1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Perhaps a reference might help. $\endgroup$ – Jason R Jul 4 '14 at 11:25
  • $\begingroup$ yeah, can't really do anything with this without context. it doesn't appear to make any sense. $\endgroup$ – robert bristow-johnson Jul 4 '14 at 12:53
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Good as always, +1 ! $\endgroup$ – jojek Jul 5 '14 at 10:26
  • $\begingroup$ @jojek: Thanks! (even though mine is a deprecated comment ...) $\endgroup$ – Matt L. Jul 5 '14 at 10:31
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Jul 4 '14 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.