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Understanding the Z-Transform 

MATLAB
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26 сен 2024

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Комментарии : 120   
@BrianBDouglas
@BrianBDouglas Год назад
If you want to play around with the DTFT MATLAB App from this video, you can find it here: github.com/aerojunkie/control-tools/tree/master/DTFT%20MATLAB%20App
@fahimrahman5196
@fahimrahman5196 11 месяцев назад
Thanks Brian
@ShahadetHosenPolash
@ShahadetHosenPolash 4 месяца назад
How you design those matlab app? How can i learn this? any tutorial video?
@nurahmedomar
@nurahmedomar Год назад
In university, we were taught how to do the DFT and Z-transform, but never taught why. And this video explains them in crystal clear. Great Job!
@HansScharler
@HansScharler Год назад
Thanks for the feedback!
@harrytsai0420
@harrytsai0420 Год назад
OMG! I've been waiting for the series of discrete-time control for nearly 5 years!
@BrianBDouglas
@BrianBDouglas Год назад
Hopefully, it's worth the wait!
@xptransformation3564
@xptransformation3564 Год назад
​@@BrianBDouglashow did you get such a perfect understanding for all this?
@erickappel4120
@erickappel4120 Год назад
I was blown away by the clearness of the explanation! Looking forward to more!
@diegoperezromero6610
@diegoperezromero6610 Год назад
This is by far the best video about the dtft and z transform i've ever seen. Very good and very detailed explanation.
@jmanius1
@jmanius1 Год назад
Brian is basically my mentor at this point
@BrianBDouglas
@BrianBDouglas Год назад
🤓 aww thanks!
@jmanius1
@jmanius1 Год назад
@@BrianBDouglas You responding has literally made my day. Thank you for your awesome videos!
@martinsanchez-hw4fi
@martinsanchez-hw4fi Год назад
Is there a list of the Matlab videos made by Brian?
@MATLAB
@MATLAB 7 месяцев назад
Hi @martinsanchez-hw4fi, You can find all of our Tech Talks on this page: bit.ly/MATLAB-Tech-Talks
@HimanshuSharma-b3q3u
@HimanshuSharma-b3q3u 3 месяца назад
bless my soul that i landed on this video. the use of complex domain in dft was a "magic" for me. today i stand near to the oz who does the "magic"
@emiliomartineziii2980
@emiliomartineziii2980 7 месяцев назад
This was truly amazing! This guy has MIT level brainpower. Like woah, his ability to explain and efficiently articulate this stuff is unreal. This channel will be my saving grace in dsp for sure! Already subscribed too. He's a legend for sure!
@BrianBDouglas
@BrianBDouglas 7 месяцев назад
🤓 haha thanks so much!
@yinkaleoogundiran1171
@yinkaleoogundiran1171 7 месяцев назад
Or MIT has Brian Douglas’ level brainpower? 😊
@clarklau1292
@clarklau1292 5 месяцев назад
I learned about DFT and z-transform many years ago, but it wasn't until I watched this video that I truly understood what the 'e' and 'z' in these two transformations actually do.
@franciskimonge1851
@franciskimonge1851 Год назад
my guy just made fourier transforms, laplace transorm and z transform all make sense , thank you
@BrianBDouglas
@BrianBDouglas Год назад
Good to hear! Thanks!
@terrycrews2281
@terrycrews2281 Год назад
been watching ur vids since 2014 ty so much for this ,brian
@mixguru9669
@mixguru9669 Год назад
My control systems professor should watch this video.
@sukursukur3617
@sukursukur3617 Год назад
I am very sure he didnt learn the background of the concept. That is why he cant convey its knowledge properly.
@Amine-gz7gq
@Amine-gz7gq Год назад
@@sukursukur3617 RU-vid FTW
@DRAMBgo
@DRAMBgo Год назад
Crazy explanation Brian, thank you!
@BingtheLizard
@BingtheLizard 9 месяцев назад
I've been able to solve a Fourier Transform in analytical form back at uni, without understanding what it was for. I've been able to implement a Discrete Fourier Transform algorithm in code to do simple frequency analysis of a signal, without understanding why the algorithm was the way it is. With your explanation, I think I finally get the "why" of the Discrete Fourier Transform, the way it uses correlation. Thanks for this explanation; I feel like one more thing is clicking into place.
@BrianBDouglas
@BrianBDouglas 9 месяцев назад
Glad it helped!
@wellid2087
@wellid2087 10 месяцев назад
One of the best introductions I have ever watched
@grxoxl
@grxoxl 7 месяцев назад
Honestly, the best explanation of the DFT I've ever seen)
@PrestonRogersJPL
@PrestonRogersJPL 9 месяцев назад
This was the best explanation on DTFT I have ever seen! Bravo!
@BrianBDouglas
@BrianBDouglas 9 месяцев назад
I appreciate that!
@Adhithya2003
@Adhithya2003 Год назад
The most intutive and lucid explanation of DTFT and Z transform I have ever seen 💖
@BrianBDouglas
@BrianBDouglas Год назад
I'm glad you liked it!
@foxlies0106
@foxlies0106 Месяц назад
thank you. very nicely presented, and helpful.
@vincentjupiterayuste2358
@vincentjupiterayuste2358 Год назад
Thank you Sir Brian
@giantbee9763
@giantbee9763 5 месяцев назад
Brian is amazing! This video is extremely clear :) One note on time delay [n-1] implying z^-1, it wasn't clear to me at first how the first example with a pulse function translated to all general cases, but it later occured to me that y^0, which in a delayed sequence matched z^-1, and every term afterward matches the original function, so hence it is the case for the general case. Such that time delay [n-2] does not imply z^-2, and so on. If I have understood correctly.
@erickappel4120
@erickappel4120 Год назад
Excelllent explanation! It is the first time someone made the DFT's relation to finite length input clear to me! Thankyou!!!
@Dinner_Cat_低能貓
@Dinner_Cat_低能貓 Год назад
I never understand why we use complex number to do the transform, until now !! TO CONSIDER THE PHASE !!
@stewartcopland7676
@stewartcopland7676 Месяц назад
Correct me if I'm wrong, but based on how you've explained it, the z transform is just a discrete version of the Laplace transform?
@Michael-ze6oz
@Michael-ze6oz 10 месяцев назад
Your explanations and teachings are awesome. This opens my eyes to so many things, and in just 1 video!. Brain, you have been blessed with an amazing gift. Thank you for sharing this with us.
@blessingshenjere484
@blessingshenjere484 4 месяца назад
Like how he explains. Like a documentary
@Volticymo
@Volticymo 10 месяцев назад
Can’t wait for z domain video
@em2129
@em2129 4 месяца назад
You are a wizard. Thank you!
@philipprottweiler2924
@philipprottweiler2924 Год назад
Awesome explanation! When will the video on the Z-Domain be published? Really looking forward to it
@BrianBDouglas
@BrianBDouglas Год назад
I hope before the end of the year. I have 4-5 more videos in my queue before it though :(
@KingBhoomie
@KingBhoomie Год назад
1:27 Input u[n] generally refers to unit step function. And Unit impulse is represented by δ[n]
@5vart5ol
@5vart5ol 11 месяцев назад
I love this, Im only 9 min in and it is all so clear to me now! Thank you. (Unless the next 10 min gives me dementia).
@BrianBDouglas
@BrianBDouglas 11 месяцев назад
Did it stick? :)
@5vart5ol
@5vart5ol 11 месяцев назад
It has indeed. Thank you.
@joelevi9823
@joelevi9823 4 месяца назад
I can't find the link to the video about the Z domain.. will be happy to get it
@hamedmajidian4451
@hamedmajidian4451 Год назад
You are the best looking forward to watching more and more!
@PrashantTiwari-i1b
@PrashantTiwari-i1b 4 месяца назад
Common man I had math exam yesterday, where were you 😢
@PankajSingh-dc2qp
@PankajSingh-dc2qp 4 месяца назад
The input signal at @ 1:31 is delta[n] and not u[n].
@Dinner_Cat_低能貓
@Dinner_Cat_低能貓 Год назад
please make more of this ! it really really help !!!!
@j.fkamaldeen
@j.fkamaldeen Год назад
Thanks alot Brian 🎉
@Pedritox0953
@Pedritox0953 Год назад
Great video!
@MarkNewmanEducation
@MarkNewmanEducation Год назад
Amazing explanation. Thanks so much for explaining things in a more intuitive way.
@joshuamasila3427
@joshuamasila3427 Месяц назад
I wonder why at 10:11, we did not make the cosine wave an imaginary signal. why choose the sine wave over the cosine wave? what would be the result if we just made the cosine wave imaginary?
@MrRaghunar
@MrRaghunar Год назад
Excellent explanation, great job!
@Eta_Carinae__
@Eta_Carinae__ Год назад
We know that we can use FT on account of e^{i /omega n} forming an orthogonal basis for all n. I'm wondering, since z^{-n} spans all complex exponentials (assuming no limit like n>0), if there are any non-zero inner products between differently-based exponential functions. I mean, just by definition you're sweeping over scalar multiples of a basis, making all elements (r e^{i /omega})^{-n} LD on (e^{i /omega})^{-n}, right?
@BalajiSankar
@BalajiSankar Год назад
Thank you.
@ernstuzhansky
@ernstuzhansky Год назад
Many thanks Brian!
@borisbvt5473
@borisbvt5473 Год назад
So, is z-transform kind of discreet Laplace Transform?
@BrianBDouglas
@BrianBDouglas 11 месяцев назад
Yep!
@rob9756
@rob9756 Год назад
Is z transform discrete Laplace transform?
@matwiz20xx
@matwiz20xx Год назад
Yes, that's right. This is why you have to see what is the application for which you are going to choose certain transform. For instance, z-transform is used in digital filter design, whereas the Laplace transform is mainly used in applications such as AC circuit analysis, and so on.
@mobilephil244
@mobilephil244 Месяц назад
This looks like exactly what the Laplace transform does. What is the difference ??????
@braedenlarson9122
@braedenlarson9122 7 месяцев назад
So cool!
@PankajSingh-dc2qp
@PankajSingh-dc2qp 4 месяца назад
There is no integrator in discrete time. Discrete time has "accumulator".
@LokeKS
@LokeKS 3 месяца назад
brilliant
@martinsanchez-hw4fi
@martinsanchez-hw4fi Год назад
Why an impulse would be integrated to a constant value of 1?
@reviewchan9806
@reviewchan9806 5 месяцев назад
If the components have both exponential and periodic signals, wouldn't that just the the Laplace transform?
@BrianBDouglas
@BrianBDouglas 5 месяцев назад
The Laplace transform is for continuous signals, the Z-transform is the discrete equivalent.
@MuhammdBilalNaz
@MuhammdBilalNaz 11 месяцев назад
is there any open source on the internet about detail on z domain
@briancoon308
@briancoon308 7 месяцев назад
Did that z domain video ever get published?
@BrianBDouglas
@BrianBDouglas 7 месяцев назад
I'm almost done with it! I had to change what I'm going to cover in it because I found a video by Youngmoo Kim that covers pretty much what I wanted to and it's brilliantly done. I'm going to call it out in my video and cover a few different things now. I'd put the link but my comment will be removed but search RU-vid for "Applied DSP No. 9: The z-Domain and Parametric Filter Design"
@kenakackrmn3898
@kenakackrmn3898 4 месяца назад
5:25 sampai 12:09
@martinsanchez-hw4fi
@martinsanchez-hw4fi Год назад
Thanks for te awesome content. Where can I find the list with Brian videos for this Channel?
@BrianBDouglas
@BrianBDouglas Год назад
I organize all of my MATLAB videos here: engineeringmedia.com/videos
@MuhammdBilalNaz
@MuhammdBilalNaz 11 месяцев назад
when the video on the z domain will be made and published
@BrianBDouglas
@BrianBDouglas 11 месяцев назад
I don't know, it keeps getting pushed back :( But I'll get it out eventually.
@khanhtruong3254
@khanhtruong3254 Год назад
Hi Brian, thanks for great explanation. At your DTFT demo 10:10, your input data signal clearly has magnitude of 1, so you choose the probing signals also have magnitude of 1 so it fits perfectly in your "dot product". If the input data signal is a combination of many frequencies with different magnitudes, it seems that we still keep the magnitude of probing signals (sine and cosine) as 1. I wonder why that still works? Shouldn't we adjust magnitude of probing signals according to the input signal?
@BrianBDouglas
@BrianBDouglas Год назад
I think I understand your question, but I might be off a bit. Let me give it a try. If a particular frequency in the input signal had an amplitude that wasn't 1 there would still be correlation between it and the probing signal with an amplitude of 1. Since we're multiplying the two signals, having a larger amplitude in one is just like multiplying the product of the two signals by a gain. So that frequency would have a larger value in the lower plot. Which is exactly what you want because you want to see "how much" of each frequency there is in the original signal. For example, if the signal is 5*sin(t) and the probing signal is sin(t), then the product of the two is 5*sin^2(t) and so the summation would be 5 times larger. Was that your question and did my answer make sense?
@binhnguyenquoc3249
@binhnguyenquoc3249 Год назад
This is how I see it, lets assume the input signal is a sum of many sinusoidal signals like x1.sin(w1.t) + x2.sin(w2.t) + x3.sin(w3.t) + .... When you have infinite data point, you can prove mathematically that with each probing signal, ONLY the exact probing signal that have the same frequency as one of the input's frequency components would have a NON-ZERO dot product (you can intuitively understand the reason why this happen by taking the integration of the product of two sinusoidal signals with different frequencies, the value would be 0), so the value of x1, x2, x3 or any amplitude of the input's frequency component is not that important (except when you need to know the amplitude of the component frequency) since the output plot would only have pulses at those exact frequencies (when you have infinite data point).
@eduardojreis
@eduardojreis 8 месяцев назад
16:50 - Where is the Z-Domain video?
@BrianBDouglas
@BrianBDouglas 8 месяцев назад
I'm just starting it now ☺ It took a while to get to it. Any particular questions you'd like answered in it while you're thinking about it? Thanks!
@eduardojreis
@eduardojreis 8 месяцев назад
@@BrianBDouglas Thank you so much for the feedback. Very much appreciated you asking! Yes! I do have questions that, if possible, I'd love to see them answers. 1) Can you give a couple more examples of how to go from the zero+poles representation to the filter equation? 2) Can you explain butter filters design and their relation to the Z-transform? What does the butter function does? 3) Can you explain `sos` for filter design? Why `sos` is more stable than the `butter` function in some cases? These are some topics I came across when first watching this video. I would find very helpful to have them addressed, but if you consider a bit out of scope for the next video you're planning that is fine. I looking forward to it anyways, your videos are great and I've learned a lot! Thank you for all the effort and quality put into them!
@hamedhojatian3539
@hamedhojatian3539 10 месяцев назад
Could it BE any more simpler to explain DFT?
@Neuromante73
@Neuromante73 Год назад
Hi Brian, since I am studying how to develop MATLAB apps, would it be possible for you to share the app that you have used during the discussion on the DTFT? Thanks
@Amine-gz7gq
@Amine-gz7gq 4 месяца назад
he put the github link in the description
@gradientenfeld
@gradientenfeld Год назад
Is this tool you showed for the DTFT somewhere available?
@BrianBDouglas
@BrianBDouglas Год назад
I just put it on GitHub. Check the pinned comment above!
@SRIGHT0
@SRIGHT0 4 месяца назад
Halo saa disuru bu guru njelasin
@40NoNameFound-100-years-ago
I am wondering why Mathworks took so long time to make such videos? We are in 2023 and version 2023 has been released. These control methods were introduced and had been modified since version 2006 .
@jimapost1496
@jimapost1496 7 месяцев назад
respect to matlab
@Amine-gz7gq
@Amine-gz7gq 4 месяца назад
For the videos, yes, but not for their software, which should be free for home users.
@Amine-gz7gq
@Amine-gz7gq Год назад
Thank you very much Brian. I love you man (I'm not gay 😂). I have a master degree in CS and system control but I never really understood system control and that's why when I have free time I enjoy watching your videos. software engineering sucks, I spend my day refactoring complex, buggy programs written by idiots and unconscious people. I'd like to do something more meaningful and intellectually stimulating like systems controls. Unfortunately, I wasn't taught the latter and maths very well, and thanks to some books and youtube videos, I'm in the process of fixing that. I have a question : Why don't we use exponential functions based on 'e' to scan/probe the sampled signal ? Is it because the signal is sampled and it doesn't make sense to use a function that expresses continuous growth/decay ? after all we can pose r = e^something.
@BrianBDouglas
@BrianBDouglas Год назад
The middle of page 608 here: www.dspguide.com/CH33.PDF explains why we use r instead of e. Hope that helps!
@kdre76
@kdre76 7 месяцев назад
L 1
@Alexis-ym9ph
@Alexis-ym9ph 11 месяцев назад
This explanation is actually pretty bad in comparison to explanation of other authors on RU-vid)
@youtube-username-placeholder
@youtube-username-placeholder 11 месяцев назад
I respectfully disagree😅
@Amine-gz7gq
@Amine-gz7gq 4 месяца назад
This subject (like many others) is so complex that a single video is not enough to explain it in depth. You need to watch several videos by different people to improve your understanding of the subject.
@marcinmartke2960
@marcinmartke2960 Год назад
boring, but not for me :)
@mightyparry
@mightyparry 11 месяцев назад
in cre di ble
@BrianBDouglas
@BrianBDouglas 10 месяцев назад
Glad you liked it!
@tidin_tss
@tidin_tss 5 месяцев назад
респект авторам, но ничего не понятно для чего его использовать вообще
@tim110-handle
@tim110-handle 8 месяцев назад
when will the z domain video come out?
@BrianBDouglas
@BrianBDouglas 8 месяцев назад
I was hoping by now! But the video got pushed for a bit to get other work done. It will be here one day though. Sorry!
@ناصرالبدراني-ش9س
this is teaching at its finest
@teebee3881
@teebee3881 Год назад
This helps so much, please make more of these
@heitorabreu3045
@heitorabreu3045 5 месяцев назад
love you Brian
@pijnappel03
@pijnappel03 8 месяцев назад
DTFT: 5:28
@joez9162
@joez9162 Год назад
@Brian I’d love to see you do an explanation of ADRC controllers!
@222_Official
@222_Official 7 месяцев назад
Amazing and very intuitive video, thanks!
@manfredbogner9799
@manfredbogner9799 8 месяцев назад
Very good
@BrianBDouglas
@BrianBDouglas 8 месяцев назад
Thanks!
@franciscorainero8673
@franciscorainero8673 Год назад
Is the shown Matlab application available on any link? (The one that calculates the FFT in min 7:50)
@BrianBDouglas
@BrianBDouglas Год назад
I just put it on GitHub. Check the pinned comment above!
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