Hello all,

Thanks for your replies. Indeed, there are some stringent conditions.
The number of equations scales as ~n^2 but the number of unknowns
(size of h1 and h2) scales as ~n.

h1[n]h2[n] = h[n]
h1[n]h2[p] + h1[p]h2[n] = 0 (for n != p)

I was wondering if this problem has been studied before, otherwise,
I'll try to create a solution from scratch.

Going into Fourier does not seem to help.

On Jan 22, 5:02 am, "sarw...@YouEyeYouSee.edu" <dvsarw...@gmail.com>
wrote:
> On Jan 22, 1:16 am, Laurent Schmalen <lo...@gmx.de> wrote:
>
>
>
> > sarw...@YouEyeYouSee.edu wrote:
> > > On Jan 18, 2:41 pm, Antoine  Bruguier <tony.brugu...@gmail.com> wrote:
>
> > >> I was looking for conditions on h such that we can find h1 and h2 such
> > >> that for every s:
> > >> (s^2) * h = (s*h1) . (s*h2)
>
> > > If the condition is required to hold for *every* input s, then
> > > it must hold when the input is the unit pulse function given
> > > by p[n] = 1 for n = 0 and p[n] = 0 otherwise.  Since p^2 = p,
> > > and p is the identity for convolution, it must be that
> > > (p^2) * h = p * h = h, and p*h1 = h1, p*h2 = h2.  Hence,
> > > h = h1.h2, that is,
>
> > >          h[n] = h1[n]h2[n] for all n.
>
> > > Since multiplication in the time domain corresponds to convolution
> > > inn the frequency domain, perhaps you need to look at spectral
> > > convolution H = (H1) * (H2) rather than spectral factorization
> > > H = (H1)(H2).
>
> > Wrong.
> > Squaring is a non-linear operation and thus your condition does NOT
> > hold. Take this little example:
>
> > h1 = [1 2 3 4];
> > h2 = [-1 -2 -3 -4];
> > s = [0 -0.1 0.2 -0.3 0.4];
> > h = h1.*h2;
> > conv(s.^2,h)
> > conv(s,h1) .* conv(s,h2)
>
> > and compare the outputs.
>
> > Regards,
> > Laurent
>
> The condition h[n] = h1[n]h2[n] for all n is NECESSARY
> if it is desired that (s^2) * h = (s*h1) . (s*h2) hold for ALL
> signals s; it is by no means SUFFICIENT.  I very much
> doubt that the decomposition sought by the OP is possible
> for all h; indeed the OP was looking for conditions on h
> that would allow for the decomposition that he wanted

On Jan 22, 1:16 am, Laurent Schmalen <lo...@gmx.de> wrote:
> sarw...@YouEyeYouSee.edu wrote:
> > On Jan 18, 2:41 pm, Antoine  Bruguier <tony.brugu...@gmail.com> wrote:
>
> >> I was looking for conditions on h such that we can find h1 and h2 such
> >> that for every s:
> >> (s^2) * h = (s*h1) . (s*h2)
>
> > If the condition is required to hold for *every* input s, then
> > it must hold when the input is the unit pulse function given
> > by p[n] = 1 for n = 0 and p[n] = 0 otherwise.  Since p^2 = p,
> > and p is the identity for convolution, it must be that
> > (p^2) * h = p * h = h, and p*h1 = h1, p*h2 = h2.  Hence,
> > h = h1.h2, that is,
>
> >          h[n] = h1[n]h2[n] for all n.
>
> > Since multiplication in the time domain corresponds to convolution
> > inn the frequency domain, perhaps you need to look at spectral
> > convolution H = (H1) * (H2) rather than spectral factorization
> > H = (H1)(H2).
>

> Wrong.
> Squaring is a non-linear operation and thus your condition does NOT
> hold. Take this little example:
>
> h1 = [1 2 3 4];
> h2 = [-1 -2 -3 -4];
> s = [0 -0.1 0.2 -0.3 0.4];
> h = h1.*h2;
> conv(s.^2,h)
> conv(s,h1) .* conv(s,h2)
>
> and compare the outputs.
>
> Regards,
> Laurent


The condition h[n] = h1[n]h2[n] for all n is NECESSARY
if it is desired that (s^2) * h = (s*h1) . (s*h2) hold for ALL
signals s; it is by no means SUFFICIENT.  I very much
doubt that the decomposition sought by the OP is possible
for all h; indeed the OP was looking for conditions on h
that would allow for the decomposition that he wanted

sarwate@YouEyeYouSee.edu wrote:
> On Jan 18, 2:41 pm, Antoine  Bruguier <tony.brugu...@gmail.com> wrote:
> 
>> I was looking for conditions on h such that we can find h1 and h2 such
>> that for every s:
>> (s^2) * h = (s*h1) . (s*h2)
> 
> If the condition is required to hold for *every* input s, then
> it must hold when the input is the unit pulse function given
> by p[n] = 1 for n = 0 and p[n] = 0 otherwise.  Since p^2 = p,
> and p is the identity for convolution, it must be that
> (p^2) * h = p * h = h, and p*h1 = h1, p*h2 = h2.  Hence,
> h = h1.h2, that is,
> 
>          h[n] = h1[n]h2[n] for all n.
> 
> Since multiplication in the time domain corresponds to convolution
> inn the frequency domain, perhaps you need to look at spectral
> convolution H = (H1) * (H2) rather than spectral factorization
> H = (H1)(H2).

Wrong.
Squaring is a non-linear operation and thus your condition does NOT
hold. Take this little example:

h1 = [1 2 3 4];
h2 = [-1 -2 -3 -4];
s = [0 -0.1 0.2 -0.3 0.4];
h = h1.*h2;
conv(s.^2,h)
conv(s,h1) .* conv(s,h2)

and compare the outputs.

Regards,
Laurent

On Jan 18, 2:41 pm, Antoine  Bruguier <tony.brugu...@gmail.com> wrote:

> I was looking for conditions on h such that we can find h1 and h2 such
> that for every s:
> (s^2) * h = (s*h1) . (s*h2)

If the condition is required to hold for *every* input s, then
it must hold when the input is the unit pulse function given
by p[n] = 1 for n = 0 and p[n] = 0 otherwise.  Since p^2 = p,
and p is the identity for convolution, it must be that
(p^2) * h = p * h = h, and p*h1 = h1, p*h2 = h2.  Hence,
h = h1.h2, that is,

         h[n] = h1[n]h2[n] for all n.

Since multiplication in the time domain corresponds to convolution
inn the frequency domain, perhaps you need to look at spectral
convolution H = (H1) * (H2) rather than spectral factorization
H = (H1)(H2).

Antoine Bruguier wrote:
> Hi Juha,
> 
> Thanks for your reply. However, I am not sure this is what I am
> looking for.
> 
> The spectral factorization finds h1 and h2 such that for every s:
> s*h = s*h1*h2
> 
> I was looking for conditions on h such that we can find h1 and h2 such
> that for every s:
> (s^2) * h = (s*h1) . (s*h2)
> 
> Let me know if I was confused and did not really understand spectral
> factorization.

Sorry, I misunderstood you question.
--
Juha

On 19 Jan., 08:26, Andor <andor.bari...@gmail.com> wrote:
> On 18 Jan., 19:08, Antoine &#4294967295;Bruguier <tony.brugu...@gmail.com> wrote:
>
> > Hello,
>
> > This is probably a classic problem, but after much Googling, I can't
> > come up with the right keywords.
>
> > In the following, * is a convolution (conv in Matlab) ^2 means squared
> > ( .^2 in Matlab) and a dot . means multiply (.* in Matlab).
>
> > I would like to be able to "factor" a filter h into two filters h1 and
> > h2 in such a way that for any signal s:
>
> > (s^2) * h = (s*h1) . (s*h2)
>
> > Any good pointers?
> > Thanks in advance
>
> The problem is underspecified. You can take h1=1 and h2 = h, for
> example.

Scratch that ...

On 18 Jan., 19:08, Antoine  Bruguier <tony.brugu...@gmail.com> wrote:
> Hello,
>
> This is probably a classic problem, but after much Googling, I can't
> come up with the right keywords.
>
> In the following, * is a convolution (conv in Matlab) ^2 means squared
> ( .^2 in Matlab) and a dot . means multiply (.* in Matlab).
>
> I would like to be able to "factor" a filter h into two filters h1 and
> h2 in such a way that for any signal s:
>
> (s^2) * h = (s*h1) . (s*h2)
>
> Any good pointers?
> Thanks in advance

The problem is underspecified. You can take h1=1 and h2 = h, for
example.

Hi Juha,

Thanks for your reply. However, I am not sure this is what I am
looking for.

The spectral factorization finds h1 and h2 such that for every s:
s*h = s*h1*h2

I was looking for conditions on h such that we can find h1 and h2 such
that for every s:
(s^2) * h = (s*h1) . (s*h2)

Let me know if I was confused and did not really understand spectral
factorization.

Antoine Bruguier wrote:
> Hello,
> 
> This is probably a classic problem, but after much Googling, I can't
> come up with the right keywords.
> 
> In the following, * is a convolution (conv in Matlab) ^2 means squared
> ( .^2 in Matlab) and a dot . means multiply (.* in Matlab).
> 
> I would like to be able to "factor" a filter h into two filters h1 and
> h2 in such a way that for any signal s:
> 
> (s^2) * h = (s*h1) . (s*h2)
> 
> Any good pointers?
> Thanks in advance

Hi, I think that the right keyword is spectral factorization or in the 
case of FIR filters you simply find the zeros of the transfer function 
and construct the subfilters by choosing some of the zeros of the 
overall filter to the H_1(z) and the remaining to the H_2(z).
--
Juha

 >> h = [1 2 2 2 1];
 >> allRoots = roots(h)
allRoots =
         0 + 1.0000i
         0 - 1.0000i
   -1.0000
   -1.0000
 >> h1 = poly([allRoots(1); allRoots(2)])
h1 =
     1.0000         0    1.0000
 >> h2 = poly([allRoots(3); allRoots(4)])
h2 =
     1.0000    2.0000    1.0000
 >> conv(h1, h2)
ans =
     1.0000    2.0000    2.0000    2.0000    1.0000