Does anyone know what the origin is of the bilinear transform? In other words, how was the bilinear transform derived?
Origin of Blinear Transform
Started by ●April 30, 2007
Reply by ●April 30, 20072007-04-30
On Apr 30, 11:27 am, Chris Barrett <"chrisbarret"@0123456789abcdefghijk113322.none> wrote:> Does anyone know what the origin is of the bilinear transform? In other > words, how was the bilinear transform derived?since it is an approximation mapping to the ideal z = e^(sT), there could be a lot of ways to get to that approximation. i know at least two ways. their in the textbooks. r b-j
Reply by ●April 30, 20072007-04-30
On Apr 30, 4:52 pm, robert bristow-johnson <r...@audioimagination.com> wrote:> On Apr 30, 11:27 am, Chris Barrett > > <"chrisbarret"@0123456789abcdefghijk113322.none> wrote: > > Does anyone know what the origin is of the bilinear transform? In other > > words, how was the bilinear transform derived? > > since it is an approximation mapping to the ideal z = e^(sT), there > could be a lot of ways to get to that approximation. > > i know at least two ways. their in the textbooks.Also at http://en.wikipedia.org/wiki/Bilinear_transform. -- Oli
Reply by ●April 30, 20072007-04-30
On Apr 30, 11:27 am, Chris Barrett <"chrisbarret"@0123456789abcdefghijk113322.none> wrote:> Does anyone know what the origin is of the bilinear transform? In other > words, how was the bilinear transform derived?Hello Chris, I recall using years ago a transform of the form ((1+x^-1)^m)(1- x^-1)^(n) for integration problems where n,m are integers. I believe this is due to Gauss. Anyway, the bilinear is a special case of this, and I'm quite sure this has been around for a long time. Now the question is likely to be who was the one to bring its use into converting S domain equations over to the z domain. I have a book with a lot of z-domain stuff in it from the 1930s. I'll have to hunt around my basement to find the book - I have thousands of them. Clay
Reply by ●April 30, 20072007-04-30
On Apr 30, 2:56 pm, Clay <phys...@bellsouth.net> wrote:> On Apr 30, 11:27 am, Chris Barrett > > <"chrisbarret"@0123456789abcdefghijk113322.none> wrote: > > Does anyone know what the origin is of the bilinear transform? In other > > words, how was the bilinear transform derived? > > Hello Chris, > > I recall using years ago a transform of the form ((1+x^-1)^m)(1- > x^-1)^(n) for integration problems where n,m are integers. I believe > this is due to Gauss. Anyway, the bilinear is a special case of this, > and I'm quite sure this has been around for a long time. Now the > question is likely to be who was the one to bring its use into > converting S domain equations over to the z domain. I have a book with > a lot of z-domain stuff in it from the 1930s. I'll have to hunt around > my basement to find the book - I have thousands of them. > > ClayI did a little digging and a transform like the above y=((1+x)^a)((1- x)^b) is due to Jacobi. It is easy to see that the bilinear is a special case of this. Clay
Reply by ●April 30, 20072007-04-30
On May 1, 3:27 am, Chris Barrett <"chrisbarret"@0123456789abcdefghijk113322.none> wrote:> Does anyone know what the origin is of the bilinear transform? In other > words, how was the bilinear transform derived?Its just Trapezoidal integration so its original goes way back. Probably Cauchy's era 18th century. Here is a simple derivation z=exp(sT)=exp(sT/2)/exp(-sT/2). Now expand num and denominator to a 1st order approx z:= (1+sT/2+..)/(1-sT/2+...) voila mes petite chiens (as Cauchy would have said) - the answer after re-arranging for s. Wang King
Reply by ●April 30, 20072007-04-30
gyansorova@gmail.com wrote:> > > voila mes petite chiens (as Cauchy would have said) - the answer after > re-arranging for s. > > > Wang King >So math has long gone to the dogs ;/
Reply by ●April 30, 20072007-04-30
gyansorova@gmail.com wrote:> On May 1, 3:27 am, Chris Barrett > <"chrisbarret"@0123456789abcdefghijk113322.none> wrote: >> Does anyone know what the origin is of the bilinear transform? In other >> words, how was the bilinear transform derived? > > Its just Trapezoidal integration so its original goes way back. > Probably Cauchy's era 18th century. > Here is a simple derivation > > > z=exp(sT)=exp(sT/2)/exp(-sT/2). > > Now expand num and denominator to a 1st order approx > > z:= (1+sT/2+..)/(1-sT/2+...) > > voila mes petite chiens (as Cauchy would have said) - the answer after > re-arranging for s.That's clear and elegant. It's the petite chien part I don't understand. Are we to think that the problem is a little bitch? Jerry -- Engineering is the art of making what you want from things you can get. ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by ●April 30, 20072007-04-30
gyansorova@gmail.com said the following on 30/04/2007 20:30:> On May 1, 3:27 am, Chris Barrett > <"chrisbarret"@0123456789abcdefghijk113322.none> wrote: >> Does anyone know what the origin is of the bilinear transform? In other >> words, how was the bilinear transform derived? > > Its just Trapezoidal integration so its original goes way back. > Probably Cauchy's era 18th century. > Here is a simple derivation > > > z=exp(sT)=exp(sT/2)/exp(-sT/2). > > Now expand num and denominator to a 1st order approx > > z:= (1+sT/2+..)/(1-sT/2+...)I've never been particularly happy with derivations like this; it seems we are matching the derivation to suit the desired result. For example, we could equally have factored z as: z = exp(sT) = exp(3sT/4)/exp(-sT/4) giving us the bilinear approximation: z ~= (1 + 3sT/4)/(1 - sT/4) -- Oli
Reply by ●April 30, 20072007-04-30
gyansorova@gmail.com said the following on 30/04/2007 20:30:> On May 1, 3:27 am, Chris Barrett > <"chrisbarret"@0123456789abcdefghijk113322.none> wrote: >> Does anyone know what the origin is of the bilinear transform? In other >> words, how was the bilinear transform derived? > > Its just Trapezoidal integration so its original goes way back. > Probably Cauchy's era 18th century. > Here is a simple derivation > > > z=exp(sT)=exp(sT/2)/exp(-sT/2). > > Now expand num and denominator to a 1st order approx > > z:= (1+sT/2+..)/(1-sT/2+...) > > voila mes petite chiens (as Cauchy would have said)I believe Cauchy would have said "mes _petits_ chiens"... -- Oli
