December 14, 2014
Very pleased to see the new book out, I've ordered it now. At the moment I'm working through the FX book - also very pleased to hear of the AU handling, which simplifies things immensely for me. So thanks for that too!
Having a bit of trouble getting the point of Equation 5.9. It's a little thing but I'd feel a lot happier continuing on if I was certain I understood it. I have no problem with the complex math basics, I've looked at those before. The text states "Equation 5.9 shows how to extract the magnitude and phase from a transfer function". Fair enough, but it's not clear to me where this is done. What are num and denom in this case, and which lines give the magnitude and phase?
That's it for now.
btw I saw in one of the other posts that there should be chapters available in the book forum, but I can't see any here...
January 28, 2017
Equation 5.9 says that the frequency response (magnitude) is found by taking the magnitude of the complex transfer function H(z). The phase response is found by taking the Argument of H(z). For digital transfer functions, H(z) may consist only of a numerator or it may be a quotient of numerator/denominator. For the numerator-only case, the magnitude and phase are simply the magnitude and argument of H(z). However, when you have both numerator and denominator, you need to take the magnitude of each and divide for the magnitude (frequency) response and for the phase you take difference of Arg(num) - Arg(denom).
Then, section 5.5 shows how to to this for the numerator-only case and 5.16 shows how to do it when you have numerator/denominator. The next examples (all fully worked out mathematically) continue in sections 5.17, 5.18, and 5.19 for the various types of filters that also are numerator/denominator examples.
January 28, 2017
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