Is there a "DIGITAL" equivalent of "Q"?
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From: Richard Owlett on 27 Apr 2010 19:59 cassiope wrote: > On Apr 26, 5:12 pm, Richard Owlett <rowl...(a)pcnetinc.com> wrote: >> Mark wrote: >> Mark wrote: >>> On Apr 26, 3:55 pm, Richard Owlett <rowl...(a)pcnetinc.com> wrote: >>>> Please note quotation marks in subject ;) >>>> Also, I'm not the oldest on group --- BUT >>>> my father operated a *LEGAL* land based spark gap xmtr >>>> All that to say that I think in "linear passive discrete" domain >>>> rather than in "digital" domain. >>>> I have a "filter" problem. >>>> I have a reasonable idea on how to implement it. >>>> *UNFORTUNATELY* requires HENRY's and FARADs ;/ >>>> I can write and solve the associated mesh equations >>>> My solution will obviously be a subset of SPICE >>>> BUT will I be able to describe either >>>> PROBLEM or SOLUTION >>>> to those educated in digital domain? >>>> {for perspective -searching this group will reveal that I once >>>> threatened to implement FFT in COBOL ;\ } >>>> 0 >>> if you provide a frequency response and phase response (if you care) >>> or a time domain response to the DSP person, they will be able to >>> design your filter.. >>> or you can try it yourself >>> http://www.mds.com/download/filterdesign.asp >>> with a free demo version >>> (very) basically, selectivity relates to the number of taps.. higher >>> selectivity = more taps... >>> Mark >>> On Apr 26, 3:55 pm, Richard Owlett <rowl...(a)pcnetinc.com> wrote: >>>> Please note quotation marks in subject ;) >>>> Also, I'm not the oldest on group --- BUT >>>> my father operated a *LEGAL* land based spark gap xmtr >>>> All that to say that I think in "linear passive discrete" domain >>>> rather than in "digital" domain. >>>> I have a "filter" problem. >>>> I have a reasonable idea on how to implement it. >>>> *UNFORTUNATELY* requires HENRY's and FARADs ;/ >>>> I can write and solve the associated mesh equations >>>> My solution will obviously be a subset of SPICE >>>> BUT will I be able to describe either >>>> PROBLEM or SOLUTION >>>> to those educated in digital domain? >>>> {for perspective -searching this group will reveal that I once >>>> threatened to implement FFT in COBOL ;\ } >>>> 0 >>> if you provide a frequency response and phase response (if you care) >>> or a time domain response to the DSP person, they will be able to >>> design your filter.. >>> or you can try it yourself >>> http://www.mds.com/download/filterdesign.asp >>> with a free demo version >>> (very) basically, selectivity relates to the number of taps.. higher >>> selectivity = more taps... >>> Mark >> Thank you for your courteous reply. >> BUT I am a "curmudgeon" >> I ask for *EXPLICIT* definition of "Q" in *DIGITAL* domain. > > Perhaps it would be useful if you would tell us how you define Q in > the analog domain (there's more than one way). I was thinking in terms of Q = {bandwidth}/{resonant frequency} Question may have become moot as Tim's answer triggered change of how I defined my goals.
From: Steve Pope on 27 Apr 2010 20:20 Richard Owlett <rowlett(a)pcnetinc.com> wrote: >I was thinking in terms of Q = {bandwidth}/{resonant frequency} I'm going to go out on a limb and say center frequency / bandwidth instead. And, even in a digital domain, it's still the same -- resonant frequency divided by the 3 dB bandwidth is still Q. Even if you're close to Fs/2, and the filter shape looks funny. Steve
From: glen herrmannsfeldt on 27 Apr 2010 22:49 Richard Owlett <rowlett(a)pcnetinc.com> wrote: (snip) > I was thinking in terms of Q = {bandwidth}/{resonant frequency} > Question may have become moot as Tim's answer triggered change of > how I defined my goals. It seems that there are different definitions for Q. Note, for example, the Wikipedia page Q_factor. Some of the difference comes from the difference between filters and oscillators, such that one is: Q = 2 pi * energy store / energy dissipated per cycle. The other is, as noted, (resonant frequency/(bandwidth), each in either cycles or radians. It seems that they are close for high Q, but not so close at lower Q. -- glen
From: Tim Wescott on 27 Apr 2010 23:38 Steve Pope wrote: > Richard Owlett <rowlett(a)pcnetinc.com> wrote: > >> I was thinking in terms of Q = {bandwidth}/{resonant frequency} > > I'm going to go out on a limb and say center frequency / bandwidth > instead. > > And, even in a digital domain, it's still the same -- resonant > frequency divided by the 3 dB bandwidth is still Q. Even if > you're close to Fs/2, and the filter shape looks funny. But in the Laplace domain Q is pretty easy to pick off of a 2nd-order characteristic polynomial; this is not so easy and direct in the z-transform world (unless you know something I don't; in which case I'd be happy if you corrected me). -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: Jerry Avins on 27 Apr 2010 23:57
On 4/27/2010 11:38 PM, Tim Wescott wrote: > Steve Pope wrote: >> Richard Owlett <rowlett(a)pcnetinc.com> wrote: >> >>> I was thinking in terms of Q = {bandwidth}/{resonant frequency} >> >> I'm going to go out on a limb and say center frequency / bandwidth >> instead. >> >> And, even in a digital domain, it's still the same -- resonant >> frequency divided by the 3 dB bandwidth is still Q. Even if >> you're close to Fs/2, and the filter shape looks funny. > > But in the Laplace domain Q is pretty easy to pick off of a 2nd-order > characteristic polynomial; this is not so easy and direct in the > z-transform world (unless you know something I don't; in which case I'd > be happy if you corrected me). Steve, I hope you can educate me. Jerry -- "I view the progress of science as ... the slow erosion of the tendency to dichotomize." --Barbara Smuts, U. Mich. ����������������������������������������������������������������������� |