[Z transform] problem with MATLAB

Hi experts, I have a question about Z-transform on MALTAB. When I convert a Laplace function F(s)=1/s to Z function, MATLAB says it is T/(z-1), but the Laplace-Z conversion table show that is z/(z-1). I know MATLAB cannot wrong because I drew a step graph of all these three functions. But all the books I found about Laplace and Z-transform also say the conversion table is right.

It's confusing me and I need an answer. Could anyone give me an explain for that? Thank in advanced.

Answer 1

[Star Strider]

The Symbolic Math Toolbox doesn’t know you’re trying to convert it from the Laplace domain to the ‘z’ domain. You have to play by its rules.

This works:

syms s t z 
T1(s) = 1/s
T2(t) = ilaplace(T1)
T3(z) = ztrans(T2)

T1(s) =
 1/s

T2(t) =
 1

T3(z) =
 z/(z - 1)

[Franklin Ngo]

Thank you for your answer. SOrry for my unclear question. I mean I use the function "c2d", and the result is totally different. As you can see:

>> T_s=tf(1,[1 0])

T_s =

1
-
s

Continuous-time transfer function.

>> c2d(T_s,0.01)

ans =

0.01
-----
z - 1

Sample time: 0.01 seconds Discrete-time transfer function.

Could you explain that for me, please.

[Star Strider]

As close as I can get is to use the Tustin approximation for ‘s’, and the Symbolic Math Toolbox for the derivation:

syms s t Ts z 
T1(s) = 1/s
Ts = sym(0.01);
T4(z) = 1/((2/Ts)*(z - 1)/(z + 1));                      % Substitute The Tustin Approximation For ‘s’
T4 = vpa(simplify(T4, 'Steps',10), 3);
pretty(T4)

    0.01
  ------- + 0.005
  z - 1.0

You may have to ask MathWorks about the exact implementation of the c2d function. It is not obvious from the documentation.

[Franklin Ngo]

OK. Thank you so much.

[Star Strider]

As always, my pleasure.

If my Answer hslped you solve your problem, please Accept it!

[Jeovane Sousa]

As I understand from here https://www.mathworks.com/help/control/ug/continuous-discrete-conversion-methods.html#bs78nig-12

Matlab uses the following method to calculate c2d by ZOH

The ZOH block generates the continuous-time input signal u(t) by holding each sample value u(k) constant over one sample period:

The signal u(t) is the input to the continuous system H(s). The output y[k] results from sampling y(t) every Ts seconds.

Resolving to :

each term will look something like that

thus,

And

Or remember the z-transform of a signal tha goes through a sample and holder, tha is:

This is what I think how c2d was implemented. No garantees it is correct or valid.

I hope I have made the correct considerations. Feel free to complement or point any mistake. Sorry about the poor English, it is not my primary language.