equivalence between Z-transform and Laplace-Transform

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&#4294967295;+2*z*wn*p+wn&#4294967295;), 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)|&#4294967295;, 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.