you explained two hours worth of lecture from my professor in 10 minutes. Amazing video
@ov3rkill
5 жыл бұрын
Can we all just agree and appreciate his pen switch skills while writing besides of course his math skills.
@bizzle9041
3 жыл бұрын
Yes we can
@taehokang2551
3 жыл бұрын
Couldn’t focus cuz of that mesmerising skilllllll
@federicopagano6590
5 жыл бұрын
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) ✔
@chinmayabehera8961
6 жыл бұрын
cz its a definite integral we don't use c here .....your the best tutor I have ever seen....THANKS
@blackpenredpen
6 жыл бұрын
chinmaya behera I am an instructor who looks young. :)
@password6975
7 жыл бұрын
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.
@djalixos4008
7 жыл бұрын
i really like your way of explaining,a big salute from Morocco ☆
@aleksgornik
2 жыл бұрын
i dont think you know how many engineering students your saving
@e.s.r5809
2 жыл бұрын
It's definitely cooler than partial fractions-- but the coolest part is not having to do partial fractions. 😉 Thank you!
@Guru_Joe_Praise2023
5 ай бұрын
In just a few minutes I have understood Convolution theorem of Laplace Transform well done sir👍
@nurten5903
Жыл бұрын
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!
@sergiorome48
2 жыл бұрын
I love these examples, so clear, so simple, so beautiful
@new-jj5il
4 жыл бұрын
Excellent explanation Without neglecting any small mathematical step... Thanks
@jobalfred9603
3 жыл бұрын
GREAT TUTOR.Youve made understand this concepts better.Highly recommend.
@bench9118
5 жыл бұрын
if i just saw this before my exam, i would have got a perfect score........ nice bro...
@skwbusaidi
5 жыл бұрын
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
@utkarshanayak1710
3 жыл бұрын
Never heard of Convolution theorem. But you explained so easily 🙏🙏❤️. Thanks #blackpenredpen Btw no +C since it is a definite integral 😎
@johnkarlorcajada3147
7 жыл бұрын
No +C because it is a definite integral.
@blackpenredpen
7 жыл бұрын
LOL! YUP!!!!
@sugarfrosted2005
6 жыл бұрын
If you want to be pedantic, it's there, it's just eaten when you substitute the end points in.
@AlgyCuber
6 жыл бұрын
and it’s an inverse laplace not an inverse derivative
@ayamohamed7468
Жыл бұрын
I have watched this video before my exam and this exact example has come and I quickly remembered watching this video .. Thank you!
@JeffNkwilimba
Жыл бұрын
I will consider you in my research report you have helped me alot in terms of calculus at the University, thanks very much ❤❤❤❤
@mariogabrielsalvatierrafra4500
7 жыл бұрын
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
@maneeshkoru2312
2 жыл бұрын
You can also use L[t.f(t)]=-dF(s)/ds, complex differentiation theorem.
@60_co_ayeshashaikh10
2 жыл бұрын
Well explained and also ur skill of switching pens is amazing, thank you for lecture😍
@parthokr
2 жыл бұрын
You are so happy when you found sin(t) as constant in v world. It made me happy too.
@stephanm.tjaden3887
5 жыл бұрын
You are awesome! You take something that seems so complicated and make it very , very easy to understand.
@Sednas
Жыл бұрын
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.
@chapahewawasam1222
3 жыл бұрын
Amazing teaching skill. Thank u so much ♥️
@levialviter2302
3 жыл бұрын
Thx a lot. You've just saved me. Stay smart.
@algion24
3 жыл бұрын
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
10 ай бұрын
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)
@FutballFocusTV
6 жыл бұрын
a big salute from Berkeley ,CA keep the good work
@blackpenredpen
6 жыл бұрын
Glad to help!
@FutballFocusTV
6 жыл бұрын
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
6 жыл бұрын
blackpenredpen my email is futtaingrp40@gmail.com
@ANANDYADAV-sc1se
2 жыл бұрын
Cool explanation , I like it
@izuchukwuokafor8130
Жыл бұрын
You are Superb Sir Blessings from the most high
@hoon8768
5 жыл бұрын
Thank you very very much!!!! From south korea
@mrinmoybhaduri9666
5 жыл бұрын
You explained it so good iam from india, thnks
@mariogabrielsalvatierrafra4500
6 жыл бұрын
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
@a.s.l711
2 ай бұрын
just how does the laplace work from 1/(s+1) becomes e^-t. what is the logic behind the conversion.
@sunset1394
5 жыл бұрын
exam tommorow and here he comes with his black pen and red pen and saves my day.
@jarikosonen4079
4 жыл бұрын
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?
@eseranceese9305
3 жыл бұрын
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
3 жыл бұрын
Uhm also what if the s on the top of it squared? Like s²/(s²+1)²?
@carultch
10 ай бұрын
@@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)
@carlos199613ful
6 жыл бұрын
Greetings from Honduras! Youre a genuis man !
@jun6003
7 жыл бұрын
thanks! it's very helpful !!
@ZARA_KEYS
3 ай бұрын
Well teaching.... Anybody in 2024??
@blackpenredpen
3 ай бұрын
Thanks!!!
@darcash1738
5 ай бұрын
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 😂
@liyanaminaj2309
2 жыл бұрын
"don't be lazy", it get me LOL
@robinrotich118
4 жыл бұрын
waauh amazing math lesson i have understood everything on convolution
@demenion3521
7 жыл бұрын
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?
@ShinSeokWoo
5 ай бұрын
Thank you exponentially !
@BK-dx3cp
2 жыл бұрын
He’s a great tutor!!
@xongram3139
4 жыл бұрын
Thank you so much....it was a 10marks question in my exam
@holyshit922
2 жыл бұрын
3:14 , integration by parts will work if you expand sin(t-v) to get two integrals
@holyshit922
7 жыл бұрын
Complex partial fraction will work We could also use differentiation
@blackpenredpen
7 жыл бұрын
yup
@andymorejon2am
6 жыл бұрын
This guy is funny af, congrats on your talent
@teo97judo
5 жыл бұрын
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
5 жыл бұрын
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)).
@santiagocas3683
4 жыл бұрын
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
10 ай бұрын
It is completely arbitrary which one gets v, and which one gets t-v, since convolution is commutative.
@emmanueljoseph8540
24 күн бұрын
Can UNIBEN STUDENTS gather here and sign attendance
@SuHAibLOL
7 жыл бұрын
integral transforms are just great
@blackpenredpen
7 жыл бұрын
Yes!
@thommythomas3123
2 жыл бұрын
good explaination
@daynaladd8894
5 жыл бұрын
Wow amazing! Thank you so much!
@himanshu11876
6 жыл бұрын
2nd method #dis function is derivative of (S^2+1)^-1,so inverse function would be multipled by t
@anishachoudhury_
6 жыл бұрын
Can u just explaim why did he use sin t-v instead of sint in 4th step
@NoobMaster-yw6eo
5 жыл бұрын
@@anishachoudhury_ it's just how the convolution is done
@NoobMaster-yw6eo
5 жыл бұрын
Hey could u write down how would u use that method on this is function cuz am kinda lost with it
@ernestamoah2612
Жыл бұрын
Thank you sir.
@ChickenJY
6 жыл бұрын
Prof, may I request a Fourier Transform/ Inverse FT videos from you ?
@منوعات-ح8ع
5 жыл бұрын
Amazing bro Thanks a lot
@queenqueen4662
6 жыл бұрын
Thank u so much sir this video helps me a lot 🙏🙏
@solinothman4094
5 жыл бұрын
I love doing math with your videos You're Amazing ❤
@suzeetasuzee9018
6 жыл бұрын
can u please upload more examples of convolution theorem.......
@IrfanNasir
6 жыл бұрын
Thank you very much sir
@manishmodak1726
5 жыл бұрын
Do it for the minus sign too without using the convolution theorem please
@maayoufamoez2217
5 жыл бұрын
all maths is here in this equation. good example thank you but i like the way you play with pens :) :)
@DedSec_7
5 жыл бұрын
But there's formula L^-1[s/(s^2+a^2)^2]=1/2a t sinat Why don't use this directly?
@blackpenredpen
5 жыл бұрын
Care to prove that formula?
@gaminghub0615
4 ай бұрын
How to do this without using convolution thm
@user-df9hk7hs1l
5 ай бұрын
Integrate denominator and take inverse to the result to get F(s) and the result is - d/ds F(s)
@sushantlakra6715
6 жыл бұрын
excellent sir ...
@flickboxextra3127
Жыл бұрын
Very good
@jerryjin5871
4 жыл бұрын
That was amazing!
@helldogforever
6 жыл бұрын
Your video helped.
@vidyatarani2554
6 жыл бұрын
Can you plz tell this question by partial fraction method . Is it possible to do this question in this method
@carultch
Жыл бұрын
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.
@fadyfahmyful
6 жыл бұрын
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
6 жыл бұрын
because we are integrating wrt v, so whatever t is just a constant, the first sin involve v so we must use cos
@angus8147
4 жыл бұрын
@@changdagong3305 you save my final term exam
@Grundini91
6 жыл бұрын
It's a definite integral, no +c
@Novak2611
3 жыл бұрын
One can simply use the derivative of Laplace transform of cos: L'(f)=-L(tf) (i am not talking about Laplace of derivative)
@mr.hridoy245
6 жыл бұрын
the lovely way to do this math, i like your your way to solve,thank you sir
@raudafaye1945
4 жыл бұрын
please sir why did u use cost???
@blackpenredpen
4 жыл бұрын
Because that’s the correct answer
@enricoperrotta5676
2 жыл бұрын
Awesome
@sydbugnano8431
3 жыл бұрын
I love how much he loves math
@adityapahalvan6484
5 жыл бұрын
Sir make the video on complex integration,contour
@hungmai7533
Жыл бұрын
thank you so much sir
@김태완-j9e
2 жыл бұрын
Is there any way not to ise Convolution theorem??
@김태완-j9e
2 жыл бұрын
Not to use
@carultch
10 ай бұрын
@@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
5 жыл бұрын
You're awesome dude
@nickdelligatti3712
6 жыл бұрын
are you wearing Supreme?
@blackpenredpen
6 жыл бұрын
: )
@mammu3635
Жыл бұрын
Find L inverse [s/(s+4)²] plzz answer for this sir I have exams this month plzz 🙏 sir
@carultch
10 ай бұрын
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)
@ronaldrosete4086
4 жыл бұрын
By the way, what's the Laplace of Y(s)/X(s)?
@carultch
10 ай бұрын
It's called the transfer function. The function that relates the output to the input of a linear time independent dynamic system, assuming X(s) represents the Laplace transform of the input, and Y(s) represents the Laplace transform of the output.
@helloitsme7553
7 жыл бұрын
no +c, cause any function has a unique La place transform and any La place transform belongs to a unique function
@blackpenredpen
7 жыл бұрын
yea!
@mutalejohn5295
Жыл бұрын
Thank you!
@saideepak3702
5 жыл бұрын
what is L inverse of 1/s(s+1)(s+2)???
@carultch
10 ай бұрын
Given: 1/(s*(s+1)*(s+2)) Use Heaviside coverup to find partial fractions: A/s + B/(s + 1) + C/(s + 2) At s = 0, A = 1/(1*2) = 1/2 At s = -1: B = 1/(-1*(-1 + 2)) = -1 At s=-2: C = 1/(-2*(-2 - 1)) = 1/2 Thus it is: 1/2/s + -1/(s + 1) + 1/2/(s + 2) Inverrt to get: 1/2 - e^(-t) + 1/2*e^(-2*t)
@diegonavia1404
6 жыл бұрын
wena hermano greetings from chile
@Yt32624
6 жыл бұрын
Instead of 1 its..... s/(s^2+4)... answer will be the same???
@carultch
Жыл бұрын
For 1/(s^2 + 4), the inverse laplace is 1/2*sin(2*t). For s/(s^2 + 4), the inverse laplace is cos(2*t) Cosine starts at the top, and has s at the top. Sine has the constant on the top.
@richellemaebaguasan3553
5 жыл бұрын
Amazing!
@blackpenredpen
5 жыл бұрын
Thank you!
@mortezamodarres2470
6 жыл бұрын
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
5 жыл бұрын
it's actually the derivative of (-1/2)/(s^2 + 1).
@carultch
8 ай бұрын
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.
@mojahedhamayel9870
4 жыл бұрын
You can tell me what equal it sin A . cosB = cos B . sin A = sin A . sin B = cos A . cos B =
@kayoderilwan9864
6 жыл бұрын
Please I tried solving using partial fraction but I was unable to do so.. Please I want to know if its possible... If yes. please solve it cause I couldn't... Thanks...
@carultch
10 ай бұрын
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.
@rahulgautam-ux9qx
4 жыл бұрын
inverse laplace transform of 1/(s^2+1)^2 please tell me solution this type of question.
@vasilisa4723
3 жыл бұрын
Hi, have you found the solution? I'm still searching it :)
@carultch
10 ай бұрын
@@vasilisa4723 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.
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