Doubt to Neso Academy: At 8:11. Why are not using the arrow symbol on the plot of x-1 where value is equal to 2? Since in all previous lectures we have represented the unit impulse signal with an arrow symbol along with weights.
inside the integral, dirac(t-tau) and x(tau) are in multiplication, how do you just put 1 in the place of dirac(t-tau)? you can't just calculate them seperately and then multiply them
The value of only the inner part of integration i.e. x(t).dirac(t-tau) is x(tau). As x(tau) is a constant, so it comes outside the integration and as we know that integration of dirac(t-tau).dt is 1. So the answer is the constant only i.e. x(tau) only. 🙂
You are the only reason I am passing Signals right now!
I just cannot be thankful enough to this UA-cam channel ...amazing work!!!
Doubt to Neso Academy: At 8:11. Why are not using the arrow symbol on the plot of x-1 where value is equal to 2?
Since in all previous lectures we have represented the unit impulse signal with an arrow symbol along with weights.
Because impluse area becomes one and now we have x(T)only which is 2
it is after this video i got the idea of convolution. thank you. much love.
Boss channel undoubtedly.
sir , thank you so much for clearing my doubts in this concept of sampling theorem
Sir your videos are really very helpful.Thank you so much
Sir, what if we want to calculate value at t=2 i.e. x(2) what will it be 1 or 2 ? Or undefined because of discontinuty or avg of 1,2 ??
fantastic... please upload sampling video as well.. thanku
simply brilliant it helped me alot
great job thank you so so much
sir plss upload all the videos as soon as possible
couldn't get last example , Anyone ? explain
the volume was very low... but rest part was very good
you have not provided videos for DTS ?
Sir I am still waiting for remaining lecture of signal and systems. .... plzzzzz provide us remaining lecture as soon as possible
I want to understand the Sampling Theorem do you make one?
excellent
where is sampling theorem lecture >??
is random process covered in any video?
please send the link
inside the integral, dirac(t-tau) and x(tau) are in multiplication, how do you just put 1 in the place of dirac(t-tau)? you can't just calculate them seperately and then multiply them
The value of only the inner part of integration i.e. x(t).dirac(t-tau) is x(tau). As x(tau) is a constant, so it comes outside the integration and as we know that integration of dirac(t-tau).dt is 1. So the answer is the constant only i.e. x(tau) only. 🙂
@@bhabanishankardas9800 well explained
nice
what if ,someone has back in lab exams , but has cleared GATE with a good AIR, will the college pass the guy from his backlog?
may be no