Wind/Solar Electrics ???

wrote in message
...
Joel Kolstad wrote:

(I can't tell you how many times I've seen people stating something like,
'The Nyquist theorem requires sampling at at least twice the highest
frequency present in the signal," when of course it says no such thing.)

What do you think it means?


Nyquist figured out that higher frequency components of the input signal
will 'alias' and you will lose the ability to tell them from lower frequency
components. In order to avoid 'losing information' and not being able to
tell whether a particular sample stream was from a low or high frequency
component, Nyquist's theorem states you must sample at least twice as fast
as the highest component present.
http://www.cs.cf.ac.uk/Dave/Multimedia/node149.html
http://www.efunda.com/designstandard...sp_nyquist.cfm

A lot of folks mistake it to think you need to sample at least twice as fast
as the 'signal of interest' also, and try to ignore high frequency
components of the input because they're 'not interested in that noise'. But
what Nyquist proved was that any frequency in the sampled signal that is
more than 1/2 the sample frequency will 'alias' and 'wrap around' and be
*indistinguisable* from a frequency component that is less than 1/2 the
sample frequency.

For example, if sampling at 1000 hz, and the sampled signal is a 900 hz
'pure sine wave', the sampled data would look *exactly* the same as if you
had sampled a 100 hz 'pure sine wave'. They would be 'indistinguisable'.
So if/when you try to convert the sampled data back to analog, how do you
know whether it should reconstruct a 100 hz wave, or 900 hz? You don't, so
you have a conundrum.

So, to avoid losing this 'information' of being able to tell if you had a
100 hz or 900 hz input, the standard thing to do is filter the input so that
there is *no* 900 hz input. Then, the resulting sample data must have come
from the 100 hz component. And if/when you want to reconstruct it, you
*know* it should form a 100 hz signal because no 900 hz signal could
possibly been present (you eliminated it before sampling).

And as Joel mentioned earlier, since most low-pass filters do not have
perfect 'cutoff' (IIRC, simple first-orders 'roll off' at something like 3
db/decade), it is more common to eliminate any frequency component that is
more than about 40% of the sampling frequency.

daestrom

On Fri, 23 Dec 2005 18:46:49 GMT, the renowned "daestrom"
wrote:


wrote in message
...
Joel Kolstad wrote:

(I can't tell you how many times I've seen people stating something like,
'The Nyquist theorem requires sampling at at least twice the highest
frequency present in the signal," when of course it says no such thing.)

What do you think it means?


Nyquist figured out that higher frequency components of the input signal
will 'alias' and you will lose the ability to tell them from lower frequency
components. In order to avoid 'losing information' and not being able to
tell whether a particular sample stream was from a low or high frequency
component, Nyquist's theorem states you must sample at least twice as fast
as the highest component present.
snip

More than twice the bandwidth.


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

"Spehro Pefhany" wrote in message
...
On Fri, 23 Dec 2005 18:46:49 GMT, the renowned "daestrom"
wrote:


wrote in message
...
Joel Kolstad wrote:

(I can't tell you how many times I've seen people stating something
like,
'The Nyquist theorem requires sampling at at least twice the highest
frequency present in the signal," when of course it says no such thing.)

What do you think it means?


Nyquist figured out that higher frequency components of the input signal
will 'alias' and you will lose the ability to tell them from lower
frequency
components. In order to avoid 'losing information' and not being able to
tell whether a particular sample stream was from a low or high frequency
component, Nyquist's theorem states you must sample at least twice as fast
as the highest component present.
snip

More than twice the bandwidth.



So, if I have a signal with a 1000 hz carrier, with a bandwidth of 50 hz,
you think I can sample it at just 150 hz and get accurate reproduction?
That's just wrong.

It is the maximum frequency component in the signal that is important. The
bandwidth is not related unless the lower edge of the band is at 0 hz
(whereupon the upper side of the band is equal to the max frequency).

daestrom

You can represent the bandwidth with double the
sampling rate as the bandwidth frequency but there is a
component missing from the sample information that has
to be known and is not part of the samples. Namely the
base frequency has to be added back into the formula.


"daestrom" wrote in
message
...

So, if I have a signal with a 1000 hz carrier, with a
bandwidth of 50 hz,
you think I can sample it at just 150 hz and get
accurate reproduction?
That's just wrong.

It is the maximum frequency component in the signal
that is important. The
bandwidth is not related unless the lower edge of the
band is at 0 hz
(whereupon the upper side of the band is equal to the
max frequency).

daestrom

In article ,
"daestrom" wrote:

"Spehro Pefhany" wrote in message
...
On Fri, 23 Dec 2005 18:46:49 GMT, the renowned "daestrom"
wrote:


wrote in message
...
Joel Kolstad wrote:

(I can't tell you how many times I've seen people stating something
like,
'The Nyquist theorem requires sampling at at least twice the highest
frequency present in the signal," when of course it says no such thing.)

What do you think it means?


Nyquist figured out that higher frequency components of the input signal
will 'alias' and you will lose the ability to tell them from lower
frequency
components. In order to avoid 'losing information' and not being able to
tell whether a particular sample stream was from a low or high frequency
component, Nyquist's theorem states you must sample at least twice as fast
as the highest component present.
snip

More than twice the bandwidth.



So, if I have a signal with a 1000 hz carrier, with a bandwidth of 50 hz,
you think I can sample it at just 150 hz and get accurate reproduction?
That's just wrong.

It is the maximum frequency component in the signal that is important. The
bandwidth is not related unless the lower edge of the band is at 0 hz
(whereupon the upper side of the band is equal to the max frequency).

daestrom



You are getting your terms confused here guys. Nyquist requires that
you input both the Center Frequency, and Bandwidth when determining
the Sampling Rate. If the sampling is done at BaseBand then only the
Bandwidth is relevent. If the sampling is not done at baseband, then
the Center Frequency, and Bandwidth are required to determine samling
rate. Example, if the Bandwith of the signal is 3Kc and the sampling is
done at BaseBand then sample rate needed would 6Kc. If the sampling is
done at 100 Mhz with the same 3Kc bandwidth, then a 200.006 Mhz sampling
rate would be required.

It is much easyier to do DSP at baseBand, than at IF Frequencies, and if
you do DSP at IF Frequencies, the lower the IF Frequency, the easyier it
is to do, and the slower the DSP has to run.

Me

If only the baseband frequency is sampled at 6kHz then
information is missing to recreate the original 100kHz
and the sampling information is insufficient to
recreate the original signal.


This is analogous to saying the number 1234 can be
represented by
(1234-234) / 1000 = 1

If I supply the number 1.0 you can regenerate the
number 1234 from it? Not true, without the rest of the
sampling information. The sample is incomplete.

Bandwidth sampling only cannot recreate the original
signal.

"Me" wrote in message
...
In article
,
"daestrom"
wrote:

"Spehro Pefhany"
wrote in message
...
On Fri, 23 Dec 2005 18:46:49 GMT, the renowned
"daestrom"
wrote:


wrote in message
...
Joel Kolstad
wrote:

(I can't tell you how many times I've seen
people stating something
like,
'The Nyquist theorem requires sampling at at
least twice the highest
frequency present in the signal," when of
course it says no such thing.)

What do you think it means?


Nyquist figured out that higher frequency
components of the input signal
will 'alias' and you will lose the ability to
tell them from lower
frequency
components. In order to avoid 'losing
information' and not being able to
tell whether a particular sample stream was from
a low or high frequency
component, Nyquist's theorem states you must
sample at least twice as fast
as the highest component present.
snip

More than twice the bandwidth.



So, if I have a signal with a 1000 hz carrier, with
a bandwidth of 50 hz,
you think I can sample it at just 150 hz and get
accurate reproduction?
That's just wrong.

It is the maximum frequency component in the signal
that is important. The
bandwidth is not related unless the lower edge of
the band is at 0 hz
(whereupon the upper side of the band is equal to
the max frequency).

daestrom



You are getting your terms confused here guys.
Nyquist requires that
you input both the Center Frequency, and Bandwidth
when determining
the Sampling Rate. If the sampling is done at
BaseBand then only the
Bandwidth is relevent. If the sampling is not done
at baseband, then
the Center Frequency, and Bandwidth are required to
determine samling
rate. Example, if the Bandwith of the signal is 3Kc
and the sampling is
done at BaseBand then sample rate needed would 6Kc.
If the sampling is
done at 100 Mhz with the same 3Kc bandwidth, then a
200.006 Mhz sampling
rate would be required.

It is much easyier to do DSP at baseBand, than at IF
Frequencies, and if
you do DSP at IF Frequencies, the lower the IF
Frequency, the easyier it
is to do, and the slower the DSP has to run.

Me

SolarFlare wrote:

If only the baseband frequency is sampled at 6kHz then
information is missing to recreate the original 100kHz
and the sampling information is insufficient to
recreate the original signal.


This is analogous to saying the number 1234 can be
represented by
(1234-234) / 1000 = 1

If I supply the number 1.0 you can regenerate the
number 1234 from it? Not true, without the rest of the
sampling information. The sample is incomplete.

Bandwidth sampling only cannot recreate the original
signal.


You've used the wrong part of 1234 for your example. The proper analogy
would be to say that 1234 can be represented by 234 in a 3 digit decimal
number system. In that case, the overflow caused by exceeding 999
results in 1234 aliasing onto 234. If you know that all your input
numbers are between 1000 and 1999, then 234 is sufficient information to
represent 1234 with no ambiguity.

The anti-alias filter on your sampling system performs the bracketing to
make sure that all the possible inputs are constrained to be within a
bandwidth of your center frequency +/- BW/2, so when sampled there is no
aliasing. In essence, that filter is the constraint that makes it work.

BTW, the same holds true for baseband sampling: The numbers in a
baseband system based on your example are assumed to be less than 1000,
so that 234 accurately represents 234. In that case if you put in 1234,
it would also map to 234 and you'd have an ambiguity. It just so
happens that in the baseband case, the representation is the same as the
original signal for signals within the bandwidth allowed by Fs/2. With
other than baseband, the representation is not the same as the number
represented, but the constraints imposed by the system allow you to
reconstruct the original value without ambiguity.

OK let's go with your analogy example of 1234 being
represnted by 234 only.

You have no way of decoding 234 into 1234 without
passing information of 1000 as your baseband info and
therefore the the number 1234 has not been successfuly
representedm as being reproduced without further
information.

Now we could further argue algorythms as part of the
information or part of the sample.


"Ray Andraka" wrote in message
news:WUdsf.34360$Mi5.17847@dukeread07...
You've used the wrong part of 1234 for your example.
The proper analogy
would be to say that 1234 can be represented by 234
in a 3 digit decimal
number system. In that case, the overflow caused by
exceeding 999
results in 1234 aliasing onto 234. If you know that
all your input
numbers are between 1000 and 1999, then 234 is
sufficient information to
represent 1234 with no ambiguity.

So, if I have a signal with a 1000 hz carrier, with a bandwidth of 50 hz,
you think I can sample it at just 150 hz and get accurate reproduction?
That's just wrong.

No, that's the whole point of this discussion.

You have to understand aliasing. The signal you want aliases
down into the baseband. Your anti-aliaising filter has
to get rid of all the junk you don't want. In this case it
includes the baseband. Since there is no baseband signal
(or other out-of-band junk) you can reconstruct the original
signal.

It's a common trick with software radios.

You do need some extra information that doesn't go in through
the A/D channel. That's the design of the system, in particular
what the anti-aliasing filter lets through.

Maybe the reason that this is so confusing is that you also need
that info the the normal/baseband case. But since that's the normal
case we don't bother mentioning it.

--
The suespammers.org mail server is located in California. So are all my
other mailboxes. Please do not send unsolicited bulk e-mail or unsolicited
commercial e-mail to my suespammers.org address or any of my other addresses.
These are my opinions, not necessarily my employer's. I hate spam.

Where is the baseband information stored if it isn't
encoded into the sampling?

"Hal Murray" wrote in message
...
So, if I have a signal with a 1000 hz carrier, with
a bandwidth of 50 hz,
you think I can sample it at just 150 hz and get
accurate reproduction?
That's just wrong.

No, that's the whole point of this discussion.

You have to understand aliasing. The signal you want
aliases
down into the baseband. Your anti-aliaising filter
has
to get rid of all the junk you don't want. In this
case it
includes the baseband. Since there is no baseband
signal
(or other out-of-band junk) you can reconstruct the
original
signal.

It's a common trick with software radios.

You do need some extra information that doesn't go in
through
the A/D channel. That's the design of the system, in
particular
what the anti-aliasing filter lets through.

Maybe the reason that this is so confusing is that
you also need
that info the the normal/baseband case. But since
that's the normal
case we don't bother mentioning it.

--
The suespammers.org mail server is located in
California. So are all my
other mailboxes. Please do not send unsolicited bulk
e-mail or unsolicited
commercial e-mail to my suespammers.org address or
any of my other addresses.
These are my opinions, not necessarily my employer's.
I hate spam.