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
ADC and undersampling
Started by ●January 27, 2004
Reply by ●February 1, 20042004-02-01
Reply by ●February 1, 20042004-02-01
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}) - FcYep, that's Eq. (1) all right.>Which can be simplified to: > >Fi = Fs + Fc - Fc > >which is just: > >Fi = FsOops, 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-]
Reply by ●February 1, 20042004-02-01
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-]
Reply by ●February 2, 20042004-02-02
Reply by ●February 3, 20042004-02-03
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-] >
Reply by ●February 3, 20042004-02-03
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. �����������������������������������������������������������������������
