Bilinear Transformation

I'm studying about digital control systems and I was given this transformation $$s=\frac{2}{T}\frac{z-1}{z+1}$$ But I don't like just accepting things so I looked it up to find the proof and I found this on wikipedia

I'm aware of the taylor and maclaurin series but I can't quite get what's happening in the second line. How is lnz equal to all that?

• Which Wikipedia article? Commented Sep 17, 2017 at 9:16
• en.wikipedia.org/wiki/Bilinear_transform Commented Sep 17, 2017 at 9:17
• It seems to be just plain silly, but perhaps that I am missing something. Commented Sep 17, 2017 at 9:26

1 Answer

From the power series of the logarithm

$$\log(1+x) = - \sum_{n>0} \frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\ldots$$

(when $\vert x \vert < 1)$, we can immediately deduce

$$\log(1-x) = - \sum_{n>0} \frac{x^n}{n}=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\ldots,$$

and substracting the two expressions we can see that only the odd degree terms survive on the right hand side:

$$\log(1+x) - \log(1-x) = 2\sum_{n>0} \frac{x^{2n-1}}{2n-1}=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right)$$

For the standard branch of the complex logarithm, we have the equality

$$\log(1+x) - \log(1-x) = \log\left(\frac{1+x}{1-x}\right).$$

Now substituting $z = \frac{1+x}{1-x}$, we get $x = \frac{z-1}{z+1}$, and we recover the expression in the second line of the image you linked.

• But why it works for all z with Re(z)>0? Commented Jun 17 at 16:51
• Got it, for any z, x< 1, so log(1+x), log(1-x) can by expressed by geometric progression Commented Jun 17 at 18:14