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

[@matthewwallace9686]

I can't believe how elegantly you just explained this. Can't be explained any more simply.

[@PunmasterSTP]

An explanation, part 2? More like "Amazing videos, and I never knew"...they were among the hidden gems of KZbin!

[@ComposingGloves]

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!

[@ahmedali6519]

This is a really good approach for understanding Z transform

[@luk45ful]

This is a really good approach for understanding Z transform!

[@lindnerlars81]

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]

It's nice to hear that Lars. Always great to hear my videos have an impact!

[@LorenMLang]

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]

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!

[@ssimm008]

glad you're back I enjoy your videos

[@mikehagerty9666]

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)|.

[@jaikumar848]

I was waiting for 2 years ....finally you uploded..you are best no doubt

[@ddorran]

Thanks - apologies for the delay.

[@adpro57]

How did you make those wonderful 3D argand plots with the unit circle highlighted in yellow?

[@ddorran]

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/

[@kayliman]

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]

-jpi/4 is correct - the angle of the complex number is -pi/4

[@darius0852]

@@ddorran but e^(-jpi/4) multiplied by power (-n) is equal to e^(jnpi/4) no? at 3.54

[@JasonConklin]

@@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]

@@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!

[@PedramNG]

Wow amazing

[@jacobhorne2763]

today I learned brits say "haych" for H, instead of "aych"

[@ddorran]

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'.

[@v.p22709]

thank you

[@AshishPatel-yq4xc]

Goes so fast, can't follow any of it

[@PunmasterSTP]

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]

I appreciate these videos but they don't seem to TEACH anything. More or less, you're simply TELLING people something.

[@fredericosevergnini8204]

To me his videos are the most important ones related to DSP on KZbin. 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]

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.

[@ahmedali6519]

This is a really good approach for understanding Z transform!