An explanation of the Z transform part 1 

Thanks Eamon. Nice to get that feedback.

Thanks York. I'm glad it helped!

@@ddorran after 5 years he's still replying. legend.

absolute legend @@upsidedowns2342

Thanks Richard!

Thanks Dakotah! Good luck with your career as an Electrical engineer!

Thanks for that!

I just came across this video in 2022, and the fact that you mentioned a year that was before the pandemic kind of struck me in an odd way. In any case, I hope that you have been doing well.

Great to get that feedback Chris. Thanks!

Thanks Tim!

Great to hear that it's helping people out Jason.

Well done to anyone who can endure these videos (it's a bit of a marathon exercise!) Hopefully they provide some insight and help develop a deeper understanding of the z-transform

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WCxM4SYsg1w.html

he was discussing the values of Z, while stability depends on values of Z^1 .... when |Z| |Z^1| >1 and vice versa

This video might help answer this ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-dEJp46SFgV4.html

I know it's been a few years but I just came across your comment and was curious. Since z is just a complex number and not a time-varying function, I don't think you'd want to put a t (or n) in it's definition. Therefore it would just be z = r*e^(jw) and not z = r*e^(jwt) or z = r*e^(jwn). We can still apply Euler's formula, since the argument of the exponential is complex, and get that z = r*e^(jw) = r*(cos(w) + j*sin(w)). However, if we're thinking about *powers* of z, where z is raised to -n, then we'd get that z^(-n) = (r*e^(jw))^(-n) = r^(-n) * (e^(jw))^(-n) = r^(-n) * e^(-jwn) = r^(-n) * (cos(-wn) + j*sin(-wn)). And since cosine is an even function and sine is odd, that would be equivalent to r^(-n) * (cos(wn) - j*sin(wn)).

See ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-2RZWvy5p3ts.html for an explanation of the statement on why "the impulse function contains all sinusoidal frequency components"

David Dorran Thanks 😀 I have seen it and I understood why we use impulse signal . When my professor was teaching about impulse response he never told us why we use 'impulse signal'. You're videos are awesome because you answer the 'whys' and don't simply jump to conclusions. 👍

You basically need the cosine and sine to capture sinusoidal oscillations of any phase shift. We can represent a sinusoidally oscillating waveform mathematically as Acos(wt + p), where p is the phase shift. Such a waveform is also equal to (Acos(p))cos(wt) - (Asin(p))sin(wt). Hope this helps.

@@ddorran oh thanks man, yea it helps :)

+AngeloYeo It'll be a while before I get a chance to do part 2 I'm afraid.

and finally you are back! :)

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WCxM4SYsg1w.html

I haven't used the CZT before. I did some reading on it when you made the post and from that would suggest that you first make sure you understand the DFT (which effectively analyses signals against a set of sinusoidal basis functions) and then try to appreciate the view of the DFT as being a sampled version of the Z-plane around the unit circle (This video might help with that - but its not explicitly clear I have to say). From my brief reading the CZT is similar to the DFT except that can be determined by sampling the Z-plane along a path that corresponds to points that ensure that the ratio of the damping factor to frequency are maintained as a constant. This corresponds to points that lie on a straight line in the s-plane if you were to map these points from the z-plane on to the s-plane. The DFT can be viewed as a special case of the CZT since the unit circle in the z-plane maps on to the imaginary axis in the s-plane i.e. a straight line but the damping factor is zero in the case of the DFT.

Professor, thanks for the fast answer. I think that the link for the video did not come. I would appreciate if you could put it again. I really need to understand how does it work. I need to use the vectors produced by CZT , DFT and maybe Walsh transform comparing the results in process that I am developing to calculate active power. Best regards.

If z is a constant 0.8 then the signal z^-n is the sequence {0.8^0 0.8^-1 0.8^-2 0.8^-3 ...} = { 1 1.2500 1.5625 1.9531 2.4414 ....}

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WCxM4SYsg1w.html

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WCxM4SYsg1w.html

e^{j*pi/2} = cos(pi/2)+j(sin(pi/2) (using eulers formula) cos(pi/2) = 0 sin(pi/2) = 1 => e^{j*pi/2} = j

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-WCxM4SYsg1w.html

Is this a joke or do you not know that in some places 'Z' is pronounced as 'zed?'

it's a bad joke clearly

I came across some Z transform videos on my way here. They were pretty useless to me as I was searching for Zed, but maybe you'll find them of interest: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-y6120QOlsfU.html