you're great!!! the way you explain things is just awesome, every word you mention is important, straight to the point. thanks,just gained my confindence in this
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (525+ videos) you might find helpful. Thanks, Adam
At 6:20, can you reduce what is in the integral because from before 0, both u(t) and u(t-Tau) make the integral zero? And from 0-t, it is non-zero because the step inputs activate each function? aka, the step functions are only there to turn the functions on.
Yes, I think what you said is essentially correct. For tau < 0, u(tau) is zero so we don't need to consider any negative tau. At tau = 0, u(tau) is the unit step function that "turns on", so this if the first point we might need to consider (and this essentially establishes the lower limit of the integral). Now, think about the unit step function u(t-tau). If t > tau, then t-tau >0 and u(t-tau) = 1. For finding the limits of the integral we're interested in the opposite case. When is t-tau < 0? When tau > t. So, for all values of tau > t, u(t-tau) = 0 and the integral is zero. However, for all values of 0 < tau < t, both u(tau) and u(t-tau) are equal to one; and that's how we decided the top limit needs to be t. Hope that helps! Adam Adam
why is it when you take the (t-tau) at 5:30 the time reversed signal gets shifted to the right and not to the left? is (t-Tau) not the same as (-tau + T)?
This is very common thing to get mixed up on. If we have a signal x(tau), then x(tau-2) is shifted to the right, and x(tau+5) is shifted to the left. However, if we've already time-reversed the signal (e.g. we are dealing with -tau as we are when doing convolution), then we shift the opposite direction. So, if we see x(2-tau) that a shift of 2 units to the right, etc. Hope that helps, Adam
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
Hello Dr. Panagos, Why can't you just do the integral from 0 to T of (tau^3)*(tau^2), why do you need to include the t term? Because isn't the overlap equal to both the functions of tau combined? and t is already included in the integral term.
We're convolving the functions in this example. When convolving, we multiply and integrate a time-reversed and time-shifted version of one of the signals. If we do what you suggest, then we wouldn't be performing convolution.
Dude, you might be the best tutor i ve ever listened, seriously! Awesome work, man, thanks a lot!!
hacayor
You're welcome, I'm glad you like the videos! Thanks for the nice comment.
what I like about your explanation is that you know engineers picks it really fast and your videos are short and the easiest to follow.
Thank you,
Thanks man! I'm taking linear systems class now. Your videos are a huge help!
Part time professor, full time engineering job, and yet here you are teaching me convolution. Technology is great isn't it?
Thanks, great video.
Thanks for the nice comment, glad you liked the video.
thank you so much!!!! the way you are teaching is so clear! you are a life saver for me!!!!
+Betul Kaplan Glad to hear that, thanks for watching!
I have been struggling to understand convolution until I found this.
Awesome explanation.
Thanks a lot.
speedblazerspz Glad the video helped, thanks for the feedback!
My proffessor never showed me this way to convolution, only the graphed boundaries, I'm glad I found your video *subbed* :^)
J1gS4wLTD Glad it helped, thanks!
This is much clearer than my native language prof. thank u so much :)!
you're great!!! the way you explain things is just awesome, every word you mention is important, straight to the point. thanks,just gained my confindence in this
Thanks for the nice comment, glad you enjoyed the video!
This is great. I was never taught to do convolution in tandem with the graphical interpretation. Enlightening... :)
Great teaching of this example. Thank you I learned from this !
awesome even after 3-years i can't pass without saying thank you
im 6 yrs late but this has helped me sm in my signals class
Glad I could help, thanks for watching!
Been searching for the right video for quite some time finally found it thank you mate, well explained!
+koyublue Glad you liked it, thanks for the nice feedback and thanks for watching!
Awesome explanation and illustrations! You've made me understand the concepts. Now I can finally start the calculating exercises, off we go!
Great, glad these videos helped. Good luck!
love your videos, reviewing these before my exam tomorrow just as a refresher and is very very useful. A+++!!!!!
Glad they're helpful for you, thanks!
YOU ARE THE BEST TEACHER
Thanks!
This saved me, thanks so much!
Glad I could help, thanks for watching. Make sure to check out my website adampanagos.org for additional content (525+ videos) you might find helpful. Thanks, Adam
Ahhh perfect, thanks so much! Will help a lot for my final 3rd year Uni exam on friday.
Glad it helped, thanks.
Adam Panagos
Adam, how would you convolve a triangle with itself? The triangle is 1+t from -1 to 0, and 1-t from 0 to 1.
Man thank you so much for this, You explain it so well and now I understand it!
Awesome, glad I could be of help. Thanks.
Very clear explanation. Thank you!
At 6:20, can you reduce what is in the integral because from before 0, both u(t) and u(t-Tau) make the integral zero? And from 0-t, it is non-zero because the step inputs activate each function? aka, the step functions are only there to turn the functions on.
Yes, I think what you said is essentially correct.
For tau < 0, u(tau) is zero so we don't need to consider any negative tau.
At tau = 0, u(tau) is the unit step function that "turns on", so this if the first point we might need to consider (and this essentially establishes the lower limit of the integral).
Now, think about the unit step function u(t-tau). If t > tau, then t-tau >0 and u(t-tau) = 1. For finding the limits of the integral we're interested in the opposite case. When is t-tau < 0? When tau > t. So, for all values of tau > t, u(t-tau) = 0 and the integral is zero. However, for all values of 0 < tau < t, both u(tau) and u(t-tau) are equal to one; and that's how we decided the top limit needs to be t.
Hope that helps!
Adam
Adam
U're the best. Thank you.
You are great man, seriously!!
Thanks!
very simple and very usefull explanation. :) Thx a lot...
Thanks!
why is it when you take the (t-tau) at 5:30 the time reversed signal gets shifted to the right and not to the left? is (t-Tau) not the same as (-tau + T)?
This is very common thing to get mixed up on.
If we have a signal x(tau), then x(tau-2) is shifted to the right, and x(tau+5) is shifted to the left.
However, if we've already time-reversed the signal (e.g. we are dealing with -tau as we are when doing convolution), then we shift the opposite direction. So, if we see x(2-tau) that a shift of 2 units to the right, etc.
Hope that helps,
Adam
Couldn't understand. x(2-tau) and x(tau-2) both shift right?
Thanks a lot for this video and others.You are so helpful thanks again :)
Thank you sir, very well explained
Thanks!
great explanation!
I’m glad you enjoyed the video! Make sure to check out my website adampanagos.org for additional content (600+ videos) you might find helpful. Thanks, Adam
thankss a lot !! It's awesome man !
Hello~Sorry~Can you explain what the u(t) is? and why they are disappeared when we change the integral zone?
The function u(t) is the "unit step function". u(t) = 0 for t < 0, and u(t) = 1 for t >= 1. Hope that helps!
patrickjmt of EE
Thanks!
thank you!!!!!very good video!
Glad you liked it, thanks!
i have alot of questions!fr expl (t_T ) isn't a line? and based on what u did sketch them like that!
Hello Dr. Panagos,
Why can't you just do the integral from 0 to T of (tau^3)*(tau^2), why do you need to include the t term? Because isn't the overlap equal to both the functions of tau combined? and t is already included in the integral term.
We're convolving the functions in this example. When convolving, we multiply and integrate a time-reversed and time-shifted version of one of the signals. If we do what you suggest, then we wouldn't be performing convolution.
you are fricking awsome
+Wilson Junior Thanks, glad you like the video.
You are awesome
Thanks!
god bless you!!
Thanks!
Why do we need to multiply by u(t) in the end? Didn't get it :/
because it is the combination of case 1 and case 2, and in case 1 the convolution was 0 for every t less than 0
thanks a lot!
Thank you,
Thanku sir