0
$\begingroup$

What's the advantage of using the Bilinear Transform?

$$H_d(z) = H_c(s)\bigg|_{s=\frac{2}{T_s}\frac{z-1}{z+1}}$$

When you can just use this equation?

$$H_d(\omega) = H_c(\Omega)\bigg|_{\Omega=\omega/T_s}$$

In other words, why does bilinear transform exist? These two equations look almost the same to me...What are the tradeoffs between using one equation verses the other? Is there a case where you would use one verses the other?

$\endgroup$
5
$\begingroup$

$$\left . H_d(z) = H_c(s) \right |_{s = \frac{2}{T\,s}\frac{z-1}{z+1}}$$ describes a transfer function in the $z$ domain that you can easily translate into a difference equation and realize in software.

$$\left . H_d(\omega) = H_c(\Omega) \right |_{\Omega = \frac{\omega}{T\,s}}$$ describes an idealized frequency response that you would like $H_d$ to have when you are done realizing it physically in software.

They're different. Note particularly the "would like to have" -- any translation from the continuous-time domain to the discrete-time domain is an approximation; part of your job is to make sure it's both practically realizable and good enough.

Note that there are other ways of approximating $H_c$ with some $H_d$ -- the bilinear transform is just one way. It has a lot of currency because it's conceptually simple, and works pretty well. It also has a lot of currency because it's easy to do with pencil, paper, and a slide rule -- today, there's numerical optimization techniques that can get you closer to some desired filtering goal, for less work (but -- I can never remember the search terms :( )

$\endgroup$
2
  • 1
    $\begingroup$ numerical optimization techniques to approximate a desired frequency response with a digital filter: The most famous one is probably the Remez exchange method. $\endgroup$ – Marcus Müller Jun 23 '20 at 19:39
  • $\begingroup$ And to add the least squares algorithm, which for most applications I deal with (optimizing overall SNR versus peak error) it usually ends up being the better choice. $\endgroup$ – Dan Boschen Jun 24 '20 at 14:49

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.