I have a system controlled by the function \$i(t)\$ and described by this time-domain equation:


Taking the Laplace transform I get this:


How can I express this like \$L(s)=H(s)I(s)\$, with \$H(s)\$ being the transfer function? I mean, what is the transfer funtion here? I got stuck because I had never seen a system in which something like \$I(s+K)\$ appeared. Can I rewrite it in terms of \$I(s)\$?

Interestingly $$\frac{I(s)-I(s+K)}{K}$$ looks like it could be \$-\frac{d}{ds}I(s)\$ in some kind of approximation but I'm still confused.

  • 1
    \$\begingroup\$ I may be wrong, but I'm not sure you did your Laplace Transform correctly... \$\endgroup\$ – James Mertz Dec 13 '12 at 6:38
  • \$\begingroup\$ Hi! Thanks for considering the question but I'm convinced the Laplace transform is correct. \$\endgroup\$ – PDRX Dec 13 '12 at 19:44

How can I express this like \$L(s)=H(s)I(s)\$?

The first thing that jumps out at me is that \$l(t)\$ is not a linear or time invariant function of \$i(t)\$.

Now, \$H(s)\$ is the transform of the impulse response \$h(t)\$ which is just \$l(t)\$ when \$i(t) = \delta(t)\$:

\$h(t) = A+\dfrac{P}{CK}\delta(t)-\dfrac{P\delta(t)-CK(O-A)}{CK}e^{-Kt} = A(1 + e^{-Kt}) - Oe^{-Kt} \$

\$H(s) = A\dfrac{2s+ K}{s(s+K)} - O\dfrac{1}{s+K}\$

But, we can write \$L(s) = H(s)I(s)\$ only if \$l(t) = h(t) * i(t)\$ which is clearly not the case here.

  • \$\begingroup\$ Thanks for the insights on the problem! (I'm not so fluent in control theory) This is a system I'm controlling with a PID loop and I was trying to analyse the steady state error to various inputs (constant, ramp, etc). But I guess unless I make it linear I can't do it. Any suggestion by the way? \$\endgroup\$ – PDRX Dec 13 '12 at 21:44

You don't need a transfer function to implement a PID loop - though you would to analyze the system. You can certainly just numerically solve this in the time domain to simulate your controller if you need to. Note that the last term in your time domain representation seems to be a transient response, decaying asymptotically to zero after about 3 or so multiples of K. Perhaps it would be fair to ignore it??

  • \$\begingroup\$ Hi!I know I don't need it to implement the PID. I was just studying how it would perform. Thanks for the comment. I thought about ignoring it first but it was not making sense. But then it did... \$\endgroup\$ – PDRX Dec 16 '12 at 16:25

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