inverse laplace of s/(s^2+1)^2, using convolution theorem

inverse laplace of s/(s^2+1)^2 using convolution theorem,
use convolution theorem to find inverse laplace transform,
blackpenredpen

[@DrLol07]

you explained two hours worth of lecture from my professor in 10 minutes. Amazing video

[@ov3rkill]

Can we all just agree and appreciate his pen switch skills while writing besides of course his math skills.

[@bizzle9041]

Yes we can

[@taehokang2551]

Couldn’t focus cuz of that mesmerising skilllllll

[@aleksgornik]

i dont think you know how many engineering students your saving

[@federicopagano6590]

We shouldn't put + C for many reasons 1°) it's a definite integral 2°) the inverse laplace transform always it's a one to one operator unique result 3°) we could think this problem like the L^-1[L{ -1/2(L {sin (t)} }'s ]= - t .f (t) then f=1/2 (sen (t)/t) ✔

[@djalixos4008]

i really like your way of explaining,a big salute from Morocco ☆

[@e.s.r5809]

It's definitely cooler than partial fractions-- but the coolest part is not having to do partial fractions. 😉 Thank you!

[@johnkarlorcajada3147]

No +C because it is a definite integral.

[@blackpenredpen]

LOL! YUP!!!!

[@sugarfrosted2005]

If you want to be pedantic, it's there, it's just eaten when you substitute the end points in.

[@AlgyCuber]

and it’s an inverse laplace not an inverse derivative

[@chinmayabehera8961]

cz its a definite integral we don't use c here .....your the best tutor I have ever seen....THANKS

[@blackpenredpen]

chinmaya behera I am an instructor who looks young. :)

[@password6975]

i learn so much from your videos, not only do they help me with the subject you are teaching, but also with my math understanding in general, you are a great teacher.

[@utkarshanayak1710]

Never heard of Convolution theorem. But you explained so easily 🙏🙏❤️. Thanks #blackpenredpen Btw no +C since it is a definite integral 😎

[@bench9118]

if i just saw this before my exam, i would have got a perfect score........ nice bro...

[@skwbusaidi]

Good . Also we can also use the fact that laplace of t f(t) = -d/ds ( F(s)) We can take f(t) = 1/2 sint

[@sergiorome48]

I love these examples, so clear, so simple, so beautiful

[@Guru_Joe_Praise2023]

In just a few minutes I have understood Convolution theorem of Laplace Transform well done sir👍

[@parthokr]

You are so happy when you found sin(t) as constant in v world. It made me happy too.

[@new-jj5il]

Excellent explanation Without neglecting any small mathematical step... Thanks

[@nurten5903]

I was struggling with a question in my textbook and decided to watch solution to another question, find this video and realized that you solved the same exact question that I was struggling with! I'm so happy thank you!

[@jobalfred9603]

GREAT TUTOR.Youve made understand this concepts better.Highly recommend.

[@maneeshkoru2312]

You can also use L[t.f(t)]=-dF(s)/ds, complex differentiation theorem.

[@ayamohamed7468]

I have watched this video before my exam and this exact example has come and I quickly remembered watching this video .. Thank you!

[@mariogabrielsalvatierrafra4500]

SUCH A GREAT EXAMPLE OF AN INVERSE LAPLACE TRANSFORMATION, I hope you can do examples of diferential ecuaions with the special functions like delta of dirac and others, i mean diferential ecuations where you have to use laplace transformation, keep doing those videos, they are so great

[@Sednas]

no, you do not need to put down +C, and that was hilarious 😂😂😂. I love your videos they are so useful but also funny sometimes.

[@a.s.l711]

just how does the laplace work from 1/(s+1) becomes e^-t. what is the logic behind the conversion.

[@JeffNkwilimba]

I will consider you in my research report you have helped me alot in terms of calculus at the University, thanks very much ❤❤❤❤

[@60_co_ayeshashaikh10]

Well explained and also ur skill of switching pens is amazing, thank you for lecture😍

[@darcash1738]

Just watching this for fun, seems pretty cool. can someone explain the step from sint * cost? why does the argument of sin become "t-v", whereas for cos it becomes simply "v"? and perhaps i should know what is being convoluted in this convolution 😂

[@stephanm.tjaden3887]

You are awesome! You take something that seems so complicated and make it very , very easy to understand.

[@sunset1394]

exam tommorow and here he comes with his black pen and red pen and saves my day.

[@algion24]

An easier way to evaluate the convolution let I = sin(t)*cos(t) = int 0 to t (sin(t-v)cos(v))dv since convolution is commutative I = int 0 to t (cos(t-v)sin(v))dv add the 2 together 2I = int 0 to t (sin(t-v)cos(v)+cos(t-v)sin(v))dv this becomes an angle sum formula for sin 2I = int 0 to t (sin(t-v+v))dv = int 0 to t (sin(t))dv = vsin(t) from 0 to t = tsin(t) - 0sin(t) = tsin(t) divide both sides by 2 I = tsin(t)/2

[@carultch]

An easier way to evaluate it without using convolution. Use the s-derivative property of the Laplace transform, where L{t*f(t)} = -d/ds F(s). Take the s-derivative of sine's Laplace transform: d/ds 1/(s^2 + 1) = -2*s/(s^2 + 1)^2 Therefore: L{t*sin(t)} = 2*s/(s^2 + 1)^2 Multiply both sides by 1/2: 1/2*L{t*sin(t)} = s/(s^2 + 1)^2 Recognize the original transform we're trying to invert in the above. Thus: L-1 {s/(s^2 + 1)^2} = 1/2*t*sin(t)

[@ANANDYADAV-sc1se]

Cool explanation , I like it

[@izuchukwuokafor8130]

You are Superb Sir Blessings from the most high

[@mariogabrielsalvatierrafra4500]

Such a great video and no we don't need to put +C because the convolution is a definite integral so the C is not necesary

[@ZARA_KEYS]

Well teaching.... Anybody in 2024??

[@blackpenredpen]

Thanks!!!

[@liyanaminaj2309]

"don't be lazy", it get me LOL

[@levialviter2302]

Thx a lot. You've just saved me. Stay smart.

[@jarikosonen4079]

The plus C could be allowed, but might depend on the initial conditions. The problem is that how to make laplace of sin(x)+2 ? It doesn't work maybe. Then if inverse laplace should give sin(x)+2, how to? This method seems to work, but requires a lot of calculation. Basically signals can be transferred both on time and signal axes. sin(x) becoming sin(x-t0)+C if transferred in both axes. One could use sin(x)+2×u(t), but that wouldn't be same as sin(x)+2, except for t>=0. If Laplace valid for t>=0 only it seems maybe also reason why u(t) should be used instead. But one could add just 2 if the offset of 2 is certain. The calculation without offset is easier maybe and doing things around the 0 instead of 2 is more practical in this math. Like solving x^2 + 3x + 1 = 0 rather than x^2 + 3x + 3 = 2. You could tune the 2nd degree polynomial equation solving to work with right side =2 rather than =0, but it would get more complicated. Can you calculate inverse laplace transform of s^2/(s^2 + w^2) as an example case also?

[@chapahewawasam1222]

Amazing teaching skill. Thank u so much ♥️

[@FutballFocusTV]

a big salute from Berkeley ,CA keep the good work

[@blackpenredpen]

Glad to help!

[@FutballFocusTV]

blackpenredpen . Thank you sir , i got this question as well . Use convolution theorem to find the inverse laplace transform of the following. f(s) = 1/s+p)(s+q) Do you have an email ?

[@FutballFocusTV]

blackpenredpen my email is [e-mail]

[@ShinSeokWoo]

Thank you exponentially !

[@demenion3521]

you should probably add an argument after the laplace transform like L{f(t)}(s). otherwise one has always to remember which variables you use normally. And also: i am used to the convolution over the whole reals. is it actually the same thing as the integral from 0 to t?

[@mrinmoybhaduri9666]

You explained it so good iam from india, thnks

[@robinrotich118]

waauh amazing math lesson i have understood everything on convolution

[@SuHAibLOL]

integral transforms are just great

[@blackpenredpen]

Yes!

[@holyshit922]

3:14 , integration by parts will work if you expand sin(t-v) to get two integrals

[@hoon8768]

Thank you very very much!!!! From south korea

[@jun6003]

thanks! it's very helpful !!

[@eseranceese9305]

Thanks! This is very helpful!. Can i ask what if the equation is inverse laplace of [ 1/(s²-9)²] is it using convolation to solve it?

[@eseranceese9305]

Uhm also what if the s on the top of it squared? Like s²/(s²+1)²?

[@carultch]

​@@eseranceese9305 A method I recommend, is to assume it is an arbitrary linear combination of t*sin(3*t), t*cos(3*t), sin(3*t), and cos(3*t). Then take the Laplace transforms of each of component function, using the s-derivative property of Laplace transforms. Set up the linear combination with undetermined coefficients, and use algebra to solve for them. L{cos(3*t)} = s/(s^2 + 9) L{sin(3*t)} = 3/(s^2 + 9) L{t*cos(3*t)} = -d/ds L{cos(3*t)}= (s^2 - 9)/(s^2 + 9)^2 L{t*sin(3*t)} = -d/ds L{sin(3*t)}= 6*s/(s^2 + 9)^2 Let the Laplace transform we're trying to invert, equal the following, and solve for A, B, C, and D: A*s/(s^2 + 9) + 3*B/(s^2 + 9) + C*(s^2 - 9)/(s^2 + 9)^2 + 6*D*s/(s^2 + 9)^2 For 1/(s^2 + 9)^2: 1/(s^2 + 9)^2 = A*s/(s^2 + 9) + 3*B/(s^2 + 9) + C*(s^2 - 9)/(s^2 + 9)^2 + 6*D*s/(s^2 + 9)^2 1 = A*s*(s^2 + 9) + 3*B*(s^2 + 9) + C*(s^2 - 9) + 6*D*s 1 = A*s^3 + 9*A*s + 3*B*s^2 + 27*B + C*s^2 - 9*C + 6*D*s A = 0 3*B + C = 0 27*B - 9*C = 1 D = 0 Solution for B&C: B = 1/54, C=-1/18 Thus: inverse Laplace of 1/(s^2 + 9)^2 = 1/54*sin(3*t) - 1/18*t*cos(3*t)

[@ernestamoah2612]

Thank you sir.

[@carlos199613ful]

Greetings from Honduras! Youre a genuis man !

[@xongram3139]

Thank you so much....it was a 10marks question in my exam

[@andymorejon2am]

This guy is funny af, congrats on your talent

[@thommythomas3123]

good explaination

[@BK-dx3cp]

He’s a great tutor!!

[@holyshit922]

Complex partial fraction will work We could also use differentiation

[@blackpenredpen]

yup

[@gaminghub0615]

How to do this without using convolution thm

[@santiagocas3683]

Broo, I got a big question, how to know, where put, t-v, for expmle, cos(t-v)sin(v), ¿Cómo sabes donde poner (t-v), ?¿ Podría haber sido cos(t-v)sin(v)? Disculpa el inglés, no soy nativo.

[@carultch]

It is completely arbitrary which one gets v, and which one gets t-v, since convolution is commutative.

[@DedSec_7]

But there's formula L^-1[s/(s^2+a^2)^2]=1/2a t sinat Why don't use this directly?

[@blackpenredpen]

Care to prove that formula?

[@HelloBangladesh389]

Thank you very much sir

[@ChickenJY]

Prof, may I request a Fourier Transform/ Inverse FT videos from you ?

[@himanshu11876]

2nd method #dis function is derivative of (S^2+1)^-1,so inverse function would be multipled by t

[@anishachoudhury_]

Can u just explaim why did he use sin t-v instead of sint in 4th step

[@NoobMaster-yw6eo]

@@anishachoudhury_ it's just how the convolution is done

[@NoobMaster-yw6eo]

Hey could u write down how would u use that method on this is function cuz am kinda lost with it

[@flickboxextra3127]

Very good

[@queenqueen4662]

Thank u so much sir this video helps me a lot 🙏🙏

[@Grundini91]

It's a definite integral, no +c

[@sydbugnano8431]

I love how much he loves math

[@solinothman4094]

I love doing math with your videos You're Amazing ❤

[@الاستاذأزهرمحمدماجستيرعلومرياض]

Integrate denominator and take inverse to the result to get F(s) and the result is - d/ds F(s)

[@Novak2611]

One can simply use the derivative of Laplace transform of cos: L'(f)=-L(tf) (i am not talking about Laplace of derivative)

[@hungmai7533]

thank you so much sir

[@manishmodak1726]

Do it for the minus sign too without using the convolution theorem please

[@daynaladd8894]

Wow amazing! Thank you so much!

[@maayoufamoez2217]

all maths is here in this equation. good example thank you but i like the way you play with pens :) :)

[@teo97judo]

Hello Steve, my name is Teo and I come all the way from Greece. I was trying to solve an inverse Laplace and googled for help so I stumbled upon this video.The problem is I can't exactly understand how to use this method on my problem. The problem is inverse Laplace of (s+3)/(s^2 + 4)^2 . I tried the partial fraction method as well but it seems that it can't be divided. I'd love to hear back from you with some help because I'm sure it will take you 5 minutes to solve it. Thank you for your time anyway.

[@DrQuatsch]

The first thing that comes to mind is separating it. L^-1{s/(s^2 + 4)^2} + 3L^-1{1/(s^2 + 4)^2}. The first part is pretty much the same as in the video, but you only have s^2 + 2^2 in the denominator, so that results in tsin(2t)/4. There's an extra factor of 1/2, because you have to match the 2 on the top for the sine part. For the other fraction you will have to calculate a convolution of two sines: 3(L^-1{1/(s^2 + 4)} * L^-1{1/(s^2 + 4)}) = 3/4(L^-1{2/(s^2 + 2^2)} * L^-1{2/(s^2 + 2^2}) = 3/4(sin(2t) * sin(2t)).

[@emmanueljoseph8540]

Can UNIBEN STUDENTS gather here and sign attendance

[@sushantlakra6715]

excellent sir ...

[@منوعات-ح8ع]

Amazing bro Thanks a lot

[@helloitsme7553]

no +c, cause any function has a unique La place transform and any La place transform belongs to a unique function

[@blackpenredpen]

yea!

[@suzeetasuzee9018]

can u please upload more examples of convolution theorem.......

[@mortezamodarres2470]

This the worst way to do this. The whole purpose of laplace transform is to avoid convolution. S/(s^2+1)^2 is the derivative of (1/2)/(s^2+1). The ILT of this expression is (1/2)sint. therefore the ILT of the original expression will be (1/2)tsintn

[@DrQuatsch]

it's actually the derivative of (-1/2)/(s^2 + 1).

[@carultch]

There is a way in general to avoid convolution, to carry out the inverse Laplace of: p(s)/(s^2 + w^2)^n Where p(s) is any polynomial, w is any real constant, and n is any integer constant. Assume the solution is a linear combination of t^k*sin(w*t) + t^k*cos(w*t), with all possible k-values from 0 to (n-1). Then construct a linear combination of corresponding Laplace transforms, each with an undetermined coefficient, to solve for algebraically. If p(s) only contains odd powers of s, then your solution will be an even function. Likewise, if p(s) only contains even powers of s, then your solution will be an odd function. These inspections allow you to eliminate half your terms, and simplify the number of coefficients to solve for.

[@mutalejohn5295]

Thank you!

[@fadyfahmyful]

why the integral of the second sin(t) is -cos(t) , I mean you did not use the same rules for both integrals of sin?

[@changdagong3305]

because we are integrating wrt v, so whatever t is just a constant, the first sin involve v so we must use cos

[@angus8147]

@@changdagong3305 you save my final term exam

[@enricoperrotta5676]

Awesome

[@john-athancrow4169]

I heard sin instead of cos and then I said, «Oh, it's COsine!»

[@helldogforever]

Your video helped.

[@harvindyadav862]

s/(s^2+a^2)^2= (-0.5)*d/ds{1/(s^2+a^2)} Now apply derivative formula.

[@mr.hridoy245]

the lovely way to do this math, i like your your way to solve,thank you sir

[@vidyatarani2554]

Can you plz tell this question by partial fraction method . Is it possible to do this question in this method

[@carultch]

Partial fractions won't help you here, because you already have the denominator as reduced as possible. Unless you explore complex roots of the denominator.

[@jerryjin5871]

That was amazing!

[@mojahedhamayel9870]

You can tell me what equal it sin A . cosB = cos B . sin A = sin A . sin B = cos A . cos B =

[@mammu3635]

Find L inverse [s/(s+4)²] plzz answer for this sir I have exams this month plzz 🙏 sir

[@carultch]

Given: s/(s + 4)^2 Add zero in a fancy way, to form a term we can cancel: (s + 4 - 4)/(s + 4)^2 = (s + 4)/(s + 4)^2 - 4/(s + 4)^2 = 1/(s + 4) - 4/(s + 4)^2 The first term inverts as e^(-4*t). The second term requires us to use the s-derivative property to unpack it. In general, L{t*f(t)} = -d/ds L{f(t)}. The expression we have to unpack, is related to the s-derivative of 1/(s + 4), so differentiate this: -d/ds 1/(s + 4) = 1/(s + 4)^2 Thus: L-1{-4/(s+ 4)^2} = -4*t*e^(-4*t) Thus, the solution is: e^(-4*t) - 4*t*e^(-4*t)

[@adityapahalvan6484]

Sir make the video on complex integration,contour

[@raudafaye1945]

please sir why did u use cost???

[@blackpenredpen]

Because that’s the correct answer

[@ibrahimadiallo9427]

good

[@김태완-j9e]

Is there any way not to ise Convolution theorem??

[@김태완-j9e]

Not to use

[@carultch]

@@김태완-j9e You can take s-derivatives of the Laplace of sine and cosine, and use the s-derivative property to find the Laplace transform of t*sin(t) and t*cos(t). Then set up a linear combination of A*cos(t) + B*sin(t) + C*t*cos(t) + D*t*sin(t). Solve for A, B, C, and D.

[@andremiller482]

You're awesome dude

[@hashem4287]

Thank you very much

[@amirrezabehzadi697]

🤯🤯🤯

[@shreetimohapatra4142]

thank uuu

[@richellemaebaguasan3553]

Amazing!

[@blackpenredpen]

Thank you!

[@diegonavia1404]

wena hermano greetings from chile

[@manishtechnicalclasses3340]

Thank full video of the poly semester exam