# Visualising a Z-transformed Transfer Function?

For designing any analog filter and various other outputs of filter we use laplace transform,I can visualise a laplace transform like for ex.
s[X(s)] can be implemented as differentiator fetched with signal x(t)while implementing differentiators we generally use capacitors or by using OPAMP similarly, [X(s)]/s can be implemented as integrators using OPAMP.

But I cannot visualize the z transform in designing of digital filters, while the implementing the digital filters we often map the s plane to z plane i cannot understand the significance

There are various method of implementing but in end we compare the s plane with z-plane , and deduce a equation where the value of z in found in place of s,

Why do we need to map the S plane to z plane?

• You don't always map the $s$ plane to the $z$ plane. You would do that if you have a continuous-time filter that you want to map to a discrete-time approximation of it. Many digital filters are designed directly in the $z$ domain. – Jason R Apr 2 '15 at 17:42
• @Jasson R yeah, i must have stated during conversion from analog to digital function. – MaMba Apr 2 '15 at 19:16

in continuous-time functions, like $x(t)$, the operation of the derivative makes sense, it is well defined. the three components to a continuous-time LTI filter are adders (two or more signals being added), scalers (a signal is simply multiplied by a constant), and integrators (the $s^{-1}$ operators). the first two components do not discriminate with regard to frequency, so with just those two components, you cannot make a "filter" that filters out some frequencies more so than others. but the integrator does act differently on sinusoids of different frequencies. higher frequencies come out of the integrator reduced in amplitude more than lower frequencies.
in discrete-time functions (or "sequences"), like $x[n]$, the derivative operator does not makes sense, it is not defined. the three components to a discrete-time LTI filter are adders (two or more signals being added), scalers (a signal is simply multiplied by a constant), and delay elements (the $z^{-1}$ operators). the first two components do not discriminate with regard to frequency, so with just those two components, you cannot make a "filter" that filters out some frequencies more so than others. but the delay element does act differently on sinusoids of different frequencies. higher frequencies come out of the delay element shifted more in phase than lower frequencies.
analog filters act on physical quantities in time. digital filters act on numbers. these numbers are samples of a continuous-time function ($x[n]=x(nT)$) and make up a sequence. there is no way to do derivatives or integrals directly. but we can delay any of these sequences by use of computer memory. that's why we talk of $H(z)$ more so than $H(s)$ when designing and implementing digital filters.
• When we look into z transform to perform delay action [z]=re^(jw) where the location of pole are give in r(angle) form ....so when z really represent delay,then why do we really need to worry about phases? – MaMba Apr 2 '15 at 19:14
• it's not about whether we worry about phase. it's about whether or not you can use differences in phase to discriminate between sinusoids of different frequencies. adding $\cos(\omega n) + \cos(\omega n + \phi_1)$ results in a different amplitude than $\cos(\omega n) + \cos(\omega n + \phi_2)$ if $\phi_1$ is different than $\phi_2$. so, like the integrator in analog filters, a delay element can be used to discriminate between frequencies and you must be able to discriminate between frequencies if you want to selectively filter out some frequencies and not others. – robert bristow-johnson Apr 2 '15 at 21:45