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I was reading this book and saw this:

enter image description here

They apply a step input R(s) to a system G(s). Now, they say that the input pole of R(s) at the origin generated a forced response of 1 in Equation 4.6. And the system pole generates the natural response which is the exponential term in Equation 4.6

My understanding was that the forced response is the response of the system due to input ONLY.

Here, they apply a input function but are now saying the output c(t) has both a forced and natural response?? Are they assuming some initial conditions in the system - otherwise, I don't understand how they are getting the exponential term!

I think they are assuming some initial conditions, since the system pole is at -a, not zero.

jaurunjljgrtutkwcy
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2 Answers2

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The forced response means that part of the response that would be zero if there was zero input. It does not mean that the response has exactly the same form as the input. So it can contain all kinds of system-dependent terms. The important point is that it is proportional to the input...but that proportionality factor is in general a system-dependent function of s.

rpm2718
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  • Hmm. But in the above case, if we had zero initial conditions in the system, would we not only get the 1 at the output? And similarly, if we applied a 0 input, would we not only get the exponential term? – jaurunjljgrtutkwcy Jun 12 '21 at 19:34
  • Typically when you are given the transfer function of a system, that is the zero-initial-condition response. So the response (including the exponential term) is what you get for zero initial conditions. If you wanted the zero-input response, we would need to know the 'zero-input response function,' which was not given in the problem. – rpm2718 Jun 12 '21 at 19:40
  • Hmm. But if the TF of the system is assuming 0 initial conditions. If we apply, no input, we should not get any output. If we apply a step input, we should get the forced response. Here, they have a TF (I assume with 0 initial conditions), they force a step input and somehow manage to get both a forced and natural response at the output? I thought natural responses are purely due to the initial conditions of the system. – jaurunjljgrtutkwcy Jun 12 '21 at 19:45
  • The natural response is what you get with only initial conditions, and no input. However, that does not mean you won't get similar terms in your driven response. The system response contributes to the forced response.....it is not simply proportional to the driving input, with no system-dependent multiplicative factors. So I think it is a bit of a confusing way that the author has chosen to call that the 'natural response.' That term would surely show up in the natural response, but that does not mean that there was an initial condition here that generated it. – rpm2718 Jun 12 '21 at 19:49
  • Yes, that was exactly my understanding of it. I understand that you can get fixed constants and exponential terms at the output of your system purely due to a forced input (depends on the system what you get at the output) - in that case both the fixed and exponential term would be classed as the forced response. My confusion was why he was splitting the output c(t) and calling one part the forced response and one the natural response, even though both are due to the forced input. – jaurunjljgrtutkwcy Jun 12 '21 at 19:52
  • I think it is more appropriate to call 1 - e^-t, both the forced response and there is a zero natural response in this case. – jaurunjljgrtutkwcy Jun 12 '21 at 19:53
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    I think you have it right....it is simply a confusing choice made by the author to label that term the 'natural response.' The exponential term is part of the forced response. – rpm2718 Jun 12 '21 at 19:54
  • Thank you for clearing this up. The book is 'Control Systems Engineering' by Nise in case you were wondering. – jaurunjljgrtutkwcy Jun 12 '21 at 19:55
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They are saying:

the input pole at the origin generated the forced response \$c_f(t)=1\$

That's because, as they are saying right above that, they are applying a step response at the input. That is the Heaviside function, whose value is zero at \$t<0\$ and one from \$t\ge 0\$. Its Laplace transform is \$1/s\$ and that's multiplied with the system's transfer function, giving (4.5).

Now, think about what the step function represents: constant value (0), a sudden jump, then again a constant value (1). Therefore the system was at rest, there was a perturbation, and then the system is forced to stay at the new value. That is the forced input.

I think they are assuming some initial conditions, since the system pole is at -a, not zero.

They are not referring to the system's pole, but at the input's: \$1/s\$ has a pole at zero. There are no initial conditions. If it helps, think of it in terms of the time domain convolution:

$$\begin{align} s(t)&=\int_{-\infty}^t{\theta(\tau)h(t-\tau)\mathrm{d}\tau} \\ &=\int_0^t{1\cdot h(t-\tau)\mathrm(d)\tau} \tag{1} \\ h(t)&=a\mathrm{e}^{-at} \tag{2} \\ s(t)&=\int_0^t{a\mathrm{e}^{-a(t-\tau)}\mathrm{d}\tau} \\ &=a\int_0^t{\mathrm{e}^{-a(t-\tau)}\mathrm{d}\tau} \\ &=a\left(\dfrac{1}{a}-\dfrac{\mathrm{e}^{-at}}{a}\right) \\ &=1-\mathrm{e}^{-at} \tag{3} \end{align}$$

Since it's zero everywhere when \$t<0\$ then it gets replaced with a constant, the value 1 for \$t\ge 0\$ (and the limits change, too). And the above just tells you that the step response is just the integral of the impulse response.

As for the forced and natural responses: the forced ersponse is the fact that the ouput converges (asymptotically) towards the imposed, or forced value, 1, and in doing so it does it with it's natural response, the exponential form given by the time constant of the system.

With these in mind, re-read everything from your picture, carefully. You'll see that everything's in there.

a concerned citizen
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