Z-Transform

On 04/25/2011 03:27 AM, dhaval_shah wrote:
> I got some idea, poles are considered bcoz real part of the poses are
> placed as a damping factor when u perform inverse Z-Transform.

You're confusing the z transform and the Laplace transform.  In the 
_Laplace_ transform the real parts of the poles determine how rapidly 
they damp out (but are different from the damping factor).  In the z 
domain it's the magnitude of the pole -- a pole with magnitude greater 
than 1 is unstable, one with magnitude equal to one is metastable, and 
one with magnitude less than one is stable.  The lower the pole 
amplitude, the more rapidly it damps out.

-- 

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

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" was written for you.
See details at http://www.wescottdesign.com/actfes/actfes.html

On Apr 21, 2:56&#4294967295;pm, Tim Wescott <t...@seemywebsite.com> wrote:
> On 04/21/2011 11:35 AM, dhaval_shah wrote:
>
> > What is the role of "zeros" in deciding the stability of the system?
> > why only "poles" are being used to decide the stability of the system?
>
> See http://www.wescottdesign.com/articles/zTransform/z-transforms.html .
>

just to add to Tim's response.

The main mathematical reason that the pole loci exclusively determine
the stability of the system is because of partial fraction expansion.

Note Eqs (17) and (20) in Tim's document.  Even though it's X(z), not
H(z) (the transfer function), the principle is the same.  The factors
in the denominator of X(z) in Eq (17) are the same as the little
denominators in the partial fraction expanded form of Eq (20).  There
is a double term (a double pole if this were H(z)), so maybe it can be
more simply explained without it.

Consider a system with transfer function (i suggest to view using
Google Groups with a fixed-width font):


              (z-q1)(z-q2) ... (z-qM)
    H(z)  =  -------------------------
              (z-p1)(z-p2) ... (z-pN)


q1, q2, ... qN are the M zeros. p1, p2, ... pN are the N poles.
Usually we assume that the number of poles is at least as many as the
number of zeros: M <= N, otherwize we can perform a form of long
division and get terms like z and z^2 outside of this big fraction.

the zeros q1, q2, ... qN and poles p1, p2, ... pN should be considered
possibly complex.  Now let's also assume (for simplicity) that none of
the poles are identical to any other poles.  The above transfer
function can be represented as a sum of partial fractions (instead of
as a product of factors with zeros divided by a product of factors
with poles):

               A_1          A_2                A_N
    H(z)  =  --------  +  --------  + ... +  --------
              z - p1       z - p2             z - pN

The denominators of each partial fraction are the same factors as
those in the denominator of the form of H(z) we started with.  The
poles are the same in both cases.  That makes sense because, if the
two forms are equivalent, when H(z) is evaluated at any pole location,
|H(z)| will blow up.  In the first form, it blows up because the
denominator of the whole transfer function goes to zero.  in the
second form (with partial fractions), it blows up because one of the
partial fractions blow up (because its denominator goes to zero).

The symbols A_1, A_2, ... A_N are *constants* that have to be just
right so that when all of these partial fractions are summed (and
placed over a common denominator), it adds up to the same H(z) as
before.  Now the zeros have some effect in the latter transfer
function (they have to), but they do not affect any partial fractions
denominator, only the numerators.

Then when we perform the inverse Z transform of this transfer
function, H(z), we get the impulse response that looks like


    h[n] = u[n]*( A_1*p1^n  +  A_2*p2^n  + ... +  A_N*pN^n )

where u[n] is the unit step function.  p1^n is an exponential
function.  in order for the impulse response to be stable and not blow
up when a unit impulse is applied to the input, then all of the pole
values must have magnitude of less than 1.

    |p1| < 1,   |p2| < 1,   ...   |pN| < 1 .

if any of those get as big as 1, the impulse response will never
settle down and decay to zero.  if any exceed 1, the exponential
functions will blow up.

that is the fundamental answer to your question.  if you're an EE
student, it's something you should learn by your 3rd year and
something similar for Laplace transforms, H(s).

r b-j