# 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:

$\Delta&space;y_{t}&space;=&space;\Psi&space;hp1(y_{t-1})&space;+&space;\beta&space;x_{t-1}$

where $hp1$ is a 1st order high pass filter with transfer function:

$hp1(z)&space;=&space;\frac{\phi&space;(1&space;-&space;z^{-1})}{1&space;-&space;\phi&space;z^{-1}}$

I was expecting the y in the above example to have an infinite final value to a step in x, because $\beta&space;x_{t-1}$ keeps feeding $\Delta&space;y$

However, the workings below give a different answer:

1. Re-writing hp1(z) in terms of its inputs only: $hp1(z)&space;=&space;\phi&space;+&space;(\phi&space;-&space;1)\sum_{i=1}^{\infty}&space;(\phi&space;z^{-1}&space;)^{i}$

2. Add $y_{t-1}$ to both sides of the system's equation: $y_t&space;=&space;y_{t-1}&space;+&space;\Psi&space;\left&space;\{&space;\phi&space;y_{t-1}&space;+&space;(\phi&space;-&space;1)\sum_{i=2}^{\infty}\phi&space;^i&space;y_{t-i}\right&space;\}+\beta&space;x_{t-1}$

3. Write the system's transfer function: $H(z)&space;=&space;\frac{\beta&space;z^{-1}}{1-(1+\Psi&space;\phi)z^{-1}&space;-&space;\Psi&space;(\phi&space;-1)\sum_{i=2}^{\infty}\phi&space;^iz^{-i}}$

4. Re-write the infinite sum in the denominator: $H(z)&space;=&space;\frac{\beta&space;z^{-1}}{1-(1+\Psi&space;\phi)z^{-1}&space;-&space;\Psi&space;(\phi&space;-1)\frac{\phi&space;^2z^{-2}}{1-\phi&space;z^{-1}}}$

5. Apply the Final Value Theorem to the response of this system to a step in x: $Final&space;Value&space;=&space;lim&space;z\mapsto1&space;\frac{z(1-z^{-1})}{(1-z^{-1})}&space;\frac{\beta&space;z^{-1}}{1-(1+\Psi&space;\phi)z^{-1}&space;-&space;\Psi&space;(\phi&space;-1)\frac{\phi&space;^2z^{-2}}{1-\phi&space;z^{-1}}}$

6. Taking the limit: $Final&space;Value&space;=&space;\frac{\beta&space;}{\Psi&space;\phi(\phi&space;-1)}$

The above suggests that the system $\Delta&space;y_{t}&space;=&space;\Psi&space;hp1(y_{t-1})&space;+&space;\beta&space;x_{t-1}$ 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 – OldSchool Nov 1 '20 at 13:55
• @OldSchool: I've added the correction of your derivation. – Matt L. Nov 1 '20 at 15:23
• Spot on. Thanks so much for taking a look. – OldSchool Nov 2 '20 at 17:55