Introduction to the convolution | Laplace transform | Differential Equations | Khan Academy

Introduction to the convolution | Laplace transform | Differential Equations | Khan Academy


In this video, I’m going to
introduce you to the concept of the convolution, one of the
first times a mathematician’s actually named something
similar to what it’s actually doing. You’re actually convoluting
the functions. And in this video, I’m not
going to dive into the intuition of the convolution,
because there’s a lot of different ways you
can look at it. It has a lot of different
applications, and if you become an engineer really of any
kind, you’re going to see the convolution in kind of a
discrete form and a continuous form, and a bunch of
different ways. But in this video I just want
to make you comfortable with the idea of a convolution,
especially in the context of taking Laplace transforms. So the convolution theorem–
well, actually, before I even go to the convolution theorem,
let me define what a convolution is. So let’s say that I have
some function f of t. So if I convolute f with g– so
this means that I’m going to take the convolution of f and
g, and this is going to be a function of t. And so far, nothing I’ve written
should make any sense to you, because I haven’t
defined what this means. This is like those SAT problems
where they say, like, you know, a triangle b means a
plus b over 3, while you’re standing on one leg or
something like that. So I need to define this
in some similar way. So let me undo this silliness
that I just wrote there. And the definition of a
convolution, we’re going to do it over a– well, there’s
several definitions you’ll see, but the definition we’re
going to use in this, context there’s actually one other
definition you’ll see in the continuous case, is the integral
from 0 to t of f of t minus tau, times g of t– let me
just write it– sorry, it’s times g of tau d tau. Now, this might seem like a very
bizarro thing to do, and you’re like, Sal, how
do I even compute one of these things? And to kind of give you that
comfort, let’s actually compute a convolution. Actually, it was hard to find
some functions that are very easy to analytically compute,
and you’re going to find that we’re going to go into a lot of
trig identities to actually compute this. But if I say that f of t, if I
define f of t to be equal to the sine of t, and I define
cosine of t– let me do it in orange– or I define g of t to
be equal to the cosine of t. Now let’s convolute
the two functions. So the convolution of f with g,
and this is going to be a function of t, it equals this. I’m just going to show you how
to apply this integral. So it equals the integral–
I’ll do it in purple– the integral from 0 to t of
f of t minus tau. This is my f of t. So it’s is going to be sine of
t minus tau times g of tau. Well, this is my g of t, so
g of tau is cosine of tau, cosine of tau d tau. So that’s the integral, and
now to evaluate it, we’re going to have to break out
some trigonometry. So let’s do that. This almost is just a very
good trigonometry and integration review. So let’s evaluate this. But I wanted to evaluate this
in this video because I want to show you that this isn’t some
abstract thing, that you can actually evaluate
these functions. So the first thing I want to
do– I mean, I don’t know what the antiderivative of this is. It’s tempting, you see a sine
and a cosine, maybe they’re the derivatives of each
other, but this is the sine of t minus tau. So let me rewrite that sine of
t minus tau, and we’ll just use the trig identity, that
the sine of t minus tau is just equal to the sine of t
times the cosine of tau minus the sine of tau times
the cosine of t. And actually, I just made a
video where I go through all of these trig identities really
just to review them for myself and actually to make a
video in better quality on them as well. So if we make this subsitution,
this you’ll find on the inside cover of any
trigonometry or calculus book, you get the convolution of f and
g is equal to– I’ll just write that f-star g; I’ll just
write it with that– is equal to the integral from 0 to t of,
instead of sine of t minus tau, I’m going to write this
thing right there. So I’m going to write the sine
of t times the cosine of tau minus the sine of tau times the
cosine of t, and then all of that’s times the
cosine of tau. I have to be careful with my
taus and t’s, and let’s see, t and tau, tau and t. Everything’s working so far. So let’s see, so
then that’s dt. Oh, sorry, d tau. Let me be very careful here. Now let’s distribute this
cosine of tau out, and what do we get? We get this is equal to– so f
convoluted with g, I guess we call it f-star g, is equal to
the integral from 0 to t of sine of t times cosine of
tau times cosine of tau. I’m just distributing
this cosine of tau. So it’s cosine squared of tau,
and then minus– let’s rewrite the cosine of t first, and I’m
doing that because we’re integrating with
respect to tau. So I’m just going to write my
cosine of t first. So cosine of t times sine of tau times
the cosine of tau d tau. And now, since we’re taking
the integral of really two things subtracting from each
other, let’s just turn this into two separate integrals. So this is equal to the integral
from 0 to t, of sine of t, times the cosine squared
of tau d tau minus the integral from 0 to t of cosine
of t times sine of tau cosine of tau d tau. Now, what can we do? Well, to simplify it more,
remember, we’re integrating with respect to– let
me be careful here. We’re integrating with
respect to tau. I wrote a t there. We’re integrating with
respect to tau. So all of these, this
cosine of t right here, that’s a constant. The sine of t is a constant. For all I know, t could
be equal to 5. It doesn’t matter that one
of the boundaries of our integration is also a t. That t would be a 5,
in which case these are all just constants. We’re integrating only with
respect to the tau, so if cosine of 5, that’s a constant,
we can take it out of the integral. So this is equal to sine of t
times the integral from 0 to t of cosine squared of tau d tau
and then minus cosine of t– that’s just a constant; I’m
bringing it out– times the integral from 0 to t of sine
of tau cosine of tau d tau. Now, this antiderivative is
pretty straightforward. You could do u substitution. Let me do it here, instead
of doing it in our heads. This is a complicated
problem, so we don’t want to skip steps. If we said u is equal to sine of
tau, then du d tau is equal to the cosine of tau, just
the derivative of sine. Or we could write that
du is equal to the cosine of tau d tau. We’ll undo the substitution
before we evaluate the endpoints here. But this was a little bit
more of a conundrum. I don’t know how to take the
antiderivative of cosine squared of tau. It’s not obvious what that is. So to do this, we’re going
to break out some more trigonometric identities. And in a video I just recorded,
it might not be the last video in the playlist, I
showed that the cosine squared of tau– I’m just using tau as
an example– is equal to 1/2 times 1 plus the cosine
of 2 tau. And once again, this is just a
trig identity that you’ll find really in the inside cover of
probably your calculus book. So we can make this substitution
here, make this substitution right there, and
then let’s see what our integrals become. So the first one over here,
let me just write it here. We get sine of t times the
integral from 0 to t of this thing here. Let me just take the 1/2 out,
to keep things simple. So I’ll put the 1/2 out here. That’s this 1/2. So 1 plus cosine of 2 tau and
all of that is d tau. That’s this integral
right there. And then we have this integral
right here, minus cosine of t times the integral from–
let me be very clear. This is tau is equal to 0
to tau is equal to t. And then this thing right
here, I did some u subsitution. If u is equal to sine of
t, then this becomes u. And we showed that du is equal
to cosine– sorry, u is equal to sine of tau. And then we showed that du is
equal to cosine tau d tau, so this thing right here
is equal to du. So it’s u du, and let’s see if
we can do anything useful now. So this integral right here, the
antiderivative of this is pretty straightforward, so
what are we going to get? Let me write this
outside part. So we have 1/2 times
the sine of t. And now let me take the
antiderivative of this. This is going to be tau plus
the antiderivative of this. It’s going to be 1/2
sine of 2 tau. I mean, we could have done
the u substitution. we could have said u is equal to
2 tau and all of that, but I think you could do that from
recognition, and if you don’t believe me, you just have to
take the derivative of this. 1/2 sine of 2 tau is the
derivative of this. You multiply, you take the
derivative of the inside, so that’s 2, so the 2 and the 1/2
cancel out, and the derivative of the outside, so
cosine of 2 tau. And you’re going to evaluate
that from 0 to t. And then we have minus
cosine of t. When we take the antiderivative
of this– let me do this on the side. So the integral of u du,
that’s trivially easy. That’s 1/2 u squared. Now, that’s 1/2 u squared, but
what was u to begin with? It was sine of tau. So the antiderivative of this
thing right here is 1/2 u squared, but u is sine of tau. So it’s going to be 1/2u, which
is sine of tau squared. And we’re going to evaluate
that from 0 to t. And we didn’t even have to do
all this u substitution. The way I often do it in my
head, I see the sine of tau, cosine of tau. if I have a function and I have
its derivative, I can treat that function just like
as if I had an x there, so it’d be sine squared of tau over
2, which is exactly what we have there. So it looks like we’re
in the home stretch. We’re taking the convolution of
sine of t with cosine of t. And so we get 1/2 sine of t. Now, if I evaluate this thing
at t, what do I get? I get t plus 1/2 sine
of 2t, that’s when I evaluated it at t. And then from that I need to
subtract it evaluated at 0, so minus 0 minus 1/2 sine of
2 times 0, which is just sine of 0. So this part right here, this
whole thing right there, what does that simplify to? Well this is 0, sine of 0
is 0, so this is all 0. So this first integral right
there simplifies to 1/2 sine of t times t plus
1/2 sine of 2t. All right, now what does this
one simplify to over here? Well, this one over here, you
have minus cosine of t. And we’re going to evaluate this
whole thing at t, so you get 1/2 sine squared of t
minus 1/2 the sine of 0 squared, which is just 0,
so that’s just minus 0. So far, everything that we have
written simplifies to– let me multiply it all out. So I have 1/2– let me just pick
a good color– 1/2t sine of t– I’m just multiplying
those out– plus 1/4 sine of t sine of 2t. And then over here I have minus
1/2 sine squared t times cosine of t. I just took the minus cosine t
and multiplied it through here and I got that. Now, this is a valid answer,
but I suspect that we can simplify this more, maybe using
some more trigonometric identities. And this guy right there
looks ripe to simplify. And we know that the sine of
2t– another trig identity you’ll find in the inside cover
of any of your books– is 2 times the sine of t
times the cosine of t. So if you substitute that there,
what does our whole expression equal? You get this first term. Let me scroll down
a little bit. You get 1/2t times the sine of
t plus 1/4 sine of t times this thing in here, so times
2 sine of t cosine of t. Just a trig identity, nothing
more than that. And then finally I have this
minus 1/2 sine squared t cosine of t. No one ever said this was
going to be easy, but hopefully it’s instructive
on some level. At least it shows you that you
didn’t memorize your trig identities for nothing. So let me rewrite the whole
thing, or let me just rewrite this part. So this is equal to 1/4. Now, I have– well let
me see, 1/4 times 2. 1/4 times 2 is 1/2. And then sine squared
of t, right? This sine times this sine is
sine squared of t cosine of t. And then this one over here is
minus 1/2 sine squared of t cosine of t. And luckily for us, or lucky
for us, these cancel out. And, of course, we had this
guy out in the front. We had this 1/2t sine
t out in front. Now, this guy cancels with this
guy, and all we’re left with, through this whole hairy
problem, and this is pretty satisfying, is 1/2t sine of t. So we just showed you that the
convolution– if I define– let me write our result. I feel like writing this
in stone because this was so much work. But if we write that f of t is
equal to sine of t, and g of t is equal to cosine of t, I
just showed you that the convolution of f with g, which
is a function of t, which is defined as the integral from 0
to t of f of t minus tau times g of tau d tau, which was equal
to– and I’ll switch colors here– which was equal to
the integral from 0 to t of sine of t minus tau times g of
tau d tau, that all of this mess, all of this convolution,
it all equals– and this is pretty satisfying– it all
equals 1/2t sine of t. And the whole reason why I went
through all of this mess and kind of bringing out the
neurons that had the trig identities memorized or having
to reproof them or whatever else is to just show you that
this convolution, it is convoluted and it seems a little
bit bizarre, but you really can take the convolutions
of actual functions and get an
actual answer. So the convolution of sine
of t with cosine of t is 1/2t sine of t. So, hopefully, you have a little
of intuition of– well, not intuition, but you at least
have a little bit of hands-on understanding of how
the convolution can be calculated.

100 thoughts on “Introduction to the convolution | Laplace transform | Differential Equations | Khan Academy

  • December 5, 2012 at 3:46 am
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    Actually, convolute is a legitimate synonym of convolve.

    Source: wolframalpha, and using google searching "define: convolute"

    Reply
  • December 15, 2012 at 4:04 pm
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    it is much better to use product to sum formula for trig : 2sinAcosB= sin(A+B)+sin(A-B)

    Reply
  • December 26, 2012 at 5:56 pm
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    In Signal and System theory when finding the zero-state response also

    Reply
  • December 30, 2012 at 10:07 am
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    very good- concepts cleared – thank u

    Reply
  • February 2, 2013 at 7:27 pm
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    wow. best video on intro to convolution that I could find! amazing!

    Reply
  • February 3, 2013 at 4:15 am
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    i just watched this video on a Saturday night and thoroughly enjoyed it, especially the ending where he was able to cancel and simplify, it's an odd sense of satisfaction.

    I need to start drinking and going out again…

    Reply
  • February 8, 2013 at 5:25 pm
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    very good comment by kaggypants

    Reply
  • April 14, 2013 at 1:49 pm
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    what tablet do you use to write with?

    Reply
  • April 24, 2013 at 6:43 am
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    con·vo·lut·ed
    /ˈkänvəˌlo͞otid/
    Adjective
    (esp. of an argument, story, or sentence) Extremely complex and difficult to follow.
    Intricately folded, twisted, or coiled.
    Synonyms
    convolute – intricate

    Reply
  • May 6, 2013 at 7:16 pm
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    nobody likes a know it all

    Reply
  • May 17, 2013 at 3:44 am
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    there was no conceptual explanation of convolution 🙁

    Reply
  • May 18, 2013 at 3:00 am
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    its maths dude!

    Reply
  • June 10, 2013 at 3:14 pm
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    i scrolled down to write exactly what you've written here.. i kept waiting for him to explain what convolution is..

    Reply
  • August 29, 2013 at 11:05 am
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    Lets hang him on a stake and burn him!!! He should never make videos that don't make 100% sense.

    Ill grab the stake. You guys grab the petrol. Khan academy director or sal. Meet us at the grand canyon. Your ass is going down!!!

    Reply
  • October 21, 2013 at 5:23 pm
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    thank you Khan !!

    Reply
  • October 30, 2013 at 3:12 am
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    Electrical engineering for the win

    Reply
  • November 3, 2013 at 10:14 pm
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    What you did was some simple maths, can you explain more what the meaning of Convolution is?

    Reply
  • November 26, 2013 at 2:15 pm
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    thank you

    Reply
  • December 13, 2013 at 3:11 am
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    isnt the integral of sin(tau) = to -cos(tau)?

    Reply
  • February 12, 2014 at 6:45 pm
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    Great introduction to convolution. My professor tends to go off on random tangents and this exercise finally gave me an idea of the nuts and bolts.

    Reply
  • February 17, 2014 at 6:15 pm
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    and look for bryan douglas   Laplace, very good!

    Reply
  • March 9, 2014 at 9:50 pm
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    τ and t and + all look so similar when written… it's like I, l, |, and 1, but less so.
    (I might stick with 𝖨, 𝓁, ¦, and 𝟣 from now on. Sorry, mobile users.)

    Reply
  • March 9, 2014 at 9:56 pm
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    Reply for a free doughnut.

    Reply
  • March 11, 2014 at 8:07 pm
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    Please someone tells me what is the software Sal is using ?

    Reply
  • April 14, 2014 at 1:01 am
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    Good explanation, thanks!

    Reply
  • April 29, 2014 at 11:11 pm
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    my brain hurts, but this is a decent explaination. thanks

    Reply
  • May 6, 2014 at 3:09 am
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    I am a ME major taking differential equations and before I started watching your lessons, I was barely passing. I now have an A+ thanks to your videos. Thanks for the help!!!

    Reply
  • June 8, 2014 at 11:20 am
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    at the very end, when writing out the answer, in the orange, you forget to write g(tau) as cos(tau). 

    Reply
  • September 17, 2014 at 7:27 am
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    your ''t'' looks like a ''+" . Apart from that, it's a very good explanation.

    Reply
  • September 30, 2014 at 3:49 am
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    good thanks

    Reply
  • October 28, 2014 at 7:09 pm
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    Sal you the real MVP

    Reply
  • November 25, 2014 at 5:08 pm
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    Thankyou so much

    Reply
  • December 4, 2014 at 6:23 pm
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    superb…made me convoluted….1/2sint..

    Reply
  • January 24, 2015 at 3:52 pm
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    Shouldn't the limits while integrating in terms of u for where u = sin(tau) be, upper limit = sin(t) , lower limit = 0 , this got me confused a bit

    Reply
  • February 11, 2015 at 5:25 am
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    why do you keep repeating the same phrase.  Once is enough

    Reply
  • February 26, 2015 at 11:44 am
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    yea but its just simple integration. what does it actually mean?

    Reply
  • March 1, 2015 at 6:22 pm
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    What kind of tool do you use for the presentation? is it a tab ?

    Reply
  • May 28, 2015 at 8:49 pm
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    I feel like this video didn't serve well as an introduction to a new concept. I think if you began the video with the proof leading up to convolution's definition, the concept will make much more sense thereafter, rather than the viewer attempting to piece together what or why you are doing what you are doing through the example.

    Reply
  • September 27, 2015 at 3:22 pm
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    how do u know which one will be t-tau? f or g?

    Reply
  • October 9, 2015 at 3:27 am
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    lol i have a test tomorrow, your 20 minute video saved me a lot of time <3

    Reply
  • February 10, 2016 at 2:49 pm
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    Why didn't you use SinA + SinB = 2.Sin((A+B)/2)Cos((A-B)/2) to substitute Sin(t-tau)Cos tau? Integration only takes 2 lines.

    Reply
  • February 27, 2016 at 6:44 am
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    Dfaq
    such a simple integration converted into bizzare

    Reply
  • March 6, 2016 at 3:54 pm
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    When you were multiplying (1/2) sin t by (1/2) sin 2t when you were distributing you inserted + where it should have been multiplication. Result should have been (1/4)(sin t)(sin 2t) but you got (1/4)sint + sin 2t

    Reply
  • April 6, 2016 at 2:39 am
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    You're my hero Sal.

    Reply
  • April 28, 2016 at 3:57 pm
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    This is the introduction to the integration rather than convolution!

    Reply
  • May 7, 2016 at 1:19 am
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    Is there any chance of expanding on this series with some applications? I'm about to finish my DE course and have found that the applications are the most difficult part.

    Reply
  • May 8, 2016 at 6:03 am
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    Thanks for the video

    Reply
  • June 5, 2016 at 2:53 am
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    what's the program used to write on PC in that way (does he use a drawing table)?

    Reply
  • June 22, 2016 at 10:16 am
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    I found this explanation a bit convoluted to be honest.

    Haha only joking.

    Convoluted, geddit?

    Reply
  • July 3, 2016 at 8:09 pm
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    You should have skipped all the algebra and trig. It is completely beside the point of convolution.

    Reply
  • July 28, 2016 at 3:12 pm
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    Khanvolution

    Reply
  • October 26, 2016 at 12:32 am
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    too long

    Reply
  • December 3, 2016 at 2:55 pm
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    8:03 Isn't it easier to use the formula of double sin instead of the replacement method? Like sin(tau)cos(tau)=sin(2tau)/2…

    Reply
  • December 9, 2016 at 5:36 pm
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    Dude, just integrate by parts twice. Done.

    Reply
  • January 13, 2017 at 1:44 pm
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    Wow. This is pie and custard to me. I thought I was clicking on a music tech video.

    Reply
  • January 19, 2017 at 11:33 pm
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    helped me a lot. thanks 😊

    Reply
  • February 13, 2017 at 12:10 am
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    Why would you choose t and tau as your variables?

    Reply
  • February 28, 2017 at 6:34 am
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    Hahahahaha guys press 7 a couple of times on your keyboard when you're on the video. TOOTYTOOTYTOOTY

    thank me later

    Reply
  • March 7, 2017 at 3:11 am
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    Very useless in terms of convolution!

    Reply
  • March 10, 2017 at 3:31 pm
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    Nice video!

    Reply
  • June 4, 2017 at 3:12 pm
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    shouldn't this be in signals and systems?

    Reply
  • June 14, 2017 at 11:28 pm
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    Ridiculously overcomplicated example with no intuition as to what convolution is/does/is useful etc.. Not up to your usual standards by a long shot.

    Reply
  • June 15, 2017 at 1:22 pm
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    a better idea is to use the product to sum formulas:

    sin A cos B = ( 1 / 2 ) * [ sin ( A + B ) + sin ( A – B ) ]

    Reply
  • August 28, 2017 at 2:52 pm
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    In this definition of convolution, the range for integration may vary. It does not have to be from 0 to t.

    Reply
  • August 28, 2017 at 4:00 pm
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    No one said this was going to be easy…..words to live by. LOL

    Reply
  • August 30, 2017 at 10:11 pm
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    Why do you call integral the anti-derivative? It is exactly that, but annoying.

    Reply
  • September 14, 2017 at 9:26 pm
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    Well explained… Thankuu..m fully satisfied

    Reply
  • October 9, 2017 at 2:50 pm
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    ok, i like your videos but this one is more like a video about integration … btw, i use maxima to do that for me 🙂

    Reply
  • November 14, 2017 at 10:10 pm
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    As others have stated I would have appreciated learning what a convolution is, how it was invented, why it is useful and some sample real-world uses cases without spending any time whatsoever on simplistic calculus mechanics.

    Reply
  • January 15, 2018 at 9:06 am
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    Hello Khan! I would´ve liked to see some examples on text exercises within this subject. Love your videos.

    Reply
  • January 20, 2018 at 8:57 pm
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    Thanks a lot Sir,
    Btw, you can just multiply and divide by 2 to Sin(tau)Cos(tau) that would convert it into sin(2tau) /2
    Now that's more easy to integrate, just write = -Cos(2tau)/(2*2) ..:)

    Reply
  • February 13, 2018 at 3:47 pm
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    So man, First question, I wonder can we do convolution to more than two functions? Second, I have a project, for now, I want to put a chirp inside of a song and detect this chirp in the song in the matlab. Should I use anything involved with convolution? Or should I use Fourier format? Please let me know!

    Reply
  • April 23, 2018 at 6:37 pm
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    At 12.37 it is faster if you know that the integral of f(x)×d/dx f(x) dx is equal to 1/2f(x)^2. You can Prove that with Integration by parts.with best regards 😉

    Reply
  • April 29, 2018 at 1:53 pm
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    is (f*g)(t) = (g*f)(t) ???

    Reply
  • May 20, 2018 at 1:19 pm
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    sin a cos a = 1/2 sin 2a !

    Reply
  • June 19, 2018 at 8:19 am
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    I feel cancelled out 😂

    Reply
  • July 26, 2018 at 7:49 am
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    Warning, triggered machine learning students with no mathematical background detected in the comment sections.

    Reply
  • July 28, 2018 at 6:25 am
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    He could have used the identity sinAcosB/2 to break to sinA+b +SinA-b

    Reply
  • October 3, 2018 at 2:02 pm
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    Yeah, this lesson taught nothing about what convolution is or what it's used for. This was just an integration exercise.

    Reply
  • November 13, 2018 at 2:31 am
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    If I said this is a convoluted explanation, would that be a bad thing?

    Reply
  • January 7, 2019 at 3:58 pm
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    Isn't the verb for convolution -> convolve rather than convolute??

    Reply
  • February 24, 2019 at 10:29 pm
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    what is toa?

    Reply
  • February 27, 2019 at 5:39 am
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    We will do the integration ourselves, tell us the idea of fooking convolution?

    Reply
  • May 30, 2019 at 10:37 am
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    Is this a video about convolution or a video about how to integrate the example function?

    Reply
  • May 31, 2019 at 10:54 pm
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    this explained nothing about what a convo is…

    Reply
  • June 26, 2019 at 3:22 pm
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    Do yourself a favour and skip the 15 minutes of integration in the middle.

    Reply
  • July 8, 2019 at 5:26 pm
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    how to solve an integral …

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  • July 9, 2019 at 1:15 pm
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    It should be about Convolution not about how to integrate

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  • July 11, 2019 at 7:49 pm
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    Thank you <3

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  • July 22, 2019 at 10:32 pm
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    can someone please for the love of god tell me what Tau represents

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  • September 6, 2019 at 3:40 pm
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    Please don't change colors so often it makes thing messy

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  • September 29, 2019 at 8:25 pm
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    Ahh, using the formula converting from the multiplication of trigonometric functions to their addition/subtraction u could get the same answer for the integration within one line.

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  • October 16, 2019 at 8:51 pm
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    https://youtu.be/8FvdFh7lnVg

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  • October 18, 2019 at 4:21 am
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    Much better than my University Engineering Prof lol. Good Gravy, university standards are terrible nowadays.

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  • October 29, 2019 at 6:05 am
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    i'm currently 17 as of now, ill see u in 3-4 years when i actually learn about this

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  • November 29, 2019 at 1:02 pm
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    You did not explain tau, I got no idea what's happening from this point on physically.

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  • January 6, 2020 at 6:55 am
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    Idk I feel like there's an Euler identity can could condense the integration down to a dense 3 lines.

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  • January 27, 2020 at 4:34 am
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    Using this in my Digital Signal Processing (Electrical Engineering) class

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