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An explanation of the Z transform part 2 - the H(z) surface 

David Dorran
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22 апр 2018

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Комментарии : 28   
@matthewwallace9686
@matthewwallace9686 4 года назад
I can't believe how elegantly you just explained this. Can't be explained any more simply.
@PunmasterSTP
@PunmasterSTP Год назад
An explanation, part 2? More like "Amazing videos, and I never knew"...they were among the hidden gems of RU-vid!
@luk45ful
@luk45ful 3 года назад
This is a really good approach for understanding Z transform!
@ComposingGloves
@ComposingGloves 6 лет назад
I just found your channel and was watching the old one where everyone was sad cause you hadn't uploaded part 2 and then I find out that 4 days ago you did just that after 2 years! Thanks so much for these clear explanations!
@ssimm008
@ssimm008 6 лет назад
glad you're back I enjoy your videos
@lindnerlars81
@lindnerlars81 4 года назад
With this video I gave the last piece of the puzzle to really understand the convergence of the z Transformed. I thank you very much for that. :)
@ddorran
@ddorran 4 года назад
It's nice to hear that Lars. Always great to hear my videos have an impact!
@jaikumar848
@jaikumar848 6 лет назад
I was waiting for 2 years ....finally you uploded..you are best no doubt
@ddorran
@ddorran 6 лет назад
Thanks - apologies for the delay.
@LorenMLang
@LorenMLang 6 лет назад
At 3:55, I think you dropped the sign on the phase angle when raising to the power of -n. I believe it should be 0.4243^-n*e^(j*pi/4n) which equals 1.66 + 1.66j for z^-1. Otherwise, great video! I'm loving this series.
@PunmasterSTP
@PunmasterSTP Год назад
I was confused and initially thought he was correct to, but trying to analyze and reply to your comment showed me the light. Thank you so much for pointing that out!
@adpro57
@adpro57 2 года назад
How did you make those wonderful 3D argand plots with the unit circle highlighted in yellow?
@ddorran
@ddorran 2 года назад
There is function in matlab and octave called mesh that I use to create 3D plots. You can see some code that might help understand how it works at dadorran.wordpress.com/2012/04/07/zpgui/
@PedramNG
@PedramNG 5 лет назад
Wow amazing
@mikehagerty9666
@mikehagerty9666 4 года назад
At [4:00], z=0.4243e^(-j*pi/4) so z^-n *should* be: z^-n = (0.4243)^-n x e^(+j*pi*n/4) . Note that you have the wrong sign on the exponent, so it should be: H(z)=1.33 + 0.83j, but of course this doesn't change the magnitude, |H(z)|.
@v.p22709
@v.p22709 4 года назад
thank you
@kayliman
@kayliman 6 лет назад
On 3:54 when you put in -n, it must than be e^+jpi/4n right? But I think it doesn't matter at the end because of the symmetry. And thanks allot for your tutorials.
@ddorran
@ddorran 6 лет назад
-jpi/4 is correct - the angle of the complex number is -pi/4
@darius0852
@darius0852 5 лет назад
@@ddorran but e^(-jpi/4) multiplied by power (-n) is equal to e^(jnpi/4) no? at 3.54
@JasonConklin
@JasonConklin 5 лет назад
@@darius0852 You're correct, there's a sign error. The purple text at 4:15 should show e^(-j*pi/4*(-n)) = 1.66+1.66j This is easily verified in Matlab: >> z = complex(0.3,-0.3) z = 0.3000 - 0.3000i >> z^-1 ans = 1.6667 + 1.6667i As you said, because this result is the complex conjugate on a symmetric surface, the final result is still correct. Not to criticize the video series, though -- this is great stuff!
@PunmasterSTP
@PunmasterSTP Год назад
@@ddorran I think the other commenters are correct and the formula for the phase angle becomes positive because you're raising it to negative n. But I don't really think that takes away from the quality of all of your videos, which is excellent!
@jacobhorne2763
@jacobhorne2763 3 года назад
today I learned brits say "haych" for H, instead of "aych"
@ddorran
@ddorran 3 года назад
I'm Irish not British. In Britain both pronunciations are used although I think 'aitch' is more commonly used. Catholics in Northern Ireland tend to use 'haitch' while protestants generally use 'aitch'.
@AshishPatel-yq4xc
@AshishPatel-yq4xc 2 года назад
Goes so fast, can't follow any of it
@PunmasterSTP
@PunmasterSTP Год назад
Yeah there was definitely a lot of information. But I think learning this type of stuff can require second (and third) watches. There's also 0.75x speed which I sometimes find helpful.
@shawnscientifica6689
@shawnscientifica6689 5 лет назад
I appreciate these videos but they don't seem to TEACH anything. More or less, you're simply TELLING people something.
@fredericosevergnini8204
@fredericosevergnini8204 4 года назад
To me his videos are the most important ones related to DSP on RU-vid. Most professors are worried about teaching the math, which paints a very abstract figure of what is going on. People learn the expressions and how to solve them, but have no clue what they actually mean or how this can be useful in real life. In videos like this the author is literally giving meaning to the concepts that are so popular in signal processing, like what is the Z transform and how to visualize it.
@PunmasterSTP
@PunmasterSTP Год назад
I think that's kind of the nature with non-interactive media like text and videos. I think interactivity like working through example problems and getting real-time feedback or discussing questions with other people in your class would probably be much more helpful.
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