jon.guerin at gmail.com
Sun Feb 6 10:09:42 EST 2005
Also, upsampling a signal by N is equivalent to:
Y(z) = X(z^N).
Y(e^jw) = X(e^jNw) | z = e^jw.
If you plot that last relationship you'll see that you get N copies of
the original spectra scaled in frequency across 0 to pi when you
upsample by N.
For example if X(e^jw) was a rectangle from 0 to pi/2 and 0 from pi/2
to pi (if we only look from 0 to pi), Y(e^jw) would have 2 rectangles;
one from 0 to pi/4 and another from pi/2 to 3pi/4 (this due to X(e^jw)
being 2pi periodic). Try upsampling a voice recording by 2, you should
be able to hear this effect.
On Sat, 05 Feb 2005 15:06:56 -0800, Norm Campbell
<norm_campbell at telus.net> wrote:
> >I'm not sure exactly what would happen to the spectrum if you simply
> >repeat the previous sample, but it's probably not as predictable as
> >inserting zeroes.
> Very predictable - it is the same as running the zero-inserted signal
> through a 1st-order filter with a zero at Nyquist (H(z) = 1 + z^-1). Thus
> you get 3 dB attenuation at Nyquist/2 (relative to DC) and infinite
> attenuation at Nyquist.
> Since you generally want to half-band lowpass filter the result anyway, you
> can use this as the first part of a half band filter, although that then
> complicates the design of the half-band filter.
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