# Are systems with more zeroes than poles really non-realizable?

I have learnt that systems whose transfer functions have more zeroes than poles become non-causal and thus non-realizable in practice for real time implementation.

But an op-amp differentiator, like the one in the figure below, is certainly realizable and has a TF of the form:

$$G(s) = -RCs$$

Couldn't we just cascade a bunch of these differentiators, at least in theory, and then generate a transfer function with more zeroes than poles?

• Real op-amps introduce poles due to their internal circuit. Commented Jan 29, 2022 at 22:41
• @ThePhoton I guess if they did not, then assuming the op-amp was ideal and the input was zero-impedance, the output would be infinite and random due to high-speed quantum fluctuations of whatever the input was connected to. Is that close to being right? Commented Jan 29, 2022 at 23:40
• because as the frequency approaches infinity, the impedance of the capacitor approaches zero and the op-amp has to output larger and larger amplitudes to cancel it through the resistor Commented Jan 29, 2022 at 23:41
• If the simplest differentiator would be possible then the MHz and upwards noise would be unbearable. OTOH, the cosmic background noise should have a very detailed map... Joking aside, even digital filters can't be zeroes, only, and if you think an FIR has no poles, think again: they're all at zero. Commented Jan 30, 2022 at 9:40
• To answer your original question, yes, you can use a delay line that is tapped, and sum the tap outputs, to make a system with that has zeros and no poles. Of course, a digital system of the same structure is commonly made., called a FIR filter. Commented Jan 30, 2022 at 19:13

But an op-amp differentiator, like the one in the figure below, is certainly realizable

Yes, it is

and has a TF of the form $$\G(s) = -RCs\$$

Generally not.

You're using the method where you set $$\V_- = V_+\$$ and solve for the transfer function. That method is only valid if the resulting circuit happens to be stable.

A more realistic method sets $$\V_o = A(s) \left(V_+ - V_- \right)\$$, calculates $$\V_-\$$ (and $$\V_+\$$ if necessary) as a function of $$\V_o\$$, and solves the resulting feedback equation. In this case, for most internally compensated, unity-gain-stable op-amps, our "more realistic" $$\A(s)\$$ is something like

$$A(s) = \frac{A_{GBW} \omega_0}{s \left(s + \omega_0 \right)}$$

where $$\A_{GBW}\$$ is the gain-bandwidth product of the op-amp in radians per second, and the manufacturer has chosen $$\\omega_0\$$ for stability, probably setting it in the range $$\\frac{A_{GBW}}{5} < \omega_0 < \frac{A_{GBW}}{2}\$$, depending on how sporting they want to be with overshoot and stability.

If you grind through all the math, you'll find out that for your circuit, with "sensible" component values (i.e. $$\R\$$ matches the op-amp's capabilities, so probably around $$\1\mathrm k \Omega \le R \le 1\mathrm M\Omega\$$, and $$\\frac{1}{RC}\$$ is well within the gain bandwidth product, your circuit is unstable, or at least shows a very strong resonance, with a little bit of differentiator action thrown in.

Trying to actually wire up that circuit and seeing the result was one of my first practical introductions to the joys of trying to make op-amps do what the simplistic theory seems to say they can.

I would like to add some "visual" information to what has already been pointed out.

A simulation, in a "real" case, will show the limits of a "differentiator" ("derivative" of an input variable).

It can be seen that, according to the choice of components, a certain range of "calculation" is reasonable (up to fo).

A margin of one "decade" of frequency is "available". To avoid any "oscillatory" risk, it is therefore advisable to "limit" drastically the frequency band at the input of a differentiator (use of a very effective low-pass) or to limit the "slew-rate" at the input.

The results also indicate the incidence of "parasitic" or "necessary" elements for the "stability" (R6 and C2) of the assembly.