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

# Does the z transform has the property?

Started by ●June 23, 2006

Reply by ●June 23, 20062006-06-23

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 Posting from Google? See http://cfaj.freeshell.org/google/ "Applied Control Theory for Embedded Systems" came out in April. See details at http://www.wescottdesign.com/actfes/actfes.html

Reply by ●June 25, 20062006-06-25

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 > > Posting from Google? See http://cfaj.freeshell.org/google/ > > "Applied Control Theory for Embedded Systems" came out in April. > See details at http://www.wescottdesign.com/actfes/actfes.html

Reply by ●June 26, 20062006-06-26

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 way, please do not answer me directly to my e-mail, rather answer via this group. Regards, Andor

Reply by ●June 26, 20062006-06-26

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 > way, please do not answer me directly to my e-mail, rather answer via > this group. > > Regards, > Andor

Reply by ●June 26, 20062006-06-26

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

Reply by ●June 26, 20062006-06-26

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

Reply by ●June 26, 20062006-06-26

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

Reply by ●June 26, 20062006-06-26

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

Reply by ●September 8, 20062006-09-08

>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 news:1151310430.436443.257480@p79g2000cwp.googlegroups.com...> 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 > way, please do not answer me directly to my e-mail, rather answer via > this group. > > Regards, > Andor >