Low pass filter is an LTI system,so its output will be convolution of input and impulse response in time domain...but in frequency domain we do multiplication instead of convolution....I mean we will multiply sampled signal in frequency domain with transfer function H omega
@@sohamroy7668 watch the previous lecture M(w) contains 1/Ts as weight so we are designing to get our m(t) back, whose amplitude is 1. we are taking a filter which has the amplitude after multiplying with the 1/Ts and the result becomes 1 so we're taking Ts
Sree Gopal If you’ve followed lecture on “Sampling Theorem” then there was a relation for Fourier Transform of s(t). S(ω) = (1/Ts)[...+M(ω+ωs)+M(ω)+ M(ω-ωs)+...] When, n = 0 M(ω) = 1 S(ω) = (1/Ts)[M(ω)] S(ω) = (1/Ts)
@@huihui666 no dude it's not arbitrary value Ts here is the sampling period and that lowpass filter gain should be Ts because only then we can recover our original signal back . In sampled signal's frequency response the peak of that triangle is at (1/Ts). but our original peak is at 1 so to recover back our signal lowpass filter gain should be Ts
@@nithinmanepalli2772How did you determine that the amplitude of sample signal spectrum is 1/Ts and that of the Original message signal spectrum is 1?
It all makes sense now, thank you.
Greatings from Turkey Sir I. learned from you a lot. This channel will always have a place in my heart.
Really crystal clear examination sir, salute
Amazing lectures sir ❤️💞thank you sir for such teachings...!!
thank you srila prabhupad , krishna and sir for explaining
Hoo nice explaination
40rad/sec is answer for homework problem
Explanation is awesome!!!!!!!!!!!😎😎
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you are superb bro . thanks lord
sir, please upload a lecture series of full digital communication
Nice video
I got the concept easily.
Thanks
Your lectures are getting me through this course!!
very very helpful video indeed! thanks a lot sir!!
thanks a lot sir🥰
Sir why is H(w)=Mr(w)/s(w)
Also why S(w) = 1/Ts for w=0
How is gain of low pass filter = Ts
Low pass filter is an LTI system,so its output will be convolution of input and impulse response in time domain...but in frequency domain we do multiplication instead of convolution....I mean we will multiply sampled signal in frequency domain with transfer function H omega
By the property can we also say mr(w)=s(t)*h(t) where h(t) is inverse Fourier transform of H(w)?
Thanku very much
It is very much cleared sir
u r the best ...
Thank you very much sir for the best explanation
Sir will you upload lectures on PSD,ESD and correlation??
is it uploaded?
Well explained...
Congratulations for 1million
Sir what if the spectrum had only the right part of the triangle I mean just from 0 to W what will it be its sampled spectrum?
Great lecture.
sir how that slope becomes 1/Ts and that mag of h(t) becomes Ts, could you please tell me that sir
Same doubt
@@sohamroy7668 watch the previous lecture M(w) contains 1/Ts as weight so we are designing to get our m(t) back, whose amplitude is 1. we are taking a filter which has the amplitude after multiplying with the 1/Ts and the result becomes 1 so we're taking Ts
Why the value of S(w) is 1/Ts , when w = 0 ?
I'm also having same doubt pls give a clarification on this.
Sree Gopal If you’ve followed lecture on “Sampling Theorem” then there was a relation for Fourier Transform of s(t).
S(ω) = (1/Ts)[...+M(ω+ωs)+M(ω)+ M(ω-ωs)+...]
When, n = 0
M(ω) = 1
S(ω) = (1/Ts)[M(ω)]
S(ω) = (1/Ts)
@@alphasatari but he said w=0 not n=0
just amazing!! thx
Best teacher!
i want to donate but I am student hope in future i could do so. thank you.
Thank you ☺️
Sir ur voice is very cute♥️
Great work
No confusion sir🙏
Thank you sir.
Thanks
thank you sir
Why the value of s(w) =1/Ts at w=0 ???
so what is omega c?
Could you explain how the amplitude of H(W) is T_s?
Ts is just an arbitrary value
^for the gain of the lowpass filter
@@huihui666 no dude it's not arbitrary value Ts here is the sampling period and that lowpass filter gain should be Ts because only then we can recover our original signal back . In sampled signal's frequency response the peak of that triangle is at (1/Ts). but our original peak is at 1 so to recover back our signal lowpass filter gain should be Ts
@@nithinmanepalli2772How did you determine that the amplitude of sample signal spectrum is 1/Ts and that of the Original message signal spectrum is 1?
You are great
Much respect to you sir 🙌🙌
How oversampling is like interpolation
sir please make video series on analog and digital communication
Excellent
Amazing explanation sir
Tqsm💖❤️🔥
#bestteacher
40pi rad per sec.
sir z transform ka lecture dalo
Hbb u
mwuthmani :-*
40π
🙏🙏🙏👏👏
S
946424
Thank you so much
#bestteacher