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

Asked by Zsolt

Zsolt (view profile)

on 25 May 2013


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:

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.



Zsolt (view profile)


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1 Answer

Answer by Azzi Abdelmalek

Azzi Abdelmalek (view profile)

on 25 May 2013
Edited by Azzi Abdelmalek

Azzi Abdelmalek (view profile)

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 (view profile)

on 25 May 2013

Thank you for fast answer. And can u tell me please if this is correct? or how can i get the same result with this form of sum?

Azzi Abdelmalek

Azzi Abdelmalek (view profile)

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