LTIfags help

>>9097496
>So the transfer function has one pole at b1, and one zero at 0, right?
Yes.
>Why is this a "one-pole" filter then, and not a "one-pole one-zero" filter?
Commonly you'd say that the order of the filter is 1. In any case, it doesn't really matter since the order of the numerator and denominator polynomials are equal after simplifying.
>Also, why do they write transfer functions of continuous systems as rational functions of "s", but for discrete systems they use rational functions of "z^-1"?
It's to distinguish the transforms used. Discrete systems use the Z-transform, whereas continuous systems use Laplace transform, which uses s. This saves some headache when you notice that you can find a function that maps one plane to another. This is sometimes used when you have a readily designed continuous filter, and want a discrete version of it, so you just "go from s-plane to z-plane" using bilinear mapping or a similar method. Study the stability areas and relation to Fourier transform to get some insight on how s and z relate.

>>9097496
>Also, why do they write transfer functions of continuous systems as rational functions of "s", but for discrete systems they use rational functions of "z^-1"... wtf? Why don't they use "z" for discrete systems too?
In addition to >>9097752 the discrete transfer function in z^-1 comes naturally after aplying the Z-Transform to sampled functions. Transfer functions in z^-1 also allow you to readily convert a discrete transfer function to a difference equation, in the same manner you get a differential equation from a continous transfer function.

>Transfer functions in z^-1 also allow you to readily convert a discrete transfer function to a difference equation, in the same manner you get a differential equation from a continous transfer function.

But isn't this true for the continuous case as well?

I mean, we write H(s) = 1 / (1 + s)
instead of H(s) = s^-1 / (s^-1 + 1).

Why is it better to use s instead of s^-1 in the continuous case, if z^-1 is preferred in the discrete case? To me they look pretty similar; i.e. integrator = s^-1, delay = z^-1
Yet for some reason the literature uses s for continuous transfer functions, and z^-1 for discrete ones as the "canonical" form...

File: discrete transfer function to difference equation (poor drawing).png (21KB, 1089x530px) Image search: [Google]

21KB, 1089x530px

>>9098383
>But isn't this true for the continuous case as well?

No. You can interpret the s variable as the derivative, in the sense that the coefficients of the denominator polynomial in s are the coefficients of the linear homogeneous differential equation, with the order of each derivative term being the same as the order of the s term.

In the discrete case, the coefficients of the difference equation are given by the coefficients of the polynomial in z^-1 because the z^-1 corresponds to the backwards shift operator. This comes from the time shift property of the Z-Transform. In your exemple, the corresponding difference equation is in pic related.