Hello,

As a result of a measurement I obtain a bandlimited signal centered at
144 kHz (12kHz Bandwidth). I was planning to use undersampling 
(Fs=192kHz) to fold this signal to 48kHz in the digital domain.
I am having a difficulty to find an ADC for this purpose. The only
ADCs I could find (I need at least 18 bit resolution) that support
Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.

Is there an ADC that could suit my application?

Thanks!

Pawel

Pawel wrote:
> Hello,
> 
> As a result of a measurement I obtain a bandlimited signal centered at
> 144 kHz (12kHz Bandwidth). I was planning to use undersampling 
> (Fs=192kHz) to fold this signal to 48kHz in the digital domain.
> I am having a difficulty to find an ADC for this purpose. The only
> ADCs I could find (I need at least 18 bit resolution) that support
> Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
> 
> Is there an ADC that could suit my application?

Why not sample a much lower frequency? 48 will not work (the 0 frequency 
ends up in themiddle of your band of interest) but something a bit lower 
(6 KHz lower or so) should work.

Take care that the quality of the sample-and-hold is more important with 
this input... the time of sampling is much more critical at high 
frequencies.


Thomas

On Tue, 27 Jan 2004 21:51:54 +0100, Zak wrote:

> Pawel wrote:
>> Hello,
>> 
>> As a result of a measurement I obtain a bandlimited signal centered at
>> 144 kHz (12kHz Bandwidth). I was planning to use undersampling 
>> (Fs=192kHz) to fold this signal to 48kHz in the digital domain.
>> I am having a difficulty to find an ADC for this purpose. The only
>> ADCs I could find (I need at least 18 bit resolution) that support
>> Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
>> 
>> Is there an ADC that could suit my application?
> 
> Why not sample a much lower frequency? 48 will not work (the 0 frequency 
> ends up in themiddle of your band of interest) but something a bit lower 
> (6 KHz lower or so) should work.
> 
> Take care that the quality of the sample-and-hold is more important with 
> this input... the time of sampling is much more critical at high 
> frequencies.
> 
> 
> Thomas

I think what he is saying is that the ADC frontend doesn't have enough
bandwidth to do under-sampling. I know some high speed ADC's are designed
for this, but I'm not too sure about these slower (192 kHz) ones.

If he went to an even slower ADC, his problems would only get worse,
presumably.

Mac

ADS1625, ADS1626, ADS8383, AD7679 ??

> 
> As a result of a measurement I obtain a bandlimited signal centered at
> 144 kHz (12kHz Bandwidth). I was planning to use undersampling 
> (Fs=192kHz) to fold this signal to 48kHz in the digital domain.
> I am having a difficulty to find an ADC for this purpose. The only
> ADCs I could find (I need at least 18 bit resolution) that support
> Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
> 
> Is there an ADC that could suit my application?
> 

Where are you getting that? The data sheet clearly states that the 0.1dB 
passband at "quad speed" sampling is 0.24 x Fs which puts you at 46KHz, 
your 48KHz is 0.25 x Fs - negligible additional attenuation 0.14dB? And 
this is the digital filter- not the analog passband.

> Why not sample a much lower frequency? (6 KHz lower or so) should
> work.

If you use very low sample rates, remember to obey the nyquist rule for
the highest (baseband?) output frequency.

Jim Adamthwaite.

Fred Bloggs <nospam@nospam.com> wrote in message news:<4017AD09.9020501@nospam.com>...
> > 
> > As a result of a measurement I obtain a bandlimited signal centered at
> > 144 kHz (12kHz Bandwidth). I was planning to use undersampling 
> > (Fs=192kHz) to fold this signal to 48kHz in the digital domain.
> > I am having a difficulty to find an ADC for this purpose. The only
> > ADCs I could find (I need at least 18 bit resolution) that support
> > Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
> > 
> > Is there an ADC that could suit my application?
> > 
> 
> Where are you getting that? The data sheet clearly states that the 0.1dB 
> passband at "quad speed" sampling is 0.24 x Fs which puts you at 46KHz, 
> your 48KHz is 0.25 x Fs - negligible additional attenuation 0.14dB? And 
> this is the digital filter- not the analog passband.

The INPUT is 144KHz

GPG wrote:
> Fred Bloggs <nospam@nospam.com> wrote in message news:<4017AD09.9020501@nospam.com>...
> 
>>>As a result of a measurement I obtain a bandlimited signal centered at
>>>144 kHz (12kHz Bandwidth). I was planning to use undersampling 
>>>(Fs=192kHz) to fold this signal to 48kHz in the digital domain.
>>>I am having a difficulty to find an ADC for this purpose. The only
>>>ADCs I could find (I need at least 18 bit resolution) that support
>>>Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
>>>
>>>Is there an ADC that could suit my application?
>>>
>>
>>Where are you getting that? The data sheet clearly states that the 0.1dB 
>>passband at "quad speed" sampling is 0.24 x Fs which puts you at 46KHz, 
>>your 48KHz is 0.25 x Fs - negligible additional attenuation 0.14dB? And 
>>this is the digital filter- not the analog passband.
> 
> 
> The INPUT is 144KHz

  The analog input is aliased into 48KHz in the digital domain, it is 
the digital domain that is filtered, not the analog, and that passband 
spec refers to the digital domain .

On 27 Jan 2004 12:21:33 -0800, pawel.kluczynski@comhem.se (Pawel)
wrote:

>Hello,
>
>As a result of a measurement I obtain a bandlimited signal centered at
>144 kHz (12kHz Bandwidth). I was planning to use undersampling 
>(Fs=192kHz) to fold this signal to 48kHz in the digital domain.
>I am having a difficulty to find an ADC for this purpose. The only
>ADCs I could find (I need at least 18 bit resolution) that support
>Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
>
>Is there an ADC that could suit my application?
>
>Thanks!
>
>Pawel

Hi,
  here are the frequency ranges (in kHz)
within which you can have your Fs sample rate:

Fs_ranges =

  150.0000 -to- 276.0000
  100.0000 -to- 138.0000
   75.0000 -to- 92.0000
   60.0000 -to- 69.0000
   50.0000 -to- 55.2000
   42.8571 -to- 46.0000
   37.5000 -to- 39.4286
   33.3333 -to- 34.5000
   30.0000 -to- 30.6667
   27.2727 -to- 27.6000
   25.0000 -to- 25.0909

Zak is right, 48 kHz won't work.

Good luck,
[-Rick-]

Rick Lyons wrote:
> On 27 Jan 2004 12:21:33 -0800, pawel.kluczynski@comhem.se (Pawel)
> wrote:
> 
> 
>>Hello,
>>
>>As a result of a measurement I obtain a bandlimited signal centered at
>>144 kHz (12kHz Bandwidth). I was planning to use undersampling 
>>(Fs=192kHz) to fold this signal to 48kHz in the digital domain.
>>I am having a difficulty to find an ADC for this purpose. The only
>>ADCs I could find (I need at least 18 bit resolution) that support
>>Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
>>
>>Is there an ADC that could suit my application?
>>
>>Thanks!
>>
>>Pawel
> 
> 
> Hi,
>   here are the frequency ranges (in kHz)
> within which you can have your Fs sample rate:
> 
> Fs_ranges =
> 
>   150.0000 -to- 276.0000
>   100.0000 -to- 138.0000
>    75.0000 -to- 92.0000
>    60.0000 -to- 69.0000
>    50.0000 -to- 55.2000
>    42.8571 -to- 46.0000
>    37.5000 -to- 39.4286
>    33.3333 -to- 34.5000
>    30.0000 -to- 30.6667
>    27.2727 -to- 27.6000
>    25.0000 -to- 25.0909
> 
> Zak is right, 48 kHz won't work.
> 
> Good luck,
> [-Rick-]
> 

Ummm- maybe you are confused by the term bandwidth. This means Fc+/-6KHz 
  and -3dB, and at quad speed this is 0.03 x Fs. Looking at the passband 
ripple graph which a nominal 0.03dB peak variation- your low sample 
rates mean the signal is spread over a correspondingly larger percentage 
of the passband- you see the *full* ripple error variation which is of 
the same order of magnitude as using the 192KHz Fs. Luck is for dummies 
and programmers, analysis is for engineers.

r.lyons@REMOVE.ieee.org (Rick Lyons) wrote in message news:<40190b23.59185953@news.west.earthlink.net>...
> 
> Hi,
>   here are the frequency ranges (in kHz)
> within which you can have your Fs sample rate:
> 
> Fs_ranges =
> 
>   150.0000 -to- 276.0000
>   100.0000 -to- 138.0000
>    75.0000 -to- 92.0000
>    60.0000 -to- 69.0000
>    50.0000 -to- 55.2000
>    42.8571 -to- 46.0000
>    37.5000 -to- 39.4286
>    33.3333 -to- 34.5000
>    30.0000 -to- 30.6667
>    27.2727 -to- 27.6000
>    25.0000 -to- 25.0909
> 
> Zak is right, 48 kHz won't work.
> 
> Good luck,
> [-Rick-]

Hello, 
Thanks for the answers!
I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
Therefore I was looking for an audio 192 kHz ADC for the application.
The problem is that they ones I could find have too small analog
passband so the udersampling trick would not work.

Regards,

Pawel

peg@slingshot.co.nz (GPG) wrote in message news:<62069f15.0401280259.e819db3@posting.google.com>...
> ADS1625, ADS1626, ADS8383, AD7679 ??

Hi!

Thanks for the info GPG!
Yes, I was condsidering to use  an ADS1626 for my application and 
just sample faster to have my 144 kHz below Nyqvist frequency. It can
be an alternative solution. But at faster sampling frequencies it has
a high power consumption, which is a limitation for me.
Lowering Fs would not help since the ADS1626 passband filter scales
with Fs so the undersampling trick would not work either, would it?

Regards,

Pawel

Fred Bloggs wrote:

> 
> 
> Rick Lyons wrote:
> 
>> On 27 Jan 2004 12:21:33 -0800, pawel.kluczynski@comhem.se (Pawel)
>> wrote:
>>
>>
>>> Hello,
>>>
>>> As a result of a measurement I obtain a bandlimited signal centered at
>>> 144 kHz (12kHz Bandwidth). I was planning to use undersampling 
>>> (Fs=192kHz) to fold this signal to 48kHz in the digital domain.
>>> I am having a difficulty to find an ADC for this purpose. The only
>>> ADCs I could find (I need at least 18 bit resolution) that support
>>> Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
>>>
>>> Is there an ADC that could suit my application?
>>>
>>> Thanks!
>>>
>>> Pawel
>>
>>
>>
>> Hi,
>>   here are the frequency ranges (in kHz)
>> within which you can have your Fs sample rate:
>>
>> Fs_ranges =
>>
>>   150.0000 -to- 276.0000
>>   100.0000 -to- 138.0000
>>    75.0000 -to- 92.0000
>>    60.0000 -to- 69.0000
>>    50.0000 -to- 55.2000
>>    42.8571 -to- 46.0000
>>    37.5000 -to- 39.4286
>>    33.3333 -to- 34.5000
>>    30.0000 -to- 30.6667
>>    27.2727 -to- 27.6000
>>    25.0000 -to- 25.0909
>>
>> Zak is right, 48 kHz won't work.
>>
>> Good luck,
>> [-Rick-]
>>
> 
> Ummm- maybe you are confused by the term bandwidth. This means Fc+/-6KHz 
>  and -3dB, and at quad speed this is 0.03 x Fs. Looking at the passband 
> ripple graph which a nominal 0.03dB peak variation- your low sample 
> rates mean the signal is spread over a correspondingly larger percentage 
> of the passband- you see the *full* ripple error variation which is of 
> the same order of magnitude as using the 192KHz Fs. Luck is for dummies 
> and programmers, analysis is for engineers.

I doubt that Rick is confused at all about acceptable sampling ranges.
The best explanation I know of the constraints on sub-band sampling is
given in the book "Understanding Digital Signal Processing" by Richard
G. Lyons (ISBN 0-201-63467-8), Section 2.3. (Don't run out to buy the
book now. There's a second edition coming out soon.) If you check with
the equations there, you'll find that Rick is right on. :-) I don't
think he directly addresses Pawel's concern, though.

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

"Pawel" <pawel.kluczynski@comhem.se> wrote in message
news:149bb2c3.0401291133.2ad6508a@posting.google.com...
> r.lyons@REMOVE.ieee.org (Rick Lyons) wrote in message
news:<40190b23.59185953@news.west.earthlink.net>...
> >
> > Hi,
> >   here are the frequency ranges (in kHz)
> > within which you can have your Fs sample rate:
> >
> > Fs_ranges =
> >
> >   150.0000 -to- 276.0000
> >   100.0000 -to- 138.0000
> >    75.0000 -to- 92.0000
> >    60.0000 -to- 69.0000
> >    50.0000 -to- 55.2000
> >    42.8571 -to- 46.0000
> >    37.5000 -to- 39.4286
> >    33.3333 -to- 34.5000
> >    30.0000 -to- 30.6667
> >    27.2727 -to- 27.6000
> >    25.0000 -to- 25.0909
> >
> > Zak is right, 48 kHz won't work.
> >
> > Good luck,
> > [-Rick-]
>
> Hello,
> Thanks for the answers!
> I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
> analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
> Therefore I was looking for an audio 192 kHz ADC for the application.
> The problem is that they ones I could find have too small analog
> passband so the udersampling trick would not work.

What happens if you sample at something just under 138kHz as Rick suggested?
You should end up with an image with bandwidth of 12kHz centered at +/-6kHz.
Then you can immediately decimate (with proper filtering) from 138 to
something like 30 like how about by a factor of 4 down to 34.5kHz?  That
leaves plenty of room for filtering between 12kHz and fs/2=17.25kHz.

The only thing the digital filter in the ADC does is keep the bandwidth
consistent with the sample rate.  It doesn't change the input bandwidth, it
controls the output bandwidth to be consistent with the output sample rate.
So, if the sample rate is 1.25MHz, the bandwidth is limited to below 625kHz
at actually 550kHz.  If the sample rate is 138kHz, it limits the output
bandwidth to (550/1250)*138 = 60.72kHz, which is what you want for that
sample rate.  The useful information is below 12kHz at this point - so you
decimate by 4 thereafter.

The other alternative is to go to a complex envelope representation by
sampling at 144kHz in quadrature.  Rick Lyon's book deals with that very
nicely.  In fact, the ADC's will do it for you by sampling at 144kHz and all
you have to do is alternately grab the output words to get the quadrature
results.  It's either that or deal with the wider bandwidth as I have done
above - and avoid having quadrature channels to deal with.  The number of
operations thereafter should be pretty close to the same either way because
either you have two channels with 1/2 the bandwidth or one channel with full
bandwidth.

I think that's right.....

Fred

Fred

"Fred Marshall" <fmarshallx@remove_the_x.acm.org> wrote in message
news:Kb-dneN_vo7nP4TdRVn-vA@centurytel.net...
>
> "Pawel" <pawel.kluczynski@comhem.se> wrote in message
> news:149bb2c3.0401291133.2ad6508a@posting.google.com...
> > r.lyons@REMOVE.ieee.org (Rick Lyons) wrote in message
> news:<40190b23.59185953@news.west.earthlink.net>...
> > >
> > > Hi,
> > >   here are the frequency ranges (in kHz)
> > > within which you can have your Fs sample rate:
> > >
> > > Fs_ranges =
> > >
> > >   150.0000 -to- 276.0000
> > >   100.0000 -to- 138.0000
> > >    75.0000 -to- 92.0000
> > >    60.0000 -to- 69.0000
> > >    50.0000 -to- 55.2000
> > >    42.8571 -to- 46.0000
> > >    37.5000 -to- 39.4286
> > >    33.3333 -to- 34.5000
> > >    30.0000 -to- 30.6667
> > >    27.2727 -to- 27.6000
> > >    25.0000 -to- 25.0909
> > >
> > > Zak is right, 48 kHz won't work.
> > >
> > > Good luck,
> > > [-Rick-]
> >
> > Hello,
> > Thanks for the answers!
> > I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
> > analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
> > Therefore I was looking for an audio 192 kHz ADC for the application.
> > The problem is that they ones I could find have too small analog
> > passband so the udersampling trick would not work.
>
> What happens if you sample at something just under 138kHz as Rick
suggested?
> You should end up with an image with bandwidth of 12kHz centered at
+/-6kHz.
> Then you can immediately decimate (with proper filtering) from 138 to
> something like 30 like how about by a factor of 4 down to 34.5kHz?  That
> leaves plenty of room for filtering between 12kHz and fs/2=17.25kHz.
>
> The only thing the digital filter in the ADC does is keep the bandwidth
> consistent with the sample rate.  It doesn't change the input bandwidth,
it
> controls the output bandwidth to be consistent with the output sample
rate.
> So, if the sample rate is 1.25MHz, the bandwidth is limited to below
625kHz
> at actually 550kHz.  If the sample rate is 138kHz, it limits the output
> bandwidth to (550/1250)*138 = 60.72kHz, which is what you want for that
> sample rate.  The useful information is below 12kHz at this point - so you
> decimate by 4 thereafter.
>
> The other alternative is to go to a complex envelope representation by
> sampling at 144kHz in quadrature.  Rick Lyon's book deals with that very
> nicely.  In fact, the ADC's will do it for you by sampling at 144kHz and
all
> you have to do is alternately grab the output words to get the quadrature
> results.  It's either that or deal with the wider bandwidth as I have done
> above - and avoid having quadrature channels to deal with.  The number of
> operations thereafter should be pretty close to the same either way
because
> either you have two channels with 1/2 the bandwidth or one channel with
full
> bandwidth.
>

Well, I didn't acknowledge here that Rick had suggested sampling at even
lower frequencies - and I should have.  The filtering necessary while
decimating will change if you do that.

Fred

Fred Bloggs wrote:
> 
> GPG wrote:
> > Fred Bloggs <nospam@nospam.com> wrote in message news:<4017AD09.9020501@nospam.com>...
> >
> >>>As a result of a measurement I obtain a bandlimited signal centered at
> >>>144 kHz (12kHz Bandwidth). I was planning to use undersampling
> >>>(Fs=192kHz) to fold this signal to 48kHz in the digital domain.
> >>>I am having a difficulty to find an ADC for this purpose. The only
> >>>ADCs I could find (I need at least 18 bit resolution) that support
> >>>Fs=192kHz had too small analog bandwidth, f.ex. cirrus CS5361.
> >>>
> >>>Is there an ADC that could suit my application?
> >>>
> >>
> >>Where are you getting that? The data sheet clearly states that the 0.1dB
> >>passband at "quad speed" sampling is 0.24 x Fs which puts you at 46KHz,
> >>your 48KHz is 0.25 x Fs - negligible additional attenuation 0.14dB? And
> >>this is the digital filter- not the analog passband.
> >
> >
> > The INPUT is 144KHz
> 
>   The analog input is aliased into 48KHz in the digital domain, it is
> the digital domain that is filtered, not the analog, and that passband
> spec refers to the digital domain .

Don't think so.
The ADC samples at about 6 MHz.
Then it filters so that aliasing will not happen when it decimates down
to
actual output rate. So 144 KHz will not alias. 

This is the fine point in oversampling converters with decimation.
Analog antialias filter need only be effective for the actual high
sample rate.

Jarmo

On 29 Jan 2004 11:33:44 -0800, pawel.kluczynski@comhem.se (Pawel)
wrote:

>r.lyons@REMOVE.ieee.org (Rick Lyons) wrote in message news:<40190b23.59185953@news.west.earthlink.net>...
>> 
>> Hi,
>>   here are the frequency ranges (in kHz)
>> within which you can have your Fs sample rate:
>> 
>> Fs_ranges =
>> 
>>   150.0000 -to- 276.0000
>>   100.0000 -to- 138.0000
>>    75.0000 -to- 92.0000
>>    60.0000 -to- 69.0000
>>    50.0000 -to- 55.2000
>>    42.8571 -to- 46.0000
>>    37.5000 -to- 39.4286
>>    33.3333 -to- 34.5000
>>    30.0000 -to- 30.6667
>>    27.2727 -to- 27.6000
>>    25.0000 -to- 25.0909
>> 
>> Zak is right, 48 kHz won't work.
>> 
>> Good luck,
>> [-Rick-]
>
>Hello, 
>Thanks for the answers!
>I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
>analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
>Therefore I was looking for an audio 192 kHz ADC for the application.
>The problem is that they ones I could find have too small analog
>passband so the udersampling trick would not work.
>
>Regards,
>
>Pawel

Hi,
  Oh shoot, I misread your first post.
Sorry.

[-Rick-]

On 29 Jan 2004 11:33:44 -0800, pawel.kluczynski@comhem.se (Pawel)
wrote:

  (snipped)
>
>Hello, 
>Thanks for the answers!
>I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
>analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
>Therefore I was looking for an audio 192 kHz ADC for the application.
>The problem is that they ones I could find have too small analog
>passband so the udersampling trick would not work.
>
>Regards,
>
>Pawel

Hi,
  I was thinkin' some more about your question, 
and darn it, you make me ask a question.

If we define:

Fc = carrier freq (Pavel's 144 kHz)
Fs = sample rate (Pavel's 198 kHz)
Fi = the positive center freq of the aliased  
       spectral replication nearest to zero Hz.

In Pavel's case:

    Fi = Fs(1 + {Fc/Fs}) -Fc          (1)

where {Fc/Fs} means the integer part of Fc/Fs.

Using that Eq. (1), Pavel's Fi (in kHz) is:

    Fi = 192(1 + 0) -144 = 48  

which is what he said.

So now here's my question:  At the risk of lookin' 
like (as Fred Sanford would say) a big dummy,
is there a way to solve the above Eq. (1) for Fs 
in terms of Fc and Fi?

For some reason (maybe Alzheimers) I can't 
figure out how to handle that {Fc/Fs} operation 
in algebra.

Thanks,
[-Rick-]

Rick Lyons wrote:

> On 29 Jan 2004 11:33:44 -0800, pawel.kluczynski@comhem.se (Pawel)
> wrote:
> 
>   (snipped)
> 
>>Hello, 
>>Thanks for the answers!
>>I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
>>analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
>>Therefore I was looking for an audio 192 kHz ADC for the application.
>>The problem is that they ones I could find have too small analog
>>passband so the udersampling trick would not work.
>>
>>Regards,
>>
>>Pawel
> 
> 
> Hi,
>   I was thinkin' some more about your question, 
> and darn it, you make me ask a question.
> 
> If we define:
> 
> Fc = carrier freq (Pavel's 144 kHz)
> Fs = sample rate (Pavel's 198 kHz)
> Fi = the positive center freq of the aliased  
>        spectral replication nearest to zero Hz.
> 
> In Pavel's case:
> 
>     Fi = Fs(1 + {Fc/Fs}) -Fc          (1)
> 
> where {Fc/Fs} means the integer part of Fc/Fs.
> 
> Using that Eq. (1), Pavel's Fi (in kHz) is:
> 
>     Fi = 192(1 + 0) -144 = 48  
> 
> which is what he said.
> 
> So now here's my question:  At the risk of lookin' 
> like (as Fred Sanford would say) a big dummy,
> is there a way to solve the above Eq. (1) for Fs 
> in terms of Fc and Fi?
> 
> For some reason (maybe Alzheimers) I can't 
> figure out how to handle that {Fc/Fs} operation 
> in algebra.
> 
> Thanks,
> [-Rick-]

A big dummy doesn't know. A big idiot thinks he does but doesn't. I'll
risk being branded with the latter title.

Solve for {Fc/Fs} first. Then, given the known Fc, see what range of Fs
makes that possible. I've introduced another variable, {Fc/Fs}, so it
may take some trial and error anyway, but at least it's directed.

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

On Sat, 31 Jan 2004 12:26:05 GMT, r.lyons@_BOGUS_ieee.org (Rick Lyons)
wrote:

>On 29 Jan 2004 11:33:44 -0800, pawel.kluczynski@comhem.se (Pawel)
>wrote:
>
>  (snipped)

  Oops, I made a typo.  I wrote:

>Fs = sample rate (Pavel's 198 kHz)

It should be

    Fs = sample rate (Pavel's 192 kHz)


Pawel, please pardon me for spelling your 
name wrong.  :-(

[-Rick-]
>
>

On Sat, 31 Jan 2004 12:26:05 +0000, Rick Lyons wrote:

> On 29 Jan 2004 11:33:44 -0800, pawel.kluczynski@comhem.se (Pawel)
> wrote:
> 
>   (snipped)
>>
>>Hello, 
>>Thanks for the answers!
>>I did not say I wanted to sample at 48 kHz. I said I have a 144 KHz
>>analog signal that I want to alias to 48 kHz by sampling at Fs=192kHz.
>>Therefore I was looking for an audio 192 kHz ADC for the application.
>>The problem is that they ones I could find have too small analog
>>passband so the udersampling trick would not work.
>>
>>Regards,
>>
>>Pawel
> 
> Hi,
>   I was thinkin' some more about your question, 
> and darn it, you make me ask a question.
> 
> If we define:
> 
> Fc = carrier freq (Pavel's 144 kHz)
> Fs = sample rate (Pavel's 198 kHz)
> Fi = the positive center freq of the aliased  
>        spectral replication nearest to zero Hz.
> 
> In Pavel's case:
> 
>     Fi = Fs(1 + {Fc/Fs}) -Fc          (1)
> 
> where {Fc/Fs} means the integer part of Fc/Fs.
> 
> Using that Eq. (1), Pavel's Fi (in kHz) is:
> 
>     Fi = 192(1 + 0) -144 = 48  
> 
> which is what he said.
> 
> So now here's my question:  At the risk of lookin' 
> like (as Fred Sanford would say) a big dummy,
> is there a way to solve the above Eq. (1) for Fs 
> in terms of Fc and Fi?
> 
> For some reason (maybe Alzheimers) I can't 
> figure out how to handle that {Fc/Fs} operation 
> in algebra.
> 
> Thanks,
> [-Rick-]


Uh, equation (1) has a problem. If you solve it for Fi, the FC  falls out.
Start by distributing the Fs back in. This gives you:

Fi = (Fs + Fs{Fc/Fs}) - Fc

Which can be simplified to:

Fi = Fs + Fc - Fc

which is just:

Fi = Fs

Or am I missing something?

Mac

On Sat, 31 Jan 2004 23:16:08 -0800, "Mac" <foo@bar.net> wrote:

  (snipped)

>Uh, equation (1) has a problem. If you solve it for Fi, the FC  falls out.

Hi,
  well, Eq, (1) is *already* solved in terms of Fi.  
Here's Eq. (1) again:

        Fi = Fs(1 + {Fc/Fs}) -Fc          (1)

>Start by distributing the Fs back in. This gives you:
>Fi = (Fs + Fs{Fc/Fs}) - Fc

Yep, that's Eq. (1) all right.

>Which can be simplified to:
>
>Fi = Fs + Fc - Fc
>
>which is just:
>
>Fi = Fs

Oops, there's the problem Mac.  The squiggly brackets 
{x} mean the integer part of x.  So {Fc/Fs} is *not* 
equal to straight Fc/Fs.  And Fs{Fc/Fs} is not equal 
to Fc.  

Now Jerry's idea of solving Eq. (1) for {Fc/Fs} first 
seems like a good idea.   Solving for {Fc/Fs}, we have:

              Fi + Fc
   {Fc/Fs} = ----------  -1.       (2)
                Fs

Which implies the ratio (Fi+Fc)/Fs must be 
an integer because {Fc/Fs} is an integer.
   
Humm, ... mumble, grumble, mumble.

Thinking about integers, solving Eq. (1) for Fs 
gives:

                Fi + Fc
        Fs = -------------- .        (3)
              {Fc/Fs} + 1

This means Fs is (Fi + Fc) divided by an integer.
However, the glitch is: Eq. (1) only applies, as 
far as I know, when {2*Fc/Fs} is an odd integer.

Shoot!  I have to think more about this. 
It looks like bandpass sampling, for me, is like 
a woman.  As soon as you think you understand 
something, you find out later that really you don't.

I'm gonna work on this during the Super Bowl.

Thanks guys,
[-Rick-]

On Sat, 31 Jan 2004 23:16:08 -0800, "Mac" <foo@bar.net> wrote:


OK, ... I think I've got it.

Pawel apparently wanted his image to be centered 
at Fs/4.  His Fs was 192 kHz, so his Fs/4 was 48 kHz.

With help from MATLAB and several bottles of my 
favorite alchoholic beverage from Holland, 
I now think Pawel can get an image centered 
at Fs/4 by choosing an Fs that satisfies:

  Fs = 4Fc/m

where m is an odd integer, and Fc is Pawel's 
144 kHz.

When m = 3, 7, 11, 15, etc.
spectral inversion will take place.

So Pawel's sample rate can be 192 kHz, 115.2 kHz,
82.28571 kHz, 64 kHz, 52.36364 kHz, 40.30792 kHz, etc.

[-Rick-]

Hello,

Thanks for all the remarks and the interesting discussion!

Pawel

Hi,

How can I determine those ranges ?

Thanks

Hern�n S�nchez


"Rick Lyons" <r.lyons@REMOVE.ieee.org> escribi� en el mensaje
news:40190b23.59185953@news.west.earthlink.net...
>
> Hi,
>   here are the frequency ranges (in kHz)
> within which you can have your Fs sample rate:
>
> Fs_ranges =
>
>   150.0000 -to- 276.0000
>   100.0000 -to- 138.0000
>    75.0000 -to- 92.0000
>    60.0000 -to- 69.0000
>    50.0000 -to- 55.2000
>    42.8571 -to- 46.0000
>    37.5000 -to- 39.4286
>    33.3333 -to- 34.5000
>    30.0000 -to- 30.6667
>    27.2727 -to- 27.6000
>    25.0000 -to- 25.0909
>
> Zak is right, 48 kHz won't work.
>
> Good luck,
> [-Rick-]
>

Hern�n S�nchez wrote:
> Hi,
> 
> How can I determine those ranges ?
> 
> Thanks
> 
> Hern�n S�nchez
> 
> 
> "Rick Lyons" <r.lyons@REMOVE.ieee.org> escribi� en el mensaje
> news:40190b23.59185953@news.west.earthlink.net...
> 
>>Hi,
>>  here are the frequency ranges (in kHz)
>>within which you can have your Fs sample rate:
>>
>>Fs_ranges =
>>
>>  150.0000 -to- 276.0000
>>  100.0000 -to- 138.0000
>>   75.0000 -to- 92.0000
>>   60.0000 -to- 69.0000
>>   50.0000 -to- 55.2000
>>   42.8571 -to- 46.0000
>>   37.5000 -to- 39.4286
>>   33.3333 -to- 34.5000
>>   30.0000 -to- 30.6667
>>   27.2727 -to- 27.6000
>>   25.0000 -to- 25.0909
>>
>>Zak is right, 48 kHz won't work.
>>
>>Good luck,
>>[-Rick-]

The only comprehensive discussion I've seen was in Rick's book 
("Understanding Digital Signal Processing" by Richard G. Lyons, ISBN 
0-201-63467-8, Section 2.3). A second edition is coming out soon.

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