Question about spectrum analysis and sampling theory

On Mon, 4 Jul 2011 13:05:25 -0700 (PDT), JoSH Lehan
<krellan@gmail.com> wrote:

>On Saturday, July 2, 2011 2:24:29 PM UTC-7, Eric Jacobsen wrote:
>> What artifacts are you talking about?   Numerically (i.e.,
>> mathematically) all of the frequencies within the supported Nyquist
>> region have essentially the same properties (if the transform is
>> linear, which it typically is for that purpose).   If there's a
>> non-rectangular window applied to the transforms that are creating the
>> waterfall display then there may be some attenuation that will be a
>> function of the window shape.
>
>Thanks.  I probably used the wrong terminology.  Digital sampling loses pre=
>cision as you approach the Nyquist frequency, degenerating completely to ju=
>st square waves once you reach the Nyquist frequency.  Let's say I'm sampli=
>ng a signal that is 1 kHz wide, and I'm sampling at 48 kHz, giving me a tot=
>al range of 0 to 24 kHz to work in (Nyquist frequency of 24 kHz).  If the s=
>ignal of interest is from 23 kHz to 24 kHz within my sampling, then it won'=
>t be as precise as if it is from somewhere in the middle, say 12 kHz to 13 =
>kHz.  That's what I meant by "artifacts".
>
>Josh

You misunderstand sampling.  There is no loss of precision as the
Nyquist frequency is approached, and the signal is recoverable as a
sine wave rather than a square wave because the support region does
not include the harmonics necessary to be a square wave rather than a
sine wave.

The support region is completely linear and uniform and the
"artifacts" you are describing are not real or are attributable to
something other than what you've described.


Eric Jacobsen
http://www.ericjacobsen.org
http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php

On Mon, 4 Jul 2011 13:08:16 -0700 (PDT), JoSH Lehan
<krellan@gmail.com> wrote:

>On Sunday, July 3, 2011 1:18:18 PM UTC-7, Andrew Holme wrote:
>> If possible, I would centre the 10 KHz wide signal around 12 KHz i.e. from 7 
>> KHz to 17 KHz.
>
>Thanks!
>
>12 kHz is half of the Nyquist frequency in this case (I'm sampling at 48 kHz, so my useful range is from 0 to 24 kHz).  In other words, if I'm using a waterfall display, this puts my signal of interest at dead center within the waterfall.
>
>I wonder what the derivation for this is, or is it just a good "rule of thumb"?
>
>Josh

It's just convenient as far as centering the signal of interest in the
waterfall display.   Sort of like aligning the steering wheel so the
spokes are comfortable for thumb rests.   What metric would a
derivation be optimizing?




Eric Jacobsen
http://www.ericjacobsen.org
http://www.dsprelated.com/blogs-1//Eric_Jacobsen.php

On Jul 4, 4:31=A0pm, eric.jacob...@ieee.org (Eric Jacobsen) wrote:
> On Mon, 4 Jul 2011 13:05:25 -0700 (PDT), JoSH Lehan
>
>
>
> <krel...@gmail.com> wrote:
> >On Saturday, July 2, 2011 2:24:29 PM UTC-7, Eric Jacobsen wrote:
> >> What artifacts are you talking about? =A0 Numerically (i.e.,
> >> mathematically) all of the frequencies within the supported Nyquist
> >> region have essentially the same properties (if the transform is
> >> linear, which it typically is for that purpose). =A0 If there's a
> >> non-rectangular window applied to the transforms that are creating the
> >> waterfall display then there may be some attenuation that will be a
> >> function of the window shape.
>
> >Thanks. =A0I probably used the wrong terminology. =A0Digital sampling lo=
ses pre=3D
> >cision as you approach the Nyquist frequency, degenerating completely to=
 ju=3D
> >st square waves once you reach the Nyquist frequency. =A0Let's say I'm s=
ampli=3D
> >ng a signal that is 1 kHz wide, and I'm sampling at 48 kHz, giving me a =
tot=3D
> >al range of 0 to 24 kHz to work in (Nyquist frequency of 24 kHz). =A0If =
the s=3D
> >ignal of interest is from 23 kHz to 24 kHz within my sampling, then it w=
on'=3D
> >t be as precise as if it is from somewhere in the middle, say 12 kHz to =
13 =3D
> >kHz. =A0That's what I meant by "artifacts".
>
> You misunderstand sampling. =A0There is no loss of precision as the
> Nyquist frequency is approached, and the signal is recoverable as a
> sine wave rather than a square wave because the support region does
> not include the harmonics necessary to be a square wave rather than a
> sine wave.
>
> The support region is completely linear and uniform and the
> "artifacts" you are describing are not real or are attributable to
> something other than what you've described.
>

my spin on this is just to remember that the basis function used in
reconstruction is sinc(t-nT), not a square and not even a sine (if you
use sines as your basis, it's a different set of coefficients than the
samples, it's the DFT of the samples).

r b-j

He's incorrect *provided* that the signal is properly bandlimited. While that's the usual assumption, I wouldn't make it here, given his other statements.

Jerry
-- 
Engineering is the art of making what you want from things you can get.

On Monday, July 4, 2011 4:31:25 PM UTC-4, Eric Jacobsen wrote:

  ...

> The support region is completely linear and uniform and the
> "artifacts" you are describing are not real or are attributable to
> something other than what you've described.

If they exist at all, they are probably aliases.

Jerry
-- 
Engineering is the art of making what you want from things you can get.

On 7/4/2011 1:05 PM, JoSH Lehan wrote:
> On Saturday, July 2, 2011 2:24:29 PM UTC-7, Eric Jacobsen wrote:
>> What artifacts are you talking about?   Numerically (i.e.,
>> mathematically) all of the frequencies within the supported Nyquist
>> region have essentially the same properties (if the transform is
>> linear, which it typically is for that purpose).   If there's a
>> non-rectangular window applied to the transforms that are creating the
>> waterfall display then there may be some attenuation that will be a
>> function of the window shape.
>
> Thanks.  I probably used the wrong terminology.  Digital sampling loses precision as you approach the Nyquist frequency, degenerating completely to just square waves once you reach the Nyquist frequency.  Let's say I'm sampling a signal that is 1 kHz wide, and I'm sampling at 48 kHz, giving me a total range of 0 to 24 kHz to work in (Nyquist frequency of 24 kHz).  If the signal of interest is from 23 kHz to 24 kHz within my sampling, then it won't be as precise as if it is from somewhere in the middle, say 12 kHz to 13 kHz.  That's what I meant by "artifacts".
>
> Josh

You've received some good comments.  I still think that you may be 
wanting an expanded display where the signal of interest takes up the 
entire width of the display.

There are a number of ways to do this:

- The obvious one (but probably not so useful) is  to just display the 
band where the signal of interest lies.  The problem with this is that 
the resolution isn't any different and the waterfall may look quite 
striped or blocky vertically unless you do some interpolation first.
And, that effect may move you to higher resolution (longer time window).

- Another method but more effort would be to use one of the various 
methods to focus on the band of interest from start to finish.  This 
will put the computing where you need it.  An arm-waving approach would 
be to:
1) bandpass filter the data.
2) downsample to a lower frequency
3) analyze that result
The same caveat about using a longer time window applies here.

I'd not particularly thought of it but good design of a waterfall is 
like good design of any video system (that's what it is after all .. 
with infinite vertical and vertical visibility limited to the last 
number of analyses) where the horizontal resolution of the signal is at 
least as good as the horizontal resolution of the display.  Otherwise 
the output will be horizontally pixelated or discretized.

Then, of course, you want the scan time (the time window) to be matched 
to the vertical scan so that pixelation in the vertical is avoided.

Fred