Thank you Mr. Morrell for these amazing, concise videos. I was completely lost at convolution, particularly with the whole Tau variable part. Your video was EXACTLY what I needed, an extremely thorough explanation of why Tau is involved and now I am able to understand the convolution process entirely. Thank you again and I am now subscribed to your amazing channel, your work is very appreciated!
cool. can I shoot one at you lot. it seems to me that convolution can be viewed graphically in 2 seperate ways. one involves the shifting of a function and computing a running integral , the other involves stacking the function (and scaling it) horizontally. has anyone else noticed this? I mean its never adressed that theres 2 graphical approaches but I find this very confusing. in dicrete time they always use the second approach but for continuous time it seems to vary depending on the mood of the instructor.
Thank you for the explanation! Since almost every example of convolution I've come across with used even functions, I never understood why h(t-tau) is used. Now I know that it is to reflect the function. I still need to build more intuition and practice, but this video gave me a good push in the right direction!!
It was a very good explanation! I'm taking the course of Automatic Control, and the convolution videos that you've posted has been a huge help. Thank you so much!
When h(t) and x(t) are both zero for values of t less than zero, the limits of integration become 0 to t. When h(t) and x(t) are non-zero for all values of t, the limits of integration are -infinity to infinity.
Prof, why are you calculating just the area underneath the function x(tao) and not under h(t-tao) as well? Aren't you multiplying the two functions first before integrating?
I believe he assumed h(t - tau) to be equal to 1 on the entire domain to make the math easier. Since it is 1, the area is always just the area under x(tau)
Not necessarily! For example, if x(t=2)=2 and h(t=2)=3, then at t=2 their product is 6. The formula states that for every t, you have to multiply the areas of both functions. Since h(t-tau) = 1, it is correct. If it was something else, then both areas should be multiplied by each other.
I have seen the convolution integral as integral from zero to t and as limit from minus infinity to infinity of x(tau)h(t - tau). Why the difference, and which is correct under which circumstances?
The thing confused me was the function h(t), now I know it acts as an indicator in this example, it has value of 0 or 1, indicating that whether we should integrate the value of f(t) in position t. In other examples, it may acts as a weight rather than an indicator, but they are the same kind of things.
THANK YOU VERY MUCH FOR EXPLAINING THE CONCEPT. Now I feel like an idiot for not seeing that Tao was a right shift (negative) or left shift (positive) along the x axis with respect to the function acted on. Now to solve some DE convolution Tao problems!
I thought I was stupid for not understanding all of this stuff in my signals class...turns out I have a horrible professor and textbook. thank you for finally making things clear. Colege would be sooo much easier if professors were actually good at teaching.
Captain Rhodes You dont know the half of it! Our text book had sections where the formulas given as the "general formula" were not correct and the real formula has to be derived as one of the exercises. and the author decided to give us the answers to each problem, however, he only gave the answer to part of the problem and did not label which part it was the answer to.
If you don't want to anger mathematicians you can just write [x*h](t), implying that to compute the result of the convolution at time t you have to use the entire signal x and the entire impulse response h.
Wow you speak german and spanish don't, i think german is pretty difficult, i only thing that i can understand is "du weiß nichts" , really you should try to post this videos in spanish, because i'm sure you would be a good help for a bunch of people... serías de mucha ayuda amigo!
I love the way you explain a very complex concept in just a simple way !!!...... Too GOOD Sir..
Thank you Mr. Morrell for these amazing, concise videos. I was completely lost at convolution, particularly with the whole Tau variable part. Your video was EXACTLY what I needed, an extremely thorough explanation of why Tau is involved and now I am able to understand the convolution process entirely. Thank you again and I am now subscribed to your amazing channel, your work is very appreciated!
Thank you! This is exactly what I was missing to link the intuitive flip-and-sweep idea to the equation in my mind.
A good way to understand that flip and shift thing is like this -
h(t-T) = h(-(T-t))
I think if you re-write it like that it makes much more sense
+Captain Rhodes thank you! you saved me from getting confused :D
+Captain Rhodes I thought exactly the same thing :)
Slightly clearer the way you wrote it, Captain.
Good lecture Darryl!
cool. can I shoot one at you lot. it seems to me that convolution can be viewed graphically in 2 seperate ways. one involves the shifting of a function and computing a running integral , the other involves stacking the function (and scaling it) horizontally. has anyone else noticed this? I mean its never adressed that theres 2 graphical approaches but I find this very confusing. in dicrete time they always use the second approach but for continuous time it seems to vary depending on the mood of the instructor.
+Captain Rhodes I didnt think graphical methods could be applied to continuous time functions?
willsoe yes they can. they are always very badly explained
Thank you for the explanation! Since almost every example of convolution I've come across with used even functions, I never understood why h(t-tau) is used. Now I know that it is to reflect the function. I still need to build more intuition and practice, but this video gave me a good push in the right direction!!
It was a very good explanation!
I'm taking the course of Automatic Control, and the convolution videos that you've posted has been a huge help.
Thank you so much!
One more step towards complete understanding. Thank you.
your engineering videos are top-notch. Thanks for explaining clearly. I wish you were my professor!
When h(t) and x(t) are both zero for values of t less than zero, the limits of integration become 0 to t. When h(t) and x(t) are non-zero for all values of t, the limits of integration are -infinity to infinity.
Thank you so much for explaining this! It's nice to now have an understanding as to why tau is used instead of just using it blindly!
Lightbulb at 8:25. Thank you so much, man.
Amazing convolution
It helped ! Hats off to you sir!
Really you make everything clear,I like that.
Finally I got it! The best explanation!
@ 8:12 why isn't the u(t) function shifted to the left by two instead so it would effectively end at -2? Wouldn't h(2-tao) be left shift by two?
Thanks,
A question, while tau domain is from minus infinity to positive infinity, why you use just positive values?
This is really helpful and detailed, thanks for this
thanks, i've had an Eureka feeling here. Now sounds clear to me. Sometimes signals has too much symbols and I feel a bit lost
Prof, why are you calculating just the area underneath the function x(tao) and not under h(t-tao) as well? Aren't you multiplying the two functions first before integrating?
I believe he assumed h(t - tau) to be equal to 1 on the entire domain to make the math easier. Since it is 1, the area is always just the area under x(tau)
Not necessarily! For example, if x(t=2)=2 and h(t=2)=3, then at t=2 their product is 6. The formula states that for every t, you have to multiply the areas of both functions. Since h(t-tau) = 1, it is correct. If it was something else, then both areas should be multiplied by each other.
great video on how it works, now i just need to find out why it's done :)
Why is the integral of the product of functions just the area under the ‘exponential’ function? Is it because the other function is always equal to 1?
Thanks ! Very Usefull
Thanks for sharing this knowledge!
the only thing i remember from class is "tau is a dummy variable!!
but your explanation makes more sense lol
Smh rote learning at best
I have seen the convolution integral as integral from zero to t and as limit from minus infinity to infinity of x(tau)h(t - tau). Why the difference, and which is correct under which circumstances?
Great explaination sir... 👌👌👌
At 7.40 when you do h(-tau+t) and shift h to the right are you assuming that t>0?
What happens when t
thanks very good explaination keep it up
Thank You! Great explanation!
The thing confused me was the function h(t), now I know it acts as an indicator in this example, it has value of 0 or 1, indicating that whether we should integrate the value of f(t) in position t. In other examples, it may acts as a weight rather than an indicator, but they are the same kind of things.
why you don't make videos for emf??
THANK YOU VERY MUCH FOR EXPLAINING THE CONCEPT. Now I feel like an idiot for not seeing that Tao was a right shift (negative) or left shift (positive) along the x axis with respect to the function acted on. Now to solve some DE convolution Tao problems!
Sir please let me know how can we change the variable from t to tau ????
So in this example what would you get for tau=2?
h(2-(3))=h(-1)=1?
You have to determine from early on who is moving and who is staying stationary (who being t or tau). That may clear things up for you here...?
I thought I was stupid for not understanding all of this stuff in my signals class...turns out I have a horrible professor and textbook. thank you for finally making things clear. Colege would be sooo much easier if professors were actually good at teaching.
college
as far as im aware, there doesnt exist a single textbook on this subject that isnt totally useless.
Captain Rhodes You dont know the half of it! Our text book had sections where the formulas given as the "general formula" were not correct and the real formula has to be derived as one of the exercises. and the author decided to give us the answers to each problem, however, he only gave the answer to part of the problem and did not label which part it was the answer to.
thats pretty standard practise actually. and dont forget authors who change notation every 5 pages or dont even tell you what the notation means lol
Captain Rhodes Yup...most textbooks aren't made for human consumption lol
thank you man
even i had same question why tou has to got to do..thanks for explanation
THANK YOU!!!!!!!!!!!!!!!!!!!!!!VERY MUCH.
amazing video
thank you :)
0:44 But convolution (*) is not multiplication (.). Reading a problem well solves half the problem ;-) Of course, tau can be t-prime (t') as well.
Wouldn't h(t - tau) simply be equal to one? Since we are just finding the area under the x(tau) part.
Why 1 though? h(t-tau) is flipping h at t.
@@petere9383
Ok, so basically τ is used as a consequence of 't' being used for the time shift
Thank you very much
the convolution the better song of the cult
Thank you!
thank you, makes a bit more sense now
THANK YOU
very beautiful
I see. Obvious when someone shows you! Thanks for the reply. :)
love it
thanks it very helpful
I wish I had you as my prof
Thankyou!!!
Thank you!!!!!
Sadly, no, I speak no Spanish. German?
If you don't want to anger mathematicians you can just write [x*h](t), implying that to compute the result of the convolution at time t you have to use the entire signal x and the entire impulse response h.
Wow you speak german and spanish don't, i think german is pretty difficult, i only thing that i can understand is "du weiß nichts" , really you should try to post this videos in spanish, because i'm sure you would be a good help for a bunch of people... serías de mucha ayuda amigo!
Permit to download
Thank you darryl
Can you explain in Spanish, please. thanks
actually tau is just a dummy variable. you can replace it with any other letter, and the result will still be the same.
haha well done, but I'm still confused.
Yo someone’s explaining signals in english for the first time?
and no indian can explain something like this
saddaam
Language is confusing
fuck I hate this, still lost.