Talk:Bilinear transform

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What is the correct formula for this transform? The current formula does not seem to agree with the "Background" paragraph. --AxelBoldt

I found two essentially equivalent definitions:

s = (2/T) * ((1 - z-1) / (1 + z-1))

and

s = (2/T) * ((z - 1) / (z + 1)).

The first one is from Feedback control of dynamic systems by Franklin et al. The second is from the help text of the bilinear operation in Octave.

What exactly do you mean by does not seem to agree? -- Ap

In the Background paragraph of the main article, it is explained that the transform maps complex numbers from the interior of the unit circle to the negative half plane. But I think the above transform maps complex numbers from the interior of the unit circle to the positive half plane. Or am I confused? --AxelBoldt

Oops, yes I am confused indeed. The above is correct, and gives values in the negative half plane. --AxelBoldt

True, both equations are the same, but I find it easier to convert the first form (above) into executable code. The second one requires a little algebra to make it useful in the real world, IMHO. Am I wrong? Madhu 23:57, 4 July 2006 (UTC)



Multiply the first equation by \frac{z}{z} and you get the second equation. They are the same. Cburnett 05:31, 28 Nov 2004 (UTC)


"When the Laplace transform of a discrete-time signal with each element of the discrete-time sequence attached to an appropriately delayed unit impulse, the result is precisely the Z transform of the discrete-time sequence with the substitution......"

-There is something wrong grammatically here. Is there a verb missing?

"is" is the verb ("the result" is the subject), but it is poorly worded. for some reason i think it's my fault. i'll work on fixing it. r b-j 23:59, 8 December 2005 (UTC)

I understand how your first approximation z=((1+sT/2)/(1-sT/2) arises and leads to the bilinear transform, but the infinite series you show for ln(z) really puzzles me. I can't concieve of any Taylor series expansion to produce it. Can you shed some light on where it comes from? Thanks. [SW] 155.104.37.17 17:19, 30 January 2007 (UTC)