Re: getting freqz from fft (complex numbers)


Andor wrote:
> Rune Allnor wrote:
> > Andor wrote:
> > > Rune Allnor wrote:
> > > > * This is the L0 or L1 norm of the spectrum, I never remember the
> > > > correct term.
> > >
> > > It should be easy to remember the correct term because an L0 norm
> > > doesn't exist.
> >
> > OK. It's been a while since I read about these kinds of things.
> > The L2 norm was the intersting one, anyway.
>
> That depends on what you are doing. I admit that the only norms I have
> seen used in practice are the 1, 2 and infinty norms.
>
>
> > > > The nice thing about norms is that if
> > > > ||x|+|y||>||x|+|z||
> > > > in one norm, it is also true in all other norms. So the
> > > > computationally
> > > > cheap norms can be used to find the extrema, and the usual
> > > > L2 norm can be used only where one wants to know the details.
> > >
> > > That won't work, Rune. Consider the two numbers z1=1.0, z2=.9 exp(i
> > > pi/4). Then
> > >
> > > ||z1||_L2 > ||z2||_L2
> > >
> > > but
> > >
> > > ||z1||_L1 < ||z2||_L1
> >
> > Ouch. It seems you are right. If so, it means the choise of norm
> > changes the topology of a normed space? At frist glance I find that
> > very hard to believe... but then, maybe not.
>
> If by topology of a complete normed space (Banach space) you mean the
> topology induced by the metric induced by the norm, then yes (I love
> maths :-). To see a difference you have to go to inifinte dimensional
> spaces however. In finite dimensional vector spaces, all norms (and
> therefore topologies induced by norms) are equivalent. I'm not sure if
> this is what you had in mind by "topology".

I have a rather naive impression of topology. It has to do with
ordering
and arrangements of elements, right? So if one normed space, say,
S1=(C^N, | |_a) is ordered such that |z1|_a>|z2|_a, and the other
space,
say S2=(C^N, | |_b), is ordered such that |z1|_b < |z2|_b for some
N-vectors z1, z2, the toplogies of the spaces S1 and S2 are different.
Right...?

> The geometry clearly is
> different - look at the complex unit "circle" in different norms:
> diamond (L1), circle (L2), square (infinity) ...

Yep, I remember that excercise from some course I took many years ago.

Rune

.

Relevant Pages

  • Re: getting freqz from fft (complex numbers)
    ... >> It should be easy to remember the correct term because an L0 norm ... seen used in practice are the 1, 2 and infinty norms. ... If by topology of a complete normed space you mean the ... > Thanks, Andor, for the correction. ...
    (comp.dsp)
  • Re: equivalent norms
    ... What is your definition of "equivalent norms"? ... What do you mean by "their balls are contained in each ... there exist d1,d2>0 such that the N1-ball of radius e and centered at ... ("General Topology", Chapter IX, Section 3.3). ...
    (sci.math)
  • Re: Norm / inner product.
    ... poincare' inequality) that two inner products are equivalent by proving ... I mean why does the equivalence of two norms imply the ... and the proof is evident (the inner products give the same topology ... the other tells us there are positive constants which bound each norm ...
    (sci.math)
  • Re: Homeomorphic non-isomorphic normed vector spaces
    ... There are norms on R^2 that give the ... same topology but are not isometric." ... Isometric, okay, but R^2 is always isomorphic to R^2. ... Julien Santini ...
    (sci.math)
  • Re: getting freqz from fft (complex numbers)
    ... Rune Allnor wrote ... and then maybe use the usual magnitude formula ... > to compute the actual magnitude for that peak. ... it is also true in all other norms. ...
    (comp.dsp)