> 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!
>
> Actually 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.
What you're doing is rejecting the discussion that I've already
presented, that when we talk about "power" in signal processing terms
we're using a metaphor for "power" in physics terms. In physics terms
it's perfectly reasonable to ask "where did the power (or energy) go?".
In DSP it's meaningless, because you've lost the connection to
physical energy.
Examples of infinite impedance ADCs (which, in physical terms, aren't
absorbing any power from the signal at all) followed by computing
systems with numerical gain, followed by DACs that coincidentally happen
to have the inverse of the gain of the ADC are interesting in their own
right, but -- beyond demonstrating how the two definitions are
significantly different -- are meaningless and trite when you are
conflating the DSP definition of "power" with the physics definition.
--
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 dbd●May 27, 20112011-05-27
On May 27, 12:14=A0am, HardySpicer <gyansor...@gmail.com> wrote:
> ...
> Nevertheless we do use the terms
> power in DSP to denote SNR ie 10log10(Po/Pin).
Never correctly.
> ALso variance is taken
> to be average power for a random signal.
>
There are classes of noise for which the numerical value of the
variance is equal to the numerical value of the average power. That
does not make it correct to take the variance to be the average power
for any random signal. The parameters have different definitions.
These may, of course, be your common practices.
Dale B. Dalrymple
Reply by Rune Allnor●May 27, 20112011-05-27
On May 27, 9:14=A0am, HardySpicer <gyansor...@gmail.com> wrote:
> On May 27, 12:18=A0pm, 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> =A0wrote:
> > >> 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) =3D 100 / (s + 10)". And such a filter can be rea=
lized
> > >>>> (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 ei=
ther
> > >>>> snicker or show you why you're wrong.
>
> > >>> I don't see any problem with such passive filter. The only hurdle c=
ould
> > >>> be a gain at S =3D 0, however even that could be realized by a piez=
o
> > >>> transoformer, for example.
>
> > >>>> Passivity isn't automatic in either the Laplace domain or the z do=
main
> > >>>> -- it's only automatic when you're using all passive components.
>
> > >>> The mistake of this thread is the omission of the fact that the pow=
er is
> > >>> a product of voltage and current, not just a voltage or just a curr=
ent.
>
> > >>> Derive the transfer functions for both voltages and currents, and t=
he
> > >>> activity or passivity of the circuit will be obvious.
>
> > >> I'm sorry, you're correct of course. =A0I should have made some qual=
ifying
> > >> 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 eve=
n 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.html
>
> Actually 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!
Wrong. That can be explained by the gain settings at the
post-DAC amplifier.
Like when you adjust volume while playing digitized music:
You don't modify the numbers stored on the CD record (or
in the .wav file) even if this would obtain the objective.
You adjust the gain at the amplifier stage after the DAC,
that drives the speakers.
Rune
Reply by HardySpicer●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.html
Actually 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
Reply by Tim Wescott●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.
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 Tim Wescott●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 Vladimir Vassilevsky●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 Tim Wescott●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 HardySpicer●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.
>
> 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 equibalent in digital. I am assuming
the analogue part maybe..