Does the steady-state value of a transfer function always converge to the derivative of the output with respect to the input?

93 Views Asked by At

I spent quite some time working with electrochemical impedance, which is no different in the math treatment than the ordinary electrical impedance. One thing that startled me, at first, was that I could always say that:

$$Z(s = 0) = \left.\frac{\mathrm{d}E}{\mathrm{d}I}\right|_{t \to \infty}$$

which is really important, because it shows that I can relate my impedance measurement at sufficiently low frequencies to the derivative of the stationary potential vs current curve.

I then continued studying other things without giving this "coincidence" much attention, since I thought it could be somehow a consequence of the laws of electricity or something like that. But then I stumbled upon a similar relation involving other transfer function and the derivatives of output with respect to input.

I started wondering if this relationship would hold true for any transfer function and tried to prove my point. So, I started this way:

$$\text{Knowing that: }H(s) = \frac{O(s)}{I(s)}$$

$$O(s) = H(s)I(s)$$

$$o(t) = \int_0^t h(\tau)i(t-\tau)\mathrm{d}\tau $$

$$\frac{\mathrm{d}o}{\mathrm{d}i} = \int_0^t h(\tau)\mathrm{d}\tau + \int_0^t \frac{\mathrm{d}h(\tau)}{\mathrm{d}i}i(t-\tau)\mathrm{d}\tau$$

$$\text{Now, if we make }t \to \infty\text{,} \left.\frac{\mathrm{d}o}{\mathrm{d}i}\right|_{t \to \infty} = \int_0^\infty h(\tau)\mathrm{d}\tau + \int_0^\infty \frac{\mathrm{d}h(\tau)}{\mathrm{d}i}i(t-\tau)\mathrm{d}\tau$$

$$\text{But we know that }F(s) = \int_0^\infty f(t)\mathrm{e}^{-st}\mathrm{d}t \implies F(0) = \int_0^\infty f(t)\mathrm{d}t$$

$$\left.\frac{\mathrm{d}o}{\mathrm{d}i}\right|_{t \to \infty} = H(0) + \mathcal{L}\left\{\frac{\mathrm{d}h(\tau)}{\mathrm {d}i}i(t-\tau)\right\}$$

So, if I have not commited any crimes during these steps, my gut feeling tells me that I'm on the right way, but I can't get rid of the second term on the right side of the last equation. I thought for a moment that this derivative could be rearranged to help me get some expression of the form $sK(s)$ and then make $s$ goes to zero, but even then it would require some well-defined characteristics of this $K(s)$ guy.

I would appreciate any comments on the general issue (does the relationship hold for any two input/output equations) and on the development that I've shown :)