Pole/Zero Plots Part 1

Sdílet
Vložit
  • čas přidán 5. 09. 2024
  • An introduction to pole/zero plots.
    This video is one in a series of videos being created to support EGR 433:Transforms & Systems Modeling at Arizona State University. Links to the other videos can be found at sites.google.co...

Komentáře • 26

  • @socencounter
    @socencounter Před 11 lety

    Thank you so much for this! This is very useful in my Analog IC design class where each pole will result in a capacitor! Now I know how to find the poles!

  • @karimkhan1312
    @karimkhan1312 Před 11 lety +1

    i am only advocate in india-- but i have under stood lecture completely

  • @MrMurtican
    @MrMurtican Před 12 lety +1

    Ẃell explained. Thank you for uploading this.

  • @sagatyba
    @sagatyba Před 12 lety +1

    thanks for the video! your mistake made me check the roots myself it was a good one! I wish you made this while I was in grad school. thank you thank you thank you *-----*

  • @mondo1019
    @mondo1019 Před 8 lety

    Thank you Thank you Thank you very well explained exactly what I needed!

  • @josepaul2000
    @josepaul2000 Před 11 lety

    Those are the roots of (s^2+4s+13). When s= either of those values you wrote, (s^2+4s+13) will equal zero.

  • @vincentedge400
    @vincentedge400 Před 11 lety +4

    can you please be so kind to explain how you reached the roots -2+j3 & -2-j3

    • @Horizon__
      @Horizon__ Před 4 lety

      By using the quadratic formula on the quadratic polynomial where j is sqrt -1

  • @alok696969
    @alok696969 Před 11 lety

    Very good lecture sir

  • @InitiateAnnihilation
    @InitiateAnnihilation Před 9 lety +3

    Could you kindly share the matlab code to plot this 3 axis diagram?

  • @circuithead94
    @circuithead94 Před 10 lety

    Nice video. I don't understand one part however. In the end the on the left of your basis set the exponential is decreasing which I feel is not consistent with my understanding of the basis set as any thing on the left of the basis set should follow an exponential growth rather than a decay. Because s= sigma + j Omega
    and e^(-sigma *t) is positive exponential growth on the left of basis set. Could you clarify this problem ? thanks

  • @dynamics529
    @dynamics529 Před 11 lety

    very well done thank you..

  • @the1618034
    @the1618034 Před 11 lety

    great series of videos these are, esp to control student. i find them easier to understand than the textbooks. i do have one question: is it possible you Sir to write down the MATLAB code for producing those beautiful 3D representation of a pole-zero plot. i would appreciate it if you do. thanks

  • @Mahmood367
    @Mahmood367 Před 8 lety +6

    how did u draw this thing with matlab??

  • @rusbelhernandez4834
    @rusbelhernandez4834 Před 5 lety

    what is the graph used for? what do you do with it?, what is need it for?

  • @davidantoniocornejoacosta5484

    thank you! can you show how to plot de 3D graphic on matlab

  • @abdallah.alammar
    @abdallah.alammar Před 5 lety

    and what is the name of the third axis?

  • @nareshmedaram7514
    @nareshmedaram7514 Před 11 lety

    Sir its really good lecture
    I do have one doubt that "what are real time applications of poles and zeros"
    please reply me sir .......Thank you.

  • @mohamedalaa4931
    @mohamedalaa4931 Před 7 lety

    could you show as the matlab code to draw this ? ... thanks

  • @nirbhaythacker6662
    @nirbhaythacker6662 Před 7 lety

    PLEASE SHARE YOUR MATLAB CODE FOR THAT PLOT, I AND SO MANY OTHER PEOPLE HAVE BEEN WANTING TO MAKE IT, BUT WE REAAAAAALLLLLLY SEEM TO BE GETTING WRONG PLOTS.

  • @grahammckay3324
    @grahammckay3324 Před 5 lety +2

    %Running on Matlab 2017
    [x,y] = meshgrid(-4: .05: 4);
    s = x + (j*y);
    z = abs(( s + 1) ./ (s.*(s.^2 + (4.*s) + 13)));
    figure;
    surf(x,y,z);
    xlabel('real');
    ylabel('imag');
    title ('Pole/Zero Plots Part 1');

  • @enbardis
    @enbardis Před 7 lety

    Why is infinity not a zero?

    • @emlmm88
      @emlmm88 Před 6 lety

      Elias Bardis Because, rigorously speaking, infinity isn't a value.

  • @sandun001
    @sandun001 Před 11 lety

    I'm still in the grad school. but I couldn't understand this yet.