Understanding the Z-Transform

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  • Опубликовано: 27 янв 2025

Комментарии • 133

  • @BrianBDouglas
    @BrianBDouglas Год назад +41

    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

  • @nurahmedomar
    @nurahmedomar Год назад +116

    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!

  • @harrytsai0420
    @harrytsai0420 Год назад +56

    OMG! I've been waiting for the series of discrete-time control for nearly 5 years!

    • @BrianBDouglas
      @BrianBDouglas Год назад +4

      Hopefully, it's worth the wait!

    • @xptransformation3564
      @xptransformation3564 Год назад +2

      ​@@BrianBDouglashow did you get such a perfect understanding for all this?

  • @erickappel4120
    @erickappel4120 Год назад +21

    I was blown away by the clearness of the explanation! Looking forward to more!

  • @diegoperezromero6610
    @diegoperezromero6610 Год назад +8

    This is by far the best video about the dtft and z transform i've ever seen. Very good and very detailed explanation.

  • @MSDhoni-pz5wc
    @MSDhoni-pz5wc Месяц назад +1

    If i were given an opportunity to add as many likes as possible, it would top the list of mine. Watched many videos but this one is straight to the point. Thank you so much to the makers of this video.

    • @MATLAB
      @MATLAB  Месяц назад +2

      I'm glad you found it helpful! Brian Douglas makes excellent videos. Look for his other videos in this channel.

  • @emiliomartineziii2980
    @emiliomartineziii2980 11 месяцев назад +3

    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!

  • @MarkNewmanEducation
    @MarkNewmanEducation Год назад +9

    Amazing explanation. Thanks so much for explaining things in a more intuitive way.

  • @jmanius1
    @jmanius1 Год назад +34

    Brian is basically my mentor at this point

    • @BrianBDouglas
      @BrianBDouglas Год назад +6

      🤓 aww thanks!

    • @jmanius1
      @jmanius1 Год назад +2

      @@BrianBDouglas You responding has literally made my day. Thank you for your awesome videos!

    • @martinsanchez-hw4fi
      @martinsanchez-hw4fi Год назад +1

      Is there a list of the Matlab videos made by Brian?

    • @MATLAB
      @MATLAB  11 месяцев назад

      Hi @martinsanchez-hw4fi, You can find all of our Tech Talks on this page: bit.ly/MATLAB-Tech-Talks

  • @PrestonRogersJPL
    @PrestonRogersJPL Год назад

    This was the best explanation on DTFT I have ever seen! Bravo!

  • @terrycrews2281
    @terrycrews2281 Год назад +2

    been watching ur vids since 2014
    ty so much for this ,brian

  • @wellid2087
    @wellid2087 Год назад

    One of the best introductions I have ever watched

  • @grxoxl
    @grxoxl 11 месяцев назад

    Honestly, the best explanation of the DFT I've ever seen)

  • @DRAMBgo
    @DRAMBgo Год назад +5

    Crazy explanation Brian, thank you!

  • @Adhithya2003
    @Adhithya2003 Год назад

    The most intutive and lucid explanation of DTFT and Z transform I have ever seen 💖

  • @Michael-ze6oz
    @Michael-ze6oz Год 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.

  • @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!!!

  • @clarklau1292
    @clarklau1292 10 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.

  • @BingtheLizard
    @BingtheLizard Год 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.

  • @PankajSingh-dc2qp
    @PankajSingh-dc2qp 8 месяцев назад +2

    The input signal at @ 1:31 is delta[n] and not u[n].

  • @HimanshuSharma-b3q3u
    @HimanshuSharma-b3q3u 7 месяцев 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"

  • @runforitman
    @runforitman 3 месяца назад

    As with everything MATLAB, very clear and concise video. Thank you!

    • @MATLAB
      @MATLAB  3 месяца назад

      Thanks for the feedback. We appreciate it!

  • @giantbee9763
    @giantbee9763 9 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.

  • @franciskimonge1851
    @franciskimonge1851 Год назад

    my guy just made fourier transforms, laplace transorm and z transform all make sense , thank you

  • @vincentjupiterayuste2358
    @vincentjupiterayuste2358 Год назад +4

    Thank you Sir Brian

  • @mixguru9669
    @mixguru9669 Год назад +6

    My control systems professor should watch this video.

    • @sukursukur3617
      @sukursukur3617 Год назад +1

      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 RUclips FTW

  • @stewartcopland7676
    @stewartcopland7676 5 месяцев назад +5

    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?

    • @thehemperor3967
      @thehemperor3967 3 месяца назад

      I thought so too at first, but for the Laplace Transformation it's F(s) = Integral( 0, +inf, f(t)*e^-s)
      Where s = gamma + i*omega, because the expanded Integral definition is Integral from 0 to infinity of f(t) * e^(-i*omega) * e^(-gamma).
      The Laplace Transform adds a real part to the complex term s, to sweep through exponentials, whereas the Z-Transform, adds another imaginary part. For me it looks like it adds the amplitude of the sinusoids (r*e(-i*omega*x), instead of adding exponentials

    • @thehemperor3967
      @thehemperor3967 3 месяца назад

      I've googled it, and it seems like it is in fact the discrete version of the Laplace Transformation, but to make it happen the added part has to be this way, dont know why at this point

  • @teebee3881
    @teebee3881 Год назад +1

    This helps so much, please make more of these

  • @222_Official
    @222_Official 11 месяцев назад

    Amazing and very intuitive video, thanks!

  • @foxlies0106
    @foxlies0106 5 месяцев назад

    thank you. very nicely presented, and helpful.

  • @joshuamasila3427
    @joshuamasila3427 5 месяцев 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?

  • @philipprottweiler2924
    @philipprottweiler2924 Год назад +3

    Awesome explanation! When will the video on the Z-Domain be published? Really looking forward to it

    • @BrianBDouglas
      @BrianBDouglas Год назад +4

      I hope before the end of the year. I have 4-5 more videos in my queue before it though :(

  • @Volticymo
    @Volticymo Год назад +1

    Can’t wait for z domain video

  • @ناصرالبدراني-ش9س
    @ناصرالبدراني-ش9س Год назад +1

    this is teaching at its finest

  • @Dinner_Cat_低能貓
    @Dinner_Cat_低能貓 Год назад

    please make more of this ! it really really help !!!!

  • @hamedmajidian4451
    @hamedmajidian4451 Год назад +1

    You are the best looking forward to watching more and more!

  • @5vart5ol
    @5vart5ol Год 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).

  • @em2129
    @em2129 8 месяцев назад

    You are a wizard. Thank you!

  • @MrRaghunar
    @MrRaghunar Год назад

    Excellent explanation, great job!

  • @Pedritox0953
    @Pedritox0953 Год назад +2

    Great video!

  • @j.fkamaldeen
    @j.fkamaldeen Год назад +1

    Thanks alot Brian 🎉

  • @joelevi9823
    @joelevi9823 8 месяцев назад

    I can't find the link to the video about the Z domain.. will be happy to get it

  • @KingBhoomie
    @KingBhoomie Год назад +2

    1:27 Input u[n] generally refers to unit step function. And Unit impulse is represented by δ[n]

  • @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 !!

  • @DhaferThyab
    @DhaferThyab 3 месяца назад +1

    You are Genius 🎉

    • @MATLAB
      @MATLAB  3 месяца назад

      Yes Brian is awesome!

  • @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).

  • @joez9162
    @joez9162 Год назад

    @Brian I’d love to see you do an explanation of ADRC controllers!

  • @Eta_Carinae__
    @Eta_Carinae__ Год назад +1

    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?

  • @eduardojreis
    @eduardojreis Год назад

    16:50 - Where is the Z-Domain video?

    • @BrianBDouglas
      @BrianBDouglas Год назад +1

      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 Год назад +2

      @@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!

  • @ernstuzhansky
    @ernstuzhansky Год назад

    Many thanks Brian!

  • @Hashim1
    @Hashim1 Месяц назад

    3:50 how did you come up with z-1/z-1??? Where did it come from, I am metaphorically mashing my head against the wall trying to understand it.

    • @fabiobiffcg4980
      @fabiobiffcg4980 24 дня назад

      1*z^-0 + 1*z^-1 + 1*z^-2 ...
      1 * (z^-0 + z^-1 + z^-2 ... )
      1 = (z-1)/(z-1) # represent 1 using a fraction where numerator = denominator
      ( (z-1)/(z-1) ) * (z^-0 + z^-1 + z^-2 ... )

  • @pijnappel03
    @pijnappel03 Год назад

    DTFT: 5:28

  • @tim110-handle
    @tim110-handle Год назад +1

    when will the z domain video come out?

    • @BrianBDouglas
      @BrianBDouglas Год назад +1

      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!

  • @borisbvt5473
    @borisbvt5473 Год назад +1

    So, is z-transform kind of discreet Laplace Transform?

  • @BalajiSankar
    @BalajiSankar Год назад +1

    Thank you.

  • @rob9756
    @rob9756 Год назад +2

    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.

  • @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!

  • @MuhammdBilalNaz
    @MuhammdBilalNaz Год назад

    is there any open source on the internet about detail on z domain

  • @heitorabreu3045
    @heitorabreu3045 9 месяцев назад

    love you Brian

  • @martinsanchez-hw4fi
    @martinsanchez-hw4fi Год назад

    Why an impulse would be integrated to a constant value of 1?

  • @Neuromante73
    @Neuromante73 Год назад +1

    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 8 месяцев назад

      he put the github link in the description

  • @briancoon308
    @briancoon308 11 месяцев назад

    Did that z domain video ever get published?

    • @BrianBDouglas
      @BrianBDouglas 11 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 RUclips for "Applied DSP No. 9: The z-Domain and Parametric Filter Design"

  • @blessingshenjere484
    @blessingshenjere484 8 месяцев назад

    Like how he explains. Like a documentary

  • @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!

  • @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

  • @PrashantTiwari-i1b
    @PrashantTiwari-i1b 8 месяцев назад

    Common man I had math exam yesterday, where were you 😢

  • @braedenlarson9122
    @braedenlarson9122 11 месяцев назад

    So cool!

  • @TestName286
    @TestName286 9 месяцев назад

    If the components have both exponential and periodic signals, wouldn't that just the the Laplace transform?

    • @BrianBDouglas
      @BrianBDouglas 9 месяцев назад +1

      The Laplace transform is for continuous signals, the Z-transform is the discrete equivalent.

  • @kenakackrmn3898
    @kenakackrmn3898 8 месяцев назад +1

    5:25 sampai 12:09

  • @manfredbogner9799
    @manfredbogner9799 Год назад

    Very good

  • @LokeKS
    @LokeKS 7 месяцев назад

    brilliant

  • @MuhammdBilalNaz
    @MuhammdBilalNaz Год назад

    when the video on the z domain will be made and published

    • @BrianBDouglas
      @BrianBDouglas Год назад

      I don't know, it keeps getting pushed back :( But I'll get it out eventually.

  • @PankajSingh-dc2qp
    @PankajSingh-dc2qp 8 месяцев назад

    There is no integrator in discrete time. Discrete time has "accumulator".

  • @hamedhojatian3539
    @hamedhojatian3539 Год назад

    Could it BE any more simpler to explain DFT?

  • @mobilephil244
    @mobilephil244 5 месяцев назад

    This looks like exactly what the Laplace transform does. What is the difference ??????

    • @perekman3570
      @perekman3570 3 месяца назад

      It's discreet as opposed to continuous.

  • @jimapost1496
    @jimapost1496 11 месяцев назад +1

    respect to matlab

    • @Amine-gz7gq
      @Amine-gz7gq 8 месяцев назад

      For the videos, yes, but not for their software, which should be free for home users.

  • @40NoNameFound-100-years-ago
    @40NoNameFound-100-years-ago Год назад

    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 .

  • @MajuBersama-f7i
    @MajuBersama-f7i 10 дней назад

    Lojic👏

  • @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 +1

      The middle of page 608 here: www.dspguide.com/CH33.PDF explains why we use r instead of e. Hope that helps!

  • @tidin_tss
    @tidin_tss 9 месяцев назад

    респект авторам, но ничего не понятно для чего его использовать вообще

  • @SRIGHT0
    @SRIGHT0 8 месяцев назад

    Halo saa disuru bu guru njelasin

  • @kdre76
    @kdre76 11 месяцев назад

    L 1

  • @mightyparry
    @mightyparry Год назад +1

    in cre di ble

  • @Alexis-ym9ph
    @Alexis-ym9ph Год назад +1

    This explanation is actually pretty bad in comparison to explanation of other authors on RUclips)

    • @youtube-username-placeholder
      @youtube-username-placeholder Год назад +2

      I respectfully disagree😅

    • @Amine-gz7gq
      @Amine-gz7gq 8 месяцев 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 +4

    boring, but not for me :)