On 24 Sep., 22:23, Rune Allnor <all...@tele.ntnu.no> wrote:> On 24 Sep, 21:54, Andor <andor.bari...@gmail.com> wrote: > > > > > > > On 24 Sep., 15:49, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > On 24 Sep, 15:42, Andor <andor.bari...@gmail.com> wrote: > > > > > On 24 Sep., 15:37, Rune Allnor <all...@tele.ntnu.no> wrote: > > > > > > On 24 Sep, 15:04, Andor <andor.bari...@gmail.com> wrote: > > > > > > > As I said, if you calculate 5 more terms in the w that I started > > > > > > below, I will tell you how to calculate w (given v1 and v2) in Matlab > > > > > > with one single command that has only 28 characters (challenge: who > > > > > > can do it in less?). > > > > > > I can do it with a 1-character command: > > > > > > %%%%%%%%%% File a.m %%%%%%%%%%%%% > > > > > > % implement computations here > > > > > > %%%%%%%%% End of a.m %%%%%%%%%%%% > > > > > > in which case the one-liner becomes > > > > > > >> a > > > > > > But that might not have been what you mean...? > > > > > Nah, only Matlab and signal processing toolbox functions are > > > > allowed :-) > > > > Then you'll have to accept my solution above: > > > I don't have the SPT. It would be much more > > > interesting with only basic matlab commands allowed. > > > Interesting perhaps, but it is much more instructive to use the > > "filter" command. > > > Consider the problem: given two vectors x and y, find w such that w*x > > = y. The solution is long division of y by x. Now remember that the > > impulse response of a rational transfer function filter is the long > > division of the numerator by the denominator polynomial. So w is > > simply the impulse response of the filter with numerator y and > > denominator x! This will, in general, be an IIR filter, thus w has > > infinite length. The neat Matlab one-liner thus is > > > w = filter(y,x,[1 zeros(1,n)]) > > > where n is the number of terms you want to compute in w. > > Why do you need the SPT for that one? > > Apart from that, your approach is sensitive to the usual > questions about poles and stability. If the denomiator > polynomial x has poles on or outside the unit circle your > approach blows up.Of course it does. The OPs denominator has all of its poles outside the unit circle. That doesn't change the fact that this impulse response w is the one and only solution to the problem of finding w sucht that w*x = y.
deconvolution problem
Started by ●September 23, 2009
Reply by ●September 24, 20092009-09-24
Reply by ●September 25, 20092009-09-25
>On 24 Sep., 22:23, Rune Allnor <all...@tele.ntnu.no> wrote: >> On 24 Sep, 21:54, Andor <andor.bari...@gmail.com> wrote: >> >> >> >> >> >> > On 24 Sep., 15:49, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> > > On 24 Sep, 15:42, Andor <andor.bari...@gmail.com> wrote: >> >> > > > On 24 Sep., 15:37, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> > > > > On 24 Sep, 15:04, Andor <andor.bari...@gmail.com> wrote: >> >> > > > > > As I said, if you calculate 5 more terms in the w that Istarted>> > > > > > below, I will tell you how to calculate w (given v1 and v2)in Matlab>> > > > > > with one single command that has only 28 characters(challenge: who>> > > > > > can do it in less?). >> >> > > > > I can do it with a 1-character command: >> >> > > > > %%%%%%%%%% File a.m %%%%%%%%%%%%% >> >> > > > > % implement computations here >> >> > > > > %%%%%%%%% End of a.m %%%%%%%%%%%% >> >> > > > > in which case the one-liner becomes >> >> > > > > >> a >> >> > > > > But that might not have been what you mean...? >> >> > > > Nah, only Matlab and signal processing toolbox functions are >> > > > allowed :-) >> >> > > Then you'll have to accept my solution above: >> > > I don't have the SPT. It would be much more >> > > interesting with only basic matlab commands allowed. >> >> > Interesting perhaps, but it is much more instructive to use the >> > "filter" command. >> >> > Consider the problem: given two vectors x and y, find w such thatw*x>> > = y. The solution is long division of y by x. Now remember that the >> > impulse response of a rational transfer function filter is the long >> > division of the numerator by the denominator polynomial. So w is >> > simply the impulse response of the filter with numerator y and >> > denominator x! This will, in general, be an IIR filter, thus w has >> > infinite length. The neat Matlab one-liner thus is >> >> > w = filter(y,x,[1 zeros(1,n)]) >> >> > where n is the number of terms you want to compute in w. >> >> Why do you need the SPT for that one? >> >> Apart from that, your approach is sensitive to the usual >> questions about poles and stability. If the denomiator >> polynomial x has poles on or outside the unit circle your >> approach blows up. > >Of course it does. The OPs denominator has all of its poles outside >the unit circle. That doesn't change the fact that this impulse >response w is the one and only solution to the problem of finding w >sucht that w*x = y. >w = filter(y,x,[1 zeros(1,n)]) is the same as deconv. deconv give the result with the rest... My question was if we can find w/ conv(w,vect1)=vect2 and eliminate the rest (rest=0)...
Reply by ●September 25, 20092009-09-25
On 25 Sep., 09:58, "sofiyya" <karimae...@gmail.com> wrote:> >On 24 Sep., 22:23, Rune Allnor <all...@tele.ntnu.no> wrote: > >> On 24 Sep, 21:54, Andor <andor.bari...@gmail.com> wrote: > > >> > On 24 Sep., 15:49, Rune Allnor <all...@tele.ntnu.no> wrote: > > >> > > On 24 Sep, 15:42, Andor <andor.bari...@gmail.com> wrote: > > >> > > > On 24 Sep., 15:37, Rune Allnor <all...@tele.ntnu.no> wrote: > > >> > > > > On 24 Sep, 15:04, Andor <andor.bari...@gmail.com> wrote: > > >> > > > > > As I said, if you calculate 5 more terms in the w that I > started > >> > > > > > below, I will tell you how to calculate w (given v1 and v2) > in Matlab > >> > > > > > with one single command that has only 28 characters > (challenge: who > >> > > > > > can do it in less?). > > >> > > > > I can do it with a 1-character command: > > >> > > > > %%%%%%%%%% File a.m %%%%%%%%%%%%% > > >> > > > > % implement computations here > > >> > > > > %%%%%%%%% End of a.m %%%%%%%%%%%% > > >> > > > > in which case the one-liner becomes > > >> > > > > >> a > > >> > > > > But that might not have been what you mean...? > > >> > > > Nah, only Matlab and signal processing toolbox functions are > >> > > > allowed :-) > > >> > > Then you'll have to accept my solution above: > >> > > I don't have the SPT. It would be much more > >> > > interesting with only basic matlab commands allowed. > > >> > Interesting perhaps, but it is much more instructive to use the > >> > "filter" command. > > >> > Consider the problem: given two vectors x and y, find w such that > w*x > >> > = y. The solution is long division of y by x. Now remember that the > >> > impulse response of a rational transfer function filter is the long > >> > division of the numerator by the denominator polynomial. So w is > >> > simply the impulse response of the filter with numerator y and > >> > denominator x! This will, in general, be an IIR filter, thus w has > >> > infinite length. The neat Matlab one-liner thus is > > >> > w = filter(y,x,[1 zeros(1,n)]) > > >> > where n is the number of terms you want to compute in w. > > >> Why do you need the SPT for that one? > > >> Apart from that, your approach is sensitive to the usual > >> questions about poles and stability. If the denomiator > >> polynomial x has poles on or outside the unit circle your > >> approach blows up. > > >Of course it does. The OPs denominator has all of its poles outside > >the unit circle. That doesn't change the fact that this impulse > >response w is the one and only solution to the problem of finding w > >sucht that w*x = y. > > w = filter(y,x,[1 zeros(1,n)]) is the same as deconv. deconv give the > result with the rest... My question was if we can find w/ > conv(w,vect1)=vect2 and eliminate the rest (rest=0)...What do you think: is there a method to divide 3 by 2 such that the rest = 0?
Reply by ●September 25, 20092009-09-25
>On 25 Sep., 09:58, "sofiyya" <karimae...@gmail.com> wrote: >> >On 24 Sep., 22:23, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> On 24 Sep, 21:54, Andor <andor.bari...@gmail.com> wrote: >> >> >> > On 24 Sep., 15:49, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> >> > > On 24 Sep, 15:42, Andor <andor.bari...@gmail.com> wrote: >> >> >> > > > On 24 Sep., 15:37, Rune Allnor <all...@tele.ntnu.no> wrote: >> >> >> > > > > On 24 Sep, 15:04, Andor <andor.bari...@gmail.com> wrote: >> >> >> > > > > > As I said, if you calculate 5 more terms in the w that I >> started >> >> > > > > > below, I will tell you how to calculate w (given v1 andv2)>> in Matlab >> >> > > > > > with one single command that has only 28 characters >> (challenge: who >> >> > > > > > can do it in less?). >> >> >> > > > > I can do it with a 1-character command: >> >> >> > > > > %%%%%%%%%% File a.m %%%%%%%%%%%%% >> >> >> > > > > % implement computations here >> >> >> > > > > %%%%%%%%% End of a.m %%%%%%%%%%%% >> >> >> > > > > in which case the one-liner becomes >> >> >> > > > > >> a >> >> >> > > > > But that might not have been what you mean...? >> >> >> > > > Nah, only Matlab and signal processing toolbox functions are >> >> > > > allowed :-) >> >> >> > > Then you'll have to accept my solution above: >> >> > > I don't have the SPT. It would be much more >> >> > > interesting with only basic matlab commands allowed. >> >> >> > Interesting perhaps, but it is much more instructive to use the >> >> > "filter" command. >> >> >> > Consider the problem: given two vectors x and y, find w such that >> w*x >> >> > = y. The solution is long division of y by x. Now remember thatthe>> >> > impulse response of a rational transfer function filter is thelong>> >> > division of the numerator by the denominator polynomial. So w is >> >> > simply the impulse response of the filter with numerator y and >> >> > denominator x! This will, in general, be an IIR filter, thus whas>> >> > infinite length. The neat Matlab one-liner thus is >> >> >> > w = filter(y,x,[1 zeros(1,n)]) >> >> >> > where n is the number of terms you want to compute in w. >> >> >> Why do you need the SPT for that one? >> >> >> Apart from that, your approach is sensitive to the usual >> >> questions about poles and stability. If the denomiator >> >> polynomial x has poles on or outside the unit circle your >> >> approach blows up. >> >> >Of course it does. The OPs denominator has all of its poles outside >> >the unit circle. That doesn't change the fact that this impulse >> >response w is the one and only solution to the problem of finding w >> >sucht that w*x = y. >> >> w = filter(y,x,[1 zeros(1,n)]) is the same as deconv. deconv give the >> result with the rest... My question was if we can find w/ >> conv(w,vect1)=vect2 and eliminate the rest (rest=0)... > >What do you think: is there a method to divide 3 by 2 such that the >rest = 0? >If we multiply with [1 0 0 0 ...] we can eliminate the rest.
