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First you need to understand why it must be the case.Tasos51 said:Summary:: A major result in control theory with no proof anywhere.
Prove that if a system's transfer function is not proper (order of numerator greater than order of denominator) then the system is not causal.
OK, but this is not a rigorous proof by any means. Further the statement "when the system has more finite zeros than poles then the overall effect is an anticipative one" needs proof.Baluncore said:First you need to understand why it must be the case.
https://lewisgroup.uta.edu/ee4314/lectures/Lecture 3.pdf
“The effects of zeros and poles of a system combine and when the system has more finite
zeros than poles then the overall effect is an anticipative one, and the system is not
causal. When a system has more poles than finite zeros (i.e. the transfer function of the
system is strictly proper) then the system is causal. If the transfer function of a system
has relative degree equal to 0 then the system is causal and there is also instantaneous
transfer between input and output.”
You did not reply to the last post in your thread from February. Were you able to look at the textbook recommended by @mpresic3 ?Tasos51 said:
mpresic3 said:I looked up in my control theory book Modern Control Theory, by William L Brogan, Third Edition. I have not explored the issue in depth but the discussion after 3.4.4. State Equations from Transfer Functions on page 88 to address your question directly. It gives the transfer function Y(s) / U(s) as the ratio of two polynomials. Please see the discussion. It is probably in the other texts too. what text are you using?
I cannot see something relevant on given reference. It would be better if the quotes are more accurate, this would save us precious time, not having to check 10 pages of text.berkeman said:You did not reply to the last post in your thread from February. Were you able to look at the textbook recommended by @mpresic3 ?
Not necessarily. I agree with @anorlunda's point that these terms seem to be definitions -- definitions don't need to be proved.Tasos51 said:OK, but this is not a rigorous proof by any means. Further the statement "when the system has more finite zeros than poles then the overall effect is an anticipative one" needs proof.
The link above is to this thread, so isn't at all helpful.Tasos51 said:
Maybe it would save you precious time in reading 10 pages, but at our expense.Tasos51 said:It would be better if the quotes are more accurate, this would save us precious time, not having to check 10 pages of text.
Agreed. However, if a system with infinite gain at infinite frequencies is not allowed then the OP makes no sense at all. The matter to be proved (or disproved?) effectively is that systems (edit: with rational system functions) with infinite gain at infinite frequency are not causal.DaveE said:Also, not really about causality (I think), but a true differentiator would require infinite gain at infinite frequency, which makes it not physically realisable.
The delta distribution and derivatives only have support at the origin, so in that sense are zero for ##t < 0## (as well as for ##t>0##) so I had assumed they are causal. If they are not considered causal then then the what the OP wants to prove is pretty trivial. You can simply do synthetic division to make the system function look something like ##H(s) = c_0\, s^p + c_1\, s^{p-1} + \cdots + c_{p-1}s + H_p(s)## where ##H_p(s)## is proper. The impulse response (just the inverse transform) then has derivatives of delta functions so would be non-causal. If that is all there is to it, then I'm surprised there has been more than one thread on this. But to me it seems weird to call the differentiator non-causal. After all, the response to an input does not occur before the input. But if that is the convention then who am I to argue? Could you provide a reference where that convention is explained?DaveE said:I don't think ## \delta(t) ## is causal. It's normally defined by the integral whose limits include t<0.
$$ \int_{-\infty}^\infty f(t)\delta(t) \, dt=f(0) $$
Or, consider it's Fourier transform, which must include t<0 components.
Yes, that makes sense. I neglected that the impulse response is defined as the response to ##\delta(t)##, so of course the output can also be non-causal.jasonRF said:I'm pretty tired so am not thinking very quickly. But even if the derivatives of the delta function aren't considered causal (which would be sufficient for 'proving' the original statement), the delta function must be. The system that is simply an ideal wire (##y=x##) obviously has the impulse response of ##\delta(t)##. Surely that is a causal, physically realizable system. Of course the system function is ##H(s)=1## so is not included in the subject of the thread.
The only thing I'm (edit: mostly) convinced of is that in an ideal world, a differentiator circuit built with ideal components (opamp, resistor, capacitor) would be causal. The fact that an ideal opamp is required doesn't bother me since virtually all of the systems we analyze with system theory are idealized. For example, V=IR doesn't actually apply for arbitrarily large currents through a resistor, and a resistor that stays linear regardless of the current through it is not physically realizable. But the mathematical models we build with the idealized components are of course very useful, even if all of them fail to be 'true' with arbitrary system inputs.DaveE said:I'm not as convinced as you about a theoretical differentiator being causal. What should it's output be at the peak of a triangle wave, where the derivative is discontinuous? Doesn't that discontinuous response require knowledge about the future value of the input? You wouldn't know you're at the peak until after it's occurred. Any delay in the response once it "saw" the peak would mean it has pole(s) in the transfer function. This is related to the problem of infinite gain required at infinite frequency.
Agreed.Tasos51 said:Furtermore, causality does not make sense for signals, only systems.
From the perspective of distribution theory (generalized functions) the support of the unit doublet is a single point - the origin. So from that perspective it can be thought of as zero for ##t \neq 0##. But as someone who knows almost nothing about control theory, I don't claim to know whether systems with such impulse responses are somehow excluded from the class of causal systems. I wouldn't be shocked to learn that they are.Tasos51 said:I think looking at the differentiator is a correct approach. So, its impulse response is the unit doublet. Causality requires it to be 0 for t<0. Is it ?
"The impulse response (just the inverse transform) then has derivatives of delta functions so would be non-causal": Well, this is not obvious to me. Am I missing something trivial ?jasonRF said:The only thing I'm (edit: mostly) convinced of is that in an ideal world, a differentiator circuit built with ideal components (opamp, resistor, capacitor) would be causal. The fact that an ideal opamp is required doesn't bother me since virtually all of the systems we analyze with system theory are idealized. For example, V=IR doesn't actually apply for arbitrarily large currents through a resistor, and a resistor that stays linear regardless of the current through it is not physically realizable. But the mathematical models we build with the idealized components are of course very useful, even if all of them fail to be 'true' with arbitrary system inputs.
Otherwise I think I have left open the possibility that, from a system theory perspective, the differentiator might be considered non-causal. Indeed, if that is the case, then as I showed in post 12 it is pretty easy to prove the statement that is the subject of this thread.
jason
This quote is taken a little out of context. Here is a larger chunk of that postTasos51 said:"The impulse response (just the inverse transform) then has derivatives of delta functions so would be non-causal": Well, this is not obvious to me. Am I missing something trivial ?
All I was saying is that a system with a transfer function that isn't proper will have derivatives of delta functions as part of the impulse response. So if systems whose impulse responses include such things are considered non-causal, then systems with an improper transfer function are non-causal.jasonRF said:The delta distribution and derivatives only have support at the origin, so in that sense are zero for ##t < 0## (as well as for ##t>0##) so I had assumed they are causal. If they are not considered causal then then the what the OP wants to prove is pretty trivial. You can simply do synthetic division to make the system function look something like ##H(s) = c_0\, s^p + c_1\, s^{p-1} + \cdots + c_{p-1}s + H_p(s)## where ##H_p(s)## is proper. The impulse response (just the inverse transform) then has derivatives of delta functions so would be non-causal.
I understand what you are saying, but when the impulse response includes generalized functions at the origin then it just isn’t obvious to me how it would be treated in the standard control theory literature. After all, the derivative of a delta is actually a functional, so maps functions to numbers. Technically, it doesn’t make sense to ask about the value at a given time. That is why in prior posts I mentioned the support instead of specifying times for which they are non-zero. So again, it is not obvious to me how folks doing control theory treat such things. If you already knew your original question was about derivatives of deltas then you should have phrased it that way; it would have saved everyone a lot of time.Tasos51 said:No they are not considered non-causal. This is to be proved, this is the whole idea. Non-causal are systems whose impule response is NOT 0 for t<0.
you have looked in a lot of control theory texts searching for a proof. Do any of those texts discuss impulse responses that include the derivatives of deltas?Tasos51 said:Summary:: A major result in control theory with no proof anywhere.
How does that work at the peak of a triangle wave? Where $$ \lim_{t \rightarrow 0^+} {tri'(t)} \neq \lim_{t \rightarrow 0^-} {tri'(t)}$$jasonRF said:Since differentiation can be computed using the limit of a backward difference it requires only present and past values
No, it's not but the network still has to change it's output then, Maybe this is the basis of a proof. The input is a continuous function with discontinuous derivatives, but the output is not a continuous function? IDK.Tasos51 said:It doesn't. The derivative is not defined at the peak.
DaveE said:edit: you can tell I'm not a mathematician by my sloppy use of "well defined", LOL.
The way I learned linearity, you would have to have valid functions f(a), f(b), and f(a+b) before you could evaluate f(a+b)=f(a)+f(b). So I would say the concept of linearity is not applicable at ##t_0##. But as you say, it all comes down to the definitions you like.jasonRF said:One more thing. If ##x_1(t)## is a function differential everywhere with derivative ##y_1(t)##, then if we use classical analysis and feed a differentiator your function plus ##x_1(t)## the output is simply 'does not exist' at ##t=t_0##, regardless of ##y_1(t_0)##. Does this violate linearity? Since we are asking the system to do things that are undefined, perhaps it doesn't matter?