An explanation of the Z transform part 1

Notes available at pzdsp.com/docs/. This is the first part of a very concise and quite detailed explanation of the z-transform and not recommended for those dealing with the z-transform for the first time.
A more gentle introduction to the z-transform can be found at • Z-Domain - Pole Zero P...

[@PunmasterSTP]

This was a truly incredible explanation, and I can't tell you how enlightening I found your animations. I'm so glad you decided to share your wisdom on KZbin, and judging from the other comments, many many people have been grateful!

[@eamonhannon1103]

David . You have a rare talent . You have a talent for communication . You take something that is complex and make it understandable . Great voice as well ,very clear .

[@ddorran]

Thanks Eamon. Nice to get that feedback.

[@riddhibora973]

I am an Electrical Engineer. Proudly say that this is the most amazing video regarding Z-transform, i ever watched. What a noble way to teach with proper fitting examples . WOW !! I know Z-transform very well, but still i can say I LEARNT A LOT :)

[@benjamindeworsop8348]

This is unbelievably useful, detailed, yet top down. You are sent from the heavens thank you!

[@NineInchFailz]

I'm about to graduate with my degree in Electrical Engineering at Oregon State University. I've had to watch hundreds(yes, literally hundreds) of videos to understand that concepts i've had to learn since i'm a slow learner, and i can say with complete confidence that this is the absolute best and most intuitive description of the Z transform i've seen.

[@ddorran]

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

[@yorkyoung4449]

No other materials explain Z transform better than this one. This really help me a lot! Thanks David!

[@ddorran]

Thanks York. I'm glad it helped!

[@upsidedowns2342]

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

[@JeremyDismukes225]

absolute legend @@upsidedowns2342

[@kpramodbaliga2152]

This is probably the best lecture series on signals and system ever put on internet. Very useful to get the intuitive knowledge of all the concepts. Please keep making more videos on Laplace, convolution and so on, which will give intuitive perspective of the equations.

[@adamwest9183]

Taught me what my professor couldn't in a whole semester. Wish i had you as my professor ;] great vids!

[@AngeloYeo]

It is the best explanation about Z-transform! thank you David

[@ywarlock]

Watch the 12 min video in 2 hours, but learn things that confused me for 5 years! Hope I can thumb up for 1000 times. Thank you for your excellent work, David!

[@ddorran]

Thanks for that!

[@TheFichek]

By far the best, impeccable explanation of z-transformation and its usage. Thank you for video. Keep up the good work! :)

[@egoadsr]

It's been a while i was looking for an explanation with visual interpretation, there is some videos about Z-transform, but yours is the most comprehensive one ;) thanks

[@Edwinthebreadwin]

I understand more now than I did in 3 months of control lectures, thank you for putting all the pieces together

[@vbris]

truly great material! no better video to explain the z transform than this one !

[@v.p22709]

Wow thank you very much. The first one and a half minute about systems stable/unstable were worth it.

[@krasnichuk]

Incredible. What a clear and comprehensive video. I learned quite a lot and really think it'll help advance my understanding of z-transforms.

[@ddorran]

Great to get that feedback Chris. Thanks!

[@muhammadnatour6087]

Great job! Very clear explanations that give a lot of intuition..in 12 mins you made me understand what my professors couldn't in a whole semester!

[@velvetsound]

Thank you so much for this video and for the series. I finally get the Z Transform. This is the best explanation yet.

[@ddorran]

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

[@ssjpsahu3726]

No words to thank you. Exceptionally good.

[@has1857]

Thanks for the great explanation. Made it a lot easier for me to understand what Z-transform is useful for.

[@esuelle]

Thank you so much. Brilliant explanation. I wish every professor could explain these concepts so well.

[@richardhayes7812]

I love it. wonderful review of the z-transform. great animations.

[@ddorran]

Thanks Richard!

[@ColorfullHD]

Crystal clear explanation. You're a hero.

[@ddorran]

Thanks Tim!

[@rxx396]

What a true gem on the Internet!

[@icytides]

Absolute perfection description of Z-transform

[@janeli1351]

beyond z-transform that i knew before. great explanation

[@OswaldChisala]

Marvelous video, David! I hadn't learnt about the Z-Transform before, but your explanation really helped get me started. Thank you, and happy 2017!

[@PunmasterSTP]

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.

[@shivanisharma-kz9pm]

It was really too much effective, thank you so much for sharing your great knowledge points with us. Really helpful.

[@izuminoo]

Thank you for your perfect English. God bless you!

[@biznock19]

Damn this is really making everything click in my head. Thanks David.

[@israelflores9286]

Excellent video! God bless you sir! I think I finally understand this!!

[@dreadlost]

Incredible video. Thank you so much!

[@tangdexian3323]

this is just simply amazing, looking forward to the part II. cheers

[@abdkrimimounir9478]

SUPPPPPPPPPPPPER ! DSP TEACHERS ALL OVER THE WORLD MUST SEE THIS VIDEO

[@lmao4982]

This is brilliant. I wish there was just a second of pause between topics and changing the visual, but nonetheless you make it extraordinarily clear (and youtube has a pause button, anyway).

[@xx903]

Great explanation, very clear and helpful. Thank you.

[@BorisGrishenco]

Thank you for this video. In this week I’m not planning to watch all of your videos ;) especially regarding Z transform and Matlab. Great job!

[@Rollmops94]

WOW. The wikipedia article on this should have a link to this.

[@jaikumar848]

Magnificent David.... ! really help me to visualize what z transform is. please release it's second part asap.. Also I want to know the practical example of discrete system that you have shown in video

[@DhaferThyab]

Tremendous video .. nice work 🎉

[@lowerpass2000]

this explanation is so good

[@SAJAN_ECE]

It is great that you came here. Believe me, this video series is one of the best explanation on z Transform available in the Internet. Please watch all the five videos. It is really worth it.

[@ddorran]

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

[@nshaiman]

Great Great piece of work on Z- Transform . Like this Video more than my whole D.S.P lectures...+++++

[@mata7648]

LIke this, very insightful. Looking forward to the part 2.

[@aleksadamjanovic7699]

Ty bro, God bless you.

[@emty6718]

Making this shit argument understandable is an act of heroism. Thanks man, you are a Hero!!!

[@VikasVJois]

Awesome video. Gives a good perspective on Z-transform. When are you planning for part 2?

[@ddorran]

kzbin.info/www/bejne/jXTbfmeJjtiaZ9k

[@abdallahgoda3977]

very helpful session , looking forward part 2 . thanks

[@sealedwings6788]

Nice vid! Nonetheless, I have one question: in the video you state that z=r*e^(jw), but (please correct me if I'm wrong) the euler identity is that e^(j*ANGLE)=fasor, so one should say that z=r*e^(j*(w*t)), where t is time. z is a fasor of radius r and angular velocity w, and that's why you can have a sinusoidal signal if you plot that on time.

[@PunmasterSTP]

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)).

[@ahmedali6519]

JUST AMAIZING

[@ahmadalghooneh2105]

Thankx David

[@RohitVerma-qu6eu]

explained very well...👍

[@MahadiHasanMeem]

Thanks so much for the amazing work. :)

[@sasona3110]

Great, Thanks

[@AdityaPrasad007]

Almost 10/10 in terms of content, crisply delivered, great ideas, you were clear on how r^-n controls the exponential growth or decay, I wish you had talked about why we have BOTH a cos and a sin in (cos(wn)-sin(wn)), together they contribute to the oscillation, one would be enough if we just wanted oscillating decay. like the second term is just phase shifted version. Maybe something to do with how together those two form an orthogonal basis? The phase shifting seems to encode the information regarding which direction of the imaginary axis that frequency is selected. It's not clear at all to me and I hope anyone who decides to use david's great video as a template decides to go a bit deeper in that. I think it might have been better to think of the complex part as a complex phasor spinning at a rate of w Hz rather than as the seperate cos or sin terms. If I did not know 3b1b I would not know what levels of animation is possible, so I have higher expectations from you. Still Thanks so much.

[@ddorran]

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.

[@AdityaPrasad007]

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

[@theAIex2]

Great explanation!!, do you have some explanation of the discrete wavelet transform using the lifting scheme? , becoming more important every year, and a lot of us would be really helped by your explanation, cheers

[@luca4229]

Maybe I am missing something: there is a theorem about transfer functions which says if all the singularities of the function lays inside the circle with |z|

[@hamdidjeridane5727]

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

[@ddorran]

This video might help answer this kzbin.info/www/bejne/mnatoWdsiKuajJY

[@johnli2026]

Amazing!

[@GSBrigida]

You are awesome ! thanks

[@arturranz]

amazing material. Waiting for part 2

[@tengoliver]

what a great video! thank you!

[@DeepakSingh-ww1dr]

Thanks alot 💖

[@LowLifeGraphicsProgrammer]

This is so good

[@BillTubbs]

Fantastic video. I've been looking for an intuitive interpretation for a while. Thanks. I'm not sure how he pronounces the 'r' variable (something like "or"). It looks like a regular 'r'... What is it?.

[@sebastianschweigert7117]

Very nice video. Thanks a lot

[@AlexAlex-fo9gt]

5:22-5:26 Did example of the signal z^-n (for z=0.9e^jx=-0.9+0j) plot correctly? z = (-0.9)^-n for n={0, 1, 2, 3, 4, ...} z = {(-0.9)^0, (-0.9)^-1, (-0.9)^-2, (-0.9)^-3, (-0.9)^-4, ...} z = (1; -1.11; 1.23; -1.37; 1,52; ...}

[@hulasanchez8003]

your lecture is realy amazing.sir what can we say about the significance of z transform in power system protecyion

[@4665726e]

Good video but I am confuzzled... at 4 minutes you say [1,1,1...] is a DC signal, which you use for the two sets of correlations. Then at just before 5 minutes you say the outcome of the correlation is 0, which indicates there is no DC component present in the signal, which I sort of took to mean, no DC input components, but that makes no sense... or does it? or did you mean the output signal?

[@AngeloYeo]

I wanna see part 2 as well... it helps me alot thanks

[@ddorran]

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

[@AngeloYeo]

and finally you are back! :)

[@ddorran]

kzbin.info/www/bejne/jXTbfmeJjtiaZ9k

[@Zuwwar]

Great video about the insight what Z-Tx is actually doing, however, I do have this question; at 11:24, the animation of the sinusoid shows an oscillating signal with some frequency when z is almost equal to -1+0j. Shouldn't the signal at this point have zero frequency because of the imaginary part equal to zero?

[@nejcdezelak6358]

Great job!

[@iraqisailor]

great explanation thank you

[@allaryyan6608]

really good video..thanks!!

[@yaghiyahbrenner8902]

Hey man thank you for this video.

[@ALLEDUCATION]

Hey where are the other part of this videos. Plz post them asap. Plz N make video on laplace transform and Fourier transform also

[@ddorran]

kzbin.info/www/bejne/jXTbfmeJjtiaZ9k

[@jsc3417]

10:23, if Z is a constant real number such as 0.8, shouldn't the 2nd graph be a straight line at 0.8? Why does it curve up towards the end?

[@ddorran]

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 ....}

[@salimrukhsar869]

great explanation

[@JohananJoysingh]

That was really good.. Appreciate the effort.. :)

[@raviteja4836]

Awesome David.Keep Going

[@knighttime19]

Fantastic !!

[@uriel_zapata]

thanks

[@andreregis13]

Professor David Dorran, I´m looking for chirp z-transform (CZT). How does it work? Where and when I need to use? Could you explain a little bit about this or indicate some book (or site) to study? Thank you very much. Best regards.

[@ddorran]

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.

[@andreregis13]

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.

[@saminder987]

very nice explanation, thank you xD

[@shreya_sinha.]

great work ! keep it up!

[@ddorran]

See kzbin.info/www/bejne/aIO9iKmvatVmqtU for an explanation of the statement on why "the impulse function contains all sinusoidal frequency components"

[@shreya_sinha.]

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. 👍

[@malakspeters6052]

great tutoring..,,,,,just one point to mention...could you please speak slowly...english is my 5th language,,,cheers

[@JeremyDismukes225]

your accent is wild, but your explanation on the topic was very helpful review for my understanding for my FE review

[@atriswar]

where is the part 2 ?????

[@jeevand1737]

Thank you sir :-)

[@FreeMarketSwine]

How is e^(j*pi/2) = j?

[@ddorran]

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

[@vign1155]

I always get teachers who are concerned in solving rather than y to solve

[@aldwinvicente3490]

Part 2 please

[@ddorran]

kzbin.info/www/bejne/jXTbfmeJjtiaZ9k

[@otherbrothersteve]

Where's part 2?

[@ddorran]

kzbin.info/www/bejne/jXTbfmeJjtiaZ9k

[@zhaozhouaghora4564]

There's just magic equations that just appear, and after that equation is said, with spurious reasons given for the pursuit of some result that seems random, a next magic equation appears. sure it is probably all logical but it just does not connect to anything practical that i can see.

[@adithyaravindra5596]

are you irish? Sound a lot like people in peaky blinders

[@Nazhull]

unipd? qualcuno?

[@joviallmountain2676]

T H A N K Y O U.

[@animatrix1851]

Stewie

[@TurbinationE]

This title is misleading. I'm just looking for a video explaining the Z Transform ffs. I don't care about this whole "Zed Transform" thing.

[@telgars]

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

[@MCSGproject]

it's a bad joke clearly

[@muntoonxt]

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: kzbin.info/www/bejne/r2eUY2OHhNGmnLc

[@moathstudying8446]

Brain Overheating!!

[@djtoddles8750]

Nice non-Indian accent, just sayin