Discrete time Final Value Theorem applied to feedback system

I wish to calculate the Final Value of systems in which a high pass filter of the output feeds back into the input.

A simple example would be:

where is a 1st order high pass filter with transfer function:

I was expecting the y in the above example to have an infinite final value to a step in x, because keeps feeding

However, the workings below give a different answer:

1. Re-writing hp1(z) in terms of its inputs only:

2. Add to both sides of the system's equation:

3. Write the system's transfer function:

4. Re-write the infinite sum in the denominator:

5. Apply the Final Value Theorem to the response of this system to a step in x:

6. Taking the limit:

The above suggests that the system has a well defined terminal value to a step in x. However I don't think that can be the case.

Where am I going wrong? Help much appreciated

I would derive the total transfer function directly in the transform domain. Your input-output equation can be written as

$$Y(z)\big(1-z^{-1}\big)=\alpha G(z)z^{-1}Y(z)+\beta z^{-1}X(z)\tag{1}$$

where $$G(z)$$ is the transfer function of the high-pass filter. From $$(1)$$ you directly obtain the transfer function

$$H(z)=\frac{Y(z)}{X(z)}=\frac{\beta z^{-1}}{1-z^{-1}\big(1+\alpha G(z)\big)}\tag{2}$$

which clearly has a pole at $$z=1$$ because $$G(1)=0$$.

The mistake in your calculation is in step 2. You increased the lower bound of the summation from $$1$$ to $$2$$ without changing the power of $$\phi$$ from $$i$$ to $$i-1$$. In the transfer function this results in a $$\phi^2$$ in the last part of the denominator instead of a $$\phi$$. The denominator should be:

$$D(z)=1 - (1+\Psi\phi)z^{-1}-\Psi(\phi-1)\frac{\phi z^{-2}}{1-\phi z^{-1}}\tag{3}$$

For $$z=1$$ this evaluates to

$$D(1)=1 - (1+\Psi\phi)-\Psi(\phi-1)\frac{\phi }{1-\phi }=1 - (1+\Psi\phi)+\Psi\phi=0\tag{4}$$

• Thanks. Keeping everything in the z domain is undoubtedly the sensible way to approach this. I'd ideally like to work out where the logic / approach in the question starts to go wrong.though Nov 1 '20 at 13:55
• @OldSchool: I've added the correction of your derivation. Nov 1 '20 at 15:23
• Spot on. Thanks so much for taking a look. Nov 2 '20 at 17:55