Z-Transform(poles, Zeros and ROC)

PrasadBC(CISC Tech) wrote:

> X(z) = 1 + az^-1+(az^-1)^2 +(az^-1)^3+.....+infinity;
> .....(eq 1)

Writing "+ infinity" to indicate infinitely many terms mixes up two 
independent notions; don't do that.

> X(z) = 1/(1-az^-1);  provided |az^-1|<1 => |z| >|a|
> ......(eq 2) 
> 
> now X(z) can be written as  X(z) = z/(z-a).            
> ......(eq 3) 
> 
> Now the question is what are the poles and zeros location of
> X(z)????????
> 
> DSP Books says Zero at z = 0      Pole at z = a from (eq 3)
> 
> but how can z=0 be substituted in the equation for X(z) =
> 1/(1-az^-1) as   0 < |a| which is not in ROC and it makes the
> original X(z) calculated using (eq 1) to go to infinity.

No. Each term in (1) is *indeterminate*, not infinite, at z = 0. 
Algebraic manipulation without regard to possible singularities -- 
such as augmenting (2) with z to get (3) -- amounts to implicit 
analytic continuation, thereby defining the value at a removable 
singularity.


Martin

-- 
Quidquid latine scriptum est, altum videtur.

> > X(z) = 1/(1-az^-1);  provided |az^-1|<1 => |z| >|a|
> > ......(eq 2)
> >
> > now X(z) can be written as  X(z) = z/(z-a).
> > ......(eq 3)
> >
equation 3 of X(z) is obtained by multiplying and dividing numerator
and denominator of X(z) of equation 2 with z.

It is the basics of algebra that you cant do it when z = 0.

1/(1-az^-1) is not equal to z/(z-a) at z = 0.

so evaluation of the poles and zeros should be done for equations
expressed in negative powers of z and not positive powers of z


X(z) = 1/(1-az^-1); doesnt have zero at z = 0, Infact it doesnt have
any zeros

In conclusion "no zeros exist outside the Region of convergence"

kee skrev:
> > > X(z) = 1/(1-az^-1);  provided |az^-1|<1 => |z| >|a|
> > > ......(eq 2)
> > >
> > > now X(z) can be written as  X(z) = z/(z-a).
> > > ......(eq 3)
> > >
> equation 3 of X(z) is obtained by multiplying and dividing numerator
> and denominator of X(z) of equation 2 with z.
>
> It is the basics of algebra that you cant do it when z = 0.
>
> 1/(1-az^-1) is not equal to z/(z-a) at z = 0.
>
> so evaluation of the poles and zeros should be done for equations
> expressed in negative powers of z and not positive powers of z
>
>
> X(z) = 1/(1-az^-1); doesnt have zero at z = 0, Infact it doesnt have
> any zeros
>
> In conclusion "no zeros exist outside the Region of convergence"

What you are say, then, is that

X(z) = 1/(1-az^-1) = 1/(1-az^-1) *1 = 1/(1-az^-1) *z/z = z/(z-a) =/=
X(z)...

Rune

On Dec 29, 11:20 am, "Rune Allnor" <all...@tele.ntnu.no> wrote:
> kee skrev:
>
>
>
> > > > X(z) = 1/(1-az^-1);  provided |az^-1|<1 => |z| >|a|
> > > > ......(eq 2)
>
> > > > now X(z) can be written as  X(z) = z/(z-a).
> > > > ......(eq 3)
>
> > equation 3 of X(z) is obtained by multiplying and dividing numerator
> > and denominator of X(z) of equation 2 with z.
>
> > It is the basics of algebra that you cant do it when z = 0.
>
> > 1/(1-az^-1) is not equal to z/(z-a) at z = 0.
>
> > so evaluation of the poles and zeros should be done for equations
> > expressed in negative powers of z and not positive powers of z
>
> > X(z) = 1/(1-az^-1); doesnt have zero at z = 0, Infact it doesnt have
> > any zeros
>
> > In conclusion "no zeros exist outside the Region of convergence"What you are say, then, is that
>
> X(z) = 1/(1-az^-1) = 1/(1-az^-1) *1 = 1/(1-az^-1) *z/z = z/(z-a) =/=
> X(z)...

At z=0, indeed.  In the same way that:

1 = 1 * z/z = z/z =/= 1 at z = 0

-- 
Oli

Oli Charlesworth wrote:
> On Dec 29, 11:20 am, "Rune Allnor" <all...@tele.ntnu.no> wrote:
> >
> > > > > X(z) = 1/(1-az^-1);  provided |az^-1|<1 => |z| >|a|
> > > > > ......(eq 2)
> >
> > > > > now X(z) can be written as  X(z) = z/(z-a).
> > > > > ......(eq 3)
> >
> > > equation 3 of X(z) is obtained by multiplying and dividing numerator
> > > and denominator of X(z) of equation 2 with z.
> >
> > > It is the basics of algebra that you cant do it when z = 0.
> >
> > > 1/(1-az^-1) is not equal to z/(z-a) at z = 0.
> >
> > > so evaluation of the poles and zeros should be done for equations
> > > expressed in negative powers of z and not positive powers of z
> >
> > > X(z) = 1/(1-az^-1); doesnt have zero at z = 0, Infact it doesnt have
> > > any zeros
> >
> > > In conclusion "no zeros exist outside the Region of convergence"What you are say, then, is that
> >
> > X(z) = 1/(1-az^-1) = 1/(1-az^-1) *1 = 1/(1-az^-1) *z/z = z/(z-a) =/=
> > X(z)...
>
> At z=0, indeed.  In the same way that:
>
> 1 = 1 * z/z = z/z =/= 1 at z = 0

but when this is a removable singularity, i think it *is* accurate and
appropriate to say something like

    " X(z) = 1/(1 - p*z^(-1))   has a zero at z = 0. "

or simply that

      X(z) = 1/(1 - p*z^(-1)) = z/(z-p) = (z-q)/(z-p)   where q = 0

they are, in fact, two different expressions of transfer function for
the same LTI system.

r b-j