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With respect to the below discussion, consider that we are talking about Continuous LTI systems characterized by constant coefficient ODEs.

Consider a cascaded system whose transfer function \$H(s)\$ is given by

$$H(s) = \frac {s+2}{(s+2)(s+3)(s+1)} =\frac 1 {(s+1)(s+3)}$$

"Recovering" the defining differential equation from this transfer function yields

$$(D^2 + 4D + 3)y = P(D)x$$

where \$P(D)\$ is unknown but irrelevant for the purposes of this discussion. What are the characteristic modes of the system?

Is it acceptable to simply say that the natural response \$y_n\$ is given as below?

$$y_n(t) = Ae^{-t} + Be^{-3t}$$

It would seem to me that the answer is no. Why can we recover an ODE from a transfer function in general? For instance, what if the cascaded system had've had \$(s-2)\$ factors rather than \$(s+2)\$ factors that cancelled (ie. one of the subsystems had a pole at \$s=2\$). In such a case, would it not be wildly incorrect to recover the ODE above since then we are ignoring an internal characteristic mode that blows up \$(e^{2t})\$?

If the procedure of recovering an ODE from a transfer function doesn't work in this blow-up case, can it work for the (s+2) case, or in general? I think not, but am looking for some guidance.

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  • \$\begingroup\$ Do you agree we can recover the ODE from cascaded transfer function when it does not have pole-zero cancellation? Do you also agree that, from a transfer function from the very first input and the very last output you cannot recover what was the transfer function of each of the systems in between that input and output? Wouldn't there be a difference between the natural response of the system as a whole and the modes/response of each subsystem? I will try to elaborate later, but I think cascaded LTI system are always well defined, so I see no problem in recovering the ODE of the overall system. \$\endgroup\$ – jDAQ Feb 27 '20 at 18:40
  • \$\begingroup\$ Hi there - yup, I absolutely agree that if the system were NOT cascaded then we could easily and acceptably (to me anyway!) recover the governing ODE. I guess my contention/question is whether we can do it in the cascaded case. I do not believe we can, since in the example I demonstrate above we would be missing an unstable characteristic mode from one of the component subsystems. We KNOW that the subsystem will blow up, therefore any overall system response that does not have this fact must be wrong. If the procedure is wrong in the case of blow up, then why shouldn't it be wrong in general? \$\endgroup\$ – 1729_SR Feb 27 '20 at 18:47
  • \$\begingroup\$ Thanks, and please do let me know if you have any light you can shed on this. I was told that, if observed externally, the cascaded system would appear to operate as per the natural response that I mentioned. But I find that hard to believe seeing as, in the blow up case, even externally we would eventually SEE the blow up. \$\endgroup\$ – 1729_SR Feb 27 '20 at 18:49
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If the poles and zeros are cancelled then the analysis will be oblivious to the internal unstable modes.

To illustrate, I am going to use the system with the unstable pole at 2 that gets cancelled. A good set of differential equations to use is any state-space realization.

I compute the state responses (using Mathematica)

Grid[{{tfm = 
    TransferFunctionModel[{{{s - 2}}, (s - 2) (s + 3) (s + 1)}, s],
   ssm = StateSpaceModel[tfm]}}, Background -> LightBlue, Frame -> All]
Grid[{Expand@StateResponse[ssm, 1, t]} // Transpose, 
 Alignment -> Left, Spacings -> {0, 2}, 
 Background -> {None, {{LightOrange, White}}}, Frame -> All]

enter image description here

The unstable mode \$e^{2 t}\$ appears in each of the state responses. However the output is \$-2 x_1+x_2\$ and the third state contributes nothing to the output. The amount of the unstable mode in the second state (\$1/15\$) is exactly twice that in the first state (\$1/30\$), and so the unstable mode does not appear in the output.

This same analysis is readily obvious in a Jordan state-space realization of the system. Now the unstable mode appears only in the third state, and the third state simply does not appear in the output.

Grid[{{tfm = 
    TransferFunctionModel[{{{s - 2}}, (s - 2) (s + 3) (s + 1)}, s],
   jssm = Last@JordanModelDecomposition[ssm]}}, 
 Background -> LightBlue, Frame -> All]
Grid[{Expand@StateResponse[jssm, 1, t]} // Transpose, 
 Alignment -> Left, Spacings -> {0, 2}, 
 Background -> {None, {{LightOrange, White}}}, Frame -> All]

enter image description here

To do a complete analysis the poles and zeros should not be cancelled. They are prone to get canceled when we use transfer functions. This does not happen with a state-space realization unless it is explicitly done.

Both the state-space realization I used above give the same characteristic polynomial. I can get the minimal polynomial explicitly.

Factor@CharacteristicPolynomial[First@Normal@#, s] & /@ {ssm, jssm, 
  MinimalStateSpaceModel@ssm}

{-(-2 + s) (1 + s) (3 + s), -(-2 + s) (1 + s) (3 + s), (1 + s) (3 + s)}

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Using as example the two following transfer functions, we will have that $$ H_1(s) = \frac{s-1}{s+5}, \; H_2(s) = \frac{s+2}{s-1}$$ will give an unstable response for \$H_2(s)\$ and a stable response for \$H_1(s)\$ resulting in the following step response two responses, one unstable and the other is stable

Yet, when we connect them in series the response from the resulting \$H_1(s)H_2(s)\$ is just a stable response which does not include, in the output, the unstable behavior of \$H_2(s)\$, if we use a state-space representation to look at both \$\begin{bmatrix} H_2(s) & H_2(s)H_1(s)\end{bmatrix}\$ we get that a stable output from the final output, but the intermediate one is unstable

although the final output gives us a stable, decaying, output the intermediate one is unstable and "blows up". So your intuition is kind of correct to point that there is something odd with the unstable pole cancelling, since you wouldn't be able to exactly place a zero on top of a pole. Yet, there is nothing wrong with it mathematically, we can see that the second transfer function has the output $$ y_2(t) = A e^{t}$$

and that the ODE for \$H_1(s)\$ would be $$ \dot{y}_1(t)+5y_1(t)= \dot{u}_1(t)- u_1(t) $$ And inputting \$ y_2(t)\$ in it we get that

$$ \dot{y}_1(t)+5 y_1(t)= \frac{d}{dt}\left(Ae^{t}\right) - \left(Ae^{t}\right) = 0.$$ Resulting in no forced response and just the homogeneous response. $$ y_1(t) = e^{-5t}y_1(0)$$

And, if we were to connect the systems in the reverse order we would get that the ODE for \$H_2(s)\$ would be $$ \dot{y}_2(t)-y_2(t)= \dot{u}_2(t)+2u_2(t),$$ $$ y_1(t) = B e^{-5t}$$ And inputting \$ y_1(t)\$ in it we get that

$$ \dot{y}_2(t)- y_2(t)= \frac{d}{dt}\left(B e^{-5t}\right) +2\left(B e^{-5t}\right) = 0,$$ $$ \dot{y}_2(t)- y_2(t)= \left(-5B e^{-5t}\right) +2\left(B e^{-5t}\right) = -3\left(B e^{-5t}\right).$$ This will have both the forced response and the homogeneous response. $$ y_2(t) = e^{t}y_2(0) + \frac{1}{2}e^{-5t}.$$

The main thing here is that for transfer function we always consider that the system had initial conditions at zero, so both orders would result in the system having the transfer function $$ H_1(s)H_2(s) = H_2(s)H_1(s) = \frac{s+2}{s+5}.$$

Long story short,

Is it acceptable to simply say that the natural response yn is given as below?

if you have those two ODEs in "series" and with initial conditions zero, yes.

In such a case, would it not be wildly incorrect to recover the ODE above since then we are ignoring an internal characteristic mode that blows up (\$e^2t\$)?

No, it will work out ok mathematically as long as you have initial conditions zero. But in practice (thinking about real systems that were cascaded) it would be a disaster, since you would only have canceled the "observability" of the unstable mode.

If the procedure of recovering an ODE from a transfer function doesn't work in this blow-up case, can it work for the (s+2) case, or in general? I think not, but am looking for some guidance.

It works, but just as the case where you have a function $$ f(x) = \frac{x(x-2)}{x-2} \neq x$$ you have to be very careful when dealing with cancellations, and point that $$ f(x) = x, \, \text{ for } x \neq 2.$$ So what you get from the reverse Laplace of a transfer function only relates the very first input and the very last output of a series of systems, with the condition that you are disregarding initial conditions of the system (and its subsystems).

Obviously you won't recover/notice any of the pole cancellations, but that does not make the ODE you got wrong, if you start from the ODE + zero initial response you do get the transfer function with the pole-zero cancelling.

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Your argument is fundamentally flawed. There is no factor in your transfer functions that gives rise to \$e^{2t}\$, there is no blow-up, whatever that is.You cannot inverse transform the individual Laplace poles and zeros and then multiply or divide them in the time domain - multiplication in the Laplace domain is convolution in the time domain. This is why we must express the transfer function in partial fractions - to decouple the factors

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  • \$\begingroup\$ You are completely missing my point here. Yes there is no blow up in the example I gave but I'm saying in another, similar example there could be. Nevertheless, none of this was the point of my question. \$\endgroup\$ – 1729_SR Feb 28 '20 at 1:30
  • \$\begingroup\$ So, give an example with 'blow-up'. \$\endgroup\$ – Chu Feb 28 '20 at 8:59
  • \$\begingroup\$ ... are you saying that the zero, (s+2), inverse transforms to \$e^{2t}\$? \$\endgroup\$ – Chu Feb 28 '20 at 12:24
  • \$\begingroup\$ No, that was a separate example. \$\endgroup\$ – 1729_SR Feb 29 '20 at 2:11

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