Z Transform Region of Convergence Explained

[@amber1862]

And it finally clicked; thank you so much! Brilliant explanation.

[@iain_explains]

Glad you found it useful.

[@abhimanyumund8679]

YOU ARE THE BEST TEACHER…it has been an eye opener for me…have seen all the lectures uploaded by you…keep doing the SUPERB work…your way of putting out waveforms and making us understand with this technique is mindblowing.

[@iain_explains]

Thanks so much. I really appreciate hearing from people who are finding the videos useful - especially people like you, who have watched all of the videos! Very impressive. I'm so glad they have been helpful.

[@prettycillium]

Thanks a lot Mr Iain, this is what I was looking for especially staying at home nowadays. I'm suprised by numbers of subscribers and views however I'm sure your contents and channel will be really valueable in the future, especially for engineers. Thanks in advance for a clear explain of z transform.

[@iain_explains]

Glad it was helpful!

[@infernal4415]

1:40 i love how he casually hides the page with paper and then releases it

[@iain_explains]

It's animations "old school" 😁 Sometimes the old ways are the best. I hadn't realised I'd been doing it "casually" though. I probably wouldn't make it as a magician. 🤣

[@eren7031]

10:36 region of convergence. Thank you for your clear explanation.

[@iain_explains]

I'm glad you found it useful.

[@vibhavshivadeker8288]

This is so beautiful!! you are the reason i can understand this beautiful thing

[@iain_explains]

That's great to hear. I'm glad you found the video helpful.

[@davidbordenkircher2179]

Best explanation I have yet to receive. Thank you! - EE Student

[@iain_explains]

Glad it was helpful!

[@iamliam1241]

Thank you so much Iain!,This is one of the best tutorial videos I've found, Your videos really help to have a deep understanding of signal analysis,I wish you could also make few videos on wireless communications and 2d dft (image processing), waiting for your next videos

[@iain_explains]

Thanks for the suggestion, I'll give it some thought.

[@stringstoparadise2392]

dear sir at 1:15 you said the expression is equivalent to taking DTFT of the signal , but as you have said in Laplace transform video that in s plane when Sigma = 0 then taking Laplace will be equivalent to Fourier transform , so here in this case also the expression will be equivalent to DTFT when we consider r = 1 in general it will be a z Transform only please correct me if I am wrong

[@iain_explains]

Yes, you've got it, that's right.

[@bourahmamasten4602]

very clear, thank you for this videos

[@qizhengzhang7286]

thank you so much! I never truly understand the ROC before seeing this video

[@iain_explains]

Glad it helped!

[@MrAmalthejus]

I wish I was taught like this in college my entire life would have been different. Thank you Sir.

[@iain_explains]

You are very welcome. I'm glad you liked the explanation. It's how I would have liked to have been taught too! 😁

[@hgtrad7655]

Thanks Lain Z transform Region of Convergence explanation very clear and synthetic, maybe you should have added a couple of examples of applications. Keep on your very good work. An advanced Electromagnetics Lecturer!

[@iain_explains]

I’m glad you liked the video.

[@michaelosinowo226]

This is amazing, very clear explanation, thanks a lot!

[@iain_explains]

Glad it was helpful! If you haven't done it already, you might like to try some of the other videos on the channel too.

[@kenc4104]

Does W=2pi on the Z plane correspond with the Fs (sampling frequency) in the time domain?

[@elfeiin]

How do I get the DFT to converge faster with fewer samples?

[@fragosoescalantemaximilian8582]

Thank you, I've never seen a explanation like this

[@iain_explains]

Glad you found it helpful.

[@lioncheng6236]

thank you s much, it helps a lot for me . but why can we multiple by r^-n and get the same frequeny as original infinty energy signal ?

[@iain_explains]

Hopefully this video will help explain: "What is the Z Transform?" kzbin.info/www/bejne/pGewemCker-igpI

[@marwanal-yoonus280]

Thank you very much for your good illustrations.

[@realislamicguidance2375]

Suppose a signal is growing exponentially. We take its Z transform & find its ROC. So what are we supposed to do practically? Are we supposed to manipulate our O/P by multiplying any signal from ROC such that our O/P converges & system is stable? And what if our signal is exponentially decaying so whats the purpose of finding ROC in this case (except for determing stability/casuality)?

[@iain_explains]

The distance between the poles and the imaginary axis affect the spectrum response. See this video for more details: "How do Poles and Zeros affect the Laplace Transform and the Fourier Transform?" kzbin.info/www/bejne/n4GXl5ahm6mBgZo

[@piggypiggy855]

Thanks a lot Mr Lain,but could you please show by doing sum calculations on exactly how to find the r.o.c please help us as r.o.c is pretty important , please upload a video that

[@iain_explains]

Thanks for the suggestion. I'll add it to my "to do" list.

[@chizobanebuwa8246]

Thank you sir. Please do a video on 'Hilbert transform' and 'Lowpass equivalent of bandpass signals'.

[@iain_explains]

Thanks for the suggestions. I've added the Hilbert transform to my "to do" list, and moved the low pass equivalent topic up the priority order.

[@chizobanebuwa8246]

Very well sir. Thank you once more.

[@realislamicguidance2375]

According to the example you gave of a microphone held too close to a speaker, its O/P grows exponentially (unstable). Lets suppose its ROC |z|>2. Choosing any value of z>2, ZT converges. Does it mean the system O/P gets stable? Do we have to multiply I/P with 3^-n for stable O/P?

[@iain_explains]

No. It means that if you modify the impulse response, by multiplying it by the function 1/r^n (for r>2 in your example) then you can take the Fourier Transform (ie. then the infinite summation converges). Then you can perform calculations in the "z/Fourier domain", and if you need to, then you can also transform back into the (discrete) time domain. If you actually want to make your system stable, then you would need to change the design of your system (eg. electric circuit) so that it has an impulse response that is stable.

[@realislamicguidance2375]

@@iain_explains Q#1. How is this information useful then? Since when we multiply impulse reponse by 1/r^n it will modify our orignal signal (impulse respone) and then its Fourier Transform would not tell information about the impulse reponse but of the modified signal. Q#2. Since we already know that our O/P is unstable, causal so what is the purpose of calculations in "z/Fourier domain"? Also tell the purpose of finding Z transform/ROC in this case?

[@ShivamSingh-xf8nb]

Thank you so much man!!! This is what i needed..god bless you♥️

[@iain_explains]

Glad it helped!

[@gill6335]

does the signal cos(wn) will also have infinite energy ? [as its periodic and is infinite for n -> infinity]

[@iain_explains]

Yes. The ROC of cos(wn)u[n] is |z|>1 (ie. not including the unit circle |z|=1).

[@War_qaa]

X(n)=(-a)^n u[-n] This is the general formula how do we solve it?

[@talicadk]

The z-domain is that the same as the frequency domain?

[@iain_explains]

It's a generalisation of the frequency domain. See this video for more details: "What is the Z Transform?" kzbin.info/www/bejne/pGewemCker-igpI

[@talicadk]

@@iain_explains thank you. :)

[@mnada72]

What is r representing in physical systems?

[@iain_explains]

The basis functions (waveforms) for the Z transformation are z^(-n), where n is the time index. For example, when r=1, these basis functions are the waveforms cos(wn)+jsin(wn), which are time domain waveforms (remember, n, is the time index), at the frequency w. So, in general, the basis functions (waveforms) are r^(-n)(cos(wn)+jsin(wn)) which are "damped" versions of cos and sin (for r>1). So, "physically", w is the frequency, and r is the damping coefficient.

[@mnada72]

@@iain_explains Thank you very much. I will review your videos again, I really began to understand signals from this channel.

[@gehadsamir5663]

THANK YOU

[@husseinalsultan]

Really amazing sir

[@iain_explains]

Thanks. Glad you found it helpful.

[@zeeshan123633]

Explanation was really good unfortunately during my graduation i didn't came across such lecture otherwise i would have topper of my batch :) 😃

[@iain_explains]

Thanks for your nice comment. Glad you liked the video.

[@zeeshan123633]

@@iain_explains i want a lecture Fourier series ...please can you make on it

[@RajeevD4138]

Thanku, u explain very well

[@iain_explains]

Glad you liked it.

[@مناورالشمري-ك7ك]

Wow 👌 that is very good sir 👏 I need some help what is the meaning of >> (Z = ρVs)

[@iain_explains]

I'm glad you liked the video. Sorry, I'm not sure what you're referring to though. Where do I mention Z = ρVs ?

[@ahmadal-tarabeen9253]

Your writing line is exactly the same of mine! I know people don't like it but its nice hhhhhhh. ;D

[@iain_explains]

Glad you like it too.

[@Alexkar98]

amazing work iain, you saved my ass!!!

[@iain_explains]

I'm glad it helped.

[@agustinlawtaro]

Excellent

[@iain_explains]

Thanks. Glad you liked it.

[@user-qv6bm5rc3j]

Yo Mr white?

[@iain_explains]

Breaking Bad or James Bond?

[@user-qv6bm5rc3j]

@@iain_explains breaking bad , you look like bryan cranston .. thanks the video u helps me alot and i have an exam today thank u again sir👍🏻