4

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 $$Y(z)(1-z^{-1}) = z^{-1}F_{\psi}\cdot Y(z) + z^{-1}F_{\phi}\cdot X(z) \tag{1}$$...


3

Complex exponentials are eigenfunctions of LTI systems because they are eigenfunctions of the convolution operator: $$\begin{align}e^{j\omega_0t}\star h(t)&=\int_{-\infty}^{\infty}h(\tau)e^{j\omega_0(t-\tau)}d\tau\\&=e^{j\omega_0t}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j\omega_0t}H(j\omega_0)\end{align}\tag{1}$$ where $h(t)$ ...


2

After taking some more time to think about it, I think the problem is finally solved. I haven't digested Tim's answer yet, but from what I can see his approach is different. Furthermore, I thought it would be nice to give a closed-form expression for the final value the system's response will reach given values for $x[-1]$ and $y[-1]$. The final value is ...


1

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 ...


1

Mathematically (and theoretically), there is no need for the exponential function to be a complex sinusoid. The math is unchanged. The problem is that practical LTI systems are not boundless nor are they acausal. So setting aside those problems, every LTI system has input/output relationship described by the convolution integral (for continuous-time) or ...


1

Complex exponential functions are most generally defined (up to a constant complex or real factor) as $t\mapsto e^{j\omega_0 t}$, $\omega_0\in \mathbb{R}$. Real exponentials are typically of the form $t\mapsto c^{ t}$, $c>0$. The latter are not a subset of the former: one reason is that complex exponentials have a modulus equal to one. This is not the ...


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