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Im sorry if it seems like a trivial question but I am confused right now. Im currently watching my professors lecture. And he basically has $\lambda(t)$ as Step response and $k(t)$ as Impulse response of an LTI system. The integral of the Impulse response gives us the Step response: $$ \lambda(t) = \int^{t}_{0} k(t) dt $$

and $$ k(t) = \lambda^{\prime}(t) $$

So to see it in action I decided to make a Simulink model of that and here it is (at least I think its good if not please tell me how to do it right):

enter image description here

As you can see those plots are not identical but I was expecting them to be (?)

Why is that?

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    $\begingroup$ Given a system H(s) as the Laplace transform of the impulse response, if you multiply that by 1/s and take the inverse Laplace Transform you will get the step response (since the integral of an impulse is a step, and the Laplace transform of an integral is 1/s $\endgroup$ – Dan Boschen Jun 5 at 21:21
  • $\begingroup$ Exactly, so why arent those 2 plots identical? $\endgroup$ – John Jun 5 at 21:26
  • $\begingroup$ Well for the top input you put in a step which has a Laplace transform equal to 1/s, but for the bottom input you put in an impulse which has a Laplace transform equal to 1 , why would they be the same? $\endgroup$ – Dan Boschen Jun 5 at 21:28
  • $\begingroup$ Oh I see, it is cascaded with 1/s—- yes I agree- it should be identical, interesting $\endgroup$ – Dan Boschen Jun 5 at 21:29
  • $\begingroup$ I know right?? Im going crazy. Is it the fault of digitization? $\endgroup$ – John Jun 5 at 21:31
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Your system H(s) is continuous

$$ H(s) = \frac{s}{s+1} $$

Using a discrete impulse does not make a lot of sense. I suspect Simulink added a "zero-order hold" or ZOH block between your discrete impulse and your system $H(s)$. As such, your impulse response will be affected by the sampling period of your discrete impulse source. If you decrease the sampling period, the second curve shoud match the step response more closely, although they will never be exactly the same.

Edit :

Decrease the width of your impulse to say 0.1 second and increase the amplitude to 10 so that the area under the curve stays 1.

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  • $\begingroup$ That makes sense. But how can I decrease that sampling period in Simulink? $\endgroup$ – John Jun 5 at 21:52
  • $\begingroup$ I don't have a Simulink license at the moment but either in the solver settings or if you click on your discrete impulse source. $\endgroup$ – Ben Jun 5 at 21:53
  • $\begingroup$ Im not sure how to do that because the sample time (as shown in the screenshot) is just 1 for both $\endgroup$ – John Jun 5 at 21:57
  • $\begingroup$ Why not use 0.1 for sample time? You will have the adjust the amplitude of the impulse to 10 however. $\endgroup$ – Ben Jun 5 at 22:19
  • $\begingroup$ You are right! It works now. Almost the same. Thank you very much. It really did confuse me after I saw them to be different $\endgroup$ – John Jun 5 at 22:21

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