Fourier Transform of Basic Signals (Sampling Function)

[@nayachen5517]

Awesome video!!! love it!

[@hikmathasanov]

Sujeet Singh, very good explanation. I was expecting we will get fourier transform of Sampling Function, Sa (t*tau/2) 2*pi/tau * rect (omega/tau)

[@gannonmalakai9706]

i realize Im quite randomly asking but do anyone know a good website to watch newly released series online ?

[@zaynarcher9793]

@Gannon Malakai Flixportal :P

[@gannonmalakai9706]

@Zayn Archer thanks, I signed up and it seems like a nice service :D I appreciate it !!

[@zaynarcher9793]

@Gannon Malakai no problem :D

[@AshishKumar-bs7um]

Conceptual & short sweet

[@Evolutsner]

Sir, I could notice a small mistake here. In the triangular waveform representation, y(t) = A rect(t/tou). To get total area = A. The amplitude of y(t) for -tou/2 to tou/2 should be A/tou. But you have taken only A. This will accordingly change the FT as A sinc (w.tou/2) [ not A.tou sinc(w.tou/2). In your previous lecture for derivation of FT of rectangular and triangular signals too, same thing you have assumed. Please check it and let me know if I am wrong. Thanks

[@manas6185]

No problem bro. Sir not consider total area A. If you want then you have said that is right. Range of t is -tou/2 to +tou/2 , then alway area*sam(w*tou/2 ) always come.

[@varshithkorukonda5926]

Sir ur simply superb👏👏

[@GauravGupta-pb8mk]

Thank you sir

[@oye_ujjwal]

why you use t=-w at 2:26 playtime,please explain

[@yashverma8266]

refer duality property

[@RomanReigns-tg5qm]

how is this function absolutely integrable because at the starting you said that we could use the formula?

[@mazensayed3250]

it's an integrable function as if you used parsavel's equation you can prove that integration of sa(wt/2) squared from -inf to +inf = 1/t

[@AshishKumar-bs7um]

NYC effort

[@siddharthrajesh8345]

Couldn't understand this topic

[@phanindrareddy4885]

One crore thanks

[@yatisinghal6440]

Where is dtft parts

[@NiteshKumar-hx2oj]

🙏🙏🙏🙏

[@vikayakumari944]

😊