# Confusion Regarding Bi Linear Transform

I was reading my book where the z-transform of a signal is derived to be ${1-e^{-2bT}z^{-1}}$ . Then it goes on to say that by applying the bilinear transform we can get

$$\frac{2(1+bT+(bT-1)z^{-1})}{T(1+z^{-1})}$$

I know a little bit about bilinear transform but this example is using it in a different context. Can someone please help me understand how the bilinear transform is being used here.

• Perhaps a reference might help. Jul 4 '14 at 11:25
• yeah, can't really do anything with this without context. it doesn't appear to make any sense. Jul 4 '14 at 12:53

As already mentioned by other people, the bilinear transform is often used to map a continuous-time system described in the $s$-domain to a discrete-time system described in the $z$-domain. However, a bilinear transform is a more general tool that can also be used to transform a discrete-time system to another discrete-time system. Since you didn't give any context I do not know about the author's motivation to do so, so I can't tell you why this is done, but I believe I can tell you how it is done (up to a constant factor):

In order to show how the discrete-time system is transformed to the other discrete-time system, I'll use the $s$-domain as an intermediate step. The function

$$1-e^{-2bT}z^{-1}$$

can be represented in the $s$-domain by

$$1-e^{-(s+2b)T}$$

because a unit delay $z^{-1}$ is exactly equivalent to $e^{-sT}$ in the continuous domain. By using a linear approximation $e^x\approx 1+x$ we get

$$1-e^{-(s+2b)T}\approx (s+2b)T$$

Now we can apply the bilinear transform

$$s=\frac{2}{T}\frac{1-z^{-1}}{1+z^{-1}}$$ which gives

$$(s+2b)T=2\left(\frac{1-z^{-1}}{1+z^{-1}}+bT\right)\tag{1}$$

The formula in your question is

$$\frac{2(1+bT+(bT-1)z^{-1})}{T(1+z^{-1})}=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}+bT\right)$$

which differs from (1) by a factor $1/T$. I can't explain this difference, but I'm pretty sure that the basic idea of the transformation is as I've described it above.

• Good as always, +1 !
– jojek
Jul 5 '14 at 10:26
• @jojek: Thanks! (even though mine is a deprecated comment ...) Jul 5 '14 at 10:31

The bi linear transform is the transform from the Laplace Transform Domain to the Z Transform.

The Laplace Transform Domain is a regular plane.
This transform transforms vertical lines in the Laplace domain into circles in the Z Domain.
Hence the Fourier Vertical Line in Laplace Domain (The Y Vertical Lines) is transformed into the unit circle in the Z Transform.

I'm a little inaccurate here, since the whole left side of the plane is transformed into the inner part of the Z Plane and the right hand of the Laplace transform it transformed to the outer part of the Z Plane (Out of the unit circle).

This transform is used to derive Digital Filters from their Analog Laplace domain representation.

• normally in DSP we think of the bilinear transform as mapping the s-plane to the z-plane in such a way that it maps the $j \Omega$ axis in the s-plane to the $e^{j \omega}$ unit circle in the z-plane. but there is a simple bilinear transform that maps z-plane to z-plane. you can use it with a digital filter to map a feature that appears at one frequency to appear at another different frequency. Jul 4 '14 at 12:57