"glen herrmannsfeldt" <gah@ugcs.caltech.edu> wrote in message 
news:B_qdnbWNK7cBLPzfRVn-gg@comcast.com...
> Fred Marshall wrote:
>
>> <tarangdadia@gmail.com> wrote in message 
>> news:1113404987.623959.199560@l41g2000cwc.googlegroups.com...
>
>>>> TO sample and be able to reconstruct the sine wave we need to sample
>>>> it at >2 x maximum frequency present.
>
>>>I think.. the better way to state this would be " to reconstruct the
>>>sine waver we need to sample it at 2X the bandwidth of the sinewave".
>>>If it is from ideal source then bandwidth tends to zero.. else if it is
>>>impure then anything greater than 2x the bw will do.
>
>> It would be a good thing if folks didn't continue to say things like:
>> "we need to sample it at 2X the bandwidth of the sinewave"
>
>> It perpetuates a common error.
>
> Isn't the bandwidth of a sine wave zero?
>
> -- glen

Yes, it is. I hope you are taking the comments and quotes in context.

I suppose that it's OK for someone to say that the bandwidth of a signal 
(which just happens to be composed of only a sinusoid at frequency f0) is 
f0 - taking that sinusoid in the context of a general lowpass signal 
bandlimited to f0.

Fred 

Fred Marshall wrote:

> <tarangdadia@gmail.com> wrote in message 
> news:1113404987.623959.199560@l41g2000cwc.googlegroups.com...

>>> TO sample and be able to reconstruct the sine wave we need to sample
>>> it at >2 x maximum frequency present.

>>I think.. the better way to state this would be " to reconstruct the
>>sine waver we need to sample it at 2X the bandwidth of the sinewave".
>>If it is from ideal source then bandwidth tends to zero.. else if it is
>>impure then anything greater than 2x the bw will do.

> It would be a good thing if folks didn't continue to say things like:
> "we need to sample it at 2X the bandwidth of the sinewave"

> It perpetuates a common error.  

Isn't the bandwidth of a sine wave zero?

-- glen

<tarangdadia@gmail.com> wrote in message 
news:1113500484.364164.184270@z14g2000cwz.googlegroups.com...
>> == 4 of 4 ==
>> Date: Wed,Apr 13 2005 10:26am
>> From: "Fred Marshall"
>> It would be a good thing if folks didn't continue to say things like:
>> "we need to sample it at 2X the bandwidth of the sinewave"
>>
>> It perpetuates a common error.  There are two ways to say it
> correctly:
>> Theoretically: "we need to sample it at *Greater Than* 2X the
> bandwidth of
>> the signal"
>
> Agreed. That was mistake on my part. In my thoughts it was " we need to
> sample it atleast at twice the bandwidth of the signal" ( but anyways
> who cares what I think ... its all about what I convey :-) ). Probably
> a little more as u have mentioned in ur post below.
>
>> In practice: "we need to sample it at a rate that is at least
>> (approximately) K*2X the bandwidth of the signal .. where K is
> perhaps 1.1
>> or 1.2 or ...."
>> The reason for the latter is that subsequent filtering is likely to
> be
>> applied and there needs to be some guard band or room for a
> transition band
>> between the signal bandwidth and the sample rate.
>
> True.
>
>>
>> If the signal is relatively narrow band then I,Q sampling at an I or
> Q rate
>> that is >1x (and thus overall >2x) the bandwidth (not related to the
> center
>> frequency) is possible.  So, that approach would apply to a signal
> that
>> approaches a pure sinusoid.  It's not OK to simply sample at >2x the
>> *bandwidth* of a narrowband signal unless some tricks are applied.
> See
>> Lyon's 2nd Ed. p474 section 13.1.2 for a trick that works well for
> this.
>
>
> Did not quite understand what you are trying to say. And do not have
> Rick Lyons book in hand. Would appreciate it if you could say what
> trick is he talking in his book.

Well, you gotta get his book!  :-)

Rick treats it as a decimation approach for downconversion of a signal 
centered at fs/4 by fs/4:
The I "modulation" is with values 1, 0, -1, 0 (which are samples of cos at 
fs/4)
The Q "modulation" is with values 0, -1, 0, 1 (which are samples of -sin)
So, there are no multiplies required at all.

Now, if you were A/D converting a known bandlimited/bandpass signal, I 
believe you could do the A/D conversion this way to generate the I/Q 
samples - then you aren't tied to a submultiple of a sampling rate.

Example:
Bandpass signal centered at 10kHz with bandwidth 1kHz.
Sample at 2.5kHz I and Q as above.  The result are I and Q samples at a 
sample rate much less than what would otherwise be >20kHz !  well, I think I 
have the numbers right....

Fred 

> == 4 of 4 ==
> Date: Wed,Apr 13 2005 10:26am
> From: "Fred Marshall"
> It would be a good thing if folks didn't continue to say things like:
> "we need to sample it at 2X the bandwidth of the sinewave"
>
> It perpetuates a common error.  There are two ways to say it
correctly:
> Theoretically: "we need to sample it at *Greater Than* 2X the
bandwidth of
> the signal"

Agreed. That was mistake on my part. In my thoughts it was " we need to
sample it atleast at twice the bandwidth of the signal" ( but anyways
who cares what I think ... its all about what I convey :-) ). Probably
a little more as u have mentioned in ur post below.

> In practice: "we need to sample it at a rate that is at least
> (approximately) K*2X the bandwidth of the signal .. where K is
perhaps 1.1
> or 1.2 or ...."
> The reason for the latter is that subsequent filtering is likely to
be
> applied and there needs to be some guard band or room for a
transition band
> between the signal bandwidth and the sample rate.

True.

>
> If the signal is relatively narrow band then I,Q sampling at an I or
Q rate
> that is >1x (and thus overall >2x) the bandwidth (not related to the
center
> frequency) is possible.  So, that approach would apply to a signal
that
> approaches a pure sinusoid.  It's not OK to simply sample at >2x the
> *bandwidth* of a narrowband signal unless some tricks are applied.
See
> Lyon's 2nd Ed. p474 section 13.1.2 for a trick that works well for
this.


Did not quite understand what you are trying to say. And do not have
Rick Lyons book in hand. Would appreciate it if you could say what
trick is he talking in his book.


Thanks.
Tarang

<tarangdadia@gmail.com> wrote in message 
news:1113404987.623959.199560@l41g2000cwc.googlegroups.com...
> >TO sample and be able to reconstruct the sine wave we need to sample
> it
>>at >2 x maximum frequency present.
>
> I think.. the better way to state this would be " to reconstruct the
> sine waver we need to sample it at 2X the bandwidth of the sinewave".
> If it is from ideal source then bandwidth tends to zero.. else if it is
> impure then anything greater than 2x the bw will do.
>
> Tarang

It would be a good thing if folks didn't continue to say things like:
"we need to sample it at 2X the bandwidth of the sinewave"

It perpetuates a common error.  There are two ways to say it correctly:
Theoretically: "we need to sample it at *Greater Than* 2X the bandwidth of 
the signal"
In practice: "we need to sample it at a rate that is at least 
(approximately) K*2X the bandwidth of the signal .. where K is perhaps 1.1 
or 1.2 or ...."
The reason for the latter is that subsequent filtering is likely to be 
applied and there needs to be some guard band or room for a transition band 
between the signal bandwidth and the sample rate.

If the signal is relatively narrow band then I,Q sampling at an I or Q rate 
that is >1x (and thus overall >2x) the bandwidth (not related to the center 
frequency) is possible.  So, that approach would apply to a signal that 
approaches a pure sinusoid.  It's not OK to simply sample at >2x the 
*bandwidth* of a narrowband signal unless some tricks are applied.  See 
Lyon's 2nd Ed. p474 section 13.1.2 for a trick that works well for this.

Regarding the square wave question, here is an interesting way to look at 
it:
We start with a square wave that we assume is of infinite extent.
We compute the Fourier Series.
We note that the Fourier Series is the same as the Fourier Transform of this 
signal.
However, the Fourier Transform is discrete because the time signal is 
periodic - thus it becomes a series and the Fourier Series formulation is 
easier to deal with and visualize.
The Fourier Transform / Series is of infinite extent in frequency - thus the 
time signal can't be sampled theoretically as the sample rate would have to 
be infinite.
A physical interpretation might be: the sample rate has to be infinite in 
order to "catch" the edges.  Any lower sample rate won't do that.

The Gibbs phenomenon is well known.  It is caused by truncating the Fourier 
Series.
The Gibbs phenomenon "ringing" can be eliminated by spectral shaping. i.e. 
lowpass filtering the square wave.  There have been analyses done that 
provide for the fastest rise time with no ringing / i.e. so the edges of the 
filtered square wave are monotonic.  Thus, it is possible to relate a 
non-ringing square wave with arbitrary rise and fall times to a bandwidth 
and filter characteristic.

The answer to the original post recognizes that it poses a trick question or 
is at least nota well-formed question:
The sample rate there is >2x the bandwidth of some postulated signal.
So, the sample rate could be very high indeed!

Then it asks if a square wave of the "same frequency" can be sampled at the 
same rate?
Well, if we assume that the sample rate is modestly >2x the original 
signal's bandwidth
and
if we assume that the square wave frequency is equal to the original 
signal's bandwidth,
then
the answer is definitely No.
However, if the sample rate is very high indeed then the answer is a 
definite Maybe depending on tolerable Gibbs phenomenon, ringing in general 
and acceptable rise and fall times of the presumed reconstructed square 
wave.  I say "presumed" because we often use the idea of reconstruction as a 
convenient way to analyze these things.

Also, it depends on the application.  If all one wants to know is if the 
square wave is high or low at any instant on some acceptable time grid then 
this is (in some sense) less demanding than recreating the edges.  But it 
does introduce the issue of ambiguity if a sample occurs at or near a 
transition.

Fred 

>TO sample and be able to reconstruct the sine wave we need to sample
it
>at >2 x maximum frequency present.

I think.. the better way to state this would be " to reconstruct the
sine waver we need to sample it at 2X the bandwidth of the sinewave".
If it is from ideal source then bandwidth tends to zero.. else if it is
impure then anything greater than 2x the bw will do.

Tarang

archilleswaterland@hotmail.com wrote:
> TO sample and be able to reconstruct the sine wave we need to sample
it
> at >2 x maximum frequency present.
>
> Can the same sampling rate be used to sample a square wave of the
same
> frequence ?


No. Mathematically, the maximum frequency present in a square wave is
infinite, so you just can't recover *perfectly* a sampled square wave.
However, in practice, you can neglect high frequency harmonics, so
sampling at about 20 * f0 (twenty times the fundamental frequency) or
more will suffice for most uses.

Hope that helps.

-- 
Manuel M.

archilleswaterland@hotmail.com wrote:
> TO sample and be able to reconstruct the sine wave we need to sample
it
> at >2 x maximum frequency present.
>
> Can the same sampling rate be used to sample a square wave of the
same
> frequence ?

No. To see why, compute the Fourier series of the square wave
train, and then reconstruct the wave by adding successively
more Fourier components.

How many overharmonics do you need to see that you have a square
wavetrain? You need to sample at more than twice the frequency of
the highest harmonic.

Rune

If you know that it is a square wave: yes.

            Regards, Wolfgang

<archilleswaterland@hotmail.com> schrieb im Newsbeitrag 
news:1113373887.037645.319150@g14g2000cwa.googlegroups.com...
> TO sample and be able to reconstruct the sine wave we need to sample it
> at >2 x maximum frequency present.
>
> Can the same sampling rate be used to sample a square wave of the same
> frequence ?
>
> thanks,
> ajw
> 

TO sample and be able to reconstruct the sine wave we need to sample it
at >2 x maximum frequency present.

Can the same sampling rate be used to sample a square wave of the same
frequence ? 

thanks,
ajw