For causal linear time-invariant systems the degree of denominator must be larger or equal than the degree of the denominator.
How can this be proven rigorously?
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1\$\begingroup\$ Think of the simple, ideal differentiator: \$s\$. What's its impulse response? Can you obtain such a response in real life? \$\endgroup\$– a concerned citizenSep 21, 2022 at 13:28
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\$\begingroup\$ This is the usual "explanation". I look for a more mathemtical srgument. \$\endgroup\$– MichaelWSep 21, 2022 at 14:28
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1\$\begingroup\$ I'm not sure about the meaning of you using quotes. Having a pure differentiator means pure derivative of the input. What if the input is discontinuous? How do you handle such case? The moment you mention causality you invite practicality, otherwise you'd be fine talking about acausal FIRs. \$\endgroup\$– a concerned citizenSep 21, 2022 at 15:07
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\$\begingroup\$ Yes, a pure differentiator cannot be realized technically. But my question is about causality, not technical feasibility. Of course it is also clear, that there is no state space model for an improper H(s). I would regard an un-causal system to have an impulse response which doesn't vanish for t<0. However, what is the impulse response of H(s) = s²/(s+1) ? It includes derivation of Dirac impulse : h(t) = -δ(t) + δ'(t) + e^(-t) but where is this un-causal? It is zero for t<0... \$\endgroup\$– MichaelWSep 22, 2022 at 10:23
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I tried to say this in my previous comment: by enforcing causality, you imply realizability thus, there can be no improper transfer function. And the "rigurous" proof:
- for a passive implementation there will always be an I/O impedance thus, there will be a pole for each zero;
- for active implementations the opamps (or any active elements) will bring their own poles into play, along with the inherent resistors for the network;
- for digital there is Nyquist.
OTOH, if you want to keep it theoretical, you can even have \$s^n\$ and nobody can stop you, not even the paper you're writing on. You can have as many Diracs, all null before \$t=0\$, and everything will work. But only on paper. Otherwise acceleration could easily be determined by a double differentiator but, in practice, not imposing a bandwidth will bring havok. So the real reason for the transfer function being proper is reality.
answered Sep 22, 2022 at 14:37
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\$\begingroup\$ Lets say we have H(s)=1/(s+4). The associated h(t) depends on the ROC and is h(t)=exp(-4t)*u(t) (CAUSAL) or h(t) = -exp(4t)*u(-t) (ANTICAUSAL). So from the transfer function alone we cannot decide whether a system is causal or non-causal. It depends always on ROC to be used. \$\endgroup\$– MichaelWSep 24, 2022 at 18:47
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\$\begingroup\$ @MichaelW Can you show how you got to the anticausal h(t) from 1/(s+4)? H(s) has a Hurwitz polynomial in the denominator, therefore the exponential and the Heaviside must come out on the positive axis. It's possible you're overthinking this. \$\endgroup\$ Sep 25, 2022 at 9:20
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\$\begingroup\$ This is an often stated correspondence which is also in my textbooks: en.m.wikipedia.org/wiki/Two-sided_Laplace_transform (see exponential decay) \$\endgroup\$– MichaelWSep 26, 2022 at 5:46
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\$\begingroup\$ @MichaelW At this point you have to decide: either you mean to follow a purely mathematical point of view, or a physically realizable one. If it's the former then what I said in my last paragraph applies, and your question no longer makes sense (you used the word causal). If it's the latter then you can't invoke a negative time scale. As I said, you're overthinking it. ;-) \$\endgroup\$ Sep 26, 2022 at 9:40