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We tend to identify systems more often by using the step response. Why? Especially when the impulse response is directly related to the transfer function?

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marked as duplicate by Olin Lathrop, Dmitry Grigoryev, PeterJ, uint128_t, CL. Sep 6 '17 at 6:45

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  • \$\begingroup\$ You are correct, in so far that the transfer function is the Laplace transform of the impulse response. However, I think we are generally more interested in how the system transfers to a new state whenever the input changes it's state, rather then the response to a temporary input disturbance, like a Dirac pulse. \$\endgroup\$ – Bart Sep 5 '17 at 10:53
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    \$\begingroup\$ Because real systems don't like infinite input voltage. \$\endgroup\$ – Dmitry Grigoryev Sep 5 '17 at 14:33
  • \$\begingroup\$ It's kind of hard to feed a system a Dirac delta in the real world, but at least you can approximate a step fairly well... \$\endgroup\$ – Mehrdad Sep 5 '17 at 22:32
  • \$\begingroup\$ @DmitryGrigoryev - But real test systems don't generate true square waves/steps. \$\endgroup\$ – Hot Licks Sep 6 '17 at 2:45
  • \$\begingroup\$ @HotLicks Depending on your definition of "true", true sine waves or true DC may also not really exist. My point is that most systems under test will like a 5V step much better than a 5kV 100ns pulse. \$\endgroup\$ – Dmitry Grigoryev Sep 6 '17 at 11:35
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Step response and transfer function are interchangeable via Laplace transform (step response -> differentiate to get impulse response -> laplace -> transfer function), but you can't run a Laplace transform on a graph pulled out of a datasheet...

For example, the user of a LDO will usually be interested in things like voltage sag/overshoot when load current steps up and down, so in this context a graph of the step response is useful.

Same if you design the control system for something like a mechanical actuator: the transient response is often more important than transfer function, you'd want to avoid overshoot, avoid excessive slew rate, have good damping, things like that. Step response shows all this in a way that is easy to understand, whereas transfer function will not (instead, it provides more insight into stability, etc).

Also testing the step response is much easier than measuring the transfer function.

And the transfer function is only valid when the system is linear, not when it is slewing. If slew limit is involved, or other large signal conditions, then step response is no longer the Laplace transform of the transfer function, because the system is no longer linear time invariant.

Opamps' datasheets usually give both, since we're both interested in the frequency response, and settling time/ringing/overshoot, clipping recovery, etc.

EDIT: I just realized you asked why step response was used instead of impulse response:

  • It is easy to generate a fast step, very difficult to generate an "infinitely short" pulse (or an approximation of).
  • The step contains much more energy and will generate a much larger response. Whereas a pulse would only generate a small blip on the output. Signal-to-noise ratio is much better with a step.
  • An impulse would not allow to test for slew rate. It would provide little information.
  • Thus the step is easier to use in practice. If you want the impulse response, do a step, then differentiate.
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    \$\begingroup\$ I don't agree with your assertion that the step response is the Laplace transform of the transfer function, but the unit impulse response is the inverse Laplace transform of the transfer function. \$\endgroup\$ – Bart Sep 5 '17 at 12:56
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An impulse, i.e. hit it with a hammer, is not very friendly, particularly in systems that have mechanical bits and pieces.

A perfect impulse cannot be generated, so you have to tailor the pulse duration to the system.

A pulse that gives meaningful response data needs to be relatively strong, and that means high amplitude (strength = area = height x duration, and duration=small, therefore height=large)

To get the step response from the impulse response you have to integrate, which means you need to know initial conditions.

An impulse response does not give DC gain information readily; a step response does.

A step isn't as violent as an impulse.

A step is a step regardless of the system dynamics.

You can get the impulse response from the step response by differentiating - and doesn't require initial conditions.

The step response is often the inherent integral of the impulse response (eg motor velocity to motor displacement) and integration has noise-rejection characteristics.

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  • \$\begingroup\$ A perfect impulse cannot be generated, that is true, but also the generation of a perfect step or ramp is not possible. But in praxis, a step aproximation is better than the aproximation of an impulse. \$\endgroup\$ – Uwe Sep 5 '17 at 15:58
  • \$\begingroup\$ Impulse ("hammer") testing is a very standard technique for mechanical systems - it is quick, and after mounting accelerometers at fixed positions on the structure you can apply the impulse at as many different positions as you want. Of course the impulses are not mathematically perfect - therefore you also record the output from a force gauge in the hammer head, so you know what impulse you actually applied in each measurement, and get the measured transfer function, not just the un-normalized impulse response. The "hammers" used can weigh from 20g to 20kg, depending what you are testing! \$\endgroup\$ – alephzero Sep 5 '17 at 22:10
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It's possible to describe an "almost perfect" step function using positive parameters δ, ε, and ε', such that:

  1. For all values of t<-δ, the output will be between -ε and +ε.
  2. For all values of t>+δ, the output will be between 1-ε and 1+ε.
  3. For all other values of t, the output will be between -ε' and 1+ε'.

Graphically, such functions may be described by saying that their output is always within an "envelope" that looks like:

 :                █
 :                █▀▀▀▀▀▀▀▀▀▀▀▀▀▀ y=1
 :                █                
 :                █                
 :▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀▀█▔▔▔▔▔▔▔▔ y=0
 :                ▀ 

Any application that would work with all functions that would meet the requirements for some particular values of δ, ε, and ε' would be satisfied by any function that met the requirements for smaller values of those parameters. The exact position of the function within the envelope may be poorly defined, but most applications wouldn't be interested in that level of detail beyond having upper limits for δ, ε, and ε'. This thus makes it easy to describe a set of functions that will be "good enough" for a particular purpose, and ensure that the behavior of a real-world system is a member of that set.

An "almost perfect" impulse function, by contrast, cannot be described in such a fashion. Behavior may be well-defined at the points where nothing is happening, but the part of the domain where behavior is least well-defined is also the part of the domain where everything interesting is happening.

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  • \$\begingroup\$ This is nice mathematics, but irrelevant in practice, since it is easy to measure the applied impulse (or step) with the same time-domain resolution as you measure the response. \$\endgroup\$ – alephzero Sep 5 '17 at 22:18
  • \$\begingroup\$ @alephzero: If an impulse is somewhere around 1usec wide and one wants to measure its total amplitude accurate to 1%, one would need to be able to resolve its width accurate to 10ns. If one looks at the step function instead, all one would need to know is the value at any time before it started to change, and at any time after it had stabilized. Performing the latter measurement with the required accuracy is apt to be much easier than doing the former. \$\endgroup\$ – supercat Sep 5 '17 at 22:26

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