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


1 Answer 1


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

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

  • $\begingroup$ 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 $\endgroup$
    – OldSchool
    Nov 1, 2020 at 13:55
  • $\begingroup$ @OldSchool: I've added the correction of your derivation. $\endgroup$
    – Matt L.
    Nov 1, 2020 at 15:23
  • $\begingroup$ Spot on. Thanks so much for taking a look. $\endgroup$
    – OldSchool
    Nov 2, 2020 at 17:55

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