# Energy

Started by October 8, 2006
```Where does the energy (or Power)  get dissipated in a digital filter?
For example, consider a simple analogue R-C low-pass filter driven by
band-limited white noise. The output power spectrum  is given by

Pout = |W(jw)|^2 * input noise power. Power is dissipated in the
resistor. So for the digital equivalent wheree does the power go?

```
```naebad wrote:
> Where does the energy (or Power)  get dissipated in a digital filter?
> For example, consider a simple analogue R-C low-pass filter driven by
> band-limited white noise. The output power spectrum  is given by
>
> Pout = |W(jw)|^2 * input noise power. Power is dissipated in the
> resistor. So for the digital equivalent wheree does the power go?

There is no dissipation because there is no actual energy in a series of
numbers. (Very good analog filters can be made with reactive elements
only. Where does the energy go with an L-C filter?)

Jerry
--
"The rights of the best of men are secured only as the
rights of the vilest and most abhorrent are protected."
- Chief Justice Charles Evans Hughes, 1927
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```
```"Jerry Avins" <jya@ieee.org> wrote in message
>> Where does the energy (or Power)  get dissipated in a
>> digital filter?
>> For example, consider a simple analogue R-C low-pass
>> filter driven by
>> band-limited white noise. The output power spectrum  is
>> given by
>>
>> Pout = |W(jw)|^2 * input noise power. Power is dissipated
>> in the
>> resistor. So for the digital equivalent wheree does the
>> power go?
>
> There is no dissipation because there is no actual energy
> in a series of numbers. (Very good analog filters can be
> made with reactive elements only. Where does the energy go
> with an L-C filter?)

Superconductivity aside,

"There's an R in the L in the lump on the bottom of the
C..."

```
```John E. Hadstate wrote:
> "Jerry Avins" <jya@ieee.org> wrote in message
>>> Where does the energy (or Power)  get dissipated in a
>>> digital filter?
>>> For example, consider a simple analogue R-C low-pass
>>> filter driven by
>>> band-limited white noise. The output power spectrum  is
>>> given by
>>>
>>> Pout = |W(jw)|^2 * input noise power. Power is dissipated
>>> in the
>>> resistor. So for the digital equivalent wheree does the
>>> power go?
>> There is no dissipation because there is no actual energy
>> in a series of numbers. (Very good analog filters can be
>> made with reactive elements only. Where does the energy go
>> with an L-C filter?)
>
> Superconductivity aside,
>
> "There's an R in the L in the lump on the bottom of the
> C..."

Even theoretically ideal Ls and Cs produce useful filters. The input
goes reactive in the stop bands, so no power gets in. There's nothing
that needs to be dissipated.

Jerry
--
"The rights of the best of men are secured only as the
rights of the vilest and most abhorrent are protected."
- Chief Justice Charles Evans Hughes, 1927
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```
```On 8 Oct 2006 16:29:26 -0700, "naebad" <minnaebad@yahoo.co.uk> wrote:

>Where does the energy (or Power)  get dissipated in a digital filter?
>For example, consider a simple analogue R-C low-pass filter driven by
>band-limited white noise. The output power spectrum  is given by
>
>Pout = |W(jw)|^2 * input noise power. Power is dissipated in the
>resistor. So for the digital equivalent wheree does the power go?
>
>

That paradox used to bug me, too.   My example was two black boxes,
one with an analog filter, and the other with a high-impedance DAC,
equivalent digital filter, and ADC.   I think in the real world it's
hard to really do this experiment effectively, but conceptually it
took me a long time to bend my brain around it.

There's still something uncomfortable about it, because I tend to
always try to keep a strong conceptual connection between the digital
processing and the analog equivalent.   Eventually, though, you have
to just make the leap to the additional level of abstraction and
realize it's just numbers floating around.

Eric Jacobsen
Minister of Algorithms, Intel Corp.
My opinions may not be Intel's opinions.
http://www.ericjacobsen.org
```
```"Jerry Avins" <jya@ieee.org> wrote in message
news:nOidnQjeIrWl6bfYnZ2dnUVZ_oudnZ2d@rcn.net...
>> "Jerry Avins" <jya@ieee.org> wrote in message
>>>> Where does the energy (or Power)  get dissipated in a
>>>> digital filter?
>>>> For example, consider a simple analogue R-C low-pass
>>>> filter driven by
>>>> band-limited white noise. The output power spectrum  is
>>>> given by
>>>>
>>>> Pout = |W(jw)|^2 * input noise power. Power is
>>>> dissipated in the
>>>> resistor. So for the digital equivalent wheree does the
>>>> power go?
>>> There is no dissipation because there is no actual
>>> energy in a series of numbers. (Very good analog filters
>>> can be made with reactive elements only. Where does the
>>> energy go with an L-C filter?)
>>
>> Superconductivity aside,
>>
>> "There's an R in the L in the lump on the bottom of the
>> C..."
>
> Even theoretically ideal Ls and Cs produce useful filters.
> The input goes reactive in the stop bands, so no power
> gets in. There's nothing that needs to be dissipated.
>

I'm don't understand a couple of things in your post.  First
of all, I'm under the impression that reactive power does
"get in" but it's reflected back out later in the cycle.
Second, I'm having trouble with the concept of producing
"useful filters" using unrealizable components.

```
```Eric Jacobsen wrote:
>
>> Where does the energy (or Power)  get dissipated in a digital filter?
>> For example, consider a simple analogue R-C low-pass filter driven by
>> band-limited white noise. The output power spectrum  is given by
>>
>> Pout = |W(jw)|^2 * input noise power. Power is dissipated in the
>> resistor. So for the digital equivalent wheree does the power go?
>>
>>
>
> That paradox used to bug me, too.   My example was two black boxes,
> one with an analog filter, and the other with a high-impedance DAC,
> equivalent digital filter, and ADC.   I think in the real world it's
> hard to really do this experiment effectively, but conceptually it
> took me a long time to bend my brain around it.
>
> There's still something uncomfortable about it, because I tend to
> always try to keep a strong conceptual connection between the digital
> processing and the analog equivalent.   Eventually, though, you have
> to just make the leap to the additional level of abstraction and
> realize it's just numbers floating around.

Much of my DSP code is in C, and debugging it can be L. I really don't
have any R's involved, in which to dissipate any energy. :-)

In most digital filters the "energy" gets stored, and eventually has to
come out. If you make a high Q filter, and put some signal through it
that pumps the filter up, that stored energy will spew out when the
stimulus is removed. Hence, the filter rings until the energy has
drained from it.

Of course, you can magically loose energy any time you want, just by
multiplying by <1.

Steve
```
```
>
> Even theoretically ideal Ls and Cs produce useful filters. The input
> goes reactive in the stop bands, so no power gets in. There's nothing
> that needs to be dissipated.
>
>

The incident power in the stop band gets REFLECTED back to the source
where it is __usually__ dissipated in the source resistance of the
source.

Mark

```
```"naebad" <minnaebad@yahoo.co.uk> writes:

> Where does the energy (or Power)  get dissipated in a digital filter?
> For example, consider a simple analogue R-C low-pass filter driven by
> band-limited white noise. The output power spectrum  is given by
>
> Pout = |W(jw)|^2 * input noise power. Power is dissipated in the
> resistor. So for the digital equivalent wheree does the power go?

I think of the situation like this. Power is a real, physical
quantity. In order for power to be generated, there must be
expenditure of real, physical joules (energy) per unit time.

In a digital signal, there is no generation of actual joules because
nothing physical is happening (at least not related to the signal - sure,
you have transistors switching and THAT requires energy).

The energy and power come into being when the digital signal is
converted to an actual physical representation.
--
%  Randy Yates                  % "Midnight, on the water...
%% Fuquay-Varina, NC            %  I saw...  the ocean's daughter."
%%% 919-577-9882                % 'Can't Get It Out Of My Head'
%%%% <yates@ieee.org>           % *El Dorado*, Electric Light Orchestra
```
```John E. Hadstate wrote:

...

> I'm don't understand a couple of things in your post.  First
> of all, I'm under the impression that reactive power does
> "get in" but it's reflected back out later in the cycle.

Look at it that way if you like. As much power comes back out the
inputout as gets in, so nothing needs to be absorbed and dissipated.

> Second, I'm having trouble with the concept of producing
> "useful filters" using unrealizable components.

Engineers deal with non-ideal material all the time. Trusses are
designed as if the joints are free to rotate, the deflection of a loaded
beam is calculated using the evidently false assumption that the beam
doesn't deflect, Digital filters are built with truncated coefficients,
and L-C filters are designed assuming ideal components. In all those
cases, the departure of the actual result from the calculated ideal is
small enough for the purpose. Are you just pulling my chain?

Jerry
--
"The rights of the best of men are secured only as the
rights of the vilest and most abhorrent are protected."
- Chief Justice Charles Evans Hughes, 1927
&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;&#4294967295;
```