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Wikipedia says in bilinear transformation from \$s\$ domain to $z$ domain relation is

$$\boxed{s \longleftarrow \frac{2}{T}\frac{z-1}{z+1}}$$

But here this relation is given like this

$$\boxed{w=\frac{z+1}{z-1}}$$

What is difference between this two?

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If you read the link which describes the second mapping $$ w = \frac{z+1}{z-1}$$

Thus inside of the unit circle in z-plane maps into the left half of w-plane and outside of the unit circle in z-plane maps into the right half of w-plane. Although w-plane seems to be similar to s-plane, quantitatively it is not same

It states that $w$ is not the $s$ plane.

Therefore, if your aim is in studying the bilinear transformation for digital filter design then you can safely use the first, well known, bilinear transformation relation, and its inverse:

$$\boxed{s \leftarrow \frac{2}{T}\frac{z-1}{z+1}}$$

Otherwise, the paper you linked is dealing with stability analysis of discrete-time systems using bilinear transform of its own selection...

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  • $\begingroup$ sir for the stabilty analysis why they use a different relation? can they use this relation $\boxed{s \leftarrow \frac{2}{T}\frac{z-1}{z+1}}$ also if the w plane is not same as the s plane they can apply RH criteria? $\endgroup$ – Rohit Nov 7 '17 at 15:37
  • $\begingroup$ They want to map poles from z-plane of discrete-time to w-plane of cont-time for stability analysis using RH method which applies to continuous-time systems. I don't know why they preferred such a mapping. Probably its warping type produces a more desired result. In the paper they describe the method. So if you want to follow that paper, then you should use their own notation. Morever you can check your claim by making the same analysis using the former mapping (let T=2) and see if they yield the same results. $\endgroup$ – Fat32 Nov 7 '17 at 16:09
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    $\begingroup$ I just used in a problem as you said they both are giving same result in terms of stability,but not sure if this can change its behaviour in other problems.anyway Thank you sir $\endgroup$ – Rohit Nov 7 '17 at 16:29

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