Re: getting freqz from fft (complex numbers)

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

> If I was right in my statement yesterday, it would mean that all
> optimization problems needed to be solved only in the least squares
> sense, since that solution would be optimal in all other senses, too.
> That's clearly not the case, so I was wrong yesterday.

I think optimizations w.r.t arbitrary norms can be achieved by iterated
WLS, by updating the weights each iteration according to some rule
determined by the required norm. I think I read that somewhere, which
doesn't necessarily make it true.

>
> Thanks, Andor, for the correction.

For comparison, the square of the L2 norm suffices, and is very cheap
to compute (as Jerry has already mentioned), so I don't think any
elaboraty tricks are needed.

Regards,
Andor

.

Relevant Pages

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