Learn how to apply the graphical "flip and slide" interpretation of the convolution integral to convolve an input signal with a system's impulse response.
In despite the clear explanation, which is great by the way, the video totally succeed due the use of a non ordinary example! Most of the books explain continuos convolution using step and exponential functions. This is the first time that I can see a convolution example using a ramp function as h(t). This is really a good job and really helped me a lot! Thank you very much!
Surely!!! Even in many first world countries they will charge you thousands of dollars to teach you, when knowledge must be free and free! Luckily that is one of the good things about my country, and that in some time I will receive my degree as an Electronic Engineer without paying anything but my taxes. Greetings from Argentina.
Memo Pony And it could be explained even shorter - At least the fact HOW convolution works - However, WHY convolution works and why it is so powerful, is quite hard to understand in my opinion - At least, I am not able to find any good information on that. So either it's too hard to understand, so people are just fine with understanding HOW it works - or I am simply too dumb to understand why convolution works and is so useful.
Out of several videos I have watched since the morning (6:00AM) until now (7:00PM) hands down to this one. It explained the concept crystal clear. Well done, sir!
The "flip" is part of the mathematical definition of convolution, and generally makes a difference in the result. An exception is when the "flipped" waveform already has even symmetry, for example a triangle ramp that goes up and comes back down at the same rate (even symmetry means you can flip the waveform about its center value and it looks the same).
At 8:13... we need the equation of the downward-sloping line between t = 1 and 2 seconds. Projecting this line back at t = zero means that the line hits the vertical axis at 4 (this is the y-intercept). Even though the line does not actually extend up to this value (because x(t) is constant in the 0 to 1 range), you need to write the function that describes x(t) during the 1 to 2 second range. Hope that helps!
Hi man, I just wanted to know, of what all to know before learning convolution pls let me know and your. Videos are really helpful for my college thank you so much ans God bless
can u please explain why did u not change the eqn while ur h(-taw) is positive slope.. for me while h(-taw), eqn of line is = (1/3)taw+1 and after shifting it while h(t-taw)= (1/3)(t-taw)+1. please correct me if i am wrong...
Tell me please, what program do you use to get the convolution when you move the inverted triangle along the axis? Is there a program in free access? Thank you in advance for the answer.
Hi, thank you for the explanation. It was would have been better if you described how to mathematically obtain the interval, rather than just showing it graphically.
Thank you very much for this video, great! I am wondering: Can you tell me please which tool you are using to show the graphical convolution? Did you program it in Mathematica or an other program? And if Mathematica: Is there already a tool to show this type of analysis?
Imagine that section from t=1 to 2 of x(t) alone, and then picture it extending in all the domain of t. This new line function would be x1(t)=-2t+4 when extended, but if you took only a piece of it, from 1 to 2, it'd still take the same equation to describe it. Hope it helps!
hi ! please tell me why we must turn around h(-t)? for what this is ? please explane..... what happened if dont this is .... ( iim sorry if i did same mistakes in my message ... just im not English speaking person)
Really appreciate your work! It has helped immensely! I have a question, if you don't mind... At 5:23, you needed to find an equation for h(t-tau). You developed this equation from the original, unmodified version of h(t), that is a ramp with negative slope. Why did you use the original h(t) instead of the modified version, h(t-tau)?
+Thomas Turkington He just came to a point, where he needed to insert the function of h. Using h(t) or the "modified version" h(t-tau) is just playing around with the argument of the function. So it's straight forward to just look at the shape of h(t) and think about the function definition, which will just give you a simple linear function as you are used to work with. Without any other constants or variables to take care of. Once you have the function of h(t), you get the function h(t-tau) by inserting "t-tau" for every occurence of "t" in h(t). That's it :)
i just had this one in my test exactly same , and i did it wrong ,despite of having watched it before because i only watched and didnt repeat it on paper.
So you must first slide, and only after you ahve slite it by t you will reverse it, right? If you do this, a very big negative t will bring your signal to the far left(first you bring it to far right, then flip it and it will go to far left) insted of brnging it to far right , is it correct?
11 years later and this is one of the best videos I have watched over the past few weeks studying for my exam.
In despite the clear explanation, which is great by the way, the video totally succeed due the use of a non ordinary example! Most of the books explain continuos convolution using step and exponential functions. This is the first time that I can see a convolution example using a ramp function as h(t). This is really a good job and really helped me a lot! Thank you very much!
that made so much more sense than the 1000$ course i paid for at university
exactly (from s.korea)
Same from Evanston
Surely!!! Even in many first world countries they will charge you thousands of dollars to teach you, when knowledge must be free and free! Luckily that is one of the good things about my country, and that in some time I will receive my degree as an Electronic Engineer without paying anything but my taxes. Greetings from Argentina.
Do you really pay to learn these kind of useless things lider shame on you
You explained what took 4 class lectures in a 11 minute video. Thank you. :D
yellow pony is best pony.
Kite Tenjo Indeed. :D
Memo Pony And it could be explained even shorter - At least the fact HOW convolution works - However, WHY convolution works and why it is so powerful, is quite hard to understand in my opinion - At least, I am not able to find any good information on that. So either it's too hard to understand, so people are just fine with understanding HOW it works - or I am simply too dumb to understand why convolution works and is so useful.
I was blown away by your example. Now this is teaching.
Superior in every way to my professors explanation today. Thanks!
Best video explaining convolution. Superb job!!!
At 5:38, the downward-sloping ramp function needs an offset (the +1) to make it have a value of 1 when time t=0 (see the given signal shape at 1:11).
Thank you so much! I've been having a hard time wrapping my mind around this graphical method. Your animation was incredibly helpful!
Man!, I wish I had you as my lecturer. You are marvellous, Keep doing that
Mensah Richard
Out of several videos I have watched since the morning (6:00AM) until now (7:00PM) hands down to this one. It explained the concept crystal clear.
Well done, sir!
Very clear explanation and nice graphical representation.
Thx! you seem a talented teacher.
This has been the most helpful video I've seen on the subject. Thank you!!
The "flip" is part of the mathematical definition of convolution, and generally makes a difference in the result. An exception is when the "flipped" waveform already has even symmetry, for example a triangle ramp that goes up and comes back down at the same rate (even symmetry means you can flip the waveform about its center value and it looks the same).
Really usefull video.
Congratulations.
Tb achei
This video is cool! we are born to recognize shape and geometry.
بارك الله فيكم وجزاكم الله خير جزاء المحسنين
Thanks, Jean, appreciate your comments!
At 8:13... we need the equation of the downward-sloping line between t = 1 and 2 seconds. Projecting this line back at t = zero means that the line hits the vertical axis at 4 (this is the y-intercept). Even though the line does not actually extend up to this value (because x(t) is constant in the 0 to 1 range), you need to write the function that describes x(t) during the 1 to 2 second range. Hope that helps!
why does the interception is 4? Shouldn't it be 2?
And I finally understand what convolution does, and the role of h(t). Thank you! I was going to memorize because I couldn't understand it!
Incredible explanation
Oh man! This is really good. Thank you NTS!
Awesome. Great presentation. Keep it up.
Hi man, I just wanted to know, of what all to know before learning convolution pls let me know and your. Videos are really helpful for my college thank you so much ans God bless
This was a beautiful explanation
thanks for the video, the flip and slide is clearly explained
can u please explain why did u not change the eqn while ur h(-taw) is positive slope.. for me while h(-taw), eqn of line is = (1/3)taw+1 and after shifting it while h(t-taw)= (1/3)(t-taw)+1. please correct me if i am wrong...
Thanks mate, I've been using this method thanks to you - seems to work everywhere haha!
what program did you use to visualize this?
This video is really helpful! Thanks for sharing!!!
Very helpful thank you! But would the same result be presented if you would'nt flip it in the beginning ?
Tell me please, what program do you use to get the convolution when you move the inverted triangle along the axis? Is there a program in free access? Thank you in advance for the answer.
Perfect explanation, that cleared up a few things. Thank you very much!
this was very helpful in understanding the concept thank you
Learning is wonderful again!
Very helpfully class
tell me what is the software you used for that graphical convolution
In which application you have plotted this graphs???
Can anyone tell me how to determine the regions for the integration? I mean based on what?
Thanks, it is a great channel, keep it going!
Great video! Thanks for posting
Great explanation, thanks. However, I still need to know how to apply this to imaging systems.
Hi, thank you for the explanation. It was would have been better if you described how to mathematically obtain the interval, rather than just showing it graphically.
Awesome demonstration! :D
Thank you very much for this video, great! I am wondering: Can you tell me please which tool you are using to show the graphical convolution? Did you program it in Mathematica or an other program? And if Mathematica: Is there already a tool to show this type of analysis?
Used Maple for the animation.
Excellent video, thank you!
WHERE HAVE YOU BEEN ALL MY LIFE????
Awesome, thank you for posting! This helped me greatly with understanding convolution! Subscribed!
at 8:10 , why is the intercept 4? I do not see it, please help !
For future students wondering how, simply use the slope formula and slope intercept form to find the equation of the line.
Imagine that section from t=1 to 2 of x(t) alone, and then picture it extending in all the domain of t. This new line function would be x1(t)=-2t+4 when extended, but if you took only a piece of it, from 1 to 2, it'd still take the same equation to describe it. Hope it helps!
can anyone xplain me when t
Thank you very much! It's been really helpful for me.
what if the regions aren't simple and obvious? How would you determine your regions for the integration then?
Sal Panahi you take longer finding them out hahaha
Or use Wolfram Mathematica :P
Why did you take inercept as 4 ?
Very interesting! Thaks a lot!
Why is it t-3 at 10:00 and not t-4? Can someone please help me? Thanks!
thank u it's very nice and helpful
How can I estiamte (without caclculations) the red curve as a product of the other two ones?
Thanks
sir how you get the ramp
very well explained!
OH my god you have saved my GPA.
Great video. Thanks.
this is the best video
Can you please do one example with a sin(pi/2 *(x-1)) and same h(t -T) in this example?
can you explain the convution in network calculus please?
hi ! please tell me why we must turn around h(-t)? for what this is ? please explane..... what happened if dont this is .... ( iim sorry if i did same mistakes in my message ... just im not English speaking person)
really an useful video
Really appreciate your work! It has helped immensely!
I have a question, if you don't mind... At 5:23, you needed to find an equation for h(t-tau). You developed this equation from the original, unmodified version of h(t), that is a ramp with negative slope. Why did you use the original h(t) instead of the modified version, h(t-tau)?
+Thomas Turkington
He just came to a point, where he needed to insert the function of h. Using h(t) or the "modified version" h(t-tau) is just playing around with the argument of the function. So it's straight forward to just look at the shape of h(t) and think about the function definition, which will just give you a simple linear function as you are used to work with. Without any other constants or variables to take care of. Once you have the function of h(t), you get the function h(t-tau) by inserting "t-tau" for every occurence of "t" in h(t). That's it :)
Thanks for this
Well explained.
Pour la première fonction comment vous avez fait l'expresion analytique il semble que je nai pas comprit du tout
thanx a lot!! the video is very helpful!!
extremely helpful! thanks!
This helped so much! Thank you!
Thanks!
@2:02, no one explains where this t comes from and why is it on the tau line.
thank you so much! it helped me a lot!
excelent ... thank you
Shit, man, you explained it perfectly. Thanks quite a lot.
where do the h go once you drop (t-T) in for t.
Why is -2t + 4 ?... i get the -2t part but i dont see how 4 is the y-intercept ...?
Thank you so much!!!
Amazing thank you
THANK YOU !
Thanks a lot!
thanks so much sir , it's helpful
Check also the following related video ua-cam.com/video/bUFv7UPavkc/v-deo.html
I picture convolution as a square with a diagonal (not the (x,x) diagonal, the other one). Fight me.
Thank you!!! :D
the second part of the input i think it's -2t+2 you made it +4
you know the slope is -2 and the x intercept is 2, so for y=mx+b we have 0=(-2*2)+b, b = 4. so y=-2x+4
oow right. thanks for helping :)
i just had this one in my test exactly same , and i did it wrong ,despite of having watched it before because i only watched and didnt repeat it on paper.
Well, you learned a lesson from this experience!
Hopefully your next exams will be better...
Good luck
رضوانه علی یاری
mistake;
when t is negative, the curve will slide to the right.
you flip before sliding. this means when t is negative it slides to left.
So you must first slide, and only after you ahve slite it by t you will reverse it, right? If you do this, a very big negative t will bring your signal to the far left(first you bring it to far right, then flip it and it will go to far left) insted of brnging it to far right , is it correct?
amazing, im not so stupid :D
the integrals are too complicated, loses focus on the actual convolution.
Oh, i am the 666th person to like this- I am from illuminati
Thank you!!