# Does the z transform has the property?

Started by June 23, 2006
```Dear ALl:

I have question with the bilateral z transform property,

D(x) is the product of  Q(x) and C(x)

After Ztransform ,

Z(D(x)) = Z(Q(x))* Z(C(x))

* is the convolution operator.

is that property hold ?

Thanks
bin

```
```yezi wrote:
> Dear ALl:
>
> I have question with the bilateral z transform property,
>
> D(x) is the product of  Q(x) and C(x)
>
> After Ztransform ,
>
> Z(D(x)) = Z(Q(x))* Z(C(x))
>
> * is the convolution operator.
>
> is that property hold ?
>
> Thanks
> bin
>
You can find this out for yourself by remembering that the z transform
is only valid for linear, time-invariant systems.  A system h is linear
if, for any two signals x1 and x2, superposition holds:

h(x1+x2) = h(x1) + h(x2).

So your system d is defined as d(x) = q(x) * c(x).

Does d(x + x) = d(x) + d(x)?  You'll find the answer is, in general,
"no", which means that your system isn't linear, which means that you
can't use the z transform to analyze it.

--

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

"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
```
```Thanks for corresponding, I have question with " complex convolution
Theorem" , which seems to me, the property holds. however for very
strict contour of integration. So would you mind explain the contour of
integration more for me. What is that mean?

Thanks
Tim Wescott wrote:
> yezi wrote:
> > Dear ALl:
> >
> > I have question with the bilateral z transform property,
> >
> > D(x) is the product of  Q(x) and C(x)
> >
> > After Ztransform ,
> >
> > Z(D(x)) = Z(Q(x))* Z(C(x))
> >
> > * is the convolution operator.
> >
> > is that property hold ?
> >
> > Thanks
> > bin
> >
> You can find this out for yourself by remembering that the z transform
> is only valid for linear, time-invariant systems.  A system h is linear
> if, for any two signals x1 and x2, superposition holds:
>
> h(x1+x2) = h(x1) + h(x2).
>
> So your system d is defined as d(x) = q(x) * c(x).
>
> Does d(x + x) = d(x) + d(x)?  You'll find the answer is, in general,
> "no", which means that your system isn't linear, which means that you
> can't use the z transform to analyze it.
>
> --
>
> Tim Wescott
> Wescott Design Services
> http://www.wescottdesign.com
>
>
> "Applied Control Theory for Embedded Systems" came out in April.
> See details at http://www.wescottdesign.com/actfes/actfes.html

```
```yezi wrote:
> Dear ALl:
>
> I have question with the bilateral z transform property,
>
> D(x) is the product of  Q(x) and C(x)
>
> After Ztransform ,
>
> Z(D(x)) = Z(Q(x))* Z(C(x))
>
> * is the convolution operator.
>
> is that property hold ?

The z-transform of the product of two time-domain sequences can be
computed with the complex convolution theorem (as you noted yourself).
You can find out about it in detailed texts about either DSP or
difference equations.

Yezi, if I am piecing your posts together correctly, you want to
compute the entropy of a discrete random variable using the complex
convolution theorem.

For discrete random variables, the z-transform equals the probability
generating function. You might have problems to determine the common
area of convergence of the two multiplicands in z domain, as the
coefficients of the z-transform of the logarithm of the discrete
probability density necessarily go to minus infinity (for random
variables with infinite domain). Only the product "x ln(x)" is well
behaved as x->0.

Why do you want to compute the entropy in such a roundabout way? By the
this group.

Regards,
Andor

```
```Thanks Andor. I am quite surprised with your excellent sagacity. You
are quite right. Acutally what the problem I am trying to figure  out
is :

I know r1(x) and r2(x), which is discrete probability density function.
The entropy of each of them is defined as before.

There is another another r3(x)= r1(x)*r2(x).  What I try to do is use
the entropy of r1(x) and r2(x) to express the entropy of r3(x).

I try to solve it in current space, however it canot work out. That is
the reason I count on the Z transform , since r3(x)=r1(x)*r2(x) has
good property in Z space. However, it turns out that because  entropy
is defined as sum(r1(x) mutiply with ln(r1(x))), I can not solve the
multiplication very well.

Till now, that is the reason I post this help. Hope my description will
clearify your question. AND I really appreciate your help. Hope To hear
more from you.

Thanks
yezi

Andor wrote:
> yezi wrote:
> > Dear ALl:
> >
> > I have question with the bilateral z transform property,
> >
> > D(x) is the product of  Q(x) and C(x)
> >
> > After Ztransform ,
> >
> > Z(D(x)) = Z(Q(x))* Z(C(x))
> >
> > * is the convolution operator.
> >
> > is that property hold ?
>
> The z-transform of the product of two time-domain sequences can be
> computed with the complex convolution theorem (as you noted yourself).
> You can find out about it in detailed texts about either DSP or
> difference equations.
>
> Yezi, if I am piecing your posts together correctly, you want to
> compute the entropy of a discrete random variable using the complex
> convolution theorem.
>
> For discrete random variables, the z-transform equals the probability
> generating function. You might have problems to determine the common
> area of convergence of the two multiplicands in z domain, as the
> coefficients of the z-transform of the logarithm of the discrete
> probability density necessarily go to minus infinity (for random
> variables with infinite domain). Only the product "x ln(x)" is well
> behaved as x->0.
>
> Why do you want to compute the entropy in such a roundabout way? By the
> this group.
>
> Regards,
> Andor

```
```yezi wrote:
> Dear ALl:
>
> I have question with the bilateral z transform property,
>
> D(x) is the product of  Q(x) and C(x)
>
> After Ztransform ,
>
> Z(D(x)) = Z(Q(x))* Z(C(x))
>
> * is the convolution operator.
>
> is that property hold ?

The only place I have seen this discussed, is the 1975 edition of
Oppenheim & Scafers "Digital Signal Processing". It's a nightmare
of complex maths, analytical functions and contour integrals in
the z plane, but they find a result. I *think* that derivation was
the reason why I bought Churchill & Brown's book on complex
maths (something in O&S 75 caused me to buy B&C, but it may
have been something else)

I say "a result" since I never attempted to have a go at the derivation

(one knows one's limits, and unlike Cauchy's, mine is finite...) and
I have no idea about whether what came out of the excercise
has anything to do with the above.

Oh well.

Rune

```
```That is quite true, I find the complex convolution therom just in the
book your mentioned. It is also disaster for me. However currently I
can not find other options to derive the relations between these 3
entropys.

:-), do we have other choices?

Rune Allnor wrote:
> yezi wrote:
> > Dear ALl:
> >
> > I have question with the bilateral z transform property,
> >
> > D(x) is the product of  Q(x) and C(x)
> >
> > After Ztransform ,
> >
> > Z(D(x)) = Z(Q(x))* Z(C(x))
> >
> > * is the convolution operator.
> >
> > is that property hold ?
>
> The only place I have seen this discussed, is the 1975 edition of
> Oppenheim & Scafers "Digital Signal Processing". It's a nightmare
> of complex maths, analytical functions and contour integrals in
> the z plane, but they find a result. I *think* that derivation was
> the reason why I bought Churchill & Brown's book on complex
> maths (something in O&S 75 caused me to buy B&C, but it may
> have been something else)
>
> I say "a result" since I never attempted to have a go at the derivation
>
> (one knows one's limits, and unlike Cauchy's, mine is finite...) and
> I have no idea about whether what came out of the excercise
> has anything to do with the above.
>
> Oh well.
>
> Rune

```
```yezi wrote:
> That is quite true, I find the complex convolution therom just in the
> book your mentioned. It is also disaster for me. However currently I
> can not find other options to derive the relations between these 3
> entropys.
>
> :-), do we have other choices?

I don't know. It depends on why you need/want to derive the result.
If one can use the complex convolution theorem to justify that
element-wise multiiplication in time domain results in a convolution
of spectrum magnitudes in z (or frequency) domain, that's enough
for me. I am happy with that the world works like that. I don't
need to prove it mathematically.

Again, it's a question of value for effort.

Rune

> Rune Allnor wrote:
> > yezi wrote:
> > > Dear ALl:
> > >
> > > I have question with the bilateral z transform property,
> > >
> > > D(x) is the product of  Q(x) and C(x)
> > >
> > > After Ztransform ,
> > >
> > > Z(D(x)) = Z(Q(x))* Z(C(x))
> > >
> > > * is the convolution operator.
> > >
> > > is that property hold ?
> >
> > The only place I have seen this discussed, is the 1975 edition of
> > Oppenheim & Scafers "Digital Signal Processing". It's a nightmare
> > of complex maths, analytical functions and contour integrals in
> > the z plane, but they find a result. I *think* that derivation was
> > the reason why I bought Churchill & Brown's book on complex
> > maths (something in O&S 75 caused me to buy B&C, but it may
> > have been something else)
> >
> > I say "a result" since I never attempted to have a go at the derivation
> >
> > (one knows one's limits, and unlike Cauchy's, mine is finite...) and
> > I have no idea about whether what came out of the excercise
> > has anything to do with the above.
> >
> > Oh well.
> >
> > Rune

```
```My oringinal problem is posted on previous:

I know r1(x) and r2(x), which is discrete probability density function.

The entropy of each of them is defined as before.

There is another another r3(x)= r1(x)*r2(x).  What I try to do is use
the entropy of r1(x) and r2(x) to express the entropy of r3(x).

I try to solve it in current space, however it canot work out. That is
the reason I count on the Z transform , since r3(x)=r1(x)*r2(x) has
good property in Z space. However, it turns out that because  entropy
is defined as sum(r1(x) mutiply with ln(r1(x))), I can not solve the
multiplication very well.

The complex convolution is just one of option solution.

Rune Allnor wrote:
> yezi wrote:
> > That is quite true, I find the complex convolution therom just in the
> > book your mentioned. It is also disaster for me. However currently I
> > can not find other options to derive the relations between these 3
> > entropys.
> >
> > :-), do we have other choices?
>
> I don't know. It depends on why you need/want to derive the result.
> If one can use the complex convolution theorem to justify that
> element-wise multiiplication in time domain results in a convolution
> of spectrum magnitudes in z (or frequency) domain, that's enough
> for me. I am happy with that the world works like that. I don't
> need to prove it mathematically.
>
> Again, it's a question of value for effort.
>
> Rune
>
> > Rune Allnor wrote:
> > > yezi wrote:
> > > > Dear ALl:
> > > >
> > > > I have question with the bilateral z transform property,
> > > >
> > > > D(x) is the product of  Q(x) and C(x)
> > > >
> > > > After Ztransform ,
> > > >
> > > > Z(D(x)) = Z(Q(x))* Z(C(x))
> > > >
> > > > * is the convolution operator.
> > > >
> > > > is that property hold ?
> > >
> > > The only place I have seen this discussed, is the 1975 edition of
> > > Oppenheim & Scafers "Digital Signal Processing". It's a nightmare
> > > of complex maths, analytical functions and contour integrals in
> > > the z plane, but they find a result. I *think* that derivation was
> > > the reason why I bought Churchill & Brown's book on complex
> > > maths (something in O&S 75 caused me to buy B&C, but it may
> > > have been something else)
> > >
> > > I say "a result" since I never attempted to have a go at the derivation
> > >
> > > (one knows one's limits, and unlike Cauchy's, mine is finite...) and
> > > I have no idea about whether what came out of the excercise
> > > has anything to do with the above.
> > >
> > > Oh well.
> > >
> > > Rune

```
```>You might have problems to determine the common
> area of convergence of the two multiplicands in z domain, as the
> coefficients of the z-transform of the logarithm of the discrete
> probability density necessarily go to minus infinity (for random
>variables with infinite domain).

Don't try to say this in one breath :-)

"Andor" <andor.bariska@gmail.com> wrote in message
> yezi wrote:
>> Dear ALl:
>>
>> I have question with the bilateral z transform property,
>>
>> D(x) is the product of  Q(x) and C(x)
>>
>> After Ztransform ,
>>
>> Z(D(x)) = Z(Q(x))* Z(C(x))
>>
>> * is the convolution operator.
>>
>> is that property hold ?
>
> The z-transform of the product of two time-domain sequences can be
> computed with the complex convolution theorem (as you noted yourself).
> You can find out about it in detailed texts about either DSP or
> difference equations.
>
> Yezi, if I am piecing your posts together correctly, you want to
> compute the entropy of a discrete random variable using the complex
> convolution theorem.
>
> For discrete random variables, the z-transform equals the probability
> generating function. You might have problems to determine the common
> area of convergence of the two multiplicands in z domain, as the
> coefficients of the z-transform of the logarithm of the discrete
> probability density necessarily go to minus infinity (for random
> variables with infinite domain). Only the product "x ln(x)" is well
> behaved as x->0.
>
> Why do you want to compute the entropy in such a roundabout way? By the