FPGA Central

Hi All,

Here is a doubt regarding Z-Transform with respect to poles and
zeros. Here is the problem

example:

x(n) = a^n * u(n)

X(z) = (n=0 to n=infinity)SUM x(n) * z^ -n;

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

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.

But ROC plots in DSP books show a zero at z=0, Are the DSP Texts
stating wrong? or is my understanding wrong?

PrasadBC(CISC Tech) skrev:
> Hi All,

> 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.

Because z^-1 means 1/z, and multiplying X(x) by z/z removes
all 1/z terms from the expression.

Rune

On Tue, 26 Dec 2006 12:46:31 -0000, Rune Allnor <[email protected]>
wrote:
>
> PrasadBC(CISC Tech) skrev:
>> Hi All,
>
>> 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.
>
> Because z^-1 means 1/z, and multiplying X(x) by z/z removes
> all 1/z terms from the expression.

That doesn't really address the question though, which was (as I
understand it) "How can convergent points exist outside the ROC?", as the
term "ROC" would imply "the set of values of z which cause X(z) to
converge". I guess the answer is that singularities can exist outside the
ROC, at which convergence still occurs, i.e. the definition of ROC is
actually "a subset of the values of z which cause X(z) to converge".

On a related note, if we look at the expanded summation form of X(z) (as
given by the OP):

X(z) = 1 + a^1.z^-1 + a^2.z^-2 + a^3.z^-3 + ... (Eq.4)

it would appear that at z=0, this should be equal to 1, and hence isn't a
zero! This problem seems to be caused by considering the general form of
the Z-transform:

inf
X(z) = SUM x[n].z^-n (Eq.5)
n=0

What is the value at z=0? This requires evaluating the summation term
when n=0, which is x[0].0^0, which would appear to be undefined. We can
get round this problem by re-defining the Z-transform as:

inf
X(z) = x[0] + SUM x[n].z^-n (Eq.6)
n=1

which in turn leads to (Eq.4). But this definition would seem to preclude
the existence of zeros at z=0 (unless x[0]=0).

In summary, I'm thoroughly confused!



--
Oli

On 26 Dec 2006 04:46:31 -0800, "Rune Allnor" <[email protected]>
wrote:

>Because z^-1 means 1/z, and multiplying X(x) by z/z removes
>all 1/z terms from the expression.

I think that you misunderstood the OP's question (either that, or I'm
reading more into it than is actually there). It looks to me like he is
addressing exactly the same issue, using exactly the same example, that
I did in this thread:

http://groups.google.com/group/comp....7a6e6a88507733

Basically, it is this: How can a transfer function have a zero at a
location outside its region of convergence? Does it make any sense at
all to discuss behavior outside the ROC?

-- Greg Berchin

Oli Charlesworth skrev:

> That doesn't really address the question though, which was (as I
> understand it) "How can convergent points exist outside the ROC?",

If that's the OP's question, I didn't catch it.

> as the
> term "ROC" would imply "the set of values of z which cause X(z) to
> converge". I guess the answer is that singularities can exist outside the
> ROC, at which convergence still occurs, i.e. the definition of ROC is
> actually "a subset of the values of z which cause X(z) to converge".

The function X(z) = z/(z-a) has exactly one point where it does
not converge. So pedantically speaking, I suppose the ROC
can be defined as "everywhere except z=a".

I think the usual terminology in context of the z transform is
that the ROC denotes a region 0 <= b < |z| < c <= inf where
no poles exist.

> In summary, I'm thoroughly confused!

Me too. After posting my first response, I thought along these
lines:

Let P be a polynomial in x,

P(x) = a_n x^n + ... + a_1 x + 1

Multiply by x^-n/x^-n = 1 and re-order coefficients:

P'(x) = x^-n +a_1 x^(n-1) + ... + a_n

Since we only multiplied by 1, P'(x)=P(x).

However, P'(0) = infinite because of the 1/x terms,
while P(0) = 1.

Where is the flaw?

And no, this is not a troll. I don't see what cancels
the effect of the 1/x terms for x = 0.

Rune

On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <[email protected]>
wrote:

>
> Oli Charlesworth skrev:
>
>> That doesn't really address the question though, which was (as I
>> understand it) "How can convergent points exist outside the ROC?",
>
> If that's the OP's question, I didn't catch it.
>
>> as the
>> term "ROC" would imply "the set of values of z which cause X(z) to
>> converge". I guess the answer is that singularities can exist outside
>> the
>> ROC, at which convergence still occurs, i.e. the definition of ROC is
>> actually "a subset of the values of z which cause X(z) to converge".
>
> The function X(z) = z/(z-a) has exactly one point where it does
> not converge. So pedantically speaking, I suppose the ROC
> can be defined as "everywhere except z=a".

I think you are confusing two things, "convergence", and "singularities"..

The function X(z) = z/(z-a) does indeed have a singularity at z=a, but it
is not meaningful to talk about "convergence" of X(z).

The term "convergence" (and therefore "ROC") can only apply to an infinite
summation; in our case it is meaningful to talk about the convergence of:

X(z) = SUM x[n].z^-n (Eq.7)

which has a clearly-defined ROC. If we only consider values within that
ROC, it is valid to specify X(z) as:

X(z) = z/(z-a) (Eq.8)

However, (Eq.8) is *not* the same as (Eq.7) outside the ROC.

Specifying a Z-transform on its own doesn't uniquely define the original
sequence. However, specifying a Z-transform *and* the ROC does uniquely
specify x[n].


> I think the usual terminology in context of the z transform is
> that the ROC denotes a region 0 <= b < |z| < c <= inf where
> no poles exist.
>
>> In summary, I'm thoroughly confused!
>
> Me too. After posting my first response, I thought along these
> lines:
>
> Let P be a polynomial in x,
>
> P(x) = a_n x^n + ... + a_1 x + 1
>
> Multiply by x^-n/x^-n = 1 and re-order coefficients:
>
> P'(x) = x^-n +a_1 x^(n-1) + ... + a_n
>
> Since we only multiplied by 1, P'(x)=P(x).
>
> However, P'(0) = infinite because of the 1/x terms,
> while P(0) = 1.
>
> Where is the flaw?

The flaw is that you multiplied by x^-n, not (x^-n/x^-n)!


--
Oli

Oli Charlesworth skrev:
> On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <[email protected]>
> wrote:
>
> >
> > Oli Charlesworth skrev:
> >
> >> That doesn't really address the question though, which was (as I
> >> understand it) "How can convergent points exist outside the ROC?",
> >
> > If that's the OP's question, I didn't catch it.
> >
> >> as the
> >> term "ROC" would imply "the set of values of z which cause X(z) to
> >> converge". I guess the answer is that singularities can exist outside
> >> the
> >> ROC, at which convergence still occurs, i.e. the definition of ROC is
> >> actually "a subset of the values of z which cause X(z) to converge".
> >
> > The function X(z) = z/(z-a) has exactly one point where it does
> > not converge. So pedantically speaking, I suppose the ROC
> > can be defined as "everywhere except z=a".
>
> I think you are confusing two things, "convergence", and "singularities".
>
> The function X(z) = z/(z-a) does indeed have a singularity at z=a, but it
> is not meaningful to talk about "convergence" of X(z).
>
> The term "convergence" (and therefore "ROC") can only apply to an infinite
> summation; in our case it is meaningful to talk about the convergence of:
>
> X(z) = SUM x[n].z^-n (Eq.7)
>
> which has a clearly-defined ROC. If we only consider values within that
> ROC, it is valid to specify X(z) as:
>
> X(z) = z/(z-a) (Eq.8)
>
> However, (Eq.8) is *not* the same as (Eq.7) outside the ROC.
>
> Specifying a Z-transform on its own doesn't uniquely define the original
> sequence. However, specifying a Z-transform *and* the ROC does uniquely
> specify x[n].

The Z transform is defined as the sum from negative infinite
to positive infinite. The Z transform (7) is well-behaved everywhere
except at z=a, for a suitable choise of summation limits. It is
guaranteed not to converge for z = a, regardless of summation
limits.

Since the ZT is linear, the ZT of one function can be separated
as the sum of ZTs two functions. Since the sign of the exponent n
is crucial for the discussion, we can make an ARBITRARY
choise to separate x[n] to one series for negative n and
another for non-negative n:

ZT{x[n]} = ZT{x1[n]} + ZT{x2[n]} (1)

where

ZT{x1[n]}^{-} = sum_{-inf}^-1 x[n] z^-n (2)
ZT{x2[n]}^{+} = sum{0}^{+inf} x[n] z^-n (3)

Exploiting the linearity of the ZT like this and employing a
divide-and-conquer strategy, one can see that it converges
everywhere except at the poles.

The convention is to define as the ROC the region where
the ZT converges under the summation limits as in (3).
This choise is CONVENIENT since we usually want to
discuss causal sequences.

Rune

On Tue, 26 Dec 2006 14:11:13 -0000, Rune Allnor <[email protected]>
wrote:

>
> Oli Charlesworth skrev:
>> On Tue, 26 Dec 2006 13:29:27 -0000, Rune Allnor <[email protected]>
>> wrote:
>> >
>> > The function X(z) = z/(z-a) has exactly one point where it does
>> > not converge. So pedantically speaking, I suppose the ROC
>> > can be defined as "everywhere except z=a".
>>
>> I think you are confusing two things, "convergence", and
>> "singularities".
>>
>> The function X(z) = z/(z-a) does indeed have a singularity at z=a, but
>> it
>> is not meaningful to talk about "convergence" of X(z).
>>
>> The term "convergence" (and therefore "ROC") can only apply to an
>> infinite
>> summation; in our case it is meaningful to talk about the convergence
>> of:
>>
>> X(z) = SUM x[n].z^-n (Eq.7)
>>
>> which has a clearly-defined ROC. If we only consider values within that
>> ROC, it is valid to specify X(z) as:
>>
>> X(z) = z/(z-a) (Eq.8)
>>
>> However, (Eq.8) is *not* the same as (Eq.7) outside the ROC.
>>
>> Specifying a Z-transform on its own doesn't uniquely define the original
>> sequence. However, specifying a Z-transform *and* the ROC does uniquely
>> specify x[n].
>
> The Z transform is defined as the sum from negative infinite
> to positive infinite. The Z transform (7) is well-behaved everywhere
> except at z=a, for a suitable choise of summation limits.

This is untrue. The Z-transform of a^n.u[n] is:

X(z) = SUM a^n.z^-n

and is not well-behaved (in the sense of convergence) for *any* value of
|z|<=|a|, not just z=a.

(This is in contrast to the expression z/(z-a), which is not well-behaved
only at z=a. Hence it is a different function, which is why we must
specify the ROC when discussing z/(z-a)).


> It is
> guaranteed not to converge for z = a, regardless of summation
> limits.

This, however, is true (with the proviso that we choose the upper limit as
+inf; for any other upper limit, it is a finite sum, and will always
"converge").


> Since the ZT is linear, the ZT of one function can be separated
> as the sum of ZTs two functions. Since the sign of the exponent n
> is crucial for the discussion, we can make an ARBITRARY
> choise to separate x[n] to one series for negative n and
> another for non-negative n:
>
> ZT{x[n]} = ZT{x1[n]} + ZT{x2[n]} (1)
>
> where
>
> ZT{x1[n]}^{-} = sum_{-inf}^-1 x[n] z^-n (2)
> ZT{x2[n]}^{+} = sum{0}^{+inf} x[n] z^-n (3)
>
> Exploiting the linearity of the ZT like this and employing a
> divide-and-conquer strategy, one can see that it converges
> everywhere except at the poles.

Again, the Z-transform does not converge *anywhere* outside the ROC, i.e..
in our example, it doesn't converge for any |z|<=|a| (except possibly at
the singular values specified as the zeros; I'm not too sure on that one!).



--
Oli

Oli Charlesworth skrev:

> Again, the Z-transform does not converge *anywhere* outside the ROC, i.e.
> in our example, it doesn't converge for any |z|<=|a| (except possibly at
> the singular values specified as the zeros; I'm not too sure on that one!).

So what is your point? That anything that happens inside the ROC
has to be discarded? If that's the case, all one would need was the
outermost pair of poles, and whatever zeros might be present
in the ROC.

Rune

On Tue, 26 Dec 2006 14:41:57 -0000, Rune Allnor <[email protected]>
wrote:

>
> Oli Charlesworth skrev:
>
>> Again, the Z-transform does not converge *anywhere* outside the ROC,
>> i.e.
>> in our example, it doesn't converge for any |z|<=|a| (except possibly at
>> the singular values specified as the zeros; I'm not too sure on that
>> one!).
>
> So what is your point? That anything that happens inside the ROC
> has to be discarded? If that's the case, all one would need was the
> outermost pair of poles, and whatever zeros might be present
> in the ROC.

I'm not sure I had a point! (and I'm not sure what you mean by "discarded")

However, to analyse stability of a causal system (causality requires that
the ROC include infinity), it is true that all you need to know are the
locations of the outermost poles, as this places a bound on the ROC (the
ROC cannot by definition include poles). If this ROC doesn't include the
unit circle, then the system is unstable.

This is the formal explanation for why poles *must* be inside the unit
circle for a causal system to be stable.



--
Oli

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