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Isnt a Step (as a Transfer function -> $ \frac{1}{s} $) and Impulse (trans fcn -> 1) therefore Impulse + $ \frac{1}{s} $ should be equal to a Step in Simulink right?

I made a Simulink model:

simulink model

But the outputted plots are not the same 🤔? Why is that? Am I wrong? Should those two blocks be essentially the same?

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The problem is that you can't use a Dirac delta impulse as an input, which would be the theoretically correct thing to do. Since you only have a discrete impulse, which has some finite value, and since Simulink probably uses something like a zero-order hold to make a continuous-time signal out of that impulse you actually use two different input signals: a step and a smoothed step.

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  • $\begingroup$ But theoretically Im correct right? Because Im refreshing some material and I just got really confused $\endgroup$ – John Dec 30 '20 at 21:35
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    $\begingroup$ @John: yes, but simulink (or anything real for that matter) won't let you do it. $\endgroup$ – Matt L. Dec 30 '20 at 21:36
  • $\begingroup$ Okay thank you :D I undestand that u cant really use something of infinite value such as a dirac impulse in real world. I was just worried that theoretically I don't understand something. But is there a way to make it work at least better than what matlab does by default? $\endgroup$ – John Dec 30 '20 at 21:38
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    $\begingroup$ @John: I'm not sure what you're trying to do, but if you want a step then just use a step. If you want to use a discrete impulse, then make sure that its amplitude is very high and also make sure that the width of the zero-order hold is very small (i.e., increase the sampling frequency). $\endgroup$ – Matt L. Dec 30 '20 at 21:41
  • $\begingroup$ Im not doing anything, just playing around. Thank you $\endgroup$ – John Dec 30 '20 at 21:48

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