Proof of the Convolution Theorem
ฝัง
- เผยแพร่เมื่อ 27 เม.ย. 2017
- Proof of the Convolution Theorem,
The Laplace Transform of a convolution is the product of the Laplace Transforms, changing order of the double integral, proving the convolution theorem,
www.blackpenredpen.com
this guy is a genius.we spent more than two weeks in lectures studying laplaces but i couldnt understand anything...the moment i watched videos from this guy hahahaha i bet im now a genius..lol...thanks lots
Munashe Kaks great thanks
So true !
I know, right? The same genius brain brain that would have realized enlightenment if he had chosen a monastic path is being used to realise mathematical equations. He is also a goo teacher. I just subscribed, maybe he has more courses coming up.
@@univuniveral9713 Brain brain u wrote 2 times
I have been trying to prove this theorem myself for about a month now because I could honestly not find proof anywhere and just seeing this video in my recommended made me smile so widely I already smashed that subscribe button. I'll be watching your videos from now on my dude. :D
Man, what a huge video... Great work! But I found it a lot simpler going "backwards", using the subtitution theorem for multivariable integration and changing the integrating region from (R+)^2 to {(x,y) | 0
Thank you! You've explained this so clearly.
probably the best channel to understand laplace thank u sir.
Such an elegant proof of Convolution theorem using the unit step function.✨️
please do more differential equations. I took that class and learned a lot in a small amount of time, and now its all gone. we spent like two weeks on all of laplace, convolution, dirac's delta function, and a whole lot of stuff that required much more than two weeks to get into.
Jerry Gutierrez will do!
Also, I finished watching your video, and I love the explanation. I was always catching up in my differential eq's class and never fully understood why the teacher did what she did, and after watching this video, I feel like the laplace of the convolution, actually makes sense, and like you, I can appreciate how something so nasty like laplace and convolution combine to make something succinct, and pretty neat to look at and think about.
You are absolutely amazing!
Omg blackpenredpen has a video about this. This guy covered so many topics
This is pure gold! This guy is a genius.
May God bless you for all you do, bro
Finally i understand it! Mindblowing video dude
Thank you so much bro!
Wow. Genius. Keep producing content like this
I found it, Convolution, Thank you... you are a blessing
this is the page with all the playlists.
blackpenredpen.com/math/DiffEq.html
Simply awesome😃
Great video once again man
Thank you for the excellent videos which taught me a great deal!! Your work is amazing and helps many people around the world to understand math better! I have a question about the converse of this theorem: does a multiplication in the t-domain correspond to a convolution in the s-domain? I think that for a Fourier transform, also the converse is true, but I'm not so sure whether this is true for the Laplace transform...
Are you going to explain the deconvolution too?
Loved it ❤
Brilliant proof!
Yay!!
Easy to understand.....
🌹🌹🌹🌹🌹🌹🌹🌹
2:55 the function graph looks like end brace you draw (using Laplace transform you should put input in braces)
P.s. yea my English is very bad
hello, do you have any videos of Fourier series?
I'm a little confused here at 6:33. Isn't u(3-v) the mirror over the vertical axis of u(v-3)?
Thank you
Thank you!
Amazing....
Could You derive invers Laplace transform formula simply from Laplace transform?
F(s)=integral(e^-st f(t))dt
->
f(t)=1/2πi integral(e^st F(s))ds
thanks mister you helps alot
You are genius........ Really.....
10:56 nice, new trick
Around 15.00, I'm confused because there were two e functions... and they merged into one function? The Laplace definition gives e^-vs(F(s)), so when you replace it back into the integral, wouldn't it give e^-s(t+v)?? Great videos! I love them and they've been so helpful for me!
Morgan Rogers I was confused at first too and was thinking the same thing as you. That e^-st is part of the definition of the Laplace transform and our f(t) from the definition is f(t-v)u(t-v) in this case. We didn’t actually do anything with the integral and just replaced it as a whole with the known result of the shift theorem, because if you wrote out the formal definition for the Laplace Transform, substituting f(t-v)u(t-v) for f(t), it would be exactly that black integral.
Respect
❤thanks sir
convoluton in the time
domain is equal to multiplication in the frequency domain.
Is this part of calculus??
What a beauty.. mathematics genius
definition of convolution is integrating from minus to plus inf.
Wait... what happened to the e^(-st) (right after the "note")?
Thanks
9:00 those open intervals after you multiply by the unit step function (USF) throw me off because I think that the original area had that "slice" but when you multiply by the USF that "slice" is no lomger accounted for. maybe the way I said it was a bit confusing, but could you tell me why?
(also, I think this is an ingenious way to approach the problem dude)
Nice
But how to solve laplace (f'(x)/g(x))
thanks
when u introduce at start u(tau-t) t acts as constant (t=a )
at start but at end in u(t-tau) you choose tau as constant (tau=a) and t as variable ,please explain this ambiguity ???
A-W-E-S-O-M-E!
11:03 Not all those functions are continuos in the interval. Namely the u(v-t) function is not. Which makes the whole thing not continuous at v=t. But I get it what you mean. The u function is not continuous, but can be integrated in that period
Alkis05 It being continuous is not relevant for the integration. You can integrate step functions over any interval.
@@angelmendez-rivera351 That is exactly what said...
I was correcting him in regard to him saying every function was continuous.
Alkis05 I’m fairly certain he meant to say all of the functions are Riemann integrable.
I don't know why, but this is still easier than algebra
1-u(tau-t)=u(t-tau) are both equal???
The Laplace Transform is itself an convolution. Moreover, convolution theorem works backward, it means that transform of convolution gives product of transforms!
What if there are constants a,b, c?
L{ c f(t)*g(t) }?
L{ a f(t)*b g(t) }?
make new functions f1(t)=c f(t)
@@krejman , thanks
Is -(u(v-t) - 1) = u(t-v)?
Yes
The u is also known as " Heaviside step function"
H(x)=d/dx(max{x,0})
Are you M.Sc or Ph.D ?
我合理地懷疑1*1*1*1=t^3/3
Just work it out. U don’t need to doubt 😆
超美
2 plate momo laga do sir ji
just make it phi bro
I expected you gave up
1:25 did my man just say Calculus 3? How the fuck is the american system so slow? Everything he has ever said is "Calculus 2" is taught in calc 1 or before here
convolution is the worst notation ever,change my mind. as someone who is almost always typing math, * meaning multiplication should be standard,its been this way since like prealgebra, so this is like changing the meaning of + to be subtraction to me. this notation is the biggest middle finger for anyone who types math all the time and honestly i came up with a better notation for that, f(t)*_{(a,b)}g(t)=∫_a^b f(u)g(t-u)du.
I cannot follow what he is doing. Will have to look at this video more than once. HE knows what he is doing, but I do not.