Lecture 5 Module 3 Convolution Example Continuous Time

[@sahanaravinder7058]

I have seen various videos in order to understand the topic .....got that into my mind after seeing the elaborate explanation in video......thankyou sir

[@mugusengajeannepo6746]

the great video i had a lot of confusions but finished in not more than 30 mins
you're life saver

[@irving9509]

Bro I can actually understand you and hear you very clearly. Legend

[@jaimeayala6969]

When you reach the time period of {2 ≤ t ≤3} wouldn't the function in your integral be just 1? Since the area of convergence/overlapping occuring is just up to the amplitude (or just a square decreasing) of x(t) = 1?

[@NatPatrick-l3w]

Thank you for the explanation, good sir

[@serignesalioumbackesourang3835]

Thank you ! very well explained

[@premat.j.2333]

your voice is similar to neso academy signals and systems teacher

[@bendustin7609]

At 20:20, your integration is incorrect. I did it many times and got the same answer, but different than yours. I also checked it with online calculators.

[@metea.890]

nice job sir shouldn't it be ((t)-(t^2/2)+(3/2))at 20:09

[@ankitbhurane9762]

Yes, you are correct! Thanks for noting that!

[@jason-92]

thanks man, i was stuck at paper like 40 min before i saw your comment.

[@papammeheia68]

Could one exolpalin to me why he didn't put 2 in equation of line is 2-t why in his integrations didn't put 2 ??

[@lisaprince1313]

in case 2 how did u get the t^2 ? it should have been taken outside the integral sign because you are integrating with respect to tau

[@benheideveld4617]

The blue hatching is often wrong. It should run all the way up to the red response function h, wherever x is 1.

[@ankitbhurane9762]

I have considered blue hatching only for the overlap between the two signals as for non-overlapping regions, it's zero.

[@shubhamkashyap6733]

thankyou very much !!!!!

[@omerahmet-r1u]

how we got the integration at 18:08 please
and thank you

[@ankitbhurane9762]

Solve it step by step: The intermediate expression before the limit substitution is:
= (t•Tau) - (Tau^2)/2
Then after substitution of limits you get the shown expression.

[@surendrabachina9681]

Y(t)=0 for t>3....I think

[@ankitbhurane5130]

Yes, you are correct! That's a typo. Please consider.🙏

[@utkarshkathuria2931]

if in the convolution I assume t=0, then how will you solve the integration?

[@ankitbhurane9762]

For t=0, both first and second sub equations are valid. i.e. 0 and t^2/2

[@Juan-iq8br]

I want to ask, what software do you use in drawing this lecture?

[@ankitbhurane9762]

Hello Juan, I use OpenBoard software. Link is here: openboard.ch/download.en.html

[@jamilajamila3658]

Please someone clearly tell me how we are getting (t-2)on the left hand side after time reversal, I can not get that. In my sense it would be (t+2) but.
Please clearly tell someone shifting process after time reversal.

[@ankitbhurane9762]

Dear Jamila,
After flipping, as the current axis is t (the point on right) so anything to the left of it would be t minus something. As the width of the signal is 2, the point on left will be t-2. Consider t as some value say 5 so t-2 will be less than that, say 3 and so on.

[@jason-92]

thank you.

[@yelinyigitoglu3375]

I think WRONG SOLVING AT 18:00

[@safeegull22]

o nooo, how you find t- tawo

[@safeegull22]

what is slope?

[@noone7692]

Here the function is h(t)=t which means slope is 45⁰ so tan(45)=1 so slope is 1

[@safeegull22]

@@noone7692 no it's not like that, after I learned from one of my friend

[@tshibengabukalonji8067]

I'm sorry to say there are a lot of mistakes in the explanations and even in the equations you gave at the end the function y(t) exists only between 0 and 1.

[@philcooper279]

Disappointing, too much playing about with the computer graphics.