On Oct 10, 5:48 am, HardySpicer <gyansor...@gmail.com> wrote:
> On Oct 10, 9:38 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
> > On 10 Okt, 03:10, HardySpicer <gyansor...@gmail.com> wrote:
>
> > > Ok suppose we have an R-C network - low pass filter with tf
>
> > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power
> > > spectrum if this is driven by zero-mean white noise as
>
> > > Phiyy(w)=mag(G(jw))^2 x power of input noise
>
> > > and the area under the PSD is the total average power.
>
> > > Now let's do the same thing digitally...and produce an equivalent
> > > digital filter G(z)
>
> > > The output periodogram is
>
> > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is
> > > the total output power.
>
> > > Now, assuming the two output power values at the output are the same,
> > > where does the missing power go? ie the output power not equal input
> > > power.
>
> > Your base assumption is wrong: Digital systems don't
> > behave like analog systems.
>
> > The physics of the analog world dictates two important
> > conditions which do not apply in the digital world:
>
> > - There is always friction present in the physical world,
> > which dissipate energy in terms of heat.
> > - One is limited to only observe real-valued signals
> > but need imaginary components to make the energy
> > computations go up.
>
> > In the digital world there are no dissipation terms,
> > only numerical accuracy issues, and one does not need
> > the complex/valued maths to make the energy computations
> > work.
>
> > In essence, the two troublesome issues from the physical
> > world, what energy is concerned, just don't exist in the
> > digital world.
>
> > Which means you will have to look elsewhere for sources
> > for discrepancy, the most common sources being certain
> > normalization factors in the most popular implementations
> > of the FFT.
>
> > Rune
>
> Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A.
> Suppose a sine-wave of unit amplitude goes in and if there are no
> other losses we get a sine wave of amplitude 0.5 out. Input power is
> 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we
> had an analogue system the power would be disipated as heat via a
> resistor divider. In the digital world the power still has to be
> disipated somewhere. If we have a gain instead of attenuation - say x2
> then this extra power flows in from the power supply in an analogue
> system and must also do in a digital system.
Replace the gain code in the DSP between the ADC and the
DAC with some random number generation code. It should
then be obvious that the power on the output side is not
really directly coupled to the power on the input side
in such a digital system.
As a famous linguist once said: don't confuse the map
with the territory.
IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M
Reply by glen herrmannsfeldt●October 11, 20072007-10-11
cs_posting@hotmail.com wrote:
(snip)
> But what about in the implementation of the math in an electronic
> computational engine?
> I had a professor who made an argument that power would ultimately be
> consumed where you "forgot" the unwanted bits, and that the true key
> to low power (consumption) digital design would be to run the
> computations backwards as well as forwards, to recreate the inputs.
It takes power (dissipation) to store data. A RAM chip, for example.
Computations that don't lose information, addition with wrap around
for example, can theoretically be done without dissipating power.
> He then went on to point out the conflict between any possible
> achievement of this and supposedly one-way operations like multiplying
> large primes... and the political implications of that. Maybe he had
> a point, or maybe he was just having fun leading us on...
There are some interested in ultra low power (less than one microwatt)
computing.
-- glen
Reply by ●October 11, 20072007-10-11
On Oct 10, 4:38 am, Rune Allnor <all...@tele.ntnu.no> wrote:
> Your base assumption is wrong: Digital systems don't
> behave like analog systems.
>
>
> In the digital world there are no dissipation terms,
> only numerical accuracy issues, and one does not need
> the complex/valued maths to make the energy computations
> work.
>
> In essence, the two troublesome issues from the physical
> world, what energy is concerned, just don't exist in the
> digital world.
In mathematics they don't exist.
But what about in the implementation of the math in an electronic
computational engine?
I had a professor who made an argument that power would ultimately be
consumed where you "forgot" the unwanted bits, and that the true key
to low power (consumption) digital design would be to run the
computations backwards as well as forwards, to recreate the inputs.
He then went on to point out the conflict between any possible
achievement of this and supposedly one-way operations like multiplying
large primes... and the political implications of that. Maybe he had
a point, or maybe he was just having fun leading us on...
Reply by glen herrmannsfeldt●October 11, 20072007-10-11
Jerry Avins wrote:
> Mark wrote:
>> the power is reflected back to the source
> Give the man a box of Snickers! There's no power to dissipate because no
> power was delivered by the source.
On average, as the message title indicates.
-- glen
Reply by Jerry Avins●October 11, 20072007-10-11
Mark wrote:
...
> the power is reflected back to the source
Give the man a box of Snickers! There's no power to dissipate because no
power was delivered by the source.
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by Mark●October 11, 20072007-10-11
On Oct 10, 2:56 pm, Jerry Avins <j...@ieee.org> wrote:
> glen herrmannsfeldt wrote:
> > cincy...@gmail.com wrote:
> > (snip on power in analog and digital signals)
>
> >> The input power is not 0.5 watts. The power doesn't actually get
> >> *delivered* to the A/D. An ideal A/D would have infinite input
> >> impedance and would receive no power, but in practice it will be some
> >> very small value.
>
> > To properly terminate the input (say it is at the end of a long
> > transmission line) it will need the appropriate load resistance.
> > That could be an external resistor or the resistor tree in the
> > A/D converter.
>
> >> There's a difference between actual power delivered
> >> to the load and the "power into 1 ohm" value. Likewise, on the output,
> >> the output power need not be 0.25 watts. If you're using the D/A to
> >> drive a 1 ohm resistor (and it can source that much current), then
> >> yes, you would have that much output power. The current would be
> >> sourced by the power supply of the output amplifier. In the digital
> >> signal processing realm, all you have are numbers (bits). Multiplying
> >> one of those numbers by a value doesn't do anything physical at all.
>
> > Now consider an analog filter. Signal coming in, Op-amp with
> > appropriate resistors, capacitors, and power supply, signal
> > going out. The incoming power is dissipated in the impedance matching
> > circuit on the input, the output power comes from the Op-amp and
> > its power source. It doesn't sound so different to me.
>
> So consider a classic cascaded pi or tee filter, perhaps with m-derived
> end sections. The only resistor is the load, and nothing reaches the
> load in the stop band. What happens in the passband is evident. What
> happens in the stopband?
>
> Jerry
> --
> Engineering is the art of making what you want from things you can get.
> =AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=
=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF=AF- Hide qu=
oted text -
>
> - Show quoted text -
the power is reflected back to the source
Mark
Reply by Jerry Avins●October 10, 20072007-10-10
glen herrmannsfeldt wrote:
> cincydsp@gmail.com wrote:
> (snip on power in analog and digital signals)
>
>> The input power is not 0.5 watts. The power doesn't actually get
>> *delivered* to the A/D. An ideal A/D would have infinite input
>> impedance and would receive no power, but in practice it will be some
>> very small value.
>
> To properly terminate the input (say it is at the end of a long
> transmission line) it will need the appropriate load resistance.
> That could be an external resistor or the resistor tree in the
> A/D converter.
>
>> There's a difference between actual power delivered
>> to the load and the "power into 1 ohm" value. Likewise, on the output,
>> the output power need not be 0.25 watts. If you're using the D/A to
>> drive a 1 ohm resistor (and it can source that much current), then
>> yes, you would have that much output power. The current would be
>> sourced by the power supply of the output amplifier. In the digital
>> signal processing realm, all you have are numbers (bits). Multiplying
>> one of those numbers by a value doesn't do anything physical at all.
>
> Now consider an analog filter. Signal coming in, Op-amp with
> appropriate resistors, capacitors, and power supply, signal
> going out. The incoming power is dissipated in the impedance matching
> circuit on the input, the output power comes from the Op-amp and
> its power source. It doesn't sound so different to me.
So consider a classic cascaded pi or tee filter, perhaps with m-derived
end sections. The only resistor is the load, and nothing reaches the
load in the stop band. What happens in the passband is evident. What
happens in the stopband?
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
Reply by glen herrmannsfeldt●October 10, 20072007-10-10
cincydsp@gmail.com wrote:
(snip on power in analog and digital signals)
> The input power is not 0.5 watts. The power doesn't actually get
> *delivered* to the A/D. An ideal A/D would have infinite input
> impedance and would receive no power, but in practice it will be some
> very small value.
To properly terminate the input (say it is at the end of a long
transmission line) it will need the appropriate load resistance.
That could be an external resistor or the resistor tree in the
A/D converter.
> There's a difference between actual power delivered
> to the load and the "power into 1 ohm" value. Likewise, on the output,
> the output power need not be 0.25 watts. If you're using the D/A to
> drive a 1 ohm resistor (and it can source that much current), then
> yes, you would have that much output power. The current would be
> sourced by the power supply of the output amplifier. In the digital
> signal processing realm, all you have are numbers (bits). Multiplying
> one of those numbers by a value doesn't do anything physical at all.
Now consider an analog filter. Signal coming in, Op-amp with
appropriate resistors, capacitors, and power supply, signal
going out. The incoming power is dissipated in the impedance matching
circuit on the input, the output power comes from the Op-amp and
its power source. It doesn't sound so different to me.
-- glen
Reply by ●October 10, 20072007-10-10
On Oct 10, 8:48 am, HardySpicer <gyansor...@gmail.com> wrote:
> On Oct 10, 9:38 pm, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
> > On 10 Okt, 03:10, HardySpicer <gyansor...@gmail.com> wrote:
>
> > > Ok suppose we have an R-C network - low pass filter with tf
>
> > > G(s)=1/(1+sT) where T=CR. We can easily calculate the output power
> > > spectrum if this is driven by zero-mean white noise as
>
> > > Phiyy(w)=mag(G(jw))^2 x power of input noise
>
> > > and the area under the PSD is the total average power.
>
> > > Now let's do the same thing digitally...and produce an equivalent
> > > digital filter G(z)
>
> > > The output periodogram is
>
> > > Phiyy(theta)=mag(G(exp(jtheta))^2 x input noise power and the area is
> > > the total output power.
>
> > > Now, assuming the two output power values at the output are the same,
> > > where does the missing power go? ie the output power not equal input
> > > power.
>
> > Your base assumption is wrong: Digital systems don't
> > behave like analog systems.
>
> > The physics of the analog world dictates two important
> > conditions which do not apply in the digital world:
>
> > - There is always friction present in the physical world,
> > which dissipate energy in terms of heat.
> > - One is limited to only observe real-valued signals
> > but need imaginary components to make the energy
> > computations go up.
>
> > In the digital world there are no dissipation terms,
> > only numerical accuracy issues, and one does not need
> > the complex/valued maths to make the energy computations
> > work.
>
> > In essence, the two troublesome issues from the physical
> > world, what energy is concerned, just don't exist in the
> > digital world.
>
> > Which means you will have to look elsewhere for sources
> > for discrepancy, the most common sources being certain
> > normalization factors in the most popular implementations
> > of the FFT.
>
> > Rune
>
> Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A.
> Suppose a sine-wave of unit amplitude goes in and if there are no
> other losses we get a sine wave of amplitude 0.5 out. Input power is
> 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we
> had an analogue system the power would be disipated as heat via a
> resistor divider. In the digital world the power still has to be
> disipated somewhere. If we have a gain instead of attenuation - say x2
> then this extra power flows in from the power supply in an analogue
> system and must also do in a digital system.
>
> Hardy
The input power is not 0.5 watts. The power doesn't actually get
*delivered* to the A/D. An ideal A/D would have infinite input
impedance and would receive no power, but in practice it will be some
very small value. There's a difference between actual power delivered
to the load and the "power into 1 ohm" value. Likewise, on the output,
the output power need not be 0.25 watts. If you're using the D/A to
drive a 1 ohm resistor (and it can source that much current), then
yes, you would have that much output power. The current would be
sourced by the power supply of the output amplifier. In the digital
signal processing realm, all you have are numbers (bits). Multiplying
one of those numbers by a value doesn't do anything physical at all.
Jason
Reply by Jerry Avins●October 10, 20072007-10-10
HardySpicer wrote:
...
> Ok let's take a simpler example - an A/D, factor of 0.5 and a D/A.
> Suppose a sine-wave of unit amplitude goes in and if there are no
> other losses we get a sine wave of amplitude 0.5 out. Input power is
> 0.5 and output power is 0.25 as measured over a 1 Ohm resistor. If we
> had an analogue system the power would be disipated as heat via a
> resistor divider. In the digital world the power still has to be
> disipated somewhere. If we have a gain instead of attenuation - say x2
> then this extra power flows in from the power supply in an analogue
> system and must also do in a digital system.
You're still barking up the wrong tree. "Power" in digital systems is a
convenient fiction. suppose instead of multiplying by 0.5, we multiplied
by 2.0. Would that be the foundation of perpetual motion? How many
joules does a bit represent?
Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯