# Average Power

Started by October 9, 2007
```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.

Hardy

```
```HardySpicer 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.

Power is a convenient fiction in digital systems. Analog filters raise a
the spectrum of noise (flat over some band of interest) delivering power
to a resistive load, Now suppose a lossless (R-L) filter interposed
between the source and load that blocks half the spectrum. The filter
doesn't get hot, so where does the blocked power go?

Jerry
--
Engineering is the art of making what you want from things you can get.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```On Oct 10, 5:15 pm, Jerry Avins <j...@ieee.org> wrote:
> HardySpicer wrote:
> > Ok suppose we have an R-C network - low pass filter with tf
>
> > G(s)=3D1/(1+sT) where T=3DCR.  We can easily calculate the output power
> > spectrum if this is driven by zero-mean white noise as
>
> > Phiyy(w)=3Dmag(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)=3Dmag(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.
>
> Power is a convenient fiction in digital systems. Analog filters raise a
> similar question not so easily answered. Suppose a broadband signal with
> the spectrum of noise (flat over some band of interest) delivering power
> to a resistive load, Now suppose a lossless (R-L) filter interposed
> between the source and load that blocks half the spectrum. The filter
> doesn't get hot, so where does the blocked power go?
>
> 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

Yes but without R the energy just gets transfered back and forth. With
an R there is always i^2R losses.

Hardy

```
```>> Analog filters raise a similar question not so easily answered.
>> The filter doesn't get hot, so where does the blocked power go?

That's easy. It gets reflected to the source if I think in terms of
S-parameters (waves), or it doesn't load the source if I think of it as an
impedance. This doesn't answer the OP's question, though...

-mn

```
```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

```
```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

```
```HardySpicer wrote:
> On Oct 10, 5:15 pm, Jerry Avins <j...@ieee.org> wrote:

>>      ... Analog filters raise a
>> similar question not so easily answered. Suppose a broadband signal with
>> the spectrum of noise (flat over some band of interest) delivering power
>> to a resistive load, Now suppose a lossless (R-L) filter interposed
>> between the source and load that blocks half the spectrum. The filter
>> doesn't get hot, so where does the blocked power go?

...

> Yes but without R the energy just gets transfered back and forth. With
> an R there is always i^2R losses.

Yes what? Transfered from where to where? If the power is delivered by
the source and doesn't reach the load, where does it go? (That's an
obfuscating question.)

Jerry
--
Engineering is the art of making what you want from things you can get.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```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.
&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;&macr;
```
```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

```
```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

```