# Re: equivalence between Z-transform and Laplace-Transform

Started by November 4, 2004
```AG wrote:

> Hi Tim,
-- snip --
>
>> The term "damping ratio" is much more slippery when you're talking
>> about discrete-time systems.  Assuming that I'm not messing up the
>> math, if you find the pole locations of your transfer function, z_0 =
>> e^{jw + q) then the "damping ratio" is
>
>
>> zeta = q / sqrt(w^2 + q^2).
>
> What I finally discovered in reading books, is that if z_0 = r*e(jw)
> (Are you sure about your definition of z_0, it looks like my r is your
> e^(q) ?), the signal is damped with a ratio 'a' at time 'k' if

Well, I'm sure that it's sound, in that you can take any arbitrary z_0
and map it using the equation.  Looking at it I realize that it should
be e^{jw - q}, however.
>
> r<e^(-ln(a)/k)   (assuming the sampling time T=1).

I suspect that this boils down to the same result as my equation, except
that your author may have been wise enough to avoid the use of the term
damping ratio for discrete-time systems.
>
> I am used to analog filter. if the filter is 1/(p&#2013266098;+2*z*wn*p+wn&#2013266098;), p
> being the operator, wn the natural pusle, and z the damping ratio.
>
Yes, that's the classic equation.

> depending on the value of z (greater or smaller than 1), the response to
> a unit step in input (1/p) will be a combination of exp() function, or
> exp() with oscillations (cos() and sin()). The fastest response is
> reached with z=1/sqrt(2) (it is the critical damping factor).
>
> I am trying to find the same parameters for the discret filter I
> mentionned in my preceding posts. I am confused with your formula.

I believe my formula is correct.  If you get the z-domain step response
of a unity-gain low pass filter you'll have (with my correction above)
either your exponentials or

h(k) = 1 - A*e^{-kq}*(cos(wk + th)),

where q and w are defined above, k is the sample index, and A and th are
some multiplier and phase shift that depends on your particular filter
transfer function.  If you crank through the math you should be able to
equate these with the continuous-time definition of the damping ratio
fairly easily.

At zeta = 1 you get w = 0, and your characteristic polynomial becomes
(z^2 - 2rz + r^2), or a double pole.  As zeta exceeds 1 you start seeing
two poles on the real line.
>
>
>> I assume that by "natural pulse" you mean impulse response;
>
> no, I mean the pulse of the response of the filter with a unit step in
> input and with a null damping. It is the 'wn' parameter mentioned above
> in my analog example.

Ah.  I've used "step response" all my life, so your term threw me.  I'll
have to remember that.
>
> Finally, in analog world, people speak about loop-noise bandwidth of a
> filter and it is defined as :
>
> B = int(|H(jw)|&#2013266098;, w=0..inifity) for the normalized H filter.
>
> Does the same thing exist in digital world ? Is it the same ?

It exists in the digital world, for essentially the same reason, except
that you have to normalize for the integration on the unit circle.  I
_think_ I'm getting my math right when I fling this off the cuff:

B = pi^-1 int(|H(e^{jw})|^2, w = 0..pi).

There are some subtleties and other gotchas there, however, not least of
which is that one doesn't often encounter purely white noise in a
sampled-time system because it's already been run through an
anti-aliasing filter before sampling.  It's best that you actually
understand the math before you apply the formula.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```