My clue is just to expand brackets and obtain coefficients of numerator and denominator of my transfer function. Am I right?
2
-
$\begingroup$ Yes, you are right, only if your filter structure is a direct form. $\endgroup$– JuanchoDec 23, 2016 at 15:33
-
$\begingroup$ going from poles and zeros to Direct Form coefficients is easy. going in the other direction is a little harder because it involves factoring polynomials. $\endgroup$– robert bristow-johnsonFeb 24, 2017 at 4:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
This might look trivial, but with the pole-zero diagram alone one can get the transfer function with a scale (gain) ambiguity.
That is, all transfer functions $(Kb,a)$, where $K$ is a constant, have the same pole-zero representation. For this reason, we usually use pole-zero-gain terminology.
