I have a system whose input-output relation is as follows

$$y(t)=x(t)+\int_{-\infty }^{t} x(\tau) \,\mathrm d \tau$$

Can I create an equivalent system by using differentiators rather than integrators?

I think something like taking derivative of both sides of the equation but improper integral makes it hard. Is there any nice way to convert this system? Thanks in advance.

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    $\begingroup$ How can a differentiator be "equivalent" to an integrator? To me it's unclear what you're trying to do. $\endgroup$ – Matt L. Mar 23 '18 at 8:31
  • $\begingroup$ Integration is the opposite of differentiation, so that will not be possible. If the integral bothers you, the function can be rewritten as $$y(t)=x(t) + x(t)*\epsilon (-t)$$ $\endgroup$ – Max Mar 23 '18 at 8:57
  • $\begingroup$ yes, you can typically create equivalent block diagram in terms of just differentiators or just integrators. In circuit theory a gyrator en.m.wikipedia.org/wiki/Gyrator can do this. An inductor can be made to behave like a capacitor. $\endgroup$ – Stanley Pawlukiewicz Mar 23 '18 at 9:42
  • $\begingroup$ @Stanley Pawlukiewicz Thanks for brief explanation and source. $\endgroup$ – Tokugava Mar 23 '18 at 9:57
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    $\begingroup$ Do you want to convert the integral equation into a differential equation? Or do you want a block diagram that uses differentiators rather than integrators? $\endgroup$ – Rodrigo de Azevedo Mar 23 '18 at 12:02

Typically you can devise equivalent block diagrams in terms of just integrators or just differentiators.

A simple example is on page 2-15 of


which should look remarkably familiar. You should be cautioned that mathematical equivalence doesn't mean that there aren't reasons where one form would be prefered over the other. OP Amp integrators have better noise characteristics. You also have to band limit an analog differentiator to make it physically realizable.

Another issue is that they will not in general have the same complexity.

When modeling ordinary differential equations systems, $$ \mathbf{M}(t) \frac{d \mathbf{y}(t)}{dt}= \mathbf{g}(t,\mathbf{y}) $$ where $\mathbf{M}(t)$ (mass matrix) is invertible is often more sparse for physical systems than $$ \frac{d \mathbf{y}(t)}{dt}= \mathbf{h}(t,\mathbf{y}) $$ which is a more general way of writing the common state space representation.

One can write : $$ \frac{d \mathbf{y}(t)}{dt}= \mathbf{M}^{-1}(t)\mathbf{g}(t,\mathbf{y}) $$ but if $\mathbf{M}(t)$ is sparse, $\mathbf{M}^{-1}(t)$ probably isn't ,which makes a big difference when the dimension of $\mathbf{y}$ is large.

The Matlab Control systems toolbox also includes state variable equations of the form

$$ \mathbf{E} \frac{d \mathbf{y}}{dt}= \mathbf{A}\mathbf{y}+ \mathbf{B}\mathbf{u}$$


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