Discover MakerZone

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn more

Discover what MATLAB® can do for your career.

Opportunities for recent engineering grads.

Apply Today

To resolve issues starting MATLAB on Mac OS X 10.10 (Yosemite) visit: http://www.mathworks.com/matlabcentral/answers/159016

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

Asked by Zsolt on 25 May 2013

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

i have found an explanation here: http://www.physicsforums.com/showthread.php?t=134657

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.

0 Comments

Zsolt

Products

No products are associated with this question.

1 Answer

Answer by Azzi Abdelmalek on 25 May 2013
Edited by Azzi Abdelmalek on 25 May 2013
Accepted answer

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 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?

Azzi Abdelmalek

Contact us