Charles Henry czhenry at gmail.com
Mon Dec 3 23:34:23 CET 2007

```On Dec 2, 2007 11:52 PM, Charles Henry <czhenry at gmail.com> wrote:
> I would consider this function and its translations to be a convenient
> basis for the set of continuous band-limited compact functions.
> It is mainly useful because it allows this sampling property.  If we
> sample the function on frequency N, we can re-create the original
> continuous function, by convolution of g(t) with a series of
> kronecker-delta functions with coefficients of the samples.

That's a mistake in reasoning, there.  This is a property of the sinc
function.  It doesn't work the same for the compact function g(t)
mentioned.

> > e.g. with a function
> > that is sin(1.5*x) over [-pi;+pi) and 0 elsewhere, the spectrum has a
> > dirac at frequency 1.5/2pi and is 0 elsewhere, right?
>
> The function you mention is not band-limited.  It has a discontinuity
> at x=pi, which means that it has infinite frequency content.

There is also a discontinuity at x=-pi.  I guess this ones needs to be computed.
F( sin(1.5x)*(-pi<x<pi) )= integral(-pi, pi,
e^(2*pi*i*f*x)*(1/2i)*(e^(1.5*i*x)-e^(-1.5*i*x)dx )
=integral(-pi, pi,  (1/2i)*(e^( (1.5+2*pi*f)*i*x)-e^( (-1.5+2*pi*f)*i*x)dx )
= -1/(2*(1.5+2*pi*f))*(e^( (1.5+2*pi*f)*i*x)+1/(2*(-1.5+2*pi*f))*(e^(
(-1.5+2*pi*f)*i*x), evaluated at x=-pi, x=pi

= -1/(3+4*pi*f)*(e^(1.5*i*pi+2*pi^2*i*f)-e^(-1.5*i*pi-2*pi^2*i*f)) +
1/(-3+4*pi*f)*(e^(-1.5*i*pi+2*pi^2*i*f)-e^(1.5*i*pi-2*pi^2*i*f))

= i/(3+4*pi*f)*(e^(2*pi^2*i*f)+e^(-2*pi^2*i*f)) +
i/(-3+4*pi*f)*(e^(2*pi^2*i*f)+e^(-2*pi^2*i*f))
= i*((-3+4*pi*f)+(3+4*pi*f))/( (4*pi*f)^2 - 9)*2*cos(2*pi^2*f)
= i*8*pi*f / ( (4*pi*f)^2 - 9) * cos(2*pi^2*f)

The spectrum has two poles at f=+/- 3/(4*pi).  The angular frequency
is 1.5 rad/sec at the poles, and the spectrum falls off asymptotically
to 1/f.

```