equivalence between Z-transform and Laplace-Transform

Started by November 2, 2004
Hi all,

I am studying a digital phase locked loop.

The closed loop filter of this loop has the following Z-Transform :

H(z) = ((K1+K2)*z^-1 - K1*z^-2) / (1 + (K1+K2-2)*z^-1 + (1-K1)*z^-2)

I would like to know the damping factor and the natural pulse of the
equivalent time continuous filter H(p).

by equivalent, I mean a filter H(p) that would give in the time domain,
the same response to an unit step than the one I would get in the
discret domain (one being the discretized version of the other).

I thought I just had to replace z by 1+p in H(z). doing the
approximation e(p)=1+p. But it doesn't seem to be the right thing to do.
(damping factor and natural pulse doesn't not corresponds each other).
Does anyone know why and could explain it ?

I hope I made it clear, and it's the right place to ask this question.

Alexandre.
AG wrote:
> Hi all, > > I am studying a digital phase locked loop. > > The closed loop filter of this loop has the following Z-Transform : > > H(z) = ((K1+K2)*z^-1 - K1*z^-2) / (1 + (K1+K2-2)*z^-1 + (1-K1)*z^-2) > > I would like to know the damping factor and the natural pulse of the > equivalent time continuous filter H(p). > > by equivalent, I mean a filter H(p) that would give in the time domain, > the same response to an unit step than the one I would get in the > discret domain (one being the discretized version of the other). > > I thought I just had to replace z by 1+p in H(z). doing the > approximation e(p)=1+p. But it doesn't seem to be the right thing to do. > (damping factor and natural pulse doesn't not corresponds each other). > Does anyone know why and could explain it ? > > I hope I made it clear, and it's the right place to ask this question. > > Alexandre.
I assume that your 'p' operator is the same as the more usual 's' operator? There are a number of approximations you can use, the most popular ones being the first difference: p = (z - 1)/Tz or p = (z - 1)/T, where T is your sample time (or one over your sample rate, if you will), and the bilinear (or Tustin) approximation: p = 2/T * (z-1)/(z+1). Both of these are _approximations_, which only hold for p (or s) that is much lower than 1/T. Your approximation is more or less a first difference except that you are not including the sample time. 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). I assume that by "natural pulse" you mean impulse response; you can get that by doing direct polynomial division in the z domain, or just simulating the system with an IIR filter (which is mathematically exactly the same thing). You can get details on this operation from "Digital Control Systems" by Houpis & Lamont, McGraw-Hill, 1985, or nearly any other good book that includes discrete-time signal processing. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Hi Tim,

Thanks for your comment, see my answers  below :

Tim Wescott wrote:
> I assume that your 'p' operator is the same as the more usual 's' operator?
yes
> > There are a number of approximations you can use, the most popular ones > being the first difference: > > p = (z - 1)/Tz or p = (z - 1)/T,
ok.
> where T is your sample time (or one over your sample rate, if you will), > > and the bilinear (or Tustin) approximation: > > p = 2/T * (z-1)/(z+1).
ok.
> Both of these are _approximations_, which only hold for p (or s) that is > much lower than 1/T. Your approximation is more or less a first > difference except that you are not including the sample time.
ok.
> 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 r<e^(-ln(a)/k) (assuming the sampling time T=1). 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. 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 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. 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 ? I will try to find your book. Thanks for the reference. Alexandre.