dvsarwate wrote:> On May 24, 2:43 pm, Jerry Avins <j...@ieee.org> asked: > > >>How long will your flashlight shine when >>powered by a pageful of sample magnitudes? > > Oh, shoot, and here I was all set to solve the > energy crisis very simply (and thus win a > Nobel prize) by just doing arithmetic left shifts > on each sample value. You spoilsport, Jerry, you!Why, governments do that all time. Just take a piece of paper and write an arbitrarily big number on it. VLV
This is daft FIR power
Started by ●May 23, 2011
Reply by ●May 25, 20112011-05-25
Reply by ●May 25, 20112011-05-25
On May 26, 1:16�am, "taks" <mtakatz@n_o_s_p_a_m.verasent.com> wrote:> >Ok fair enough. Daft question. So where does the excess power go in a > >digital filter > >when you attenuate? > > You need to keep in mind that once you have digital data, the relative > power levels are fixed (noise and signal.) �No matter how you scale the > data (gain or attenuation,) the same information is still there (other than > the effects of quantization - if you scale by very large or very small > numbers you may lose information.) �There is no "power" in this signal any > more, it is just numbers. �These numbers are somewhat arbitrary, though > ordered (in amplitude and likely time) based on whatever was originally > digitized. > > MarkNo power but when you output from the D/A it becomes a signal again.
Reply by ●May 25, 20112011-05-25
>No power but when you output from the D/A it becomes a signal again. >Immaterial. You're simply sending numbers to a D/A. The D/A does not change anything material w.r.t. the numbers that you created (other than adding its own distortion) and its operation is largely independent of what you have done to create or modify those numbers. I'm really not sure what your point is anymore. Mark
Reply by ●May 26, 20112011-05-26
On May 26, 8:43�am, "taks" <mtakatz@n_o_s_p_a_m.verasent.com> wrote:> >No power but when you output from the D/A it becomes a signal again. > > Immaterial. �You're simply sending numbers to a D/A. �The D/A does not > change anything material w.r.t. the numbers that you created (other than > adding its own distortion) and its operation is largely independent of what > you have done to create or modify those numbers. > > I'm really not sure what your point is anymore. > > MarkIf you took a signal + noise and measured its power, then filtered it (analogue low pass unity pass gain) and measured the power again it would be less than the input power. Clearly power has been dissipated in the circuit. There must be an equibalent in digital. I am assuming the analogue part maybe..
Reply by ●May 26, 20112011-05-26
On 05/26/2011 11:10 AM, HardySpicer wrote:> On May 26, 8:43 am, "taks"<mtakatz@n_o_s_p_a_m.verasent.com> wrote: >>> No power but when you output from the D/A it becomes a signal again. >> >> Immaterial. You're simply sending numbers to a D/A. The D/A does not >> change anything material w.r.t. the numbers that you created (other than >> adding its own distortion) and its operation is largely independent of what >> you have done to create or modify those numbers. >> >> I'm really not sure what your point is anymore. >> >> Mark > > If you took a signal + noise and measured its power, then filtered it > (analogue low pass unity pass gain) and measured the power again it > would be less than the input power. Clearly power has been dissipated > in the circuit. There must be an equivalent in digital. I am assuming > the analogue part maybe..If you take a signal in the real world and applied it to a filter, then every erg of energy that was in the original signal would either be reflected back the way it came, burned up as heat, or emitted from the filter along some route other than the main signal path. That's physics, because thermodynamics holds sway you and energy can't come from nowhere. There really isn't an analog to that in the digital world. What your original example was more akin to was an active analog filter, with at least one gain element. In the digital world "power" and "energy" aren't real concepts -- they're metaphors for what _would_ be happening to the signal if it _were_ in the real world. In the real world, you could say "I have a filter, whose transfer function is H(s) = 100 / (s + 10)". And such a filter can be realized (or at least approximated). But if you then go on to say "and it's passive" then the half the engineers in the room would start to either snicker or show you why you're wrong. Passivity isn't automatic in either the Laplace domain or the z domain -- it's only automatic when you're using all passive components. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 26, 20112011-05-26
Tim Wescott wrote:> In the real world, you could say "I have a filter, whose transfer > function is H(s) = 100 / (s + 10)". And such a filter can be realized > (or at least approximated). But if you then go on to say "and it's > passive" then the half the engineers in the room would start to either > snicker or show you why you're wrong.I don't see any problem with such passive filter. The only hurdle could be a gain at S = 0, however even that could be realized by a piezo transoformer, for example.> Passivity isn't automatic in either the Laplace domain or the z domain > -- it's only automatic when you're using all passive components.The mistake of this thread is the omission of the fact that the power is a product of voltage and current, not just a voltage or just a current. Derive the transfer functions for both voltages and currents, and the activity or passivity of the circuit will be obvious. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●May 26, 20112011-05-26
On 05/26/2011 12:18 PM, Vladimir Vassilevsky wrote:> > > Tim Wescott wrote: > > >> In the real world, you could say "I have a filter, whose transfer >> function is H(s) = 100 / (s + 10)". And such a filter can be realized >> (or at least approximated). But if you then go on to say "and it's >> passive" then the half the engineers in the room would start to either >> snicker or show you why you're wrong. > > I don't see any problem with such passive filter. The only hurdle could > be a gain at S = 0, however even that could be realized by a piezo > transoformer, for example. > >> Passivity isn't automatic in either the Laplace domain or the z domain >> -- it's only automatic when you're using all passive components. > > The mistake of this thread is the omission of the fact that the power is > a product of voltage and current, not just a voltage or just a current. > > Derive the transfer functions for both voltages and currents, and the > activity or passivity of the circuit will be obvious.I'm sorry, you're correct of course. I should have made some qualifying comment about everything going into equal impedances, or being normalized, etc. None the less -- the concept of "power" in the digital domain loses a great deal of meaning in the transition from the real world, and even in the real world -- as I have demonstrated and you have pointed out -- there are pitfalls to effectively keeping track of it. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 26, 20112011-05-26
On May 26, 4:02�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On 05/26/2011 12:18 PM, Vladimir Vassilevsky wrote: > > > > > > > > > > > > > Tim Wescott wrote: > > >> In the real world, you could say "I have a filter, whose transfer > >> function is H(s) = 100 / (s + 10)". And such a filter can be realized > >> (or at least approximated). But if you then go on to say "and it's > >> passive" then the half the engineers in the room would start to either > >> snicker or show you why you're wrong. > > > I don't see any problem with such passive filter. The only hurdle could > > be a gain at S = 0, however even that could be realized by a piezo > > transoformer, for example. > > >> Passivity isn't automatic in either the Laplace domain or the z domain > >> -- it's only automatic when you're using all passive components. > > > The mistake of this thread is the omission of the fact that the power is > > a product of voltage and current, not just a voltage or just a current. > > > Derive the transfer functions for both voltages and currents, and the > > activity or passivity of the circuit will be obvious. > > I'm sorry, you're correct of course. �I should have made some qualifying > comment about everything going into equal impedances, or being > normalized, etc. > > None the less -- the concept of "power" in the digital domain loses a > great deal of meaning in the transition from the real world, and even in > the real world -- as I have demonstrated and you have pointed out -- > there are pitfalls to effectively keeping track of it.... power comes out of the barrel of a gun ... Mao Tse Tung It is important to use well defined terms. Jerry -- Engineering is the art of making what you want from things you can get.
Reply by ●May 26, 20112011-05-26
On 05/26/2011 03:44 PM, Jerry Avins wrote:> On May 26, 4:02 pm, Tim Wescott<t...@seemywebsite.com> wrote: >> On 05/26/2011 12:18 PM, Vladimir Vassilevsky wrote: >> >> >> >> >> >> >> >> >> >> >> >>> Tim Wescott wrote: >> >>>> In the real world, you could say "I have a filter, whose transfer >>>> function is H(s) = 100 / (s + 10)". And such a filter can be realized >>>> (or at least approximated). But if you then go on to say "and it's >>>> passive" then the half the engineers in the room would start to either >>>> snicker or show you why you're wrong. >> >>> I don't see any problem with such passive filter. The only hurdle could >>> be a gain at S = 0, however even that could be realized by a piezo >>> transoformer, for example. >> >>>> Passivity isn't automatic in either the Laplace domain or the z domain >>>> -- it's only automatic when you're using all passive components. >> >>> The mistake of this thread is the omission of the fact that the power is >>> a product of voltage and current, not just a voltage or just a current. >> >>> Derive the transfer functions for both voltages and currents, and the >>> activity or passivity of the circuit will be obvious. >> >> I'm sorry, you're correct of course. I should have made some qualifying >> comment about everything going into equal impedances, or being >> normalized, etc. >> >> None the less -- the concept of "power" in the digital domain loses a >> great deal of meaning in the transition from the real world, and even in >> the real world -- as I have demonstrated and you have pointed out -- >> there are pitfalls to effectively keeping track of it. > > ... power comes out of the barrel of a gun ... Mao Tse Tung > > It is important to use well defined terms.And he defined it well! -- Tim Wescott Wescott Design Services http://www.wescottdesign.com Do you need to implement control loops in software? "Applied Control Theory for Embedded Systems" was written for you. See details at http://www.wescottdesign.com/actfes/actfes.html
Reply by ●May 27, 20112011-05-27
On May 27, 12:18�pm, Tim Wescott <t...@seemywebsite.com> wrote:> On 05/26/2011 03:44 PM, Jerry Avins wrote: > > > > > On May 26, 4:02 pm, Tim Wescott<t...@seemywebsite.com> �wrote: > >> On 05/26/2011 12:18 PM, Vladimir Vassilevsky wrote: > > >>> Tim Wescott wrote: > > >>>> In the real world, you could say "I have a filter, whose transfer > >>>> function is H(s) = 100 / (s + 10)". And such a filter can be realized > >>>> (or at least approximated). But if you then go on to say "and it's > >>>> passive" then the half the engineers in the room would start to either > >>>> snicker or show you why you're wrong. > > >>> I don't see any problem with such passive filter. The only hurdle could > >>> be a gain at S = 0, however even that could be realized by a piezo > >>> transoformer, for example. > > >>>> Passivity isn't automatic in either the Laplace domain or the z domain > >>>> -- it's only automatic when you're using all passive components. > > >>> The mistake of this thread is the omission of the fact that the power is > >>> a product of voltage and current, not just a voltage or just a current. > > >>> Derive the transfer functions for both voltages and currents, and the > >>> activity or passivity of the circuit will be obvious. > > >> I'm sorry, you're correct of course. �I should have made some qualifying > >> comment about everything going into equal impedances, or being > >> normalized, etc. > > >> None the less -- the concept of "power" in the digital domain loses a > >> great deal of meaning in the transition from the real world, and even in > >> the real world -- as I have demonstrated and you have pointed out -- > >> there are pitfalls to effectively keeping track of it. > > > ... power comes out of the barrel of a gun ... Mao Tse Tung > > > It is important to use well defined terms. > > And he defined it well! > > -- > > Tim Wescott > Wescott Design Serviceshttp://www.wescottdesign.com > > Do you need to implement control loops in software? > "Applied Control Theory for Embedded Systems" was written for you. > See details athttp://www.wescottdesign.com/actfes/actfes.htmlActually I am being Devil's Advocate here a bit because I find it interesting. Now if you take a string of numbers we call a digital signal and divide them by two of course no power is lost as we are dealing with numbers only. However, if I read a signal into a computer and attenuate by two and send it out, then the power of my output signal has indeed changed. I think this can be explained by the fact that it is not the same signal that comes out! In other words it is a reconstruction or clone of the orginal scaled down. No power is dissipated at all since in the ideal case let us say the ADC has infinite input impedance. Nevertheless we do use the terms power in DSP to denote SNR ie 10log10(Po/Pin). ALso variance is taken to be average power for a random signal. Hardy
