| Author |
Message |
Att
Guest
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 Posted:
Mon Jan 03, 2005 1:41 am Post subject:
Re: Noise floor / FFT relationship |
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The references below look like they would be very good.
Another reference more specific to SNR and frequency measurements is
published by
Bruel and Kjaer. It is called "Frequency Analysis" (ISBN 87 87355 07 8)
Mine cost $25 in 1998, hardbound 342 pages, and I felt like I got a steal on
it.
Has extensive discussion of both analog and digital measurement techniques
and instrument usage.
Dave Shaw
"Jon Harris" <goldentully@hotmail.com> wrote in message
news:33ph9hF43vd1cU1@individual.net...
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Rick Lyons
Guest
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| Posted:
Mon Jan 03, 2005 7:06 pm Post subject:
Re: Noise floor / FFT relationship |
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On Fri, 31 Dec 2004 15:12:23 -0800, Tim Wescott
<tim@wescottnospamdesign.com> wrote:
| Quote: | robert bristow-johnson wrote:
in article 10tbj4acvnai99b@corp.supernews.com, Tim Wescott at
tim@wescottnospamdesign.com wrote on 12/31/2004 17:02:
Robert Lyons's "Understanding Digital Signal Processing" is
highly recommended on this group.
it's Richard (Rick) Lyons.
good book.
Thanks Robert.
Sorry Rick.
And yes, it's a good book.
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Hi Tim & R B-J,
a compliment from DSP gurus like you
two guys means a lot to me.
Thanks,
[-Rick-] |
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robert bristow-johnson
Guest
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| Posted:
Tue Jan 04, 2005 1:30 am Post subject:
Re: Noise floor / FFT relationship |
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in article 41d9512c.274384000@news.sf.sbcglobal.net, Rick Lyons at
r.lyons@_BOGUS_ieee.org wrote on 01/03/2005 09:06:
| Quote: | On Fri, 31 Dec 2004 15:12:23 -0800, Tim Wescott
tim@wescottnospamdesign.com> wrote:
robert bristow-johnson wrote:
in article 10tbj4acvnai99b@corp.supernews.com, Tim Wescott at
tim@wescottnospamdesign.com wrote on 12/31/2004 17:02:
Robert Lyons's "Understanding Digital Signal Processing" is
highly recommended on this group.
it's Richard (Rick) Lyons.
good book.
Thanks Robert.
Sorry Rick.
And yes, it's a good book.
Hi Tim & R B-J,
a compliment from DSP gurus like you
two guys means a lot to me.
|
an old issue: we still need to talk about the DFT of all ones where N = an
even number. there was something not so kosher in your treatment of it as i
recall, but i have to familiarize myself with it again. i think there is a
phase problem or something because if N is even, you really cannot have a
symmetric spectrum about DC. there will always be at least one more "one"
on either the left or the right.
--
r b-j rbj@audioimagination.com
"Imagination is more important than knowledge." |
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Tim Wescott
Guest
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| Posted:
Tue Jan 04, 2005 2:47 am Post subject:
Re: Noise floor / FFT relationship |
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robert bristow-johnson wrote:
| Quote: | in article 41d9512c.274384000@news.sf.sbcglobal.net, Rick Lyons at
r.lyons@_BOGUS_ieee.org wrote on 01/03/2005 09:06:
On Fri, 31 Dec 2004 15:12:23 -0800, Tim Wescott
tim@wescottnospamdesign.com> wrote:
robert bristow-johnson wrote:
in article 10tbj4acvnai99b@corp.supernews.com, Tim Wescott at
tim@wescottnospamdesign.com wrote on 12/31/2004 17:02:
Robert Lyons's "Understanding Digital Signal Processing" is
highly recommended on this group.
it's Richard (Rick) Lyons.
good book.
Thanks Robert.
Sorry Rick.
And yes, it's a good book.
Hi Tim & R B-J,
a compliment from DSP gurus like you
two guys means a lot to me.
an old issue: we still need to talk about the DFT of all ones where N = an
even number. there was something not so kosher in your treatment of it as i
recall, but i have to familiarize myself with it again. i think there is a
phase problem or something because if N is even, you really cannot have a
symmetric spectrum about DC. there will always be at least one more "one"
on either the left or the right.
|
Well, you can have a symmetric spectrum if you remember that the DFT
spectrum is periodic -- the Fs/2 bin is also the -Fs/2 bin, so its
contents are automatically symmetric around 0.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com |
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Rune Allnor
Guest
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| Posted:
Tue Jan 04, 2005 2:55 am Post subject:
Re: Noise floor / FFT relationship |
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Tim Wescott wrote:
| Quote: | robert bristow-johnson wrote:
in article 41d9512c.274384000@news.sf.sbcglobal.net, Rick Lyons at
r.lyons@_BOGUS_ieee.org wrote on 01/03/2005 09:06:
On Fri, 31 Dec 2004 15:12:23 -0800, Tim Wescott
tim@wescottnospamdesign.com> wrote:
robert bristow-johnson wrote:
in article 10tbj4acvnai99b@corp.supernews.com, Tim Wescott at
tim@wescottnospamdesign.com wrote on 12/31/2004 17:02:
Robert Lyons's "Understanding Digital Signal Processing" is
highly recommended on this group.
it's Richard (Rick) Lyons.
good book.
Thanks Robert.
Sorry Rick.
And yes, it's a good book.
Hi Tim & R B-J,
a compliment from DSP gurus like you
two guys means a lot to me.
an old issue: we still need to talk about the DFT of all ones
where N = an
even number. there was something not so kosher in your treatment
of it as i
recall, but i have to familiarize myself with it again. i think
there is a
phase problem or something because if N is even, you really cannot
have a
symmetric spectrum about DC. there will always be at least one
more "one"
on either the left or the right.
Well, you can have a symmetric spectrum if you remember that the DFT
spectrum is periodic -- the Fs/2 bin is also the -Fs/2 bin, so its
contents are automatically symmetric around 0.
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I have to agree with Tim here. The issue of whether N is even or odd
boils down to whether there is a coefficient exactly at Fs/2 or not.
< running frantically for cover to avoid the imminent crossfire >
Rune |
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Jon
Guest
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| Posted:
Tue Jan 04, 2005 7:58 am Post subject:
Re: Noise floor / FFT relationship |
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How about this for a simpler question? The noise floor of 16-bit
audio is said to have a 96 (or 98) dB dynamic range. How would the
noise floor have to be measured to get the value -96 dBFS out of it? Is
that equivalent to an RMS value? Then I can measure real devices and
compare to theoretical minimum...
And how would this measurement method compare to the heating power (or
RMS voltage measurement) of an analog piece of equipment?
Eric Jacobsen wrote:
| Quote: | FFTs have processing gain proportional to the relationship between the
input bandwidth and the bandwidth of a single FFT bin.
So a strict answer to your question depends on the length of the FFT
as well as the nature of any windowing used, additional averaging,
etc.
|
Ah. I see that now. Because each point in the FFT behaves as a
little bandpass filter, so if there are more points, they are narrower
bandwidth and each measure a lower level. (Right?) So to measure noise
floor directly from the FFT would require every FFT value, knowledge of
windowing function including width and overlap, and FFT size. To get an
RMS value of the noise floor itself with no signal present would just
require RMSing all the values and converting to dBFS. (You could also
use the FFT to filter out the test signal, inverse FFT, and measure
noise floor the same way, right?)
Tim Wescott wrote:
| Quote: | Because somebody's confused. If you RMS a voltage, or a current, or a
string of numbers, then you're finding the DC value that would burn up
the same amount of power when applied to a given resistor -- so 10VDC
applied to a 10 ohm resistor delivers 10W, and so does 10Vrms.
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Yes, but the RMS of the string of numbers isn't the power. It's just
the steady DC value that would dissipate the same average amount of
power. Average power is still a square and resistor multiplication
away. I'm not sure what "power" means in the digital world, I guess.
| Quote: | But there are drumbeats, which have a very high peak/RMS ratio. This is
your problem. A 1V RMS signal could be anything from 1.414*sin(wt) to a
train of pulses that are 100us wide and 32V high. There -- no Diracs,
but things that could be recorded if you had enough dynamic range, and
which have _identical_ RMS voltages.
|
Ok, but the dynamic range measurement of an audio system would be the
ratio between RMS value of a maxed out sine wave and RMS value of the
noise floor (which is probably close to white noise), right? How does
this kind of dynamic range compare to the kind used in the theoretical
96 dB for 16-bit calculated from the quantization noise?
(Thanks everyone else for books and direct FFT instructions.) |
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Jon Harris
Guest
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| Posted:
Wed Jan 05, 2005 12:00 am Post subject:
Re: Noise floor / FFT relationship |
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"Jon" <u035m4i02@sneakemail.com> wrote in message
news:2AoCd.22904$EL5.2537@trndny09...
| Quote: | How about this for a simpler question? The noise floor of 16-bit
audio is said to have a 96 (or 98) dB dynamic range. How would the
noise floor have to be measured to get the value -96 dBFS out of it? Is
that equivalent to an RMS value? Then I can measure real devices and
compare to theoretical minimum...
|
You describe the procedure yourself at the bottom of this message. See my
comments there.
| Quote: | Tim Wescott wrote:
Because somebody's confused. If you RMS a voltage, or a current, or a
string of numbers, then you're finding the DC value that would burn up
the same amount of power when applied to a given resistor -- so 10VDC
applied to a 10 ohm resistor delivers 10W, and so does 10Vrms.
Yes, but the RMS of the string of numbers isn't the power. It's just
the steady DC value that would dissipate the same average amount of
power. Average power is still a square and resistor multiplication
away. I'm not sure what "power" means in the digital world, I guess.
But there are drumbeats, which have a very high peak/RMS ratio. This is
your problem. A 1V RMS signal could be anything from 1.414*sin(wt) to a
train of pulses that are 100us wide and 32V high. There -- no Diracs,
but things that could be recorded if you had enough dynamic range, and
which have _identical_ RMS voltages.
|
All this is true, but I'm not sure it is relevant here. Dynamic range
measurement of an audio system is always made using sinusoidal inputs (or no
inputs).
| Quote: | Ok, but the dynamic range measurement of an audio system would be the
ratio between RMS value of a maxed out sine wave and RMS value of the
noise floor (which is probably close to white noise), right? How does
this kind of dynamic range compare to the kind used in the theoretical
96 dB for 16-bit calculated from the quantization noise?
|
Technically, the dynamic range of a digital system (e.g. a DAC) is measured by
comparing the full scale sine wave RMS vs. the RMS level with a sine wave
at -60dB (relative to full scale). The -60dBFS tone is compensated for by
either placing an analog notch filter at the frequency in question (typically
1kHz) placed before the RMS level detector or subtracting out its contribution.
One reason it is measured this way because some DACs "cheat" and have a hardware
mute function that kicks in when they see all zeros on the input, artificially
lowering the noise floor in that special case. What you describe is usually
called the signal-to-noise ratio. However, in practice, unless you have one of
those "cheating" DACs, the two measurements are often nearly identical (within a
dB or so).
The Metzler text I recommended earlier states, "The signal at -60dB assures that
the digital converter continues to operate, generating quantization noise and
distortion." I think this ties in to the issue of comparing a practical device
to a theoretical 16-bit device you've been asking about. |
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Jon
Guest
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| Posted:
Wed Jan 05, 2005 4:48 am Post subject:
Re: Noise floor / FFT relationship |
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Jon Harris wrote:
| Quote: | lowering the noise floor in that special case. What you describe is usually
called the signal-to-noise ratio. However, in practice, unless you have one of
those "cheating" DACs, the two measurements are often nearly identical (within a
dB or so).
|
Yes it is usually described as SNR, but I thought they were identical
for a digital system. Dynamic range is ratio of noise floor to full
scale sine, and SNR is ratio of noise floor to arbitrary "nominal"
signal, which for digital seems to be defined as full scale sine. So
they are identical by my definitions. Are my definitions wrong?
| Quote: | The Metzler text I recommended earlier states, "The signal at -60dB assures that
the digital converter continues to operate, generating quantization noise and
distortion." I think this ties in to the issue of comparing a practical device
to a theoretical 16-bit device you've been asking about.
|
Yes, that makes sense. That would also measure harmonic distortion if
it was present, right?
--
Include "newsgroup" in the subject line to reply by email (or get dumped
with the spam). |
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Jon Harris
Guest
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| Posted:
Wed Jan 05, 2005 6:37 am Post subject:
Re: Noise floor / FFT relationship |
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"Jon" <u035m4i02@sneakemail.com> wrote in message
news:6XFCd.10068$PY6.4111@trndny02...
| Quote: | Jon Harris wrote:
lowering the noise floor in that special case. What you describe is usually
called the signal-to-noise ratio. However, in practice, unless you have one
of
those "cheating" DACs, the two measurements are often nearly identical
(within a
dB or so).
Yes it is usually described as SNR, but I thought they were identical
for a digital system. Dynamic range is ratio of noise floor to full
scale sine, and SNR is ratio of noise floor to arbitrary "nominal"
signal, which for digital seems to be defined as full scale sine. So
they are identical by my definitions. Are my definitions wrong?
|
The definitions are the same (or very similar), but the method of measurement
differs for a digital system. Like I stated before, they _usually_ give the
same results (the "cheating" DAC being the main exception I'm aware of).
| Quote: | The Metzler text I recommended earlier states, "The signal at -60dB assures
that
the digital converter continues to operate, generating quantization noise
and
distortion." I think this ties in to the issue of comparing a practical
device
to a theoretical 16-bit device you've been asking about.
Yes, that makes sense. That would also measure harmonic distortion if
it was present, right?
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Yes. But in most systems, the harmonic distortion components of a -60dB signal
are going to be way under the noise floor unless it was grossly non-linear. |
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Jon
Guest
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| Posted:
Thu Jan 06, 2005 7:59 am Post subject:
Re: Noise floor / FFT relationship |
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Eric Jacobsen wrote:
| Quote: | the sum of the noise in each bin. So the longer the FFT the lower
the noise floor will be for the same input statistics. People
exploit this all the time to reveal low-level signals or spurious
responses that are otherwise difficult to see.
|
So wait a second... What if you took an infinite-point FFT?
*Can opener and squishy sounds*
:-)
--
Include "newsgroup" in the subject line to reply by email (or get dumped
with the spam). |
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Jon Harris
Guest
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| Posted:
Thu Jan 06, 2005 7:59 am Post subject:
Re: Noise floor / FFT relationship |
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"Jon" <u035m4i02@sneakemail.com> wrote in message
news:bV1Dd.28893$EL5.11209@trndny09...
| Quote: | Eric Jacobsen wrote:
the sum of the noise in each bin. So the longer the FFT the lower
the noise floor will be for the same input statistics. People
exploit this all the time to reveal low-level signals or spurious
responses that are otherwise difficult to see.
So wait a second... What if you took an infinite-point FFT?
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Come back after you've finished calculating it, and I'll tell you them! :-) |
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Jon
Guest
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| Posted:
Fri Jan 07, 2005 6:50 am Post subject:
Re: Noise floor / FFT relationship |
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Jon Harris wrote:
| Quote: | So wait a second... What if you took an infinite-point FFT?
Come back after you've finished calculating it, and I'll tell you them! :-)
|
Yeah, yeah... I am obviously talking about a computer with infinite
processing speed. C'mon. :-P
So what about a 2^1,000,000,000,000,000,000,000,000-point FFT, then?
The "noise floor" line just keeps dropping? Exposing what?
(I don't even remember how to take, for example, a 256-point FFT of a
64-point signal.) :-(
--
Include "newsgroup" in the subject line to reply by email (or get dumped
with the spam). |
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Rick Lyons
Guest
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| Posted:
Fri Jan 07, 2005 7:28 pm Post subject:
Re: Noise floor / FFT relationship |
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On Mon, 03 Jan 2005 15:30:07 -0500, robert bristow-johnson
<rbj@audioimagination.com> wrote:
(snipped)
| Quote: |
Hi Tim & R B-J,
a compliment from DSP gurus like you
two guys means a lot to me.
an old issue: we still need to talk about the DFT of all ones where N = an
even number. there was something not so kosher in your treatment of it as i
recall, but i have to familiarize myself with it again. i think there is a
phase problem or something because if N is even, you really cannot have a
symmetric spectrum about DC. there will always be at least one more "one"
on either the left or the right.
--
r b-j rbj@audioimagination.com
|
Hi,
Not knowing what "kosher" means, I'm not sure what to say.
As you know, the DFT of an all-ones time sequence is a
sequence starting with a non-zero valued sample followed by
samples that are all zero-valued.
Although my understanding of symmetry has changed
recently. I now believe that there are two types of
symmetry.
[-Rick-] |
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Rick Lyons
Guest
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| Posted:
Fri Jan 07, 2005 7:31 pm Post subject:
Re: Noise floor / FFT relationship |
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On Sat, 01 Jan 2005 06:57:05 GMT, eric.jacobsen@ieee.org (Eric
Jacobsen) wrote:
(snipped)
| Quote: |
FFTs have processing gain proportional to the relationship between the
input bandwidth and the bandwidth of a single FFT bin. This was
mentioned in another reply to the effect that the total noise power is
the sum of the noise in each bin. So the longer the FFT the lower
the noise floor will be for the same input statistics. People
exploit this all the time to reveal low-level signals or spurious
responses that are otherwise difficult to see.
|
Hi Eric,
you're not kiddin' !!
The SETI researchers implement multimillion-point
FFTs in order to detect an itsy bitsy glob of
RF energy from little green men out there.
[-Rick-] |
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Eric Jacobsen
Guest
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| Posted:
Fri Jan 07, 2005 11:41 pm Post subject:
Re: Noise floor / FFT relationship |
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On Fri, 07 Jan 2005 01:50:54 GMT, Jon <u035m4i02@sneakemail.com>
wrote:
| Quote: | Jon Harris wrote:
So wait a second... What if you took an infinite-point FFT?
Come back after you've finished calculating it, and I'll tell you them! :-)
Yeah, yeah... I am obviously talking about a computer with infinite
processing speed. C'mon. :-P
So what about a 2^1,000,000,000,000,000,000,000,000-point FFT, then?
The "noise floor" line just keeps dropping? Exposing what?
|
If the number of samples keeps increasing along with the FFT length,
then, yes, the noise floor keeps dropping.
| Quote: | (I don't even remember how to take, for example, a 256-point FFT of a
64-point signal.) :-(
|
Zero pad.
Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org |
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