:03
to the negative T and don't forget we're in the T world and don't forget the chandu the derivative of e to the negative T is e to the negative T and the channel says x negative s because negative is the constant with the teeth just like that all right now first part is just this times that so this is equal to this times that which is e to the negative T times f of T this is the first
2:12:33
part of the answer but remember this is a definite integral so go ahead and plug in plug in from zero to infinity yeah and then we are - - we still have to do the part right here and this is still going to give us and integral so we have to actually do - times - oh it's actually a plus so I just put down right - times - this - ingre first times that - simplify becomes a plus very cool anyway
2:13:10
and I'll put everything except the integral so you have the integral and then I'll put this and that is e to the negative T and then we have F of T DT and this is the comment from zero to infinity and you might be you know worried about that didn't put on yes well I can put on s here passing see asymmetry world s it's just a constant allow me to put on s in the front right so have a look the hardest part to teach the past tense point is - radius anyway could I ask in the front so
2:13:48
that's the s in blue now we know a few things first when we plug in infinity to this right here because in the beginning of the video which is the pollak towards our course all assume that f is with exponential order so that means if you put infinity to this it would be zero so I will just have to write this down again for this down f of T suppose it has exponential water so when we do that the first part will be 0 right so that's nice then of course that's the first part
2:14:29
we are going to - put a 0 into here and there so we have the rest is just e to the negative we are doing that in the humor so you put the zero into the T so it's negative s times zero and then times f of zero like so right and then we add again here is dice but what's this part actually although this is yes here we go what's this part this is our definition of the Laplace transform of D to F very nice huh this right here is nicely equal to capital F of s of course you
2:15:15
can if you like if you'll read a Laplace of F of T but yeah you prove that anyway and we're almost done just like clean things up notice this is just what one right and this is minus F of zero so be sure that you do the following and how do I write it or I would write it down like that so okay so our fitting everything here finally we have the F oh yes - E - f so you have to refer back to what's your little F it's alright so you just look like a functions
2:15:53
and usually you do this when you're solving a differential equation initial value problem actually so usually this would be given and you have F of 0 that it's for this and we are done very nice right and again F of s if you like this is the Laplace transform of little F up like that
2:16:21
all right so that's it now after the first the robot here what should we do sure let's go ahead and do our second derivative hmm the C number 20 a plus transform I want to figure out what's the second derivative we can do this again and just have to integrate No twice so you have three rows of the stuff but life doesn't have to be that hard sometimes we use we have that already so use that make good use of our results so this is how we are going to do it so I put this down in
2:17:06
blue first this is what we know right so have a look here we know that the Laplace transform and F prime of T let me write it down like this this is the same as saying we have F of T right and then we differentiate that that's how we get priority good and this is equal to well you see F of S is equal to that so let me put this down right here for the F of s so this is equal to s in the front and then F of s is the same as that so that's why properties wrote it down
2:17:49
over there and then the input here is f of T very cool and then in the end here we have little F of zero that's a - little F of zero just like this this is what we know and now let's make the best use of this now what's the connection between second truck in the first derivative yes you just differentiate the first derivative again so have a good happy dog so now if you want to get the first if you want to get a second term over here let me write it down better for you guys
2:18:26
sorry now so let me just literally put on now if you have Laplace of the second the reality of T like this right this is nicely equal to the Laplace transform you just have to be appreciate the first derivative so as you can see all I did is just rewrite this f double Prime the same as that Prime and the Panda again yeah now check this out I thought he little puppy little T e to F of zero this this and now they are the same time now my kind right here is f prime so next oh
2:19:13
I have to do is the following here as two FS and then I will just do the Laplace and then this is just going to be this yeah so put down F prime of T very nicer and then minus instead of F will put on F prime so we have F prime of zero like that and this is so wonderful have a look this this and that are the same very very nice isn't it now do we know what this is of course we do is the same as that so we know the following I'll just put this down for you guys this is equal
2:20:02
to the following s in the front but for all this part we can just write it like that which is with a parenthesis and then we have the s capital f of s minus zero so this is for that so f of zero
2:20:26
yeah and then in the end we have the minus F prime of zero now finally we just do some work and usually we put down the half point first before that and that's why I did actually I didn't do that nevermind so I'll just write down everything for you guys you distribute the s you have s square f of s minus s little F of 0 and then minus F prime of 0 like this Congrats this is our Laplace transform of the second derivative of some motion very quiet
2:21:10
okay so twenties are us transportin and then four more to go
2:21:24
so earlier we did a first and also the second derivative and now what should we do yes we shall do the integral so number twenty-one right here the best for right here so number 21 let's go ahead and do the Laplace transform and instantly recap oh is the integral going from some number to T right some was responsible to T so you have to tweak a body this is the integral going from zero to T and then here I'm using C as my Tommy variable so I have F of T DT and then that's
2:22:19
something cool to what I don't know yeah can we use this again this again I think so because the rub t think the cooking cancel half a dog so that's to what we did earlier so let me raise down here for you guys this is what we know earlier when we do the Laplace transform of the derivative of some function and I wrote it down like this right this right here is equal to s times the Laplace transform of F of T and a minus F of 0 like that yeah now have a look this is my input
2:23:17
let's go ahead and put this into here here and also here how do I do that I will show you don't worry so our function here I will change a little bit because the notation is going to be this is the F already so that's not good I mean actually put down G if you don't mind right because in our case here this right here I'll just cut out to be G of T and again this is not a capital F whatsoever because capital F is reserved for the output of the Laplace of the two F so that's
2:23:58
why I'm using G so in this case I will just say the missive I can put that down so in this case I'll just say that right G of B that and again you into a quick wipe a little F is plugged into your parking key the output is it with T so G of T and you get the integral from 0 to t f of v TV and of course that go ahead and plug in to everything so we are looking at this right here G of T is
2:24:32
just an integral so it's just what we are going to do integral from 0 to t f of P P P prime that and then close that and then s naught plus transform of the integral going from 0 to T and then we have F of V DV good - and I'm sorry this right here I'm using G G so this radius also be G and should be G of 0 ok the reason now try this down I'm per proton G of T like this on purpose it's because G
2:25:16
of 0 it's this well all we have to do is putting 0 into the T you see here is where that he is I just have to dance you were right here so G of 0 it's actually integral from zero to zero of that F of P TV and that's nice T equal to zero so this is the gentleman D equal to zero which is very nice right so just wanna show you that this part goes away because of how we will how do you take it a
2:25:52
review of this the rub deep and integral cancel by the fundamental theorem of calculus part one so I have a look the left-hand side they can so you put a T right here though how can you put it inside here though so we actually end up with F of T and that's it right that's it and then on the right hand side of course you have the s times the Laplace transform of the thing that we are looking for so we have the integral going from 0 to t for P P like this and finally what can
2:26:35
we do great onion so of course I want to figure out this all i have to do is divide the s sample size so finally Laplace transform of the integral from 0 to t f of T DT this is equal to the Laplace transform of F of T and then on the outside right on the outside the tip ID pass so you can put this on OS and of course if you ally you can write us f of s divided by s after you whichever one
2:27:12
that you like so one way or the other just like that very cool now next it's another Laplace of intergroup bias with a star that's a new kind it's a special kind of integral how the tongue fiducia
2:27:32
pop quiz for you guys do you guys know what does the word Laplace me if you know leave a comment down below right and question number two what does do a composition me if you know if I comment down below alright so number 22 right here here we go here we have the Laplace transform and input here is what we call the it convolution namely F of T and you do the star right here and with another
2:28:07
function GOP like that now here is the idea this is actually now regular multiplication if you have a Laplace transform two functions the functions has to be special enough like the ones who have done either e to the AG were T to some power so you can doubt with nice formulas but if you just have a general function stack the plastron spoon of L and T time Sankey whatsoever there's no proper rule for the Laplace unfortunately however if you have the composition inside
2:28:43
it's a different story you actually end up with something extremely nice in this case not sure I guess what is but of course the first thing is to tell you get the definition of the convolution so write this down right here this is Cody come for just this is called account for Lucian and again F of P star GOP by the way that's called a star that's why I say stop this right here thank you
2:29:19
going from 0 to T so it's kind of similar to the one with the early ok but not exact inside yeah F of T minus the dummy variable T and then motor time this is the regular multiplication and then G of B TV yeah so just for fun for you guys right just for fun question for you guys what's one dot one question - what's one start one huh leave a comment down below and let me know if you know yes for that in fact yeah if you have seen my videos in the past thank you so much alright so I
2:30:06
did that in the past off I would include a link to the video in the description for your convenience oh my god man it's not funny because Laplace is integral and now we're putting it not interpret there we'll try our best first Laplace transform is the integral going from 0 to infinity e to the negative as yeah and then we'll put this inside and I might as well just put on the definition
2:30:37
right here already so I will have integral going from 0 to T and then we have F of t minus V and multiplied by G of T DT so this is the composition and then on the outside we still have the TT
2:31:02
it's okay it's just stopping they're gonna see what we can do first thing first when you go from zero to T about Hyup idea we don't we don't like that we prefer to have red handles right so now we can just change the water of the integration that's much better if you want to change integration right now leave a comment down below how you care but I'm not gonna do that now I will lie to her this going from zero to infinity as well because though that can actually end up with
2:31:33
a really nice that plus for the function f and also for the function G you will see now how can I make the happen though hmm one hint right here is that we have f of t minus V have we seen that G minus V s input already yes we have where yes the unit step function so the see what we can do with unit step function right well in our case here if you would like you can let's see how should I
2:32:09
let me just focus on the red part right here so I am just going to graph this power here just a report I so I will just do it in Graham blue okay so I will keep it as a picture right here suppose here we have f of t minus T times G of P so this is our function in red and here remember we are in the v-world sophie is the fee is the variable i don't know how it looks like let me just give
2:32:44
you a sketch so maybe it looks like this I have no idea really all right so let's say this is how the function oxide now here is the deal I would like to multiply by the unit step function well V is the variable t is the constant and let me just remind you guys when we have the unit step function this time table because they both look like they're opposed to me so I've already done like this when we have number sorry let me do it like this first home we have a variable
2:33:21
- hashtags the number I just easier that way right this right here produce the following why ever done and number is right this is why I've read that number is anything before the number is zero anything after that it jumps to one I thought this is how it goes if the in size path deep is one so variable greater than the number is one so just like that however if you have the unit step function if you have the number minus the variable this will keep you the other direction so what
2:34:03
that means is that anything so that's a here's the number here's a numbering read anything before here's one add anything after that is negative one so be really careful with this now here is the deal pay attention to the red integral we are just going from zero to T so with this picture right here we just want to go from zero to T and again T is our number in this situation and that's the
2:34:40
integral that's the area for the integral that's all we need with only anything for work so we actually had to multiply by this kind of situation but it's actually so wonderful because you notice that he is the number B is the variable in the v-world like you're right so have a look if we have this so I'll just put it down blue if we have F of T minus V G of B and we multiplied by U of T
2:35:12
minus D and again this right here is the variable and this is the number so when I multiply this with the function here what happened what happens is that anything before in main tanks so the blue portion is no domain names and and you see after that becomes zero and zero times that will just keep your zero like this so the blue portion is for the blue function right here so what I'm gonna
2:35:45
do is I can multiply by U of T minus V and then to get the same area we can't you just go from zero to infinity everybody happy so now this idea right here really it really well okay so let me erase this because we will push any more space all right first thing first copy down the things that we have so this is integral 0 to infinity here is e to the negative T and then this is the integral of
2:36:17
going from zero to something let me not put on yet I already start get F of t minus V G of T and then let me multiply this by U of T minus V because I'm doing this from you front because I multiplied by U of T minus V front here none worth zero so I will change that to infinity so that's really very nice and then just continue we have the TV and then we have that TT and everybody happy right now
2:36:53
perhaps we'll just clean things up a little bit I'll do this in black and red I will first put this inside and then we'll switch to water in the gross I will write on this two steps right first integral in black and the integral in red they are both going from zero to infinity I'm just going to put this inside so we have the e to the negative as T and then we have the red part which is the F
2:37:23
of t minus V G of T and then of course that keep the blue blue of course you of t minus V and then we have the TV and then we have that all right I'm just putting everything inside and now I will switch the water of the integration well what's that being said the Ranger cuoco first bottom that heat at the integral in black will call next and then what still write down everything in second actually no let's let's do this in your head right so let me put the brain tech one outside
2:38:03
so we have this and we have the BP on the outside right and then I will have the T T so we have the integral from zero to this now we have everything right here you can put it down right here but in the t world hey gia V is a constant so we can put that to the front this one has to stay because he has that he this and that they also have to stay because he has that he so I can put a G of V in
2:38:31
the front right here sorry G of T first to your fee and then the integral from zero to infinity and then the rest I was just ready time black e to the negative T and then this is f of t minus V and then u of t minus V like this then I'm doing the TT first and then the P after that just like that very cool unit step function and I think we have done this already yes have a look here we have the integral in red yeah that's zero to infinity and this is G of P good what is this
2:39:25
this is the Laplace of this function isn't it yes yes yes so I can write everything down right here as the Laplace transform of F of P minus P U of T minus P and then on outside we have the T P I should stop using my pen otherwise my hands are just going to get really dirty so this is what we have very nice huh all right so in the front we still have the bread integral going from zero to infinity and we have G of T do another class of this yes we do well in this case though V is
2:40:15
a constant black in the black right here T is the variable so V is the key is the constant so it's pretty tricky so be sure you just kind of keep all this little things in mind so we have the F of T minus P yeah it's the constant earlier we were doing f of t minus a times U of T minus area and how it get is the following we get e to the negative P and then you have the s yeah and then we have the rest it's just F of s right because the input match so just have this right here and
2:41:07
again I'm just using feet in 38 so same thing and then on the outside we still have the TV so nice huh because now F of s is just a constant in v-world I know it's just really crazy but it's really fun you can put the F of s all the way in the front and then you look at that real part so here we have the integral going from 0 to infinity and allow me to write this down first which is e to the negative V s and then this right here is actually the mean red yes Espie yeah let me
2:41:40
read yes sv and then write down G of P and then we have TV and do we recognize so this is V V V match and then it goes from 0 to infinity this right here is precisely the Laplace transform of the G function so finally we can just write down F of s and regular multiplication of capital G of s like this this is so so so wonderful and of course if you like and ready dance the Laplace transform of D thought he little F of T did hold multiplication right just a regular modification
2:42:20
and compares your type you like and this is just a plot transform of little G of like that so one or the other right just like this so so so cool this is one of the most wonderful Laplace transform and yeah so so so nice right all right and again don't forget to comment what's one star one right
2:42:57
all right so let me show you guys eat samples okay I'll just let me finish number 23 I'll keep you guys tweet samples okay what's the Laplace transform of sine Riggleman regular multiplication 1/2 cosine T versus the other one Laplace transform of sine T star which is the composition and this is cosine T right just like that in fact I'm not going to say this is question number 23 this is just a Honorable Mention right so let me just say this is the Honorable Mention
2:43:48
right so Laplace transform sign he co-signed this is so not pretty sorry sign he co-signed he / sister all right which one is easier this one they mean - this one first this right here it's a convolution we can use the convolution theorem that we proved the earlier this is the same as a to the Laplace transform of the first function and then you multiply by the Laplace transform of the second function which is cosine T like this what's the first one well well this is just 1
2:44:29
over s square plus 1 squared what's the second one it's just estimate op over s square plus 1 squared final answer s on the top we are multiplying so it's parentheses s square plus 1 and then square that one square is 1 so that's right at the end put on square then how can you do this this is the regular mode two-pager it's not easy but it's also still doable so notice that this is just a sound
2:44:58
right here for you guys when you have sine cosine you can look at the top angle formula sine of 2t this is just to sign cosines right just like that very good now here we have just one front so I can justify deposits by two so here is the deal I will just write it down so this right here it's a same as saying the Laplace transform I would divide it by two so I will have 1 over 2 and then we have
2:45:36
sine of 2t just like that it's the same so use identities sometimes now 1/2 is 1/2 good and then sine of 2t well we do this a few times already yong-ho top right so we have the two on the top and then over s square plus 2 square like that hey guess what listen that can so pretty cool so final answer we get 1 over s square plus 4 cool all right so this is how you could use the convolution and you have to just really make sure if your competition will regular multiplication and all
2:46:28
that good stuff now just two more to go so let's go ahead 23 Laplace transform and now here I have square root of x okay this right here it does have a lot of fast transform and I'll show you guys what the answer is you can use the definition and then you can do some use substitution and then you can use the Gaussian integral and so on so on so on but I will take a shortcut right here right so
2:47:01
change I can just show you how can extend the idea of the factorial so know the following remember we know that the Laplace transform of T to some power n this right here it's equal to well pay attention to what n is this right here is equal to n factorial that divided by as to the plus one power like this and the truth is this right here doesn't have to be just 0 1 2 3 4 and so on you can
2:47:37
actually have a lot more numbers this right here if you like you can have the Laplace transform of T to the Arts power I'm using all right here and we will say this is our factorial over s to the R plus 1 in this case though R has to be greater than negative 1 and what I'm going to do is actually the following not yet I'm sorry all right that's that's that's put atoms are it or has to
2:48:11
be negative 1 greater than negative 1 right so no no no no all right these are not almost identical that for we are going to say we can actually use what we call the gamma function to extend the idea right here so that's what we have and another notice let me just show you this right here oh yeah so this is called a gamma function let me just put this down gum function and it's like hangman of course like this gamma of some input R this right here is equal to the following a plus
2:48:58
is what is integral what gamma it's also integral you have the integral going faster protein nothing I'm not making this up I don't know who did well anyway nobody did it legitimate math so you start with T / stop and then you to R and then you have minus 1 and then you do e to the negative like this so this is what we call the gamma function and the good thing is that you can say the gamma
2:49:28
of and you can also have the PI function but let me just use comma right here the gamma of our plus 1 which is precise you put our plus 1 here and just work that out and you can just say this right here is equal to n factorial R factorial the mere creature spread is done 0 to infinity and let me just you just triggers T to R plus 1 minus 1 just are ya and then e to the negative T DT and again this particular one it's also called a PI function and just cap the PI of our right so
2:50:11
I just don't know it's accounting class the PI but this is not a PI ap it's just a capital PI function and the very nice thing is that you end up with r factorial but is very nice and again in order for this to work you will show technically say or has to be greater than negative 1 ok so just to make you slightly more legit by the way you can do the following our factorial is equal to this so you can register as gamma of R plus 1 and then over as to the R plus 1 like so very nice
2:50:50
right and the best part is that you can use this as your new definition to extend the concept of factorial and you can end up with the following if you have RS equal to 1/2 right then you get 1/2 factorial and again you can just put a 1/2 in here and that's pretty much going to be that that's pretty much how you're going to start with the Laplace transform the square root if you want to do the definition but I have to have a video on this so you can go ahead and check that out
2:51:27
it's a famous result if you have seen me already 1/2 factorial is nicely equal to square root of pi over 2 so this is a very nice result it used the gamma function so with this being said this is how we can do it now I'll just put it down here the Laplace transform of the square root of x sorry no not squared of X I'm not using X I'm sorry I squared yeah this right here of course it's the
2:52:06
same as saying the Laplace transform of T to the one-half power and we can just put a 1/2 into here if you wish so in other world can say this is the 1/2 factorial again 90 you choose cents you have to use a gamma function to actually work that out divided by dou s xx 1/2 plus 1 and now based on what I told you guys right here this right here 1/2 factorial is just nice to equal to square root
2:52:39
of PI over 2 yeah so cool PI over 2 and then s 1/2 plus 1 it needs no introduction is just s to the 3 half like this so this is it yeah this is it so for this one let me just show you guys what the gamma function is and then it actually works out very nicely pretty cool stuff just like that 3 hours not so bad oh my life goes to run a marathon 26.2 miles under 4 hours and my latest one which
2:53:22
is which happened in two weeks ago right there a marathon the meadow I showed you guys last time take a guess my time that take a guess my mirrors on time I post my I post a picture on my Instagram all that so you guys know you know it but if you don't you can take this all right question number 20 what is this really question number 24 and then Laplace transform off I start with an L so
2:53:57
of course now and haha an actual log of T crazy stuff huh so let's have a look this is going to look like the Interpol conference row to infinity e to the negative T and then we have to have the Ellen you might do it a few ways you can do this from scratch with integration like this and working with you working out power to it with the gamma function that I just showed you guys earlier
2:54:41
and man I forgot a comma function it's somewhere anyway that here we go so this is what we know so they mean to us put this down right here and let me actually just start with what we had earlier we know that let me put it down right here any more space it should be okay we know that just I vomited earlier the Laplace transform of T to the art power I will put this down in red why not
2:55:21
I mean black why not so like this right this right here is our factorial over s to the art class one power but our factorials SMS gamma of R plus 1 so Buddhist honest gamma of our plus 1 over s to the R plus 1 isn't it nice and right here of course we can recall the definition of the class transform this right here is the integral going from 0 to infinity e to the negative T and then we have P to
2:55:55
the earth power and then we have and that's equal to this of course so gamma of R plus 1 over s to the R plus 1 power good stuff now compare this and that and to squeeze out an LNG the thing about the relative because with the a temp out really similar to this already right earlier if you want to squeeze out at he we can just differentiate this with respect to s so can squeeze out a G but
2:56:26
now I'll just squeeze out there and T I'm not okay and this part anymore actually have to look at this part because when we have T to the art power if we differentiate this with respect to R then we can squeeze out Ln T and that's very very very nice so let me write it down for you guys then just put on some note right here yeah so recall that if you differentiate your function as a DDX
2:56:58
some number for the base that jet pays to X power if this is the variable then you get B to the x times and B so just keep that in mind so it's tying my I will go ahead and look at this and I will go ahead and differentiate this with respect to R yeah just like that and training for this show put on the parsha OPA I will do that inside anyway so here we go I'm going to differentiate both sides this is the key function in terms of our Isis the constant anyway so when I put a sign
2:57:34
inside that's the time change that partial zero to infinity and then we have the partial with respect to R and then the inside is e to the neck yes t e to the R and then we have the T and then again the right hand side were just go ahead and do our usual derivative let's say I do s is the constant we know already anyway here we have the gamma of R plus 1 over s to our class 1 all right here we
2:58:03
go this is just a constant so it will stay we will have the integral going from 0 to infinity this is staying so it's just e to the negative T yep then literati with respect to R of T we are we get this so I'll put everything down red we get e to the R times Ln P like that and then we have the PT very good now the right hand side just go ahead and do the derivative by using the quotient rule because
2:58:37
I don't the party on the top they both have R so now here's the deal on the bottom I will have to square it so as to the R plus 1 and then I will square that and then on the top I will put on the first function on the bottom function so it's s to the R plus 1 times the derivative of the top and for now I will just write down gamma prime of R plus 1 what's the chendo says the channel says x
2:59:08
dt roof inside but the derivative of our class was just 1 so it doesn't matter then you continue - the top function which is oh man gamma of R plus 1 times the derivative of the bottom with respect to R yeah so you first repeat as to the R plus 1 and times the root of this which is again just 1 so it doesn't matter so this right here is what we have this is what we have now compare this and that
2:59:42
what can we do I can just make R equal to 0 don't want your by yet yeah so I would just say that R equal to 0 in that case the left hand side give us the integral from 0 to infinity just e to the negative as T and then Ln T DT which is exactly what we are trying to get now on the bottom here we have s to the 0 plus 1 to s 1 and then to a second power oh no we have s square on the bottom
3:00:16
good put your own here we have s to the first good and then we have the the relative of the gamma function at 1 and then minus gamma 1 oops just make it one more kiss tick and then over n times s to the first power like this so that's pretty good now what's gamma prime of 1 and what's gamma 1 okay so let's see here note so remember gamma off and because we're talking about functions so
3:00:59
perhaps I'll talk about X yeah our wiper actually that's just put the arc because that's what we do so gamma of our car yeah earlier though we said gamma of R plus 1 is equal to R factorial right and you can see that I can just push the Rho into that so this is actually pretty that's that's to this first no let's just do this first why not sometimes anyway I'm sorry note gamma of R the definition of gamma is going to be the integral from 0 to infinity T to the R minus 1 and then oh
3:01:46
yeah and then e to the negative T DT like this all right so this is what we have a belief that's yes good all right so gamma of 1 in fact this is how we can argue that this is zero factorial how can a good stuff actually one you can put on one right here this is integral from 0 to infinity T to the 1 minus 1 power so that will be 0 actually and then you do e to the negative power and you can do this in your head integrating e to a negative Y because this is just
3:02:30
one anyway and then you could negative e to the negative T plus infinity plug in 0 worked out you actually end up one so this right here is nice - equal to one and in fact that zero factorial now let's look at a gamma prime of R so to do that I would actually have to do the derivative right here and again we will technically how to differentiate this with respect to R so use the like nice rule right so we actually end up with the following so the action went down
3:03:06
here so gamma prime of R equals okay still have the integral going from 0 to infinity this right here stays because teachers are constant in the art world we are doing this with respect to art so you have this and we did that earlier so if you depreciate that you actually get this repeat which is T to the R minus 1 times L and T you actually have this and then you have this right here stays which is e to the negative T like this right so if you have the gamma prime of
3:03:47
1 when R is equal to 1 you will end up with the following put a 1 in here G to the 0 is just 1 so it's nothing there and you actually get integral from 0 to infinity I'm so sorry gamma prime of 1 this right here is the integral going from 0 to infinity this right here is just hit with 0 which is 1 and you end up with L and G times e to the negative T and you just want to know what the answer to this is this right here so cool this is also
3:04:25
called gamma this is the oldest constant gamma right and let me just show you this is approximately 0.577 and in fact I'm not going to do this integral right here for you guys I'm sorry but of course you guys can go ahead and check out dr. PI and check out dr. Kyon's video I will have this video in the description for your convenience so although this right here it's actually a special number called gamma if you haven't seen
3:05:04
that before very cool stuff so right here I will actually come here and tell you this right here is equal to gamma like that very very good stuff right and that's pretty much it I'm afraid I'm making a small-time mistake let's see okay I should have
3:05:42
hmm
3:05:48
Ahana this rainy should be negative karma akarma that's see see this is what I have to look at doctor pants video sometimes that all right let's see if it makes sense if you have zero this is negative infinity times one so it's negative and then went to infinity does zero so when he is one that would be zero so you have negative part and then you have some not so big may pass the parcel I believe okay I believe this right here should be negative gamma otherwise it won't work so it
3:06:36
has to be negative gamma let me just double check to see if I have any mistakes or not yes I really believe this ratio pian negative gamma and make a gamma can we see okay so this negative gamma is just like a constant just like pi water II all right or Euler's constant yeah and I will come back here and then actually this is negative gamma right finally ladies and gentlemen as you can see the left-hand side is equal to the hands of obviously but of course we have the s
3:07:18
s then s we can cancel them out so you can see that on the top you have negative gamma minus 1 over cancelled out so you have s like that oh one more thing I forgot I'm so sorry I'm very sorry I'm very sorry all right so you have all this to little row but if that's too easy with quotient rule the bottom function square and the top is the bottom function times the derivative of the top minus the top function times DT really at the bottom derivative of s to the R plus 1 with
3:07:59
respect to R as our plus one times Ln s times s why is that what we did earlier test just about with you are you so in firms re this right here which will have another L&S after that yes so I'm missing that the own s well anyway the rest is OK right here and again this s dies and that s quick can reduce you finally care negative gamma minus 1 Ln s sauce - yes all for s side uh-huh so already done negative gamma minus 1 s oo / s like this tan tan tan tan tan so done
3:08:46
yes not so bad all right so this right here is it $24 transforms for you guys right so that's pretty much it and in fact I will have to ask you guys the following now so problem just as we got this so I leave this to you guys question for you guys this is again the Honorable Mention question what's the Laplace transform of 1 minus cosine T over T right what's Laplace transform of that and then after this can you also find me the integral going from 0 to infinity of
3:09:26
1 minus cosine T over T e to the T all right so please do that and let's see if you guys can and hopefully this right here 3 T negative gamma and if so that would be great all right so that's it and hopefully you guys all at this and by the way I should say s is greater than 0 somewhere but if because otherwise you will converge fine so anyway this right here is it and as usual I wanted to put on my marathon medal this is the one I got from 2017 this register
3:10:13
both sites by sketcher in my hand it's dirty very dirty dirty dirty right okay so and you're stuck every time I put on my metal so that's it hopefully you just find this video to be helpful and yeah let me know how you guys to doing other stuff which you get the best which you guys safe and healthy right so as always that's it
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