How can initial conditions be taken into account to calculate a system's terminal value using the final value theorem or some other technique I wish to calculate the Final Value of the following system given initial conditions not at rest, and assuming the X will not change from its initial condition value.
The transfer function is
$$H(z) = \frac{z^{-1} \phi / (1 - \phi z^{-1})}{1 - z^{-1} \psi / (1 - \psi z^{-1})}$$
It arises from the following feedback system
$$Y(z)(1-z^{-1}) = z^{-1}F_{\psi}\cdot Y(z) + z^{-1}F_{\phi}\cdot X(z)$$
where $F_{\alpha}(z)$ is a 1st order high pass filter with parameter $\alpha$ of the form
$$F_{\alpha}(z) = \frac{\alpha (1 - z^{-1})}{1 - \alpha z^{-1}}$$
When the initial conditions are rest, then the final value of the system response to a step in x from $0$ to $x[0]$ is readily obtained using the Final Value Theorem:
$$fv = \lim_{z\rightarrow 1} z(1-z^{-1})H(z)\frac{x[0]}{1 - z^{-1}}$$
which in this case resolves to
$$\frac{x[0] \cdot \phi / (1 - \phi )}{1 - \psi / (1 - \psi )}$$
However this is not where the system will settle if initial conditions are non zero.
How does one solve for the final value when initial conditions are not zero?
Ah ha!  There's a lot of obfuscation in the problem as stated, plus a bit of over-emphasis on the final value theorem.

$$H(z) = \frac{z^{-1} \phi / (1 - \phi z^{-1})}{1 - z^{-1} \psi / (1 - \psi z^{-1})} \tag 0$$

OK, so far so good.

It arises from the following feedback system
$$Y(z)(1-z^{-1}) = z^{-1}F_{\psi}\cdot Y(z) + z^{-1}F_{\phi}\cdot X(z) \tag{1}$$
where $F_{\alpha}(z)$ is a 1st order high pass filter with parameter $\alpha$ of the form
$$F_{\alpha}(z) = \frac{\alpha (1 - z^{-1})}{1 - \alpha z^{-1}} \tag{2}$$

This was bothering me yesterday, but I didn't notice the essential problem.  The general form of the filter described in (2) is first-order, and the delay operators ($z^{-1}$) on the right side of (1) potentially add a second pole.
The original transfer function (0) is in a screwy form, but it appears to be second-order.  So the whole problem manages to worm its way through the pole-zero cancellation check by virtue of being stated in such a non-standard way.  This question deserves to be an extra-credit problem in a control systems class, or problem 10 out of 10 in a final by a particularly sadistic teacher, or perhaps a routine problem if the instructor believes in homework sets that no one will be able to fully solve.
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 $y_k$ 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.
So, first, shine some light on this obfuscated* problem.  I'll do that by substituting (2) into (1):
$$Y(z)(1-z^{-1}) = 
     z^{-1}\frac{\psi(1-z^{-1})}{1 - \psi z^{-1}}Y(z) + 
     z^{-1}\frac{\phi(1-z^{-1})}{1 - \phi z^{-1}}X(z) \tag 3$$
Now the pole-zero cancellation becomes painfully evident: we're expressing a filter of the form $1 - z^{-1}$ on the left-hand side, but when we divide through by $1 - z^{-1}$, it no longer appears in the equation.
Oh joy.
So, drawing this out as a feedback loop (and changing from $z^{-1}$ to $z$ notation, because I'm a control systems guy):

Notation is going to get a bit awkward here.  Bear with me.  The transfer function in (2) can be realized as a state-space system (where $x$ is a state variable, not the system input):
$$\begin{aligned}
x_k = \alpha x_{k-1} + (\alpha - 1)u_k \\
y_k = -x_k + u_k
\end{aligned}
\tag 4
$$
So (awkwardly) noting that $x_{1,k}$ is the first element of the state vector $\mathbf{x}$ at time $k$, but is not the system input $x_k$, then the 'top' input to the summing junction is (4) with $\alpha = \phi$ and an extra delay:
$$\begin{aligned}
\begin{bmatrix}x_{1,k} \\ x_{2,k}\end{bmatrix} = 
\begin{bmatrix} \phi &amp; 0 \\ 0 &amp; 0\end{bmatrix} 
  \begin{bmatrix}x_{1,k-1} \\ x_{2,k-1}\end{bmatrix} + 
\begin{bmatrix}\phi - 1 \\ 1\end{bmatrix} x_k \\
u_{1,k} = x_{2, k-1} - x_{1,k-1}
\end{aligned}
\tag 5
$$
The output of the system, $y_k$, is itself just a state variable (oh joy -- not sarcastically this time).  So just let $$y_k = x_{3, k} \tag 6$$.
The filter in the feedback path is again described by (4), with $\alpha = \psi$ and an extra delay:
$$\begin{aligned}
x_{4,k} = \psi x_{4,k-1} + (\psi - 1)x_{3, k} \\
u_{2,k} = x_{4,k-1} - x_{3, k-1}
\end{aligned}
\tag 7
$$
Finally, the integrator in the forward portion of the loop is:
$$x_{2,k} = x_{2,k-1} + u_{1,k} - u_{2,k} \tag 8$$
This goes together into a fourth-order system:
$$\begin{aligned}
\mathbf{x}_k = \begin{bmatrix}
\phi &amp; 0 &amp; 0 &amp; 0\\
0 &amp; 0 &amp; 0 &amp; 0 \\
-1 &amp; 1 &amp; 0 &amp; -1 \\
0 &amp; 0 &amp; \psi - 1 &amp; \psi
\end{bmatrix}
\mathbf{x}_{k-1} + \begin{bmatrix}\phi - 1 \\ 1 \\ 0 \\ 0\end{bmatrix}x_k \\
y_k = \begin{bmatrix}0 &amp; 0 &amp; 1 &amp; 0\end{bmatrix}
\end{aligned}
\tag 9
$$
Recast that into $\mathbf{x}_k = A \mathbf{x}_{k-1} + B\,u_k,\,y_k = C\mathbf{x}_k$, and you can test the $A$ and $C$ matrices to see that it is observable, at least for some combinations of $\phi$ and $\psi$ (it is, I checked).  Since there's pole-zero cancellation, we can assume that it isn't controllable (you can check that for yourself).
Because it's observable, we can consider the system
$$\mathbf{x}_k = A \mathbf{x}_{k-1} + \mathbf{x}_0 \delta(k)$$
The response of this system to any given $\mathbf{x}_0$ is $$Y(z) = C(\mathbf{I}z - A)^{-1}\mathbf{x}_0$$  Now you can find the system response for each initial value in $\mathbf{x}$ using the final value theorem.
* I looked it up -- &quot;obfuscate&quot; literally means &quot;to darken, to put into shadow&quot;.  So shining a light on the thing really is the antonym of &quot;obfuscate&quot;.