Discrete transfer fcn 1/z-1 (Z-transform) How it Works?

15 views (last 30 days)
Hi,
i would like to ask you for little explanation of a simple model.
In this model is one Constant block with value = 1.
The second block is discrete transfer fcn with Z-Transform in this form : 1/z-1
time is set to 2 sec and the sample time of trans. fcn blok to 1. the result is 2 but why? how it works.. this concrete function 1/z-1
and in my view it works like this: 1*1^(-1) + 1*^(-2) = 2 (but im not sure, plz answer if its wrong or correct)
my next question is, can be z=1 , i think "z" in form at the page i pasted? or what the "z" is?
thank you very much. i hope its easy but i realy wondering about it a lot.

Accepted Answer

Azzi Abdelmalek
Azzi Abdelmalek on 25 May 2013
Edited: Azzi Abdelmalek on 25 May 2013
If the input of your system is u[n] and its output is Y[n] then
% H(z)=Y(z)/U(z)
% 1/(z-1)=Y(z)/U(z)
% which gives Y(z)(z-1)=U(z)
% zY(z)-Y(z)=U(z)
% In the temporal domain
% y[n+1]-y[n]=u[n]
% y[n+1]=y[n]+u[n]
% finaly: y[n]=y[n-1]+u[n-1]
% In your example
% y[1]=y[0]+u[0]= 0+1
% y[2]=y[1]+u[1]=1+1=2
% y[3]=y[2]+u[2]=2+1=3
% and so on
  1 Comment
Zsolt
Zsolt on 25 May 2013
Edited: Zsolt on 25 May 2013
Thank you for fast answer. And can u tell me please if this is correct? http://i.imgur.com/GH6MlRZ.png or how can i get the same result with this form of sum?

Sign in to comment.

More Answers (0)

Categories

Find more on Programming in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!