z-transform of unit step function?
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Diamond
on 26 May 2014
Commented: Fardin Ahasan Maraz
on 14 Nov 2021
i have this n-domain function H[n]= (0.5^n)*u[n] and i need to find z transformation of this function. i've tried this one:
>> syms n z
>> y = ((0.5)^n)*heaviside(n);
>> yz = ztrans (y,n,z)
yz =
1/(2*z - 1) + 1/2
isn't supposed to be 1/(-0.5^z-1)+1 ? is there any solution to solve this z transform?
Accepted Answer
Geoff Hayes
on 27 May 2014
I don't have the Symbolic Toolbox but I think that the provided answer is correct. If
x[n] = 0.5^n * u[n]
=> x[n] = {…,0,0,0.5^0 * 0, 0.5^1 * 1, 0.5^2 * 1,…}
then the z-transform is
sum(n=-Inf:+Inf)(0.5^n)*(z^-n)
= sum(n=0:+Inf)(0.5^n)*(z^-n) % ignore all n<0
= 0.5 + sum(n=1:+Inf)(0.5^n)*(z^-n)
= 0.5 + sum(n=1::+Inf)(0.5/z)^n
= 0.5 + (0.5/z)/(1-(0.5/z)) % assuming abs(0.5/z)<1
= 0.5 + 0.5/(z-0.5)
= 1/2 + (1/2)/(z-1/2)
= 1/(2*z - 1) + 1/2
2 Comments
Geoff Hayes
on 27 May 2014
As Azzi mentions below, MATLAB's heaviside defines H(0) to be 0.5, whereas other definitions of the same function define H(0) to be 1. (See http://en.wikipedia.org/wiki/Heaviside_step_function#Zero_argument for details.)
The formula in the third row of your table assumes that H(0)=1 and can be derived similar to above.
I think that it is fine to define u[n] with heaviside(n) as long as your are aware (and state) that heaviside(0) = 0.5.
More Answers (3)
Amine BENZOUBIR
on 22 Oct 2019
use this commande to change the origine to 1
v=1;
sympref('HeavisideAtOrigin', v);
0 Comments
Azzi Abdelmalek
on 27 May 2014
Edited: Azzi Abdelmalek
on 27 May 2014
The heaveaside function of Matlab is defined with heaviside(0) equal to 0.5
If you look at the table using another definition of heaviside (e(0)=1), you will find the z-transform of a^n is z/(z-a) . The heaviside defined in Matlab can be written as
heaviside(n)=e(n)-delta(n) (delta is Kronecker function), the z-transform is z/(z-a)-0.5
In your case replace a by 0.5
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