[PD] tabread4~~
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.
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