Fourier Transform of Basic Signals (Signum Function)

[@mohammedalfaifi8165]

6:00 -----------------> Substituting a with 0 will be easier

[@zetaconvex1987]

Once again, I'd like to thank you for this great series of lectures.

[@Dh_okudama]

you are the best FT ever

[@Alabdforallah]

U r the best teacher ever thanks for every thing

[@rd1abrar924]

Excellent video 😍

[@ajayhari3018]

Sir... how to plot the magnitude Spectra and phase Spectra of signam function

[@awet_hadush22]

I would like to say thanks alot,

[@chinkigautam6782]

at 4:45 shouldn't there e a (-) sign also with e?

[@波嘎-i6g]

I have a same question

[@filankestayn]

@@波嘎-i6g it does not matter a goes to zero after the limit

[@galibms3457]

no because our goal is to converge the line, that second equation is considered when t

[@HarishKumar-dn2ve]

Hello sir please complete the course ASAP so that we can prepare for gate sir

[@blehblehbleh7099]

how was the gate exam?

[@NatureLover-oq6uc]

Thenkew brother for your priceless efforts....tomorrow morning is my exam and just learning the concepts now🥲

[@dhfm-tl4bi]

u r great sir

[@HaiderAli-co9jl]

It would be easier with differentiation in time domain property

[@akellamadhav3923]

could you explain how?

[@azadshaikh6089]

is this method only called exponential approximation?

[@srinuvadthya4879]

Superb sir

[@MeH75M]

Why the convergence of signum function takes place in exponential form?

[@khushmeetsingh1520]

How are u taking the value of converging waveform as e-at .u(t)

[@marvinkamande93]

I think it's because the equation is actually -u(-t)=-e^at*u(-t) and so the two negatives cancel each other out

[@sayanghosh316]

Since u(t) is a non converging part so,to make it converging(decaying)we have to involve a exponential part e^-at with lim a tends to 0 same goes for u(-t)

[@phanindrareddy4885]

Superb sirrrrrrr

[@adamjouamaa1286]

Thank U

[@chao-mingchang1990]

In case this derivation is correct, it would lead to the Fourier transform of the unit step function be 1/jw only, which contradicts the fact, unfortunately.

[@saiharish7975]

Sir i need the problem of the discrete time Fourier transform of the signal sin nΠ/2*(u(n))

[@Fffffffffffffffgggg]

Other easier solution : sgn(x) = u(t) +1 And do the F.T. To the right side

[@yashsharma-yc8sl]

If we are doing same using Fourier Transform formula , we are getting the same result , why ??

[@mayanksisodiya8567]

Sir, please also draw spectrum of fourier transfrom of.. Basic signal

[@dadannagariniveditha]

Y j^2 is 1

[@434ecechandrasekhar8]

Sir why sgn(t) = u(t) - u(t) ? Why cant we write u(t) + u(t)? Plz tell sir🤞

[@rajashri8925]

sgn(t) = u(t) - u(-t) , this will curve the sgn curve, if you try to plot u(t)+u(t) you will get u(t) signal but with 2 times the amplitude. which will not be sgn(t)

[@ranjeet5806]

@@rajashri8925 yeah but if I do a -1 to that 2*u(t) i will get sgn(t)

[@nuayt]

Thank you for sharing, but it bothers me that here we can use 1/(a+jw) as the Fourier Transform of [lim a->0] u(t)exp(-at) in order to calculate the transform of sign(t), but not in lecture 190. Because on the lecture 190 that proves the Fourier Transform of u(t), you use the signum function as a shortcut instead of transforming [lim a->0] u(t)exp(-at) and accepting this approximation directly. Why is that equality valid in this proof, but not valid in lecture 190? Direct links to the lectures used as reference would be helpful. Thanks again.

[@rajatganguly4032]

Nice one neso academy. Lekin mere alsi dosto , pahelehi a=0 mar dena jaldi ho jaega...

[@ankitanandy4805]

Thanks

[@srilakshmil3944]

I think it will be very easy if we integrate over the given time period sir....

[@rajashri8925]

Yes. You will get the same answer but conceptually it will be incorrect. Since you only apply FT when it is absolutely integrable.