0
$\begingroup$

Why is it that we Z-transform a difference equation to get a the transfer function of an digital filter?

How come a digital filter is given in the Z-domain, and what is the Z-domain?

And for that sake, why do analog filters operate in the S-domain, and what is the S-domain?

$\endgroup$
4
$\begingroup$

The S-transform allows you to deal with differential equations in an algebraic manner - so they become easier to solve. Since continuous/analog filters consist of integrators and differentiators the S-transform is therefore a natural way to deal with these systems.

The z-transform provides an algebraic way of dealing with finite difference systems and therefore it is a natural way to deal to discrete-time systems i.e. digital.

In the S-transform, setting $s=j\omega$ results in the Fourier transform. For continuous time systems we are interested in whether poles are on the right or left side of the $j\omega$ axis, because that determines the stability.

For the Z-transform, setting $z=e^{j\omega}$ results in the Discrete-Time Fourier transform. To determine stability, we are interested in whether the poles are inside or outside the unit circle.

So the S and Z domains are similar - they allows you solve continuous and discrete time systems, respectively, using algebra.

$\endgroup$
2
$\begingroup$

Applying a filter to a signal is convolution. For discrete signals, convolution equals polynomial or power (Laurent) series multiplication. The Z "transform" just formally puts filter coefficients and signal samples in power resp. Laurent series so that their multiplication can be interpreted back as the application of the filter to the signal.

There are certain important filters whose coefficient sequence is connected to a geometric sequence, so that the power/Laurent series with those coefficients has, again formally, a closed form as rational function.


Added: The real transform behind the formal Z transform is the Fourier series. Any stable signal or filter, i.e., with exponentially decaying coefficiens to both sides, and some signals that are not that good have a convergent Fourier series.

Now you might know that there is some funny business with the function space and pointwise convergence etc., the important part however is that the converse direction, recovering the coefficients from the periodic function, is much less problematic. So you can think of a signal being encoded by a periodic (square integrable function) $X(ω)$ as

$$x_n=\frac1{2π}\int_{-π}^π X(ω)e^{-i\,nω}\,dω$$

In terms of information content, $(x_n)$ and $X(ω)$ are equivalent. The application of a filter $f_k$ with Fourier series $F(ω)$ to encode it then has (depending on convention) the Fourier series

$$\sum_n (f*x)_ne^{iωn}=\sum_n \sum_k f_ke^{iωk}\,x_{n-k}e^{iω(n-k)}=F(ω)X(ω)$$

So to get the code of the filtered signal, you just multiply the codes of the filter and the signal pointwise.

By that design, any rational function $h(z)$ that acts as a Z transform is always the Z transform of a stable signal, the encoded coefficients are obtained by evaluating $X(ω)=h(e^{iω})$ over the unit circle,

$$x_n=\frac1{2π}\int_{-π}^π h(e^{iω})e^{-i\,nω}\,dω=\frac1{2πi}\int_{|z|=1} \frac{h(z)}{z^{n+1}}\,dz$$

so that the resulting coefficients are those for the Laurent series relative to a narrow annulus around the unit circle. end Added



The Laplace or s transform does something similar for continuous signals. Convolution in the time picture is the same as pointwise multiplication in the Fourier picture. In the beginning and formally, the s transform is just a Fourier transform hiding the complex unit. There is more to it, the domains of Fourier and Laplace transforms each have parts that are not contained in the other.

The power of the Laplace transform stems from the fact that it can perfectly express solutions of linear (ordinary and partial) differential equations with constant coefficients. Among solutions of such ODE are the polynomials, exponential and trigonometric functions and products thereof.

$\endgroup$
  • $\begingroup$ Ok.. But could you tell more about the domain it apears in.. the thing that confuses me is why is it we are using a unity circle in a argang diagram to determine stability, and how the s-plane is connected to to those?? what domain is the S-plane, and what domain is the z-plane.. $\endgroup$ – user7277 Jan 7 '14 at 15:20

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.