the first channel on youtube which starts from the basics honestly , You are very genius. Sir, pls upload all the lectures in signals and systems 10. Discrete Fourier transform and Fast Fourier transform 11. Digital filters 12. Discrete cosine transform Please accept my sincere appreciation and respect
Guys in the comment section who keep asking for more and quicker please stop. You've got to respect Neso for it's time he is giving away for us. We've got to appreciate what we're provided with! Thank you Neso Academy for all your videos ❤
I'm not attending my the gate academy coaching classes and totally relying on your videos . please keep them coming. Complete signals and systems as soon as possible. And upload all the remaining videos before 1st july. God bless you Mr.Neso.
@@noha284 its a property of geometric series.. when you add up to infinite terms and apply formula, then the mod of common ratio is less than 1.. watch previous lecture to understand.
As you can see the terms are fractions hence when they are summated they will always converge to some finite number since the condition for the z transform to exist is less than infinity which is a finite number.
Neso , sometimes I just think, about how many students have been saved by this man!. 🔥
U r really the saviour of students especially ece students
cse too bro
the first channel on youtube which starts from the basics honestly , You are very genius.
Sir, pls upload all the lectures in signals and systems
10. Discrete Fourier transform and Fast Fourier transform
11. Digital filters
12. Discrete cosine transform
Please accept my sincere appreciation and respect
Guys in the comment section who keep asking for more and quicker please stop. You've got to respect Neso for it's time he is giving away for us. We've got to appreciate what we're provided with!
Thank you Neso Academy for all your videos ❤
I'm not attending my the gate academy coaching classes and totally relying on your videos .
please keep them coming.
Complete signals and systems as soon as possible.
And upload all the remaining videos before 1st july.
God bless you Mr.Neso.
I passed my mids from this man , now tomorrow will be my final and I have learnt many from this man's videos of Signals and Systems. Thank You Neso.
The great channel for engineers....thnku so much sir 🙏🙏😊😊😊😊😊
Watch at 2X speed. Will be better for you.
6:53 why does [2(r^-1)] have to be less than 1 and greater than -1?
I think it says that only when [2(r^-1)] is greater than -1 and less than 1 then the z transform exists
But how do you know it’s -1 and 1, where did these numbers come from
@@noha284 its a property of geometric series.. when you add up to infinite terms and apply formula, then the mod of common ratio is less than 1.. watch previous lecture to understand.
keep them coming!! thnk you!
Thank you so much!!!!! Helped me a lot for my semester exam tomorrow!
Keep up the good work!!! Wishing you with loads of luck!
how |e power -j(omega)n|=1??
euler form e^(-jwn)=cos(wn)-jsin(wn)
So |e^(-jwn)|=sqrt(Cos^2 wn+Sin^2 wn )=1
i think this is the ANS
Sir in how many videos will you complete this chapter
sir,e to the power minus jwn ka value 1 kyu ho jata hai????? please explain.
oo sir mil gaya ans...
euler form e^(-jwn)=cos(wn)-jsin(wn)
So |e^(-jwn)|=sqrt(Cos^2 wn+Sin^2 wn )=1
ye hoga na sir???
Well said manisha das
Tq
What will be for x(n) = (1/3)^n.u(-n)?
But why you take r=1 for dtft??
Same qs for me
How e-jwn is 1??
The absolut value is equal to 1.
e^j(-wn) = cos(wn) - jsin(wn) [Euler's formula]
|e ^j(-wn) | = |cos(wn) - j sin(wn)| = |cos(wn)^2 + sin(wn)^2 | = | 1 | = 1.
@@josecarlosferreira4942 THANK YOU BRO
Why is e ^(-jwn) =1 ? 2:52
modulus of e^(-jwn) is always 1 as e^(-jwn)=cos(jwn) - i*sin(jwn
modulus of that complex number is always 1
How the expanded term is less than infinite? At n = infinite there is a chance of getting infinite value, no?
1+ x + x^2 + x^3 + ... till infinity, is finite if -1 (2), like an exponentially increasing curve.
As you can see the terms are fractions hence when they are summated they will always converge to some finite number since the condition for the z transform to exist is less than infinity which is a finite number.
Why are we letting r = 1 arbitrarily in the beginning?
Residue theorem
show that an=(1/6)5^n is a solution of nonhomogeneous recurrence relation an+2 -5an+1-6an=2^n....plz sir solved this problem. ....
Srrrrrr opamp ko complete kya jayega kya analog mai ya sirf utna hi hai.... Pls help srr
Watch op amp vdos by t Natarajan sir on nptel
Voice in the video is very low.
Upload videos with louder voice.
Existence
how exp(-jwn)=1 ?
modulus of e^(-jwn) is always 1 as e^(-jwn)=cos(jwn) - i*sin(jwn
modulus of that complex number is always 1
Because mod is sq root a^2+b^2 for a complex number a+ib. Here it will be cos^2jwn +sin^2n which is 1
it's transform not trans form ffs