I have two zeros at $z=-1$ and two complex conjugate poles at $z=A\cos\theta\pm jA\sin\theta$

This gives me the next transfer function

$$H(z)=\frac{1+2z^{-1}+z^{-2}}{1-2A\cos\theta z^{-1}+A^2z^{-2}}$$

To get filter's DC gain I'm trying to substitute $z$ with $1$:


Following graphical method described here (gain as a product of lengths of vectors from every zero to given point divided by product of lengths of vectors from every pole to given point) I've got next formula:


Where is my mistake? Why I've lost square root in the first case?


I think you lost the square root somewhere in your graphical method: Considering the nominator (product of lengths of vectors from every zero given point), I end up with


i.e. you have two vectors of length 2 (two zeros at z=-1).

Then, for the denominator you have

$$ D = \sqrt{(A\sin\theta)^2+(A\cos\theta-1)^2}^2=A^2-2A\cos\theta+1 $$

And finally $G=N/D$ yields the same result as your analytical solution.

| improve this answer | |
  • $\begingroup$ Oh, that was very stupid mistake. Instead of squaring the square root I've multiplied it by two. $\endgroup$ – e_asphyx Dec 25 '16 at 16:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.