Convolution of Square with Rectangle

[@MinChitXD]

Man, the part that you actually had the extra paper to drag it along the way at the bottom for the impulse response absolutely helps me understand this concept within the range of your videos without needing to find and read some painful paper work on Google, really appreciate the way you deliver your message

[@iain_explains]

I'm so glad you found it helpful.

[@roysignior4309]

really you did help a lot of students, half of my class failed to pass this course and now all of them are sitting here with you

[@iain_explains]

I am really glad the videos are helping you and your classmates. Please let me know if there are any topics you'd like me to cover, that aren't already on my channel (the full list is on my webpage: iaincollings.com )

[@drehkick32]

Finally, I have truly understood convolution. Thank you very very much. Liked and subscribed.

[@iain_explains]

Glad it helped!

[@KisoreMuthiahAnnamalai]

I couldn't understand the part where A becomes 2A after convolution in output @timestamp 8:00. Can anyone help me with this??

[@iain_explains]

A doesn't become 2A. The value 2A is the value of the integral. The integral adds up the area under the function. In this case, the function is x(tau)h(t-tau). The height of that function is A over the interval of tau between -1 and 1, and it's zero outside that. So the area equals A times 2 (since the interval between -1 and 1 is of length 2). Hopefully this video helps: "Convolution in 5 Easy Steps" kzbin.info/www/bejne/l37EqGd7qtKqr6c

[@KisoreMuthiahAnnamalai]

@@iain_explains Thank you professor. Your explanations are truly great. I just love it.

[@sahilanand2360]

What an amazing way to explain. Finally understood this concept. I have my end sems in 2 days and will update you how it went. Thanx for the awesome explanation sir!

[@iain_explains]

That's great to hear. Good luck with your exams!

[@katherinegoh7759]

Thank you for the awesome explanation. Now i see hope in my signals test tomorrow

[@iain_explains]

Best of luck!

[@salmansabbir-u5e]

Sir, your way of teaching is really good,its also help us to understood the convolution.

[@iain_explains]

Glad to hear that

[@hothifashurbaji9616]

the right video to understand convolution! Thanks for the effort and dedication.

[@iain_explains]

I'm so glad you liked it.

[@belbel9225]

Thanks for the explanation! 😊

[@iain_explains]

Glad it was helpful!

[@PameBc]

Thanks a lot!! I've been trying to understand convolution and you've helped me a lot

[@iain_explains]

That's great to hear. Have you watched my two videos that explain what the convolution equation is actually doing? They might help with general intuition. "How to Understand Convolution" kzbin.info/www/bejne/rmSplZdsi8R7pc0 and "What is Convolution? And Two Examples where it arises" kzbin.info/www/bejne/jmPGe2usdshjg7c

[@user-zj9pq5xc7x]

too good. thank you for such an easy explanation

[@iain_explains]

Happy to help

[@黃品元-f7d]

thank u for saving my final exam

[@iain_explains]

I'm glad the video helped. Have you seen my latest video for students preparing for exams? "Essentials of Signals & Systems: Part 1" kzbin.info/www/bejne/qKiWhmttlruZnsU

[@dirac9]

Such clear explanations. Thank you!

[@iain_explains]

Glad the video helped.

[@hussain7md166]

Big thanks to the well explained lecture ❤

[@iain_explains]

I'm glad you found it useful.

[@dmitrikazantsev3692]

I am sorry I am lost a bit, maybe you could give me a hand please. At 5:10. Let's consider tau+2 = -1 (tau = -3), that moment when h(t-tau) rectangular response touches x(t) rectangular signal. We know, that x(t) rectangular signal can be looked at as endless number of delta functions with area = "A" each, basically x(t) rectangle consists of them. At that (tau = -3) moment there will be minimum (just one delta function) overlap between two functions with rectangular shapes: x(t) and h(t). In this case first delta function of x(t) = A (it is always =A) will generate full 4-time_units-long response with amp=A. Value of first result component x(touch_moment)*h( t + touch_moment) = A*1 = A. This is how I understand delta function works. If this is correct, than result graph at the bottom should start from A, not 0. If this is not correct, than why in "Convolution Equation Explained" kzbin.info/www/bejne/iJ7IgXqBjLSGg7M video we started drawing result y(t) graph from x(0)*h(0) = "non-zero value" (which was not marked), and which faded in time? This is to my mind the similar situation when h(t) response is modulated with x(0) delta, and h(t) basically just repeated itself while being modulated with x(0) value. Those two cases look similar to me but display different behavior. In one case y(t) result increases proportionally x(t) and h(t) overlap area which starts from zero, in another case y(t) result is drawn with no connection to x(t) and h(t) overlap, and there cannot be any overlap between h(t) shape and delta function x(t). Hopefully it makes sense. Thank you for the video!

[@iain_explains]

In answer to the first part of your question, yes, the square function of width T can be thought of as an infinite set of delta functions packed in next to each other, over the time period T where x(t)=A. But each of those "infinite" delta functions will have an area that approaches 0, as the number of delta functions approaches infinity, in such a way that the total area under all the delta functions equals AT. So, when there is the "first overlap" (with just the "first delta function"), the area will be zero.

[@iain_explains]

Hopefully that also gives you insights into the second part of your question. In that other video you mentioned, the more delta functions there are, the smaller their area will be, and the smaller their corresponding response will be (even though I didn't draw it that way - because it was too hard to try to explain that aspect in a single diagram - and most people don't realise about that technical detail - and the overall insight is enough to provide the intuition I was wanting to convey). You've thought about it more deeply, which is great!!

[@dmitrikazantsev3692]

Oh, yes, it makes sense now! Your both answers shifted my “missing link” from understanding of convolution operation to understanding of delta function, which is easier, so it's a good news :) . I see my problem with delta function: I made a famous mistake your warned about in one of videos - I treated delta as having amplitude of A (height), while it has AREA of A, and if its width t->0, then its height (amplitude) -> infinity, with const t*h=A. Having said that, if we “squeeze” unlimited deltas inside rectangular shape, their amplitude will be limited by the signal’s amplitude, and with width t->0 its area ->0 cannot maintain A any longer. In the same time speaking about "isolated" delta function, we still mean good-old delta function with t->0, h->infinity, and area=A. Thank you for solving this, I couldn’t get out this trap without your help. If you approve my brainstorm, I will be sleeping like a baby :)

[@iain_explains]

Yes, that’s right. You got it! 😁

[@dmitrikazantsev3692]

Great! Thank you.

[@vedantshinde6511]

These are super helpful man.Thank ya

[@iain_explains]

Glad you found them useful.

[@mohamudrashid4528]

why 2A instead of 1A for the height

[@iain_explains]

The "height" is the value of the integral, which in other words is the area under the curve, which in this case is the area under the overlapping rectangle, which in this case is (base x height) = 2 x A when they both overlap fully (ie. the base is between -1 and 1, which is a distance/length of 2)

[@cs74546]

thank you so much for this precious video

[@iain_explains]

Thanks for your nice comment. I'm glad yo found the video helpful.

[@hritijrana1409]

'linear increase in area' was a little confusing.overall a good explanation

[@iain_explains]

Glad you liked it.

[@kyf6705]

You are GOAT, but why didn’t use integral

[@iain_explains]

I'm glad you liked the video. I did use the integral - I just didn't write out all the maths steps. The integral effectively takes the area under the curve, and in this case the area under the curve is easy to calculate, since the curve is just a rectangular function. If you'd like other examples, see my website, that has a full listing of the videos I've made: www.iaincollings.com/signals-and-systems

[@elmou7651]

thank you, well explained.

[@iain_explains]

Thanks. Glad you found it useful.

[@alfderbabybenz7092]

how is this only 360p?

[@iain_explains]

Sorry. I made this video in 2018, when high definition wasn't such a "thing" on KZbin, and I wasn't sure how big a file I was allowed to upload.

[@zfolwick]

some more pixels would help.

[@iain_explains]

Sorry about that. I made this video in 2018, when I didn't know what file sizes I was allowed to upload to KZbin.

[@jbergamp]

Greek T is pronounced tao.. like ciao in italian. You are eating 'a' . Thanks for the video

[@iain_explains]

And you should use a capital letter when you name a language. ie. "Italian" ... you're focusing on the wrong thing if you're worrying about criticising my pronunciation! Better to focus on your understanding of the content.