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According to this question and answer the following system cannot be adequately captured by a z-transform transfer function.

$$y[n] = y[n-1] + F_{\psi}(y[n-1)) + F_{\phi}(x[n-1])$$ where $F_{\alpha}(z)$ is a first order high pass filter of the form $$F_{\alpha}(z) = \frac{\alpha (1 -z^{-1})}{1-\alpha z^{-1}} $$

The answer states that

The problem is that there's a pole-zero cancellation that snuck right by me and everyone else. It's evident in the left-hand side of (1), where the derivative of yk is the subject of the equation.

So the reason that you can't solve this problem as stated using the final value theorem is that you cannot adequately represent the system using a transfer function. There may be some way to save this within transfer function notation, but I just tried and failed at the first step, so I'm going to do it in state-space

What limitations of the z-transform (or others) require that this system be analyzed using alternative methods? What features of systems in general pose the same difficulty and why?

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  • $\begingroup$ I think this is a controllability/observability issue. I might have some time to analyze it more properly. $\endgroup$ – Ben Nov 25 '20 at 16:23
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A transfer function describes an LTI system. As such, the given system can be described by a transfer function. However, if there are non-zero initial conditions, the system is no longer linear because there's a contribution in the output that does not depend on the input signal but only on the initial conditions. Consequently, the transfer function cannot directly be used to compute the system's response if there are non-zero initial conditions.

Nevertheless, the (unilateral) $\mathcal{Z}$-transform can still be used to compute the system's response, even with non-zero initial conditions by transforming the difference equation and using

$$\mathcal{Z}\big\{ y[n-k]\big\}=z^{-k}Y(z)+\sum_{m=0}^{k-1}z^{-m}y[m-k],\qquad k>0\tag{1}$$

EXAMPLE: Let's use a simple example with a similar pole-zero cancellation as in the original problem to illustrate the point. Consider a system described by

$$y[n]-y[n-1]=\alpha \big(x[n]-x[n-1]\big)\tag{2}$$

The corresponding transfer function is

$$H(z)=\frac{Y(z)}{X(z)}=\frac{\alpha(1-z^{-1})}{1-z^{-1}}=\alpha\tag{3}$$

Clearly, $y[n]=\alpha x[n]$ is a solution of $(2)$. It is also the only solution if we require the system to be linear. However, it is not the only solution if we allow non-linear systems because there are infinitely many solutions of the form

$$y[n]=\alpha x[n]+c\tag{4}$$

with an arbitrary constant $c$. Note that these solutions cannot be inferred from the transfer function $(3)$.

Let's now use the $\mathcal{Z}$-transform to solve $(2)$ with initial conditions $y[-1]\neq 0$ and $x[-1]=0$. Transforming $(2)$ using $(1)$ gives

$$Y(z)(1-z^{-1})-y[-1]=\alpha X(z)(1-z^{-1})$$

which results in the following $\mathcal{Z}$-transform of the output:

$$Y(z)=\alpha X(z)+\frac{y[-1]}{1-z^{-1}}\tag{5}$$

In the time domain this becomes

$$y[n]=\alpha x[n]+y[-1]u[n]\tag{6}$$

where $u[n]$ is the unit step. Eq. $(6)$ is just a causal version of $(4)$.

This shows that the $\mathcal{Z}$-transform can be used to compute the system's response with non-zero initial conditions, even though the transfer function alone is inadequate for solving the problem.

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    $\begingroup$ IMHO the key here is that the $\mathcal{Z}$-transform can be used, but transfer functions cannot. Note that you'd have the same problem with a continuous-time system with pole-zero cancellations. $\endgroup$ – TimWescott Nov 23 '20 at 17:25
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    $\begingroup$ @TimWescott: That's right. I've tried to clarify this in my answer. $\endgroup$ – Matt L. Nov 23 '20 at 17:55
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    $\begingroup$ I think it is covered by this paper from Kalman. Section 9 especially. pdfs.semanticscholar.org/5881/… Quote from Kalman : In the process of algebraic manipulations, some transfer functions may have (exactly or very nearly) common factors in the numerator and denominator, which are then canceled. This is an indication that a part of the dynamics of the system is not represented by the transfer function $\endgroup$ – Ben Nov 25 '20 at 16:22
  • $\begingroup$ @Ben: Very good, Kalman's papers are usually good references! $\endgroup$ – Matt L. Nov 25 '20 at 16:32
  • $\begingroup$ @Ben. Nice reference. Thanks so much! $\endgroup$ – OldSchool Nov 25 '20 at 16:32

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