[music-dsp] Derivation of the Tustins method (was Re: Simulating Valve Amps)
Ok, so what I'm really asking is why did someone (Tustin?) decide to make this substitution? exp (sT) = exp (sT/2) / exp (-sT/2) which can be written: exp (sT/2 - (-sT/2)) On 23 June 2014 23:58, Andrew Simper a...@cytomic.com wrote: Ok, but where does On 23 June 2014 22:59, robert bristow-johnson r...@audioimagination.com wrote: On 6/23/14 10:50 AM, Andrew Simper wrote: Ok, I'm still stumped here. Can someone please show me a reference to how the bi-linear transform is created without using trapezoidal integration? not that you wanna hear from me, but the usual derivation in textbooks goes something like this: z = e^(sT) = e^(sT/2) / e^(-sT/2) =approx (1 + sT/2)/(1 - sT/2) (hence the bilinear approximation) solving for s gets s =approx 2/T * (z-1)/(z+1) Ok, but what are the origins of that approximation? Where did it actually come from. Looking at it in hindsight is fine, but doesn't tell me anything. -- dupswapdrop -- the music-dsp mailing list and website: subscription info, FAQ, source code archive, list archive, book reviews, dsp links http://music.columbia.edu/cmc/music-dsp http://music.columbia.edu/mailman/listinfo/music-dsp
Re: [music-dsp] Derivation of the Tustins method (was Re: Simulating Valve Amps)
Here is a reply from Ivan to the old thread, that I am including here in this new thread: On 24 June 2014 00:25, Ivan Cohen ivan.co...@orosys.fr wrote: Not sure about what you mean here, but to get these approximations, you use the Taylor series of exp(x) and ln(x) for x - 0 : exp(x) = sum_(k=0 to N) x^k / k ! exp(x) = 1 + x + x^2/2! + x^3/3! + ... ln(x) = 2 * sum(k=0 to N) 1 / (2k+1) ((x - 1) / (x + 1))^(2k-1) ln(x) = 2 ( (x - 1)/(x+1) + 1/3 ((x - 1)/(x + 1))^3 + ... ) So, if we do the normal substitution, we can write this : z = exp (sT) z = exp (sT/2) * exp (sT/2) = exp (sT/2) / exp (-sT/2) z = (1 + sT/2 + (sT/2)² / 2... ) / (1 - sT/2 + (-sT/2)² / 2 + ...) With the inverse mapping : s = 1/T * ln(z) s = 2/T * ( (z - 1)/(z+1) + 1/3 (z-1)^3 /(z+1)^3 + ...) At the first order you get the approximations z = (1 + sT/2) / (1 - sT/2) or s = 2/T (z -1)/(z+1) Ivan COHEN http://musicalentropy.wordpress.com I'm after the reason behind the normal substitution, what prompted someone (Tustin?) to come up with that? -- dupswapdrop -- the music-dsp mailing list and website: subscription info, FAQ, source code archive, list archive, book reviews, dsp links http://music.columbia.edu/cmc/music-dsp http://music.columbia.edu/mailman/listinfo/music-dsp
