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I am thinking about these terms in the context of linear control.

A transfer function is proper if the degree of the numerator is not greater than the degree of the denominator. I've read often that improper transfer functions are "not causal". I also often see the word "unrealizable" used often in this context.

If a control transfer function I've designed is improper, does that mean it is "causal" and/or "unrealizable"? What is the difference between these terms? What do they mean in practice?

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  • $\begingroup$ This question is related. $\endgroup$
    – Matt L.
    Commented Mar 2, 2017 at 15:52
  • $\begingroup$ Thank you for the link. I buy that proper systems (and only proper systems) are causal. I suppose I am interested in whether applying the label "causal" gives us any other useful information about the system (beyond the definition of causality). "Realizability" seemed like an obvious candidate which is why I mentioned it. $\endgroup$
    – Max
    Commented Mar 2, 2017 at 15:56
  • $\begingroup$ An improper transfer-function can be realizable, and casual. E.g., what is the relative degree of a PID control law? $\endgroup$
    – Arnfinn
    Commented Mar 3, 2017 at 9:08
  • $\begingroup$ But real PD control laws don't actually have infinite high-frequency gain, right? It seems like any time you would actually try to implement (for example) a derivative term of the form $K(T_ds+1)$, there's really a high-frequency pole that we just don't account for in the model, i.e., $G_c = K(T_ds+1)/(\tau s+1)$. It's just that the high frequency pole at $1/\tau$ is negligible so we don't include it in the model, but it really is there and it makes the transfer function proper. Similar to how lumped parameter models break down at high frequencies. Is this the case? $\endgroup$
    – Max
    Commented Mar 3, 2017 at 13:44
  • $\begingroup$ Well, yes, in practice it is impossible to implement a differentiator, and it is also a bad idea to try to approximate one to a high degree in most cases, but mathematically speaking it is still realizable and causal. $\endgroup$
    – Arnfinn
    Commented Mar 4, 2017 at 5:38

1 Answer 1

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Causality is a necessary condition for realizability. Stability (or, at least, marginal stability) is also important for a system to be useful in practice.

For linear time-invariant (LTI) systems, which are fully characterized by their transfer function, we get realizability constraints on the transfer function. For continuous-time LTI systems, if we work at frequencies for which the lumped element model is valid, we require the system's transfer function to be rational for the system to be realizable. Also for discrete-time LTI systems we require rationality of the transfer function, which implies that the system can be realized by adders, multipliers, and delay elements.

For an LTI system to be causal and stable, its poles must lie in the left half-plane (for continuous-time systems), or inside the unit circle (discrete-time systems). From this it follows that the rational transfer function of an LTI system must be proper, otherwise you would get one or more poles at infinity, causing the system to be unstable (or non-causal).

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    $\begingroup$ So to be clear, if we say a transfer function is not realizable, that mean in practice something like "there is no physical system corresponding to this transfer function"? $\endgroup$
    – Max
    Commented Mar 2, 2017 at 16:47
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    $\begingroup$ @Max: Yes, not even an "ideal" physical system (i.e., with zero tolerances, no quantization, etc.). $\endgroup$
    – Matt L.
    Commented Mar 2, 2017 at 17:53

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