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I was reading about second order systems and forced/natural responses:

enter image description here

I just want to explain my understanding of what generates the forced and natural responses and want to ask if it is correct:

  • The input poles of the input function decide the form of the forced response - step, etc.
  • The system poles decide the form of the natural response.
  • The amplitude of the forced and natural responses are dictated by ALL of the poles - both input and system.

I don't know if my explanation is correct because if you consider a system that is completely underdamped, that will just continue to oscillate at the output when a step input. In that case, we never even see the form of the forced response at the output (another step) or is the forced response there but just overshadowed by the natural oscillation?

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  • \$\begingroup\$ Applying a step input to a non-damped system is like powering up an oscillator. You see the forced response immediately as before, the system was flat silent. And it's the same for any other system. \$\endgroup\$
    – Janka
    Jun 26, 2021 at 14:09
  • \$\begingroup\$ Oh.. That makes a lot of sense actually. Just the fact that we see oscillations after applying the step and having nothing before is the forced response itself. \$\endgroup\$ Jun 26, 2021 at 14:11
  • \$\begingroup\$ It's the main difference between the Fourier and Laplace transformations. The Fourier transformation is about the steady state when all effects from single events sink into the noise, and the Laplace transformation is especially about the single events. \$\endgroup\$
    – Janka
    Jun 26, 2021 at 14:34
  • \$\begingroup\$ "In that case, we never even see the form of the forced response at the output". We do! The resulting oscillation is actually shifted up by the step part! \$\endgroup\$
    – AJN
    Jun 26, 2021 at 15:02
  • \$\begingroup\$ @jaurunjljgrtutkwcy You really got some nice, succinct answers here! +1 to all. \$\endgroup\$
    – jonk
    Jun 26, 2021 at 17:29

3 Answers 3

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In that case, we never even see the form of the forced response at the output.

We do! The resulting oscillation is actually shifted up by the step part! Notice that the response is always above zero. this is because the step shifted the sinusoid part of the oscillation (which is otherwise symmetric about zero).

is the forced response there but just overshadowed by the natural oscillation?

Yes.

sys = tf([100], [1, 0.1, 100])
step(sys, 10);

step response of under damped system

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  • \$\begingroup\$ Ahh. That plot explains it really well. So the natural oscillation is just centered around the forced response. A side question - is there any way to view just the natural response at the output? I mean, we could put some initial conditions into the system but that's the same as applying a forced input poles right since a system TF is defined with zero initial conditions? \$\endgroup\$ Jun 26, 2021 at 15:11
  • \$\begingroup\$ IMO, You are right. If by view, you mean in a simulation, I guess you could plot the part of the inverse transform corresponding to the natural response (and suppress the other term). In a real system: you can probably get similar effect either by setting an initial condition and zero input or an impulse input. \$\endgroup\$
    – AJN
    Jun 26, 2021 at 15:18
  • \$\begingroup\$ The natural response is the response to initial conditions. The forced response is the response to input signals or 'forcing functions'. \$\endgroup\$
    – Chu
    Jun 27, 2021 at 7:57
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The input poles of the input function decide the form of the forced response - step, etc.

If by "form" you mean whether it's constant or damped sinusoid or whatever -- yes. But the input function can have zeros, too, and those do shape the response. For instance, an input function of the form \$X(s) = \frac{a s + b}{s^2}\$ will have a step of amplitude \$a\$ plus a ramp with a slope of \$b\$.

The system poles decide the form of the natural response

Yes.

The amplitude of the forced and natural responses are dictated by ALL of the poles - both input and system.

Poles and zeros

if you consider a system that is completely underdamped, that will just continue to oscillate at the output when a step input.

You will see the input in the output, but a "forever" natural response could be there, too.

But note: if you're overly clever, the "forever" natural response won't be forever. Here's a "low pass" filter with zero damping, $$H = \frac{\omega_0^2}{s^2 + \omega_0^2}$$ where \$\omega_0 = 5(2\pi)\$. It's excited by a finite-duration pulse that someone just happened to arrange to be exactly five cycles of the filter's natural response.

It even has a Laplace transform -- $$X(s) = \frac{1 - e^{- s}}{s}$$

enter image description here

In time-domain terms, the reason that the natural response goes away is because the input signal acts like a step at \$t = 0\$, added to a negative step at \$t = 1\$. Because the "negative step" is an integer number of cycles after the initial step, it's response cancels out the natural response.

In frequency-domain terms, the reason that there's no persistent natural response is because \$X\$ has an infinite number of zeros on the imaginary axis at \$\omega = 2 \pi n\$. One of these falls at \$\omega = 0\$, which is why there's no persisting DC response to the input; another falls at \$\pm\omega_0\$, which is why there's no persisting natural response.

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The natural response of a system is the response to initial conditions; the forced response is the response to an applied signal or signals.

If a system has only initial conditions and no input signal, the response is the natural response.

If a system has zero initial conditions and one or more inputs are applied, the response is the forced response.

If a system has initial condition(s) and input signal(s), the response will contain both a natural response and a forced response. In this case the residues at the system poles will have contributions from the natural response and the forced response.

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  • \$\begingroup\$ I understand but in this case when a step input is applied, the system will respond with it's natural response even though there are no initial conditions? \$\endgroup\$ Jun 27, 2021 at 15:54
  • \$\begingroup\$ I think in the s-domain, having initial conditions or a step input are modelled the same. \$\endgroup\$ Jun 27, 2021 at 16:02
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    \$\begingroup\$ @jaurunjljgrtutkwcy, No. When a step is applied the system will respond with its forced response to a step function. And no again, initial conditions are NOT the same as applying a step function, that's why they have different names. Also, if a system has initial conditions then it cannot be represented by a transfer function. A transfer function requires zero initial conditions. \$\endgroup\$
    – Chu
    Jun 27, 2021 at 16:48
  • \$\begingroup\$ Ok I am confused. I understand what you are saying but if I consider an undamped system with zero initial conditions and I apply a step input to that system, the system will oscillate at it's natural frequency, so that will be the natural response or is that the forced response? \$\endgroup\$ Jun 27, 2021 at 17:31
  • \$\begingroup\$ If you have a look at the image I have in the original post, they show exactly that. Applying a step input (1/s) to a transfer function (assuming zero ICs) and they group the output into a forced response and a natural response. How is the step input in this case generating both a natural and forced response? \$\endgroup\$ Jun 27, 2021 at 17:34

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