Z transfer fcn under simulink and cmd line

I create a simple cmd from S to Z domain as below, and the result is matched with my understanding；

clear

rth1=0.0268;

tau1=1066.6;

Ts=0.05;

tf1=tf(rth1,[tau1 1])

dtf1=c2d(tf1,Ts)

figure;

step(tf1);

figure

step(dtf1);

but when i create this under simulink, the result is not matched, as you can see the Z transfer fcn output is quite different with the S transfer fcn.

Is there anyone can help me?

[Jon]

Please attach you Simulink model and any scripts or other functions needed to run your example - thanks

[Jon]

Please see my answer below, I was able to work out what the issue was without your code, so no need to attach it now

Answer 1

[Jon]

This appears to be due to a round off error in the denominator of the discrete transfer function input to Simulink.

Try for example with the denominator set to

[1 -0.999949]

instead of

[1 -1]

and you will see that the step response of the continuous and discrete approximation are now much closer

[Jon]

with the denominator set to

[1 -0.999953]

they are right on top of each other

[Jon]

If in your MATLAB command window you change the formatting to long and look at the denominator of your dtf1 you will see something like this:

format long
>> dtf1.Denominator

ans =

  1×1 cell array

    {[1 -0.999953123168882]}

So in fact the pole is not at 1, which would make it an integrator as you saw in your Simulink but rather just inside the unit circle

[Jon]

The underlying issue here is that your sample time is very small 0.05s relative to the time constant of over 1066s for your continuous system, so you are greatly oversampling the system. This makes the last term in your denominator very close but not actually equal to -1

[Bruce]

Thanks a lot, Jon. Your answer and explaination is quite clear, thanks again!

[Jon]

Your welcome, it was an interesting problem